ELECTROSTATICS

17. Properties :
(i) Electric lines of force start from a positive charge and end on a negative charge.
(ii) No two lines of force can intersect each other. If they does so then at the point of intersection two tangents could be drawn, which gives two directions of electric field at the same point, which is impossible.
(iii) The tangent drawn at any point on line of force gives the direction of force acting on a positive charge placed at that point.
(iv) These lines have a tendency to contract in length like a stretched elastic string. This actually explains attraction between opposite charges. (v) These lines have a tendency to separate from each other in the direction perpendicular to their length. This explains repulsion between like charges.
(vi) Intensity of electric field is given by the number of electric lines of force in a unit area at that point.
(vii) Lines of force of a uniform field are parallel and at equal distance.
(viii) Unit positive charge gives 4 π K 4 π K (4pi)/(K)\frac{4 \pi}{\mathrm{K}}4πK lines in a medium of dielectric constant K K K\mathrm{K}K.
(ix) Important : Electric lines of force can never enter the conductor, because inside the conductor the intensity of electric field is zero.
(x) Important: Lines of force leaves the surface of conductor normally.
Illustration :
Which is true?
(1) E A < E B > E C E A < E B > E C E_(A) < E_(B) > E_(C)E_{A}<E_{B}>E_{C}EA<EB>EC
(2) E A > E B > E C E A > E B > E C E_(A) > E_(B) > E_(C)E_{A}>E_{B}>E_{C}EA>EB>EC
(3) E A > E B < E C E A > E B < E C E_(A) > E_(B) < E_(C)E_{A}>E_{B}<E_{C}EA>EB<EC
(4) E A < E B < E C E A < E B < E C E_(A) < E_(B) < E_(C)E_{A}<E_{B}<E_{C}EA<EB<EC
Sol. (2) Number of electric lines of force in unit area is maximum at A A AAA and least at C C CCC.
so E A > E B > E C E A > E B > E C E_(A) > E_(B) > E_(C)E_{A}>E_{B}>E_{C}EA>EB>EC
Illustration :
A metal sphere is placed in an uniform electric field which one is a correct electric line of force?
Sol. (4) Only 4 is normal to the conducting surface.

18. ELECTRIC POTENTIAL :

Work done in bringing a unit positive charge from infinity to any point keeping kinetic energy constant is termed as potential at that point i.e. if
W = W = W=\mathrm{W}=W= work done in bringing a positive charge q 0 q 0 q_(0)\mathrm{q}_{0}q0 from infinity to that point, then, V = W q 0 V = W q 0 V=(W)/(q_(0))V=\frac{W}{q_{0}}V=Wq0
(i) Electric potential at infinity is taken to be zero.
(ii) It is not path dependent quantity if simply depends upon the starting and end points.
(iii) It is a scalar quantity.
(iv) Unit : Volt or Joule/Coulomb
(v) Dimension : [ M 1 L 2 T − 3 A − 1 ] M 1 L 2 T − 3 A − 1 [M^(1)L^(2)T^(-3)A^(-1)]\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right][M1 L2 T−3 A−1]
(vi) Potential due to a positive charge is positive and potential due to a negative charge is negative, here potential being positive and negative implies whether work is done on the charge or done by the charge respectively. (vii) Potential due to a point charge Q Q QQQ at a distance r r rrr is V = 1 4 π ϵ 0 q r V = 1 4 π ϵ 0 q r V=(1)/(4piepsilon_(0))(q)/(r)V=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}}{r}V=14πϵ0qr
⇒ V α 1 r ⇒ V α 1 r =>Valpha(1)/(r)\Rightarrow \mathrm{V} \alpha \frac{1}{r}⇒Vα1r
(viii) Total potential at a point due to a group of charges is scalar sum of individual potentials. V p = V 1 + V 2 + … . V n V p = V 1 + V 2 + … . V n V_(p)=V_(1)+V_(2)+dots.V_No\mathrm{V}_{\mathrm{p}}=\mathrm{V}_{1}+\mathrm{V}_{2}+\ldots . \mathrm{V}_{\mathrm{n}}Vp=V1+V2+….Vn
(ix) Electric field is gradient of electric potential at that point. E = − d v d r E = − d v d r E=-(dv)/(dr)E=-\frac{d v}{d r}E=−dvdr
The negative sign implies that direction of electric field is in the direction of decreasing potential.
(x) Work done in bringing a charge Q Q QQQ from infinity to that point is
W = Q V W = Q V W=QV\mathrm{W}=\mathrm{QV}W=QV where V V V\mathrm{V}V is potential at that point.
(xi) Potential of earth is taken to be zero.
Illustration :
Can metal sphere of 1 c m 1 c m 1cm1 \mathrm{~cm}1 cm radius held a charge of 1 coulomb? [Air gets ionized at the Electric field of E max = 3 × 10 6 E max = 3 × 10 6 E_(max)=3xx10^(6)E_{\max }=3 \times 10^{6}Emax=3×106 volt / m ] / m {://m]\left./ \mathrm{m}\right]/m]
Sol. No. The potential of a metal sphere of radius 1 c m 1 c m 1cm1 \mathrm{~cm}1 cm is given by
V = 1 4 π ϵ 0 q r = 9 × 10 9 × 1 1 × 10 − 2 = 9 × 10 11 V = 1 4 π ϵ 0 q r = 9 × 10 9 × 1 1 × 10 − 2 = 9 × 10 11 V=(1)/(4piepsilon_(0))(q)/(r)=9xx10^(9)xx(1)/(1xx10^(-2))=9xx10^(11)V=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}=9 \times 10^{9} \times \frac{1}{1 \times 10^{-2}}=9 \times 10^{11}V=14πϵ0qr=9×109×11×10−2=9×1011
The potential is much greater then needed to ionise the air and hence the charge leakes to surrounding air. [Air gets ionized at the potential of 3 × 10 6 3 × 10 6 3xx10^(6)3 \times 10^{6}3×106 volt]
Illustration :
Infinite number of same charge q q qqq are placed at x = 1 , 2 , 4 , 8 … x = 1 , 2 , 4 , 8 … x=1,2,4,8dotsx=1,2,4,8 \ldotsx=1,2,4,8…. What is the potential at x = 0 x = 0 x=0x=0x=0 ?
Sol. V = 1 4 π ε 0 ( q 1 + q 2 + q 4 + q 8 + … . . ) V = 1 4 π ε 0 q 1 + q 2 + q 4 + q 8 + … . . quad V=(1)/(4piepsi_(0))((q)/(1)+(q)/(2)+(q)/(4)+(q)/(8)+dots..)\quad V=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{1}+\frac{q}{2}+\frac{q}{4}+\frac{q}{8}+\ldots ..\right)V=14πε0(q1+q2+q4+q8+…..)
= q 4 π ε 0 1 ( 1 − 1 2 ) = 2 q 4 π ε 0 = q 2 π ε 0 [ ∴ a + a r + … … ∞ = a 1 − r a r < 1 = q 4 π ε 0 1 1 − 1 2 = 2 q 4 π ε 0 = q 2 π ε 0 ∴ a + a r + … … ∞ = a 1 − r a r < 1 {:[=(q)/(4piepsi_(0))(1)/((1-(1)/(2)))=(2q)/(4piepsi_(0))=(q)/(2piepsi_(0))],[[:.a+ar+dots dots oo=(a)/(1-r_(a))quad r < 1:}]:}\begin{gathered} =\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\left(1-\frac{1}{2}\right)}=\frac{2 q}{4 \pi \varepsilon_{0}}=\frac{q}{2 \pi \varepsilon_{0}} \\ {\left[\therefore a+a r+\ldots \ldots \infty=\frac{a}{1-r_{a}} \quad r<1\right.} \end{gathered}=q4πε01(1−12)=2q4πε0=q2πε0[∴a+ar+……∞=a1−rar<1
Illustration :
If the alternative charges are unlike, then what will be the potential ?
Sol. Then , V = 1 4 π ∈ 0 ( q 1 − q 2 + q 4 − q 8 + … … ∞ ) V = 1 4 π ∈ 0 q 1 − q 2 + q 4 − q 8 + … … ∞ V=(1)/(4piin_(0))((q)/(1)-(q)/(2)+(q)/(4)-(q)/(8)+dots dots oo)V=\frac{1}{4 \pi \in_{0}}\left(\frac{\mathrm{q}}{1}-\frac{\mathrm{q}}{2}+\frac{\mathrm{q}}{4}-\frac{\mathrm{q}}{8}+\ldots \ldots \infty\right)V=14π∈0(q1−q2+q4−q8+……∞)
= 1 4 π ϵ 0 [ ( q 1 + q 2 + … . . ∞ ) − ( q 2 + q 8 + … ∞ ) ] = q 4 π ϵ 0 [ 1 1 − 1 4 − 1 2 ( 1 1 − 1 4 ) ] = 1 4 π ϵ 0 2 q 3 [ ∴ a + a r + a r 2 … ∞ = a 1 − r r < 1 ] = 1 4 π ϵ 0 q 1 + q 2 + … . . ∞ − q 2 + q 8 + … ∞ = q 4 π ϵ 0 1 1 − 1 4 − 1 2 1 1 − 1 4 = 1 4 π ϵ 0 2 q 3 ∴ a + a r + a r 2 … ∞ = a 1 − r r < 1 {:[=(1)/(4piepsilon_(0))[((q)/(1)+(q)/(2)+dots..oo)-((q)/(2)+(q)/(8)+dots oo)]],[=(q)/(4piepsilon_(0))[(1)/(1-(1)/(4))-(1)/(2)((1)/(1-(1)/(4)))]=(1)/(4piepsilon_(0))(2q)/(3)],[[:.a+ar+ar^(2)dots oo=(a)/(1-r)r < 1]]:}\begin{aligned} & =\frac{1}{4 \pi \epsilon_{0}}\left[\left(\frac{q}{1}+\frac{q}{2}+\ldots . . \infty\right)-\left(\frac{q}{2}+\frac{q}{8}+\ldots \infty\right)\right] \\ & =\frac{q}{4 \pi \epsilon_{0}}\left[\frac{1}{1-\frac{1}{4}}-\frac{1}{2}\left(\frac{1}{1-\frac{1}{4}}\right)\right]=\frac{1}{4 \pi \epsilon_{0}} \frac{2 q}{3} \\ & {\left[\therefore a+a r+a r^{2} \ldots \infty=\frac{a}{1-r} r<1\right]} \end{aligned}=14πϵ0[(q1+q2+…..∞)−(q2+q8+…∞)]=q4πϵ0[11−14−12(11−14)]=14πϵ02q3[∴a+ar+ar2…∞=a1−rr<1]
The work done in taking a charge from one point to the other in an electric field is called the potential difference between two points.
Thus, if w w w\mathrm{w}w be work done in moving a charge q 0 q 0 q_(0)\mathrm{q}_{0}q0 from B B B\mathrm{B}B to A A A\mathrm{A}A, then the potential difference is given by-
V A − V B = W ( B → A ) q 0 V A − V B = W ( B → A ) q 0 V_(A)-V_(B)=(W_((BrarrA)))/(q_(0))\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{W}_{(\mathrm{B} \rightarrow \mathrm{A})}}{\mathrm{q}_{0}}VA−VB=W(B→A)q0
(i) Unit of potential difference is volt.
(ii) This is a scalar quantity.
(iii) Potential difference does not depend upon Co-ordinate system.
(iv) Potential difference does not depend upon the path followed. This is, because electric field is a conservative force field and work done due to conservative force field does not depend upon path followed.
Illustration :
In the following fig. Along which path the work done will be maximum in carrying a charge from A A AAA to B B BBB in the presence of any another charge
Sol. Same for all the path
[Because the work done doesn't depend upon the path]

19. Illustration :

A charge 20 μ C 20 μ C 20 mu C20 \mu C20μC is situated at the origin of X X XXX-Y plane. What will be potential difference between points ( 5 a , a ) ( 5 a , a ) (5a,a)(5 a, a)(5a,a) and ( − 3 a , 4 a ) ( − 3 a , 4 a ) (-3a,4a)(-3 a, 4 a)(−3a,4a).
Sol. Distance between ( 0 , 0 ) & ( 5 a , a ) ( 0 , 0 ) & ( 5 a , a ) (0,0)&(5a,a)(0,0) \&(5 a, a)(0,0)&(5a,a),
r 1 = 25 a 2 + 0 = 5 a ∴ V 1 = k q 5 a r 1 = 25 a 2 + 0 = 5 a ∴ V 1 = k q 5 a {:[r_(1)=sqrt(25a^(2)+0)=5a],[:.V_(1)=(kq)/(5a)]:}\begin{aligned} r_{1} & =\sqrt{25 \mathrm{a}^{2}+0}=5 a \\ \therefore V_{1} & =\frac{\mathrm{kq}}{5 \mathrm{a}} \end{aligned}r1=25a2+0=5a∴V1=kq5a
Distance between
( 0 , 0 ) & ( − 3 a , 4 a ) r 2 = 9 a 2 + 16 a = 5 a ( 0 , 0 ) & ( − 3 a , 4 a ) r 2 = 9 a 2 + 16 a = 5 a (0,0)&(-3a,4a)r_(2)=sqrt(9a^(2)+16 a)=5a(0,0) \&(-3 a, 4 a) r_{2}=\sqrt{9 a^{2}+16 a}=5 a(0,0)&(−3a,4a)r2=9a2+16a=5a
V 2 = k q 5 a V 2 = k q 5 a V_(2)=(kq)/(5a)V_{2}=\frac{\mathrm{kq}}{5 \mathrm{a}}V2=kq5a
∴ V 1 − V 2 = 0 ∴ V 1 − V 2 = 0 :.quadV_(1)-V_(2)=0\therefore \quad V_{1}-V_{2}=0∴V1−V2=0

20. Relationship between electric potential and intensity of electric field :

(i) V A = − ∫ ∞ A E → ⋅ d r → , V A = V A = − ∫ ∞ A   E → ⋅ d r → , V A = quadV_(A)=-int_(oo)^(A) vec(E)* vec(dr),V_(A)=\quad V_{A}=-\int_{\infty}^{A} \vec{E} \cdot \overrightarrow{d r}, V_{A}=VA=−∫∞AE→⋅dr→,VA= electric potential at point A A AAA.
(ii) Potential difference between two points in an electric field is given by negative value of line integral of electric field i.e.
V B − V A = − ∫ A B E → ⋅ d r → V B − V A = − ∫ A B   E → ⋅ d r → V_(B)-V_(A)=-int_(A)^(B) vec(E)* vec(dr)\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-\int_{\mathrm{A}}^{\mathrm{B}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dr}}VB−VA=−∫ABE→⋅dr→
(iii) E → = − ∇ V = − grad ⁡ V E → = − ∇ V = − grad ⁡ V vec(E)=-gradV=-grad V\overrightarrow{\mathrm{E}}=-\nabla \mathrm{V}=-\operatorname{grad} \mathrm{V}E→=−∇V=−grad⁡V
∇ = ( ∇ = ( grad=(\nabla=(∇=( gradient ) = ( ∂ ∂ x i ^ + ∂ ∂ y j ^ + ∂ ∂ x k ^ ) ) = ∂ ∂ x i ^ + ∂ ∂ y j ^ + ∂ ∂ x k ^ )=((del)/(delx)( hat(i))+(del)/(dely)( hat(j))+(del)/(delx)( hat(k))))=\left(\frac{\partial}{\partial \mathbf{x}} \hat{\mathbf{i}}+\frac{\partial}{\partial \mathbf{y}} \hat{\mathbf{j}}+\frac{\partial}{\partial \mathbf{x}} \hat{\mathbf{k}}\right))=(∂∂xi^+∂∂yj^+∂∂xk^)
E x = − ∂ V ∂ x , E y = − ∂ V ∂ y , E z = − ∂ V ∂ z E x = − ∂ V ∂ x , E y = − ∂ V ∂ y , E z = − ∂ V ∂ z E_(x)=-(delV)/(delx),E_Yes=-(delV)/(dely),E_(z)=-(delV)/(delz)\mathrm{E}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}}, \mathrm{E}_{\mathrm{y}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{y}}, \mathrm{E}_{\mathrm{z}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{z}}Ex=−∂V∂x,Ey=−∂V∂y,Ez=−∂V∂z
(iv) If v v vvv is a function of r r rrr only, then E = − d V d r E = − d V d r E=-(dV)/(dr)E=-\frac{d V}{d r}E=−dVdr
(v) For a uniform electric field, E = − Δ V Δ r E = − Δ V Δ r E=-(DeltaV)/(Deltar)\mathrm{E}=-\frac{\Delta \mathrm{V}}{\Delta \mathrm{r}}E=−ΔVΔr and it's direction is along the decrease in the value of V V V\mathrm{V}V.
Illustration :
Electric potential for a point ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z) is given by V = 4 x 2 V = 4 x 2 V=4x^(2)V=4 x^{2}V=4x2 volt. Electric field at point ( 1 , 0 , 2 ) ( 1 , 0 , 2 ) (1,0,2)(1,0,2)(1,0,2) is -
Sol. E = − d V d x = − 8 x E = − d V d x = − 8 x quad E=-(dV)/(dx)=-8x\quad E=-\frac{\mathrm{dV}}{\mathrm{dx}}=-8 xE=−dVdx=−8x
E at ( 1 , 0 , 2 ) = − 8 V / m ( 1 , 0 , 2 ) = − 8 V / m (1,0,2)=-8V//m(1,0,2)=-8 \mathrm{~V} / \mathrm{m}(1,0,2)=−8 V/m
∴ ∴ :.\therefore∴ Magnitude of E E EEE
= 8 V / m = 8 V / m =8V//m=8 \mathrm{~V} / \mathrm{m}=8 V/m direction along − x − x -x-x−x axis.
Illustration :
Potential in the x − y x − y x-yx-yx−y plane is given as V = 5 ( x 2 + x y ) V = 5 x 2 + x y V=5(x^(2)+xy)V=5\left(x^{2}+x y\right)V=5(x2+xy) volts. Find the electric field at the point ( 1 , − 2 ) ( 1 , − 2 ) (1,-2)(1,-2)(1,−2).
Sol. E x = − ∂ V ∂ x = − ( 10 x + 5 y ) = − 10 + 10 = 0 , E y = − ∂ V ∂ y = − 5 x = − 5 V / m E x = − ∂ V ∂ x = − ( 10 x + 5 y ) = − 10 + 10 = 0 , E y = − ∂ V ∂ y = − 5 x = − 5 V / m quadE_(x)=-(delV)/(delx)=-(10 x+5y)=-10+10=0,E_Yes=-(delV)/(dely)=-5x=-5V//m\quad E_{x}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}}=-(10 x+5 y)=-10+10=0, E_{y}=-\frac{\partial \mathrm{V}}{\partial \mathrm{y}}=-5 x=-5 \mathrm{~V} / \mathrm{m}Ex=−∂V∂x=−(10x+5y)=−10+10=0,Ey=−∂V∂y=−5x=−5 V/m
∴ E → = ( − 5 j ^ ) V / m ∴ E → = ( − 5 j ^ ) V / m :. vec(E)=(-5 hat(j))V//m\therefore \overrightarrow{\mathrm{E}}=(-5 \hat{\mathrm{j}}) \mathrm{V} / \mathrm{m}∴E→=(−5j^)V/m
Illustration :
Graph shows variation of potential v / s v / s v//sv / sv/s distance. Calculate electric field at various points.
E = − d v d r = − E = − d v d r = − E=(-dv)/(dr)=-\mathrm{E}=\frac{-\mathrm{dv}}{\mathrm{dr}}=-E=−dvdr=− slope = − tan ⁡ θ = − tan ⁡ θ =-tan theta=-\tan \theta=−tan⁡θ
distance E ( V / m ) E ( V / m ) E(V//m)\mathrm{E}(\mathrm{V} / \mathrm{m})E(V/m)
S = 1 S = 1 S=1\mathrm{S}=1S=1 -1
Sol. r = 1.5 r = 1.5 quadr=1.5\quad \mathrm{r}=1.5r=1.5 -1
r = 2.5 r = 2.5 r=2.5\mathrm{r}=2.5r=2.5 0
r = 5 r = 5 r=5\mathrm{r}=5r=5 1
distance E(V//m) S=1 -1 Sol. quadr=1.5 -1 r=2.5 0 r=5 1 | | distance | \(\mathrm{E}(\mathrm{V} / \mathrm{m})\) | | :---: | :---: | :---: | | \(\mathrm{S}=1\) | -1 | | | Sol. \(\quad \mathrm{r}=1.5\) | -1 | | | | \(\mathrm{r}=2.5\) | 0 | | \(\mathrm{r}=5\) | 1 | |
Illustration :
An oil drop 'B' has charge 1.6 × 10 − 19 C 1.6 × 10 − 19 C 1.6 xx10^(-19)C1.6 \times 10^{-19} \mathrm{C}1.6×10−19C and mass 1.6 × 10 − 14 k g 1.6 × 10 − 14 k g 1.6 xx10^(-14)kg1.6 \times 10^{-14} \mathrm{~kg}1.6×10−14 kg. If the drop is in equilibrium position, then what will be the potential diff. between the plates.[The distance between the plates is 10 m m ] 10 m m ] 10mm]10 \mathrm{~mm}]10 mm]
Sol. For equilibrium, electric force = = === weight of drop
⇒ q E = m g ⇒ q E = m g =>qE=mg\Rightarrow q E=m g⇒qE=mg
⇒ q ⋅ V d = m g ⇒ q ⋅ V d = m g =>q*(V)/(d)=mg\Rightarrow q \cdot \frac{\mathrm{V}}{\mathrm{d}}=m g⇒q⋅Vd=mg
⇒ V = m g d q ⇒ V = m g d q =>quad V=(mgd)/(q)\Rightarrow \quad V=\frac{\mathrm{mgd}}{\mathrm{q}}⇒V=mgdq
= 1.6 × 10 − 14 × 9.8 × 10 × 10 − 3 1.6 × 10 − 19 = 1.6 × 10 − 14 × 9.8 × 10 × 10 − 3 1.6 × 10 − 19 =(1.6 xx10^(-14)xx9.8 xx10 xx10^(-3))/(1.6 xx10^(-19))=\frac{1.6 \times 10^{-14} \times 9.8 \times 10 \times 10^{-3}}{1.6 \times 10^{-19}}=1.6×10−14×9.8×10×10−31.6×10−19
V = 10 4 V = 10 4 V=10^(4)V=10^{4}V=104 volt
[When a charged particle is in equilibrium in electric field, the following formula is often used q E = m g ] q E = m g ] qE=mg]q E=m g]qE=mg]
Illustration :
Compare field, potential and surface charge density at A , B , C A , B , C A,B,CA, B, CA,B,C.
Sol. Surface of a metal is an equipotential surface (EPS) so charge on irregular shaped metal distributes to create same potential on surface
⇒ V A = V B = V C ⇒ V A = V B = V C =>V_(A)=V_(B)=V_(C)\Rightarrow V_{A}=V_{B}=V_{C}⇒VA=VB=VC Now radius of curvature (R) for straight line = ∞ = ∞ =oo=\infty=∞
R C > R B > R A R C > R B > R A R_(C) > R_(B) > R_(A)R_{C}>R_{B}>R_{A}RC>RB>RA Now Q A R A = Q B R B = Q C R C Q A R A = Q B R B = Q C R C (Q_(A))/(R_(A))=(Q_(B))/(R_(B))=(Q_(C))/(R_(C))\frac{\mathrm{Q}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{A}}}=\frac{\mathrm{Q}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\mathrm{Q}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{C}}}QARA=QBRB=QCRC or ( Q ∝ R ) ( Q ∝ R ) (Q prop R)(Q \propto R)(Q∝R)
Electric Field E ∝ Q R 2 E ∝ Q R 2 Eprop(Q)/(R^(2))\mathrm{E} \propto \frac{\mathrm{Q}}{\mathrm{R}^{2}}E∝QR2 or E ∝ 1 R ⇒ E A > E B > E C E ∝ 1 R ⇒ E A > E B > E C Eprop(1)/(R)=>E_(A) > E_(B) > E_(C)\mathrm{E} \propto \frac{1}{\mathrm{R}} \Rightarrow E_{A}>E_{B}>E_{C}E∝1R⇒EA>EB>EC
surface charge density σ ∝ Q R 2 ⇒ σ A > σ B > σ C σ ∝ Q R 2 ⇒ σ A > σ B > σ C sigma prop(Q)/(R^(2))=>sigma_(A) > sigma_(B) > sigma_(C)\sigma \propto \frac{\mathrm{Q}}{\mathrm{R}^{2}} \Rightarrow \sigma_{A}>\sigma_{B}>\sigma_{C}σ∝QR2⇒σA>σB>σC
So, if there is any possibility of charge leakage, it starts from point A. (corona discharge) Illustration :
Find V A , V B , V C V A , V B , V C V_(A),V_(B),V_(C)V_{A}, V_{B}, V_{C}VA,VB,VC, and If V A = V C V A = V C V_(A)=V_(C)V_{A}=V_{C}VA=VC then what is the required condition ?
Sol. V A = σ a ϵ 0 − σ b ϵ 0 + σ c ϵ 0 = σ ϵ 0 ( a − b + c ) V A = σ a ϵ 0 − σ b ϵ 0 + σ c ϵ 0 = σ ϵ 0 ( a − b + c ) quadV_(A)=(sigmaa)/(epsilon_(0))-(sigmab)/(epsilon_(0))+(sigmac)/(epsilon_(0))=(sigma)/(epsilon_(0))(a-b+c)\quad \mathrm{V}_{\mathrm{A}}=\frac{\sigma \mathrm{a}}{\epsilon_{0}}-\frac{\sigma \mathrm{b}}{\epsilon_{0}}+\frac{\sigma \mathrm{c}}{\epsilon_{0}}=\frac{\sigma}{\epsilon_{0}}(\mathrm{a}-\mathrm{b}+\mathrm{c})VA=σaϵ0−σbϵ0+σcϵ0=σϵ0(a−b+c)
V B = σ a 2 ϵ 0 b − σ b ϵ 0 + σ c ϵ 0 V C = σ a 2 ϵ 0 c − σ b 2 ϵ 0 c + σ c ϵ 0 = σ ϵ 0 ( a 2 − b 2 + c 2 c ) V B = σ a 2 ϵ 0 b − σ b ϵ 0 + σ c ϵ 0 V C = σ a 2 ϵ 0 c − σ b 2 ϵ 0 c + σ c ϵ 0 = σ ϵ 0 a 2 − b 2 + c 2 c {:[V_(B)=(sigmaa^(2))/(epsilon_(0)(b))-(sigmab)/(epsilon_(0))+(sigmac)/(epsilon_(0))],[V_(C)=(sigmaa^(2))/(epsilon_(0)c)-(sigmab^(2))/(epsilon_(0)c)+(sigmac)/(epsilon_(0))=(sigma)/(epsilon_(0))((a^(2)-b^(2)+c^(2))/(c))]:}\begin{aligned} & \mathrm{V}_{\mathrm{B}}=\frac{\sigma \mathrm{a}^{2}}{\epsilon_{0} \mathrm{~b}}-\frac{\sigma \mathrm{b}}{\epsilon_{0}}+\frac{\sigma \mathrm{c}}{\epsilon_{0}} \\ & \mathrm{~V}_{\mathrm{C}}=\frac{\sigma \mathrm{a}^{2}}{\epsilon_{0} \mathrm{c}}-\frac{\sigma \mathrm{b}^{2}}{\epsilon_{0} \mathrm{c}}+\frac{\sigma \mathrm{c}}{\epsilon_{0}}=\frac{\sigma}{\epsilon_{0}}\left(\frac{\mathrm{a}^{2}-\mathrm{b}^{2}+\mathrm{c}^{2}}{\mathrm{c}}\right) \end{aligned}VB=σa2ϵ0 b−σbϵ0+σcϵ0 VC=σa2ϵ0c−σb2ϵ0c+σcϵ0=σϵ0(a2−b2+c2c)
⇒ ⇒ =>\Rightarrow⇒ Now If V A = V C ⇒ ( a − b + c ) = a 2 − b 2 + c 2 c ⇒ c ( a − b ) = a 2 − b 2 ⇒ c = a + b V A = V C ⇒ ( a − b + c ) = a 2 − b 2 + c 2 c ⇒ c ( a − b ) = a 2 − b 2 ⇒ c = a + b V_(A)=V_(C)=>(a-b+c)=(a^(2)-b^(2)+c^(2))/(c)=>c(a-b)=a^(2)-b^(2)=>c=a+bV_{A}=V_{C} \Rightarrow(a-b+c)=\frac{\mathrm{a}^{2}-\mathrm{b}^{2}+\mathrm{c}^{2}}{\mathrm{c}} \Rightarrow c(a-b)=a^{2}-b^{2} \Rightarrow c=a+bVA=VC⇒(a−b+c)=a2−b2+c2c⇒c(a−b)=a2−b2⇒c=a+b

21. Equipotential Surface :

(a)
(b)
( c ) ( c ) (c)(\mathrm{c})(c)
( d ) ( d ) (d)(\mathrm{d})(d)
(i) These are the imaginary surface (drawn in an electric field) where the potential at any point on the surface has the same value.
(ii) No two equipotential surfaces ever intersects.
(iii) Equipotential surfaces are perpendicular to the electric field lines.
(iv) Work done in moving a charge from a one point to the other on an equipotential surface is zero irrespective of the path followed and hence there is no change in kinetic energy of the charge. (v) Component of electric field parallel to equipotential surface is zero.
(vi) Nearer the equipotential surfaces, stronger the electric field intensity.
Illustration :
Some equipotential surfaces are shown in fig. What is the correct order of electric field intensity?
Sol. E B > E C > E A E B > E C > E A E_(B) > E_(C) > E_(A)E_{B}>E_{C}>E_{A}EB>EC>EA, because potential gradient at B B BBB is maximum.

22. POTENTIAL ENERGY OF CHARGED PARTICLE IN ELECTRIC FIELD :

(i) Work done in bringing a charge from infinity to a point against the electric field is equal to the potential energy of that charge.
(ii) Potential energy of a charge of a point is equal to the product of magnitude of charge and electric potential at that point i.e.P.E. = q V = q V =qV=\mathrm{qV}=qV
(iii) Work done in moving a charge from one point to other in an electric field is equal to change in it's potential energy i.e. work done in moving Q Q QQQ from A A AAA to B = q V B − q A B = q V B − q A B=qV_(B)-q_(A)B=q V_{B}-q_{A}B=qVB−qA
= U B − U A = U B − U A =U_(B)-U_(A)=\mathrm{U}_{\mathrm{B}}-\mathrm{U}_{\mathrm{A}}=UB−UA
(iv) Work done in moving a unit charge from one point to other is equal to potential difference between two points.

23. Potential Energy of System :

(i) The electric potential energy of a system of charges is the work that has been done in bringing those charges from infinity to near each other to form the system.
(ii) If a system is given negative of it's potential energy, then all charges will move to infinity. This negative value of total energy is called the binding energy.

24. Work Done in An Electric Field :

(i) If electric potential at a point is V V V\mathrm{V}V then potential energy (PE) of a charge placed at that point will be qv.
(ii) Work done in moving a charge from A A A\mathrm{A}A to B B B\mathrm{B}B is equal to change in P E P E PE\mathrm{PE}PE of that charge, W A B = W A B = W_(AB)=\mathrm{W}_{\mathrm{AB}}=WAB= work done from A A A\mathrm{A}A to B B B\mathrm{B}B
= P E B − P E A = q ( V B − V A ) = P E B − P E A = q V B − V A =PE_(B)-PE_(A)=q(V_(B)-V_(A))=\mathrm{PE}_{\mathrm{B}}-\mathrm{PE}_{\mathrm{A}}=\mathrm{q}\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)=PEB−PEA=q(VB−VA)
(iii) Work done in moving a charge along a closed surface in an electric field is zero.
(iv) Total energy remains constant in an electric field i.e. K E A + P E A = K E B + P E B K E A + P E A = K E B + P E B KE_(A)+PE_(A)=KE_(B)+PE_(B)\mathrm{KE}_{\mathrm{A}}+\mathrm{PE}_{\mathrm{A}}=\mathrm{KE}_{\mathrm{B}}+\mathrm{PE}_{\mathrm{B}}KEA+PEA=KEB+PEB K E = K i n e t i c K E = K i n e t i c KE=Kinetic\mathrm{KE}=\mathrm{Kinetic}KE=Kinetic energy P E = P E = PE=\mathrm{PE}=PE= Potential energy
(v) A free charge moves from higher PE to lower PE state in an electric field. Hence
(a) a + ve charge will move from higher potential to lower potential while,
(b) a-ve charge will move from lower potential to higher potential
(c) Work done for displacement through r → r → vec(r)\vec{r}r→ for a charge experiencing a force
F → = W = F → ⋅ r → F → = W = F → ⋅ r → vec(F)quad=W= vec(F)* vec(r)\overrightarrow{\mathrm{F}} \quad=\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{r}}F→=W=F→⋅r→
Illustration :
A charge Q Q QQQ is placed at the centre of a circle of a radius ' r r rrr '. What will be the work done in taking a charge q q qqq from A A AAA to diametrically opposite point B B BBB ?
Sol. Potential energy of q q qqq at A A AAA
= U A = 1 4 π ϵ 0 Q q r = U A = 1 4 π ϵ 0 Q q r =U_(A)=(1)/(4piepsilon_(0))(Qq)/(r)=U_{A}=\frac{1}{4 \pi \epsilon_{0}} \frac{Q q}{r}=UA=14πϵ0Qqr
PE of q q qqq at B B BBB
= U B = 1 4 π ϵ 0 Q q r = U B = 1 4 π ϵ 0 Q q r =U_(B)=(1)/(4piepsilon_(0))(Qq)/(r)=U_{B}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{Qq}}{r}=UB=14πϵ0Qqr
∴ ∴ :.\therefore∴ Work done = U B − U A = 0 = U B − U A = 0 =U_(B)-U_(A)=0=U_{B}-U_{A}=0=UB−UA=0
Illustration:
Three charges Q , + q Q , + q Q,+qQ,+qQ,+q and + q + q +q+q+q are placed at the vertices of a right angled isosceles triangle as shown in the figure. The net electrostatic energy of
the configuration is zero., if Q Q QQQ is equal to :
Sol. Net electrostatic energy
U = k Q q a + k Q q 2 a + k q q a For U = 0 ; Q q a + Q q 2 a + q q a = 0 ⇒ Q q ( 1 a + 1 2 a ) = − q 2 a U = k Q q a + k Q q 2 a + k q q a  For  U = 0 ; Q q a + Q q 2 a + q q a = 0 ⇒ Q q 1 a + 1 2 a = − q 2 a {:[U=(kQq)/(a)+(kQq)/(sqrt2a)+(kqq)/(a)],[" For "U=0;(Qq)/(a)+(Qq)/(sqrt2a)+(qq)/(a)=0],[=>quad Qq((1)/(a)+(1)/(sqrt2a))=(-q^(2))/(a)]:}\begin{aligned} & \mathrm{U}=\frac{\mathrm{kQq}}{\mathrm{a}}+\frac{\mathrm{kQq}}{\sqrt{2} \mathrm{a}}+\frac{\mathrm{kqq}}{\mathrm{a}} \\ & \text { For } U=0 ; \frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\sqrt{2} \mathrm{a}}+\frac{\mathrm{qq}}{\mathrm{a}}=0 \\ & \Rightarrow \quad Q q\left(\frac{1}{\mathrm{a}}+\frac{1}{\sqrt{2} \mathrm{a}}\right)=\frac{-\mathrm{q}^{2}}{\mathrm{a}} \end{aligned}U=kQqa+kQq2a+kqqa For U=0;Qqa+Qq2a+qqa=0⇒Qq(1a+12a)=−q2a[
\Rightarrow \quad Q\left(\frac{\sqrt{2}+1}{\sqrt{2} \mathrm{a}}\right)=\frac{-\mathrm{q}}{\mathrm{a}} \quad \Rightarrow \quad Q=-q\left(\frac{\sqrt{2}}{\sqrt{2}+1}\right) \quad=-q\left(\frac{2}{2+\sqrt{2}}\right)
]
Illustration:
What is work done by the electrostatic field when we put the four charges together, as shown in the figure. Each side of the square has a length a. Initially charges were at infinity.
Sol. U i = 0 U i = 0 U_(i)=0\mathrm{U}_{\mathrm{i}}=0Ui=0 [Where charges are separated by infinite distance]
U f = 1 4 π ε 0 ( − 4 q 2 a + q 2 2 a + ( − q ) 2 2 a ) [for 6 pairs of charges] U f = 1 4 π ε 0 − 4 q 2 a + q 2 2 a + ( − q ) 2 2 a  [for  6  pairs of charges]  U_(f)=(1)/(4piepsi_(0))((-4q^(2))/(a)+(q^(2))/(sqrt2a)+((-q)^(2))/(sqrt2a))quad" [for "6" pairs of charges] "\mathrm{U}_{\mathrm{f}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{-4 \mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{q}^{2}}{\sqrt{2} \mathrm{a}}+\frac{(-\mathrm{q})^{2}}{\sqrt{2} \mathrm{a}}\right) \quad \text { [for } 6 \text { pairs of charges] }Uf=14πε0(−4q2a+q22a+(−q)22a) [for 6 pairs of charges] 
Work done by field
− Δ U = U i − U f = 1 4 π ε 0 q 2 a ( 4 − 2 a ) − Δ U = U i − U f = 1 4 π ε 0 q 2 a 4 − 2 a -DeltaU=U_(i)-U_(f)=(1)/(4piepsi_(0))(q^(2))/(a)(4-(sqrt2)/(a))-\Delta \mathrm{U}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{a}}\left(4-\frac{\sqrt{2}}{\mathrm{a}}\right)−ΔU=Ui−Uf=14πε0q2a(4−2a)
Illustration :
Two (-ve) charge, each of magnitude q q qqq are situated at 2 r 2 r 2r2 r2r distance apart. A ( + v e ) A ( + v e ) A(+ve)A(+v e)A(+ve) charge q q qqq is lying at the centre between them. The potential energy of the system is U 1 U 1 U_(1)U_{1}U1. If the two nearest charges are mutually interchanged and the potential energy becomes U 2 U 2 U_(2)U_{2}U2, then U 1 U 2 U 1 U 2 (U_(1))/(U_(2))\frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}U1U2 will be:
Sol. U 1 = − k q 2 r + − k q 2 r + k q 2 2 r = − 3 k q 2 2 r U 1 = − k q 2 r + − k q 2 r + k q 2 2 r = − 3 k q 2 2 r quadU_(1)=(-kq^(2))/(r)+(-kq^(2))/(r)+(kq^(2))/(2r)=(-3kq^(2))/(2r)\quad \mathrm{U}_{1}=\frac{-\mathrm{kq}^{2}}{\mathrm{r}}+\frac{-\mathrm{kq}^{2}}{\mathrm{r}}+\frac{\mathrm{kq}^{2}}{2 \mathrm{r}}=\frac{-3 \mathrm{kq}^{2}}{2 \mathrm{r}}U1=−kq2r+−kq2r+kq22r=−3kq22r
U 2 = − k q 2 r + k q 2 r − k q 2 2 r = − k q 2 2 r so U 1 U 2 = 3 U 2 = − k q 2 r + k q 2 r − k q 2 2 r = − k q 2 2 r  so  U 1 U 2 = 3 U_(2)=(-kq^(2))/(r)+(kq^(2))/(r)-(kq^(2))/(2r)=(-kq^(2))/(2r)quad" so "(U_(1))/(U_(2))=3\mathrm{U}_{2}=\frac{-\mathrm{kq}^{2}}{\mathrm{r}}+\frac{\mathrm{kq}^{2}}{\mathrm{r}}-\frac{\mathrm{kq}^{2}}{2 \mathrm{r}}=\frac{-\mathrm{kq}^{2}}{2 \mathrm{r}} \quad \text { so } \frac{\mathrm{U}_{1}}{\mathrm{U}_{2}}=3U2=−kq2r+kq2r−kq22r=−kq22r so U1U2=3
Illustration :
If electric potential energy of given system is positive then prove that 2 Q > 3 q 2 Q > 3 q 2Q > 3q2 Q>3 q2Q>3q
Sol. U ( U ( U(U(U( system ) = ) = )=)=)= Sum of potential energy of all pairs
= k ( Q ) ( 2 Q ) a − k 2 Q q a − k Q q a = k Q a ( 2 Q − 3 q ) = k ( Q ) ( 2 Q ) a − k 2 Q q a − k Q q a = k Q a ( 2 Q − 3 q ) =(k(Q)(2Q))/(a)-(k2Qq)/(a)-(kQq)/(a)=(kQ)/(a)(2Q-3q)=\frac{\mathrm{k}(\mathrm{Q})(2 \mathrm{Q})}{\mathrm{a}}-\frac{\mathrm{k} 2 \mathrm{Qq}}{\mathrm{a}}-\frac{\mathrm{kQq}}{\mathrm{a}}=\frac{\mathrm{kQ}}{\mathrm{a}}(2 \mathrm{Q}-3 \mathrm{q})=k(Q)(2Q)a−k2Qqa−kQqa=kQa(2Q−3q)
Given U U UUU is positive means U > 0 U > 0 U > 0U>0U>0
⇒ k Q a ( 2 Q − 3 q ) > 0 ⇒ 2 Q > 3 q ⇒ k Q a ( 2 Q − 3 q ) > 0 ⇒ 2 Q > 3 q =>(kQ)/(a)(2Q-3q) > 0=>2Q > 3q\Rightarrow \frac{\mathrm{kQ}}{\mathrm{a}}(2 \mathrm{Q}-3 \mathrm{q})>0 \Rightarrow 2 Q>3 q⇒kQa(2Q−3q)>0⇒2Q>3q Illustration :
For the given figure charge q 1 = 2 × 10 − 8 C q 1 = 2 × 10 − 8 C q_(1)=2xx10^(-8)Cq_{1}=2 \times 10^{-8} \mathrm{C}q1=2×10−8C and q 2 = − 0.4 × 10 − 8 C q 2 = − 0.4 × 10 − 8 C q_(2)=-0.4 xx10^(-8)Cq_{2}=-0.4 \times 10^{-8} \mathrm{C}q2=−0.4×10−8C. A charge q 3 q 3 q_(3)q_{3}q3 of 0.2 × 10 − 8 C 0.2 × 10 − 8 C 0.2 xx10^(-8)C0.2 \times 10^{-8} \mathrm{C}0.2×10−8C is taken along arc of circle from C C CCC to D D DDD then potential energy of charge q 3 q 3 q_(3)q_{3}q3 is decreased by ----- percent
Sol. EPE of q 3 q 3 q_(3)q_{3}q3 at C = q 3 ( k q 1 0.8 + k q 2 1 ) C = q 3 k q 1 0.8 + k q 2 1 C=q_(3)((kq_(1))/(0.8)+(kq_(2))/(1))\mathrm{C}=\mathrm{q}_{3}\left(\frac{\mathrm{kq}_{1}}{0.8}+\frac{\mathrm{kq}_{2}}{1}\right)C=q3(kq10.8+kq21)
E P E E P E EPEE P EEPE of q 3 q 3 q_(3)q_{3}q3 at D = q 3 ( k q 1 0.8 + k q 2 0.2 ) D = q 3 k q 1 0.8 + k q 2 0.2 D=q_(3)((kq_(1))/(0.8)+(kq_(2))/(0.2))\mathrm{D}=\mathrm{q}_{3}\left(\frac{\mathrm{kq}_{1}}{0.8}+\frac{\mathrm{kq}_{2}}{0.2}\right)D=q3(kq10.8+kq20.2)
or EPE at C C CCC and D D DDD are 3.78 × 10 − 7 J 3.78 × 10 − 7 J 3.78 xx10^(-7)J3.78 \times 10^{-7} J3.78×10−7J and 0.9 × 10 − 7 J 0.9 × 10 − 7 J 0.9 xx10^(-7)J0.9 \times 10^{-7} J0.9×10−7J.
% % %\%% decrement = 2.88 3.78 × 100 = 76 % = 2.88 3.78 × 100 = 76 % =(2.88)/(3.78)xx100=76%=\frac{2.88}{3.78} \times 100=76 \%=2.883.78×100=76% (decreases, 76%)
MOTION OFA CHARGED PARTICLE INAN ELECTRIC FIELD :
(i) Charged particle experience force in an electric field.
(ii) Magnitude of force on a charge q q qqq in an electric field E E EEE is F = q E F = q E F=qEF=q EF=qE
(iii) Direction of force on a positive charge is same as direction of electric field while it is opposite to direction of electric field in case of negative charge.
Illustration :
A particle having a charge of 1.6 × 10 − 19 C 1.6 × 10 − 19 C 1.6 xx10^(-19)C1.6 \times 10^{-19} \mathrm{C}1.6×10−19C enters midway between the plates of a parallel plate capacitor. The initial velocity of particle is parallel to the plates. A potential difference of 300 volts is applied to the capacitor plates. If the length of the capacitor plate is 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm and they are separated by 2 c m 2 c m 2cm2 \mathrm{~cm}2 cm. Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of particle is 12 × 10 − 24 k g 12 × 10 − 24 k g 12 xx10^(-24)kg12 \times 10^{-24} \mathrm{~kg}12×10−24 kg.
Sol. The situation is shown in fig .
Here E = Potential difference d E =  Potential difference  d E=(" Potential difference ")/(d)E=\frac{\text { Potential difference }}{\mathrm{d}}E= Potential difference d
= 300 2 / 100 = 15000 V m = 300 2 / 100 = 15000 V m =(300)/(2//100)=15000(V)/(m)=\frac{300}{2 / 100}=15000 \frac{\mathrm{V}}{\mathrm{m}}=3002/100=15000Vm
As the particle does not come, its maximum deflection y = 1 c m = 10 − 2 m y = 1 c m = 10 − 2 m y=1cm=10^(-2)my=1 \mathrm{~cm}=10^{-2} \mathrm{~m}y=1 cm=10−2 m
We know that y = 1 2 ⋅ q E m ( x u ) 2 y = 1 2 ⋅ q E m x u 2 y=(1)/(2)*(qE)/(m)((x)/(u))^(2)\mathrm{y}=\frac{1}{2} \cdot \frac{\mathrm{qE}}{\mathrm{m}}\left(\frac{\mathrm{x}}{\mathrm{u}}\right)^{2}y=12⋅qEm(xu)2
u 2 = 1 2 ⋅ q E m y ⋅ x 2 1 2 ( 1.6 × 10 − 19 ) ( 15000 ) ( 12 × 10 − 24 ) ( 10 − 2 ) ( 1 10 ) 2 = 10 8 ∴ u = 10 4 m / s u 2 = 1 2 ⋅ q E m y ⋅ x 2 1 2 1.6 × 10 − 19 ( 15000 ) 12 × 10 − 24 10 − 2 1 10 2 = 10 8 ∴ u = 10 4 m / s {:[u^(2)=(1)/(2)*(qE)/(my)*x^(2)],[(1)/(2)((1.6 xx10^(-19))(15000))/((12 xx10^(-24))(10^(-2)))((1)/(10))^(2)=10^(8)],[:.u=10^(4)m//s]:}\begin{aligned} & \mathrm{u}^{2}=\frac{1}{2} \cdot \frac{\mathrm{qE}}{\mathrm{my}} \cdot \mathrm{x}^{2} \\ & \frac{1}{2} \frac{\left(1.6 \times 10^{-19}\right)(15000)}{\left(12 \times 10^{-24}\right)\left(10^{-2}\right)}\left(\frac{1}{10}\right)^{2}=10^{8} \\ & \therefore \mathrm{u}=10^{4} \mathrm{~m} / \mathrm{s} \end{aligned}u2=12⋅qEmy⋅x212(1.6×10−19)(15000)(12×10−24)(10−2)(110)2=108∴u=104 m/s
When an electron enters normally to electric field, it's path becomes parabolic.

25. ELECTROSTATICS

26. ELECTRIC FLUX :

⇒ ⇒ =>quad\Rightarrow \quad⇒ It is denoted by ' ϕ ϕ phi\phiϕ '.
⇒ ⇒ =>quad\Rightarrow \quad⇒ It is a scalar quantity.
⇒ ⇒ =>quad\Rightarrow \quad⇒ It is defined as the total number of lines of force passing normally through a curved surface placed in the field.
⇒ ⇒ =>quad\Rightarrow \quad⇒ It is given by the dot product of E → E → vec(E)\vec{E}E→ and normal infinitesimal area d s → d s → vec(ds)\overrightarrow{d s}ds→ integrated over a closed surface.
d ϕ = E → ⋅ d s → ϕ = ∮ E → ⋅ d s → = ∮ E d s cos ⁡ θ ( ∮ = Close integral ) d ϕ = E → ⋅ d s → ϕ = ∮ E → ⋅ d s → = ∮ E d s cos ⁡ θ ( ∮ =  Close integral  ) {:[dphi= vec(E)* vec(ds)],[phi=oint vec(E)* vec(ds)=ointEdscos thetaquad(oint=" Close integral ")]:}\begin{aligned} & \mathrm{d} \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}} \\ & \phi=\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\oint \mathrm{Eds} \cos \theta \quad(\oint=\text { Close integral }) \end{aligned}dϕ=E→⋅ds→ϕ=∮E→⋅ds→=∮Edscos⁡θ(∮= Close integral )
where θ = θ = theta=\theta=θ= angle between electric field and normal to the area
(a)
(b)
⇒ ⇒ =>quad\Rightarrow \quad⇒ (a) if θ = 0 ∘ , ϕ = Eds θ = 0 ∘ , ϕ = Eds theta=0^(@),phi=Eds\theta=0^{\circ}, \phi=\operatorname{Eds}θ=0∘,ϕ=Eds (maximum)
(b) if θ = 90 0 , ϕ = θ = 90 0 , ϕ = theta=90^(0,)phi=\theta=90^{0,} \phi=θ=900,ϕ= zero
⇒ ⇒ =>quad\Rightarrow \quad⇒ Unit : (a) Newton - metre 2 / 2 / ^(2)//^{2} /2/ coulomb
(b) Volt - meter
⇒ ⇒ =>quad\Rightarrow \quad⇒ Dimension : [ M L 3 T − 3 A − 1 ] M L 3 T − 3 A − 1 [ML^(3)T^(-3)A^(-1)]\left[\mathrm{M} \mathrm{L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right][ML3 T−3 A−1]
⇒ ⇒ =>quad\Rightarrow \quad⇒ Flux due to a positive charge goes out of the surface while that due to negative charge comes into the surface.
⇒ ⇒ =>quad\Rightarrow \quad⇒ Value of electric flux is independent of shape and size of the surface.
⇒ ⇒ =>quad\Rightarrow \quad⇒ If only a dipole is present in the surface then net flux is zero.
⇒ ⇒ =>quad\Rightarrow \quad⇒ Net flux of a surface kept in a uniform electric field is zero.
⇒ ⇒ =>quad\Rightarrow \quad⇒ Net flux from a surface is zero does not imply that intensity of electric field is also zero.

27. GAUSS'S LAW :

This law states that electric flux ϕ E ϕ E phi_(E)\phi_{\mathrm{E}}ϕE through any closed surface is equal to 1 / ε 0 1 / ε 0 1//epsi_(0)1 / \varepsilon_{0}1/ε0 times the net charge ' q q qqq ' enclosed by the surface i.e
ϕ E = ∮ E → ⋅ d s → = q ε 0 ϕ E = ∮ E → ⋅ d s → = q ε 0 phi_(E)=oint vec(E)* vec(ds)=(q)/(epsi_(0))\phi_{\mathrm{E}}=\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{\mathrm{q}}{\varepsilon_{0}}ϕE=∮E→⋅ds→=qε0
The closed surface can be hypothetical and then it is called a Gaussian surface. If the closed surface enclosed a number of charges q 1 , q 2 … … … . q n q 1 , q 2 … … … . q n q_(1),q_(2)dots dots dots.q_Noq_{1}, q_{2} \ldots \ldots \ldots . q_{n}q1,q2……….qn etc., then
ϕ = ∮ E → ⋅ d s → = Σ q ε 0 = ( q 1 + q 2 + … ⋅ q n ) ε 0 ϕ = ∮ E → ⋅ d s → = Σ q ε 0 = q 1 + q 2 + … ⋅ q n ε 0 phi=oint vec(E)* vec(ds)=(Sigma q)/(epsi_(0))=((q_(1)+q_(2)+dots*q_No))/(epsi_(0))\phi=\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{\Sigma q}{\varepsilon_{0}}=\frac{\left(q_{1}+q_{2}+\ldots \cdot q_{n}\right)}{\varepsilon_{0}}ϕ=∮E→⋅ds→=Σqε0=(q1+q2+…⋅qn)ε0

28. Flux is :

⇒ ⇒ =>quad\Rightarrow \quad⇒ Independent of distances between charges inside the surface and their distribution.
⇒ ⇒ =>quad\Rightarrow \quad⇒ Independent of shape , size and nature of surface.
⇒ ⇒ =>quad\Rightarrow \quad⇒ Dependent on charges enclosed by surface, their nature and on the medium.
⇒ ⇒ =>quad\Rightarrow \quad⇒ Net flux due to a charge outside the surface will be zero.
⇒ ⇒ =>quad\Rightarrow \quad⇒ If Σ Q = 0 Σ Q = 0 Sigma Q=0\Sigma Q=0ΣQ=0, then ϕ = 0 ϕ = 0 phi=0\phi=0ϕ=0 but it is not necessary that E = 0 E = 0 E=0E=0E=0
(i) A charge q q qqq is placed at the centre of a cube, then
(a) Total flux through cube = q ε 0 = q ε 0 =(q)/(epsi_(0))=\frac{q}{\varepsilon_{0}}=qε0
(b) Flux through each surface = q 6 ε 0 = q 6 ε 0 =(q)/(6epsi_(0))=\frac{q}{6 \varepsilon_{0}}=q6ε0
(ii) A charge q q qqq is placed at the centre of a face of a cube , then total flux through cube
= q 2 ε 0 = q 2 ε 0 =(q)/(2epsi_(0))=\frac{q}{2 \varepsilon_{0}}=q2ε0
Assume ...... A second cube can be assumed adjacent to the first cube, total flux through both cubes = q ε 0 = q ε 0 =(q)/(epsi_(0))=\frac{q}{\varepsilon_{0}}=qε0, So flux through each cube = q 2 ε 0 = q 2 ε 0 =(q)/(2epsi_(0))=\frac{q}{2 \varepsilon_{0}}=q2ε0
(iii) Now, q q qqq is placed at a corner then the flux will be q 8 ε 0 q 8 ε 0 (q)/(8epsi_(0))\frac{q}{8 \varepsilon_{0}}q8ε0 Illustration :
A hemispherical surface of radius R R RRR is kept in a uniform electric field E such that E E EEE is parallel to the axis of hemi-sphere, Net flux from the surface will be -
Sol. ϕ = ∮ E → ⋅ d s → = E ⋅ π R 2 ϕ = ∮ E → ⋅ d s → = E ⋅ π R 2 quad phi=oint vec(E)* vec(ds)=E*piR^(2)\quad \phi=\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=E \cdot \pi R^{2}ϕ=∮E→⋅ds→=E⋅πR2.
= ( E ) = ( E ) =(E)=(E)=(E) (Area of surface perpendicular to E)
= E ⋅ π R 2 = E ⋅ π R 2 =E*piR^(2)=E \cdot \pi R^{2}=E⋅πR2
Illustration :
A rectangular surface of length 4 m 4 m 4m4 \mathrm{~m}4 m and breadth 2 m 2 m 2m2 \mathrm{~m}2 m is kept in an electric field of 20 N / c 20 N / c 20N//c20 \mathrm{~N} / \mathrm{c}20 N/c. Angle between the surface and electric field is 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘. What is flux thought this surface?
Angle between surface and E → E → vec(E)\overrightarrow{\mathrm{E}}E→ is given to be 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘. This is not the ' θ θ theta\thetaθ ' used in our formula, ' θ θ theta\thetaθ ' is the angle between normal to surface and E → E → vec(E)\overrightarrow{\mathrm{E}}E→. So here θ = 90 − 30 = 60 ∘ θ = 90 − 30 = 60 ∘ theta=90-30=60^(@)\theta=90-30=60^{\circ}θ=90−30=60∘
Sol. ϕ = E A cos ⁡ θ = 20 × 8 cos ⁡ 60 = 80 V − m ϕ = E A cos ⁡ θ = 20 × 8 cos ⁡ 60 = 80 V − m phi=EA cos theta=20 xx8cos 60=80V-m\phi=E A \cos \theta=20 \times 8 \cos 60=80 \mathrm{~V}-\mathrm{m}ϕ=EAcos⁡θ=20×8cos⁡60=80 V−m
Ex. Flux entering a closed surface is 2000 V 2000 V 2000V2000 \mathrm{~V}2000 V-m. Flux leaving that surface is 6000 V 6000 V 6000V6000 \mathrm{~V}6000 V-m. Find the charge inside surface.
Sol. Net flux = ϕ out − ϕ in = ϕ out  − ϕ in  =phi_("out ")-phi_("in ")=\phi_{\text {out }}-\phi_{\text {in }}=ϕout −ϕin 
ϕ = ( 6000 − 2000 ) = 4000 V − m ϕ = ( 6000 − 2000 ) = 4000 V − m phi=(6000-2000)=4000 V-m\phi=(6000-2000)=4000 V-mϕ=(6000−2000)=4000V−m
ϕ = q ε 0 q = ( 4000 ) ( 8.85 × 10 − 12 ) = 0.36 μ c ϕ = q ε 0 q = ( 4000 ) 8.85 × 10 − 12 = 0.36 μ c phi=(q)/(epsi_(0))quad q=(4000)(8.85 xx10^(-12))=0.36 mu c\phi=\frac{\mathrm{q}}{\varepsilon_{0}} \quad q=(4000)\left(8.85 \times 10^{-12}\right)=0.36 \mu cϕ=qε0q=(4000)(8.85×10−12)=0.36μc
Ex. A charge Q Q QQQ is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and the Gauss's law, find the flux of the electric field due to this charge through the surface of the hemisphere
Sol. Let us imagine another hemispherical surface over identical given one.
Both being symmetric with respect to Q Q QQQ, hence flux will be same through both the hemisphere ( ϕ 1 = ϕ 2 ) ϕ 1 = ϕ 2 (phi_(1)=phi_(2))\left(\phi_{1}=\phi_{2}\right)(ϕ1=ϕ2).
Net flux ϕ 1 + ϕ 2 = q n c ϵ 0 = Q ϵ 0 ϕ 1 + ϕ 2 = q n c ϵ 0 = Q ϵ 0 quadphi_(1)+phi_(2)=(q_(nc))/(epsilon_(0))=(Q)/(epsilon_(0))\quad \phi_{1}+\phi_{2}=\frac{\mathrm{q}_{\mathrm{nc}}}{\epsilon_{0}}=\frac{\mathrm{Q}}{\epsilon_{0}}ϕ1+ϕ2=qncϵ0=Qϵ0
⇒ ϕ 1 = Q 2 ∈ 0 ⇒ ϕ 1 = Q 2 ∈ 0 =>quadphi_(1)=(Q)/(2in_(0))\Rightarrow \quad \phi_{1}=\frac{\mathrm{Q}}{2 \in_{0}}⇒ϕ1=Q2∈0

29. APPLICATION OF GAUSS'S LAW :

Electric field due to a charged conducting sphere/ Hollow conducting or insulating sphere.
Case: 1 OP = r ≥ R = r ≥ R =r >= R=r \geq R=r≥R
E → = k q r 2 r ^ ( σ = surface charge density ) E → = k q r 2 r ^ ( σ =  surface charge density  ) {:[ vec(E)=(kq)/(r^(2)) hat(r)],[(sigma=" surface charge density ")]:}\begin{aligned} \vec{E} & =\frac{k q}{r^{2}} \hat{r} \\ (\sigma & =\text { surface charge density }) \end{aligned}E→=kqr2r^(σ= surface charge density )
Case: 2 r = R 2 r = R 2quadr=R2 \quad \mathrm{r}=\mathrm{R}2r=R
E → = k q R 2 r ^ E → = k q R 2 r ^ vec(E)=(kq)/(R^(2)) hat(r)\overrightarrow{\mathrm{E}}=\frac{\mathrm{kq}}{\mathrm{R}^{2}} \hat{\mathrm{r}}E→=kqR2r^
Case: 3 r < R 3 r < R 3quad r < R3 \quad r<R3r<R
E → = 0 E → = 0 vec(E)=0\overrightarrow{\mathrm{E}}=0E→=0

30. Graph :

31. Electric potential :

Case : 1 r > R 1 r > R 1r > R1 \mathrm{r}>\mathrm{R}1r>R
V out = 1 4 π ε 0 Q r V out  = 1 4 π ε 0 Q r V_("out ")=(1)/(4piepsi_(0))(Q)/(r)\mathrm{V}_{\text {out }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{r}}Vout =14πε0Qr
Case: 2 r = R 2 r = R 2r=R2 \mathrm{r}=\mathrm{R}2r=R
V surface = 1 4 π ε 0 Q R V surface  = 1 4 π ε 0 Q R V_("surface ")=(1)/(4piepsi_(0))(Q)/(R)\mathrm{V}_{\text {surface }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}}Vsurface =14πε0QR
Case: 3 r < R 3 r < R 3quad r < R3 \quad r<R3r<R
V in = 1 4 π ε 0 Q R V in  = 1 4 π ε 0 Q R V_("in ")=(1)/(4piepsi_(0))(Q)/(R)\mathrm{V}_{\text {in }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}}Vin =14πε0QR

32. Graph :

33. Electric field due to solid insulating sphere :

A charge given to a solid insulating sphere is distributed equally throughout its volume.

34. Electric Field:

Case: 1 r > R 1 r > R 1r > R1 \mathrm{r}>\mathrm{R}1r>R (point is outside the sphere)
E → = 1 4 π ε 0 Q r 2 r ^ E → = 1 4 π ε 0 Q r 2 r ^ vec(E)quad=(1)/(4piepsi_(0))(Q)/(r^(2)) hat(r)\overrightarrow{\mathrm{E}} \quad=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{r}^{2}} \hat{r}E→=14πε0Qr2r^
Case: 2 r = R 2 r = R 2r=R2 \mathrm{r}=\mathrm{R}2r=R (point is at the surface)
E → = 1 4 π ε 0 Q R 2 r ^ = E max = E surface E → = 1 4 π ε 0 Q R 2 r ^ = E max = E surface  vec(E)=(1)/(4piepsi_(0))(Q)/(R^(2)) hat(r)=E_(max)=E_("surface ")\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}^{2}} \hat{\mathrm{r}}=\mathrm{E}_{\max }=\mathrm{E}_{\text {surface }}E→=14πε0QR2r^=Emax=Esurface 
Case: 3 r < R 3 r < R 3r < R3 \mathrm{r}<\mathrm{R}3r<R (point is inside the sphere)
E → = 1 4 π ε 0 Q R 3 r r ^ = k Q R 3 r r ^ E i n ∝ r E → = 1 4 π ε 0 Q R 3 r r ^ = k Q R 3 r r ^ E i n ∝ r {:[ vec(E)=(1)/(4piepsi_(0))(Q)/(R^(3))r hat(r)=(kQ)/(R^(3))r hat(r)],[E_(in)propr]:}\begin{aligned} & \overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}^{3}} \mathrm{r} \hat{r}=\frac{\mathrm{kQ}}{\mathrm{R}^{3}} \mathrm{r} \hat{\mathrm{r}} \\ & \mathrm{E}_{\mathrm{in}} \propto \mathrm{r} \end{aligned}E→=14πε0QR3rr^=kQR3rr^Ein∝r
at r = 0 , E = 0 r = 0 , E = 0 r=0,E=0\mathrm{r}=0, \mathrm{E}=0r=0,E=0

35. Graph :

36. Electric Potential due to solid insulating sphere :

Case: 1 r > R 1 r > R 1quadr > R1 \quad \mathrm{r}>\mathrm{R}1r>R
V out = 1 4 π ε 0 Q r V out  = 1 4 π ε 0 Q r V_("out ")=(1)/(4piepsi_(0))(Q)/(r)\mathrm{V}_{\text {out }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{r}}Vout =14πε0Qr
Case : 2 r = R 2 r = R 2quad r=R2 \quad r=R2r=R
V surface = 1 4 π ε 0 Q R V surface  = 1 4 π ε 0 Q R V_("surface ")=(1)/(4piepsi_(0))(Q)/(R)\mathrm{V}_{\text {surface }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}}Vsurface =14πε0QR
Case : 3 r < R 3 r < R 3quad r < R3 \quad r<R3r<R
V in = 1 4 π ε 0 Q ( 3 R 2 − r 2 ) 2 R 3 V centre = 3 2 1 4 π ε 0 Q R ( Imp) V centre = 3 / 2 V surface V in  = 1 4 π ε 0 Q 3 R 2 − r 2 2 R 3 V centre  = 3 2 1 4 π ε 0 Q R (  Imp)  V centre  = 3 / 2 V surface  {:[V_("in ")=(1)/(4piepsi_(0))(Q(3R^(2)-r^(2)))/(2R^(3))],[V_("centre ")=(3)/(2)(1)/(4piepsi_(0))(Q)/(R)quad(" Imp) "],[V_("centre ")=3//2V_("surface ")]:}\begin{aligned} & \mathrm{V}_{\text {in }}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}\left(3 \mathrm{R}^{2}-\mathrm{r}^{2}\right)}{2 \mathrm{R}^{3}} \\ & \mathrm{~V}_{\text {centre }}=\frac{3}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}} \quad(\text { Imp) } \\ & \mathrm{V}_{\text {centre }}=3 / 2 \mathrm{~V}_{\text {surface }} \end{aligned}Vin =14πε0Q(3R2−r2)2R3 Vcentre =3214πε0QR( Imp) Vcentre =3/2 Vsurface 
Graph :
Electric field due to infinitely long charge :
(i) E E EEE at point p p ppp
E → = λ 2 π ε 0 r r ^ or E = λ 2 π ε 0 r E → = λ 2 π ε 0 r r ^  or  E = λ 2 π ε 0 r vec(E)=(lambda)/(2piepsi_(0)r) hat(r)quad" or "quad E=(lambda)/(2piepsi_(0)r)\vec{E}=\frac{\lambda}{2 \pi \varepsilon_{0} r} \hat{r} \quad \text { or } \quad E=\frac{\lambda}{2 \pi \varepsilon_{0} r}E→=λ2πε0rr^ or E=λ2πε0r
(ii)
Electric field at a point due to an infinite sheet of charge :
(i) If σ = σ = sigma=\sigma=σ= surface charge density. Intensity at points near to the sheet = E → = σ 2 ε 0 r ^ = E → = σ 2 ε 0 r ^ = vec(E)=(sigma)/(2epsi_(0)) hat(r)=\vec{E}=\frac{\sigma}{2 \varepsilon_{0}} \hat{r}=E→=σ2ε0r^
(ii) Direction of electric field is perpendicular to the sheet of charge.
(iii) Intensity of electric field does not depend upon the distance of points from the sheet for the points in front of sheet i.e.. There is an equipotential region near the charged sheet.

37. Electric field due to infinite charged metal sheet :

(i) Intensity at points near the sheet
E → = σ ε 0 r ^ E → = σ ε 0 r ^ vec(E)=(sigma)/(epsi_(0)) hat(r)\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_{0}} \hat{\mathrm{r}}E→=σε0r^
where
σ = σ = sigma=\sigma=σ= surface charge density
Electric field due to charged ring: Q Q QQQ charge is distributed over a ring of radius R R RRR :
(i) Intensity of electric field at a distance x x xxx from the centre of ring along it's axis -
E = 1 4 π ε 0 Q x ( R 2 + x 2 ) 3 / 2 E = 1 4 π ε 0 Q cos ⁡ θ r 2 and it's direction will be along the axis of the ring E = 1 4 π ε 0 Q x R 2 + x 2 3 / 2 E = 1 4 π ε 0 Q cos ⁡ θ r 2  and it's direction will be along the axis of the ring  {:[E=(1)/(4piepsi_(0))(Qx)/((R^(2)+x^(2))^(3//2))],[E=(1)/(4piepsi_(0))(Q cos theta)/(r^(2))" and it's direction will be along the axis of the ring "]:}\begin{aligned} & E=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q x}{\left(R^{2}+x^{2}\right)^{3 / 2}} \\ & E=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q \cos \theta}{r^{2}} \text { and it's direction will be along the axis of the ring } \end{aligned}E=14πε0Qx(R2+x2)3/2E=14πε0Qcos⁡θr2 and it's direction will be along the axis of the ring 
(ii) Intensity will be maximum at a distance R / 2 R / 2 R//sqrt2R / \sqrt{2}R/2 from the centre and
E max = 2 3 3 ⋅ 1 4 π ε 0 ⋅ Q R 2 E max = 2 3 3 ⋅ 1 4 π ε 0 ⋅ Q R 2 E_(max)quad=(2)/(3sqrt3)*(1)/(4piepsi_(0))*(Q)/(R^(2))\mathrm{E}_{\max } \quad=\frac{2}{3 \sqrt{3}} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Q}}{\mathrm{R}^{2}}Emax=233⋅14πε0⋅QR2
(iii) Electric potential at a distance x x x\mathrm{x}x from centre, V = 1 4 π ε 0 Q ( x 2 + R 2 ) V = 1 4 π ε 0 Q x 2 + R 2 V=(1)/(4piepsi_(0))(Q)/(sqrt((x^(2)+R^(2))))\mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\sqrt{\left(\mathrm{x}^{2}+\mathrm{R}^{2}\right)}}V=14πε0Q(x2+R2)

38. CONDUCTORS :

(a) Inside a conductor, electrostatic field is zero : In the previous chapter, we have already discussed that "when there is no electric current inside or on the surface of a conductor, the electric field inside the conductor is everywhere zero".
(b) At the surface of a charged conductor, electrostatic field must be normal to the surface at every point : If the field E → E → vec(E)\overrightarrow{\mathrm{E}}E→ is not normal tothe surface, it will have a nonzero component along the surface. Hence the free charge on the surface will move due to electrostatic force on it. But free charge on the surface in electrostatics remains at rest. So the electrostatic field at the surface of a charged conductor must be normal to the surface.
(c) The interior of a conductor can have no excess charge in the static situation : in the previous chapter we have already discussed it.
(d) Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface : Since the field E = 0 E = 0 E=0E=0E=0 inside the conductor and its has no nonzero component along the surface, so the potential gradient d V d r = 0 d V d r = 0 (dV)/(dr)=0\frac{\mathrm{dV}}{\mathrm{dr}}=0dVdr=0 because − d V d r − d V d r -(dV)/(dr)-\frac{\mathrm{dV}}{\mathrm{dr}}−dVdr is equal tot he component of the field along dr. If we take dr along the surface or inside the surface, then − d V d r = 0 − d V d r = 0 -(dV)/(dr)=0-\frac{\mathrm{dV}}{\mathrm{dr}}=0−dVdr=0
⇒ V ⇒ V =>V\Rightarrow \mathrm{V}⇒V is constant (e) Electric field at the surface of a charged conductor : E → = σ ε 0 n ^ E → = σ ε 0 n ^ vec(E)=(sigma)/(epsi_(0)) hatNo\vec{E}=\frac{\sigma}{\varepsilon_{0}} \hat{\mathrm{n}}E→=σε0n^ where σ σ sigma\sigmaσ is the surface charge density which may be positive or negative and n ^ n ^ hatNo\hat{\mathrm{n}}n^ is the unit vector normal to the surface in the outward direction.
(f) Electrostatic Shielding : In an electrostatic situation, if a conductor contains a cavity and if no charge is present inside the cavity, then there can be no net charge anywhere on the surface of the cavity. This means without being electrocuted. This is known as electrostatic shielding. Remember the following facts.
(i) There is no net charge anywhere on the surface of the cavity. i.e., σ = 0 σ = 0 sigma=0\sigma=0σ=0
(ii) The potential inside (or on the surface of) the cavity
= = === The potential inside the material of the conductor
= = === The potential on the surface of the conductor = V 0 = V 0 =V_(0)=\mathrm{V}_{0}=V0 (constant)
(iii) The electric field in (or on the surface of) the cavity
= = === The electric field inside material of the conductor
= = === zero
But the electric field on the surface of a charged conductor is not zero.
[Figure shows an uncharged conductor in an external electric field E ∝ 0 E ∝ 0 Eprop0\mathrm{E} \propto 0E∝0. The free electrons in the conductor distribute themselves on the surfaces as shown in the figure so that the net electric field inside the conductor becomes zero and the net field at the surface is perpendicular to the surface.]
(iv) Except for spherical surfaces, the charge is not distributed uniformly on the surface of a conductor. At the sharp poitns or edges, the surface charge density ( σ ) ( σ ) (sigma)(\sigma)(σ) is very high and hence the electric field ( E = σ ε 0 ) E = σ ε 0 (E=(sigma)/(epsi_(0)))\left(\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}\right)(E=σε0) becomes very strong. The air around such sharp points may become ionised, producing the corona discharge in which the charge jumps from th conductor to the air because of the dielectric breakdown of the air.
Figure shows a charged conductor with nonuniform distribution of charge. σ 1 σ 1 sigma_(1)\sigma_{1}σ1 and σ 2 σ 2 sigma_(2)\sigma_{2}σ2 are the surface charge densities at the portions where radii of curvature are r 1 r 1 r_(1)r_{1}r1 and r 2 r 2 r_(2)r_{2}r2 then
σ 1 σ 2 = r 2 r 1 E 2 = σ 2 ε 0 σ 1 σ 2 = r 2 r 1 E 2 = σ 2 ε 0 (sigma_(1))/(sigma_(2))=(r_(2))/(r_(1))quadE_(2)=(sigma_(2))/(epsi_(0))\frac{\sigma_{1}}{\sigma_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}} \quad \mathrm{E}_{2}=\frac{\sigma_{2}}{\varepsilon_{0}}σ1σ2=r2r1E2=σ2ε0
(a) The corona discharge from the pointed end of the conductor at the top of a building protect the building from lightning strikes.

39. ELECTROSTATICS

(b) It is electrostatic shielding that protects a person from lightning strikes if he is in a car.
(c) Figure (a) shows an uncharged conducting plate placed in a region without electric field.
Figure (b) shows an uncharged conducting plate placed in a region having electric field E. Due to electric force on the free electrons in the conductor, they redistribute themselves till the electric field produced due to the redistribution neutralizes the original field. Hence the net field inside the conductor becomes zero.
(a)
(b)
(d) Figure shows a metallic slab between two charged plate. The field E E E\mathrm{E}E due to the charged plate is directed toward right and the field E E E\mathrm{E}E due to the induced charge in the slab is directed towards left, and hence the net field inside the slab becomes zero.
(e) Figure (a) shows a conducting shell with charge q.
In figure (b), the conducting shell with charge q q qqq has been surrounded by another larger conducting shell which is uncharged. Charges − q − q -q-\mathrm{q}−q and + q + q +q+\mathrm{q}+q are induced on the its inner and outer surfaces but net charge in the outer shell is stell zero.
In the figure (c), the outer shell of figure (b) is earthed. The free charge on the outer surface goes to the earth but the inner charge remains bounded to the charge on the inner shell so that the potential of the outer shell connected to the earth becomes zero.
[Remember that an earthed conductor is always at zero potential due to flow of charge either from the earth to the conductor or from the conductor to the earth]
(a)
(b)
( c ) ( c ) (c)(\mathrm{c})(c) (f) Figure (a), shows two concentric conducing shells. Some charge q 1 q 1 q_(1)\mathrm{q}_{1}q1 is given to the outer shell. No charge is developed on the inner shell.
(a)
(b)
In the figure (b), the inner shell is earthed and hence some charge q 2 q 2 q_(2)\mathrm{q}_{2}q2 is developed on it so that its potential becomes zero.
On the surface of the inner shell, the net potential is 1 4 π ε 0 ( q 2 R 2 + q 1 R 1 ) = 0 1 4 π ε 0 q 2 R 2 + q 1 R 1 = 0 (1)/(4piepsi_(0))((q_(2))/(R_(2))+(q_(1))/(R_(1)))=0\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{2}}{\mathrm{R}_{2}}+\frac{\mathrm{q}_{1}}{\mathrm{R}_{1}}\right)=014πε0(q2R2+q1R1)=0
or, q 2 = − q 1 R 2 R 1 q 2 = − q 1 R 2 R 1 q_(2)=-q_(1)(R_(2))/(R_(1))\mathrm{q}_{2}=-\mathrm{q}_{1} \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}q2=−q1R2R1
(g) Figure (a) is an uncharged metallic solid sphere of radius R.
(a)
(b)
(c)
In figure (b), a positive point charge Q Q QQQ has been placed at a distance r r rrr from the centre of the sphere. The point charge Q Q Q\mathrm{Q}Q exerts force on the electrons in the sphere and hence the free electrons redistribute themselves so that the left half is negatively charged and the right half is positively charge. The charge distribution is non-uniform.
At the centre of the sphere, the field due to the point charge Q Q QQQ is E P = 1 4 π ε 0 Q r 2 E P = 1 4 π ε 0 Q r 2 E_(P)=(1)/(4piepsi_(0))(Q)/(r^(2))E_{P}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}EP=14πε0Qr2, toward right. Since the field inside the conductor is zero, the field due the induced charge on the sphere is equal in magnitude of E P E P E_(P)E_{P}EP but opposite in direction.
i.e., E i = Q 4 π ε 0 r 2 E i = Q 4 π ε 0 r 2 E_(i)=(Q)/(4piepsi_(0)r^(2))\mathrm{E}_{\mathrm{i}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}Ei=Q4πε0r2
or,
The potential at any point of the sphere = = === potential at its centre = 1 4 π ε 0 Q r = 1 4 π ε 0 Q r =(1)/(4piepsi_(0))(Q)/(r)=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{r}}=14πε0Qr
In the figure (c), the sphere is earthed, so V = 0 V = 0 V=0V=0V=0 hence
1 4 π ε 0 ( Q r + Q i R ) = 0 1 4 π ε 0 Q r + Q i R = 0 (1)/(4piepsi_(0))((Q)/(r)+(Q_(i))/(R))=0\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{Q}}{\mathrm{r}}+\frac{\mathrm{Q}_{\mathrm{i}}}{\mathrm{R}}\right)=014πε0(Qr+QiR)=0 or, Q i = − Q R / r Q i = − Q R / r Q_(i)=-QR//r\mathrm{Q}_{\mathrm{i}}=-\mathrm{QR} / \mathrm{r}Qi=−QR/r
That the induced charge Q i Q i Q_(i)Q_{i}Qi is non-uniformly distributed on the surface of the sphere

40. ELECTRIC DIPOLE :

(i) A system consisting of two equal and opposite charges separated by a small distance is termed an electric dipole.
Example : N a + C l − , H + C l N a + C l − , H + C l Na^(+)Cl^(-),H^(+)Cl\mathrm{Na}^{+} \mathrm{Cl}^{-}, \mathrm{H}^{+} \mathrm{Cl}Na+Cl−,H+Cl - etc.
(ii) DIPOLE MOMENT : The product of the magnitude of charges and distance between them is called the dipole moment.
(a) This is a vector quantity which is directed from negative to positive charge.
(b) Unit : Coulomb - metre (C-M)
(c) Dimension : [ M 0 L 1 T 1 A 1 ] M 0 L 1 T 1 A 1 [M^(0)L^(1)T^(1)A^(1)]\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1} \mathrm{~A}^{1}\right][M0 L1 T1 A1]
(d) It is denoted by p → p → vec(p)\vec{p}p→ that is p → = q d → p → = q d → vec(p)=q vec(d)\vec{p}=q \vec{d}p→=qd→
Electric field at any point due to dipole :
E = E r 2 + E θ 2 = P 4 π ε 0 r 3 1 + 3 cos 2 ⁡ θ E = E r 2 + E θ 2 = P 4 π ε 0 r 3 1 + 3 cos 2 ⁡ θ E=sqrt(E_(r)^(2)+E_(theta)^(2))=(P)/(4piepsi_(0)r^(3))sqrt(1+3cos^(2)theta)E=\sqrt{E_{r}^{2}+E_{\theta}^{2}}=\frac{P}{4 \pi \varepsilon_{0} r^{3}} \sqrt{1+3 \cos ^{2} \theta}E=Er2+Eθ2=P4πε0r31+3cos2⁡θ
(ii) Angle between the resultant E → E → vec(E)\vec{E}E→ and r → r → vec(r)\vec{r}r→, α α alpha\alphaα is given by α = tan − 1 ⁡ ( E θ E r ) = tan − 1 ⁡ ( 1 2 tan ⁡ θ ) α = tan − 1 ⁡ E θ E r = tan − 1 ⁡ 1 2 tan ⁡ θ alpha=tan^(-1)((E_(theta))/(E_(r)))=tan^(-1)((1)/(2)tan theta)\alpha=\tan ^{-1}\left(\frac{E_{\theta}}{E_{r}}\right)=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)α=tan−1⁡(EθEr)=tan−1⁡(12tan⁡θ)
(iii) If θ = 0 θ = 0 theta=0\theta=0θ=0, i.e point is on the axis -
E axis = 2 × 1 4 π ε 0 ⋅ P → r 3 = 2 K P r 3 E axis  = 2 × 1 4 π ε 0 ⋅ P → r 3 = 2 K P r 3 E_("axis ")=2xx(1)/(4piepsi_(0))*( vec(P))/(r^(3))=(2KP)/(r^(3))\mathrm{E}_{\text {axis }}=2 \times \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\overrightarrow{\mathrm{P}}}{\mathrm{r}^{3}}=\frac{2 \mathrm{KP}}{\mathrm{r}^{3}}Eaxis =2×14πε0⋅P→r3=2KPr3[
\theta=0, \text { i .e . along the axis. }
]
(iv) If θ = 90 ∘ θ = 90 ∘ theta=90^(@)\theta=90^{\circ}θ=90∘, i.e. point is on the line bisecting the dipole perpendicularly
E equator = − 1 4 π ε 0 ⋅ P → r 3 = − K P → r 3 E equator  = − 1 4 π ε 0 ⋅ P → r 3 = − K P → r 3 E_("equator ")=(-1)/(4piepsi_(0))*( vec(P))/(r^(3))=(-K vec(P))/(r^(3))\mathrm{E}_{\text {equator }}=\frac{-1}{4 \pi \varepsilon_{0}} \cdot \frac{\overrightarrow{\mathrm{P}}}{\mathrm{r}^{3}}=\frac{-\mathrm{K} \overrightarrow{\mathrm{P}}}{\mathrm{r}^{3}}Eequator =−14πε0⋅P→r3=−KP→r3
(v) So, E axis = 2 E equator E axis  = 2 E equator  E_("axis ")=2E_("equator ")\mathrm{E}_{\text {axis }}=2 \mathrm{E}_{\text {equator }}Eaxis =2Eequator  (for same r r r\mathrm{r}r )
(vi) E axis = 1 4 π ε 0 ⋅ 2 Pr ( r 2 − ℓ 2 ) 2 E axis  = 1 4 π ε 0 ⋅ 2 Pr r 2 − ℓ 2 2 quadE_("axis ")=(1)/(4piepsi_(0))*(2Pr)/((r^(2)-ℓ^(2))^(2))\quad \mathrm{E}_{\text {axis }}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 \operatorname{Pr}}{\left(r^{2}-\ell^{2}\right)^{2}}Eaxis =14πε0⋅2Pr(r2−ℓ2)2
E equator = 1 4 π ε 0 ⋅ 2 P ( r 2 + ℓ 2 ) 3 / 2 E equator  = 1 4 π ε 0 ⋅ 2 P r 2 + ℓ 2 3 / 2 E_("equator ")=(1)/(4piepsi_(0))*(2P)/((r^(2)+ℓ^(2))^(3//2))\mathrm{E}_{\text {equator }}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 \mathrm{P}}{\left(\mathrm{r}^{2}+\ell^{2}\right)^{3 / 2}}Eequator =14πε0⋅2P(r2+ℓ2)3/2
where P = q ⋅ ( 2 ℓ ) P = q ⋅ ( 2 ℓ ) P=q*(2ℓ)P=q \cdot(2 \ell)P=q⋅(2ℓ)
V = 1 4 π ε 0 ⋅ q ( 2 ℓ ) cos ⁡ θ r 2 V = 1 4 π ε 0 ⋅ P cos ⁡ θ r 2 = 1 4 π ε 0 ⋅ P → ⋅ r → r 3 V = 1 4 π ε 0 ⋅ q ( 2 ℓ ) cos ⁡ θ r 2 V = 1 4 π ε 0 ⋅ P cos ⁡ θ r 2 = 1 4 π ε 0 ⋅ P → ⋅ r → r 3 {:[V=(1)/(4piepsi_(0))*(q(2ℓ)cos theta)/(r^(2))],[V=(1)/(4piepsi_(0))*(Pcos theta)/(r^(2))=(1)/(4piepsi_(0))*( vec(P)* vec(r))/(r^(3))]:}\begin{aligned} & \mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}(2 \ell) \cos \theta}{\mathrm{r}^{2}} \\ & \mathrm{~V}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{P} \cos \theta}{\mathrm{r}^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^{3}} \end{aligned}V=14πε0⋅q(2ℓ)cos⁡θr2 V=14πε0⋅Pcos⁡θr2=14πε0⋅P→⋅r→r3
where θ θ theta\thetaθ is the angle between P → P → vec(P)\vec{P}P→ and r → r → vec(r)\vec{r}r→.
(viii) If θ = 0 , V axis = q d 4 π ε 0 ⋅ r 2 θ = 0 , V axis  = q d 4 π ε 0 ⋅ r 2 theta=0,V_("axis ")=(qd)/(4piepsi_(0)*r^(2))\theta=0, \mathrm{~V}_{\text {axis }}=\frac{\mathrm{qd}}{4 \pi \varepsilon_{0} \cdot r^{2}}θ=0, Vaxis =qd4πε0⋅r2
(ix) If θ = 90 ∘ , V equator = 0 θ = 90 ∘ , V equator  = 0 theta=90^(@),V_("equator ")=0\theta=90^{\circ}, \mathrm{V}_{\text {equator }}=0θ=90∘,Vequator =0 Illustration :
What is the dipole moment
of the system shown in figure
Sol. There are two dipoles of P → = q ( a ) P → = q ( a ) vec(P)=q(a)\overrightarrow{\mathrm{P}}=\mathrm{q}(\mathrm{a})P→=q(a)
so P n e t = 3 p = 3 q a P n e t = 3 p = 3 q a P_(net)=sqrt3p=sqrt3qa\mathrm{P}_{\mathrm{net}}=\sqrt{3} \mathrm{p}=\sqrt{3} \mathrm{qa}Pnet=3p=3qa
ELECTROSTATICS

41. Electric Dipole In an Electric Field - Uniform Electric Field :

(i) When an electric dipole is placed in an uniform electric field, A torque acts on it which subjects the dipole to rotatory motion. This τ τ tau\tauτ is given by τ = P E sin ⁡ θ τ = P E sin ⁡ θ tau=PE sin theta\tau=P E \sin \thetaτ=PEsin⁡θ or
τ = P → × E → τ = P → × E → tau= vec(P)xx vec(E)\tau=\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}τ=P→×E→
(ii) Potential energy of the dipole

42. Cases :

(a) If θ = 0 ∘ θ = 0 ∘ theta=0^(@)\theta=0^{\circ}θ=0∘, i.e. P → ∥ E → , τ = 0 P → ∥ E → , τ = 0 vec(P)|| vec(E),tau=0\vec{P} \| \vec{E}, \tau=0P→∥E→,τ=0 and U = − P E U = − P E U=-PEU=-P EU=−PE, dipole is in the minimum potential energy state and no torque acting on it and hence it is in the stable equilibrium state.
(b) For θ = 180 ∘ θ = 180 ∘ theta=180^(@)\theta=180^{\circ}θ=180∘, i.e. P → P → vec(P)\vec{P}P→ and E → E → vec(E)\vec{E}E→ are in opposite direction, then τ = 0 τ = 0 tau=0\tau=0τ=0 but U = P E U = P E U=PEU=P EU=PE which is maximum potential energy state. (unstable equilibrium)
(c) θ = 90 0 θ = 90 0 theta=90^(0)\theta=90^{0}θ=900, i.e. P → ⊥ E → P → ⊥ E → vec(P)_|_ vec(E)\overrightarrow{\mathrm{P}} \perp \overrightarrow{\mathrm{E}}P→⊥E→, then
τ = P E ( maximum) and U = 0 τ = P E (  maximum) and  U = 0 tau=PE(" maximum) and "U=0\tau=\mathrm{PE}(\text { maximum) and } \mathrm{U}=0τ=PE( maximum) and U=0
(a) There is no net force acting on the dipole in a uniform electric field.
(b) Dipole can only perform rotatory motion.
(c) If dipole is placed in a nonuniform electric field, it performs rotatory as well as translatory motion because now a net force also acts on the dipole along with the torque. (important)
Work done in rotating on electric dipole in an electric field :
(i) To rotate the dipole by an angle θ θ theta\thetaθ from the state of stable equilibrium W = P E ( 1 − cos ⁡ θ ) W = P E ( 1 − cos ⁡ θ ) W=PE(1-cos theta)\mathrm{W}=\mathrm{PE}(1-\cos \theta)W=PE(1−cos⁡θ).
(ii) Work done in rotating the dipole from θ 1 θ 1 theta_(1)\theta_{1}θ1 to θ 2 θ 2 theta_(2)\theta_{2}θ2 in an uniform electric field
W = P E ( cos ⁡ θ 1 − cos ⁡ θ 2 ) W = P E cos ⁡ θ 1 − cos ⁡ θ 2 W=PE(cos theta_(1)-cos theta_(2))\mathrm{W}=\mathrm{PE}\left(\cos \theta_{1}-\cos \theta_{2}\right)W=PE(cos⁡θ1−cos⁡θ2)
(iii) Work done in rotating the dipole through 180 ∘ 180 ∘ 180^(@)180^{\circ}180∘ from stable equilibrium state
W = 2 P E = 2 ( potential energy) W = 2 P E = 2 (  potential energy)  W=2PE=2(" potential energy) "\mathrm{W}=2 \mathrm{PE}=2(\text { potential energy) }W=2PE=2( potential energy) 
Illustration :
An electron and a proton are placed at distance of 1 Å 1 Å 1"Å"1 \AA1Å. What will be dipole moment of so formed dipole?
Sol. p = q d = 1.6 × 10 − 19 × 1 × 10 − 10 p = q d = 1.6 × 10 − 19 × 1 × 10 − 10 quad p=qd=1.6 xx10^(-19)xx1xx10^(-10)\quad p=q d=1.6 \times 10^{-19} \times 1 \times 10^{-10}p=qd=1.6×10−19×1×10−10
= 1.6 × 10 − 29 = 1.6 × 10 − 29 =1.6 xx10^(-29)=1.6 \times 10^{-29}=1.6×10−29 coulomb metre
Illustration:
A dipole of dipole moment P P PPP lies in a uniform electric field E E EEE such that dipole direction is along field. If dipole is rotated through 180 ∘ 180 ∘ 180^(@)180^{\circ}180∘ such that dipole direction becomes opposite to the field direction. Find the work done by the electrostatic field.
Sol.
U i = − P → ⋅ E → = − P E cos ⁡ 0 ∘ = − P E U f = − P ⋅ E ⋅ cos ⁡ ( 180 ∘ ) = P E U i = − P → ⋅ E → = − P E cos ⁡ 0 ∘ = − P E U f = − P ⋅ E ⋅ cos ⁡ 180 ∘ = P E {:[U_(i)=- vec(P)* vec(E)=-PEcos 0^(@)=-PE],[U_(f)=-P*E*cos(180^(@))=PE]:}\begin{aligned} \mathrm{U}_{\mathrm{i}}=-\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{E}} & =-\mathrm{PE} \cos 0^{\circ}=-\mathrm{PE} \\ \mathrm{U}_{\mathrm{f}} & =-\mathrm{P} \cdot \mathrm{E} \cdot \cos \left(180^{\circ}\right)=\mathrm{PE} \end{aligned}Ui=−P→⋅E→=−PEcos⁡0∘=−PEUf=−P⋅E⋅cos⁡(180∘)=PE
work done by the field = − Δ U = U i − U f = − 2 P E = − Δ U = U i − U f = − 2 P E =-DeltaU=U_(i)-U_(f)=-2PE=-\Delta \mathrm{U}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=-2 \mathrm{PE}=−ΔU=Ui−Uf=−2PE

43. FORCE ON THE SURFACE OF A CHARGED CONDUCTOR :

(i) If surface charge density on a surface is σ σ sigma\sigmaσ, then electric field intensity at a point near this surface is σ ε 0 σ ε 0 (sigma)/(epsi_(0))\frac{\sigma}{\varepsilon_{0}}σε0.
(ii) When a conductor is charged then it's entire surface experiences an outward force perpendicular to the surface.
(iii) The force per unit area of the charged surface is called as the electrical pressure , P electrical = σ 2 2 ε 0 N / m 2 P electrical  = σ 2 2 ε 0 N / m 2 P_("electrical ")=(sigma^(2))/(2epsi_(0))N//m^(2)\mathrm{P}_{\text {electrical }}=\frac{\sigma^{2}}{2 \varepsilon_{0}} \mathrm{~N} / \mathrm{m}^{2}Pelectrical =σ22ε0 N/m2
(iv) The direction of this force is perpendicular to the surface.

44. Energy associated with the electric field :

(i) The energy stored per unit volume around a point in an electric field E E E\mathrm{E}E is given by
U = 1 2 ε 0 E 2 U = 1 2 ε 0 E 2 U=(1)/(2)epsi_(0)E^(2)\mathrm{U}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}U=12ε0E2
This is also called energy density.
(ii) If in place of vacuum some medium is present then U = 1 2 ε 0 ε r E 2 U = 1 2 ε 0 ε r E 2 U=(1)/(2)epsi_(0)epsi_(r)E^(2)\mathrm{U}=\frac{1}{2} \varepsilon_{0} \varepsilon_{\mathrm{r}} \mathrm{E}^{2}U=12ε0εrE2
(iii) For the electric field around a charged conducting sphere U = 1 8 π ε 0 ⋅ q 2 R U = 1 8 π ε 0 ⋅ q 2 R U=(1)/(8piepsi_(0))*(q^(2))/(R)U=\frac{1}{8 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{R}U=18πε0⋅q2R
Where q = q = q=\mathrm{q}=q= charge on sphere
R = R = R=\mathrm{R}=R= radius of sphere

45. Drop of a charged liquid :

If n n n\mathrm{n}n identical drops each having a charge q q q\mathrm{q}q and radius r r r\mathrm{r}r coalesce to form a single large drop of radius R R R\mathrm{R}R and charge Q Q Q\mathrm{Q}Q, then
(a) Charge will be conserved i.e. n q = Q n q = Q nq=Q\mathrm{nq}=\mathrm{Q}nq=Q
(b) Volume will be conserved i.e.
n ⋅ 4 3 π r 3 = 4 3 π R 3 or R = n 1 / 3 r n ⋅ 4 3 π r 3 = 4 3 π R 3  or  R = n 1 / 3 r n*(4)/(3)pir^(3)=(4)/(3)piR^(3)" or "R=n^(1//3)r\mathrm{n} \cdot \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3} \text { or } \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r}n⋅43πr3=43πR3 or R=n1/3r
(c) Potential of each small drops = V = 1 4 π ε 0 ⋅ q r = V = 1 4 π ε 0 ⋅ q r =V=(1)/(4piepsi_(0))*(q)/(r)=\mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}}=V=14πε0⋅qr
(d) Potential of large drop = V ′ = V ′ =V^(')=\mathrm{V}^{\prime}=V′
V ′ = 1 4 π ε 0 Q R ⇒ V ′ = n 2 / 3 V V ′ = 1 4 π ε 0 Q R ⇒ V ′ = n 2 / 3 V V^(')=(1)/(4piepsi_(0))(Q)/(R)=>V^(')=n^(2//3)V\mathrm{V}^{\prime}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}} \Rightarrow \mathrm{V}^{\prime}=\mathrm{n}^{2 / 3} \mathrm{~V}V′=14πε0QR⇒V′=n2/3 V
(4)
(e) Electric field at surface of small drop = E = E =E=\mathrm{E}=E
E = 1 4 π ε 0 ⋅ q r 2 E = 1 4 π ε 0 ⋅ q r 2 E=(1)/(4piepsi_(0))*(q)/(r^(2))\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}E=14πε0⋅qr2
(f) Electric field at surface of large drop = E ′ = E ′ =E^(')=\mathrm{E}^{\prime}=E′
E ′ = 1 4 π ε 0 ⋅ Q R 2 E ′ = n 1 / 3 E . E ′ = 1 4 π ε 0 ⋅ Q R 2 E ′ = n 1 / 3 E . {:[E^(')=(1)/(4piepsi_(0))*(Q)/(R^(2))],[E^(')=n^(1//3)E.]:}\begin{aligned} & E^{\prime}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R^{2}} \\ & E^{\prime}=n^{1 / 3} \mathrm{E} . \end{aligned}E′=14πε0⋅QR2E′=n1/3E.

46. VAN DE GRAAFF GENERATOR :

In 1929, Rober J. Van de Graaff designed a mechine which could build up high voltages of the order of a few million volt. This machine acts on the principle of corona discharge.
It consists of large metal spherical shell supported on an insulating column. A long narrow belt made of insulating material passes over the two pulleys shown in the figure.
The positive charge is sprayed at the lower position of the belt throught he corona discharge by the lower metal brush with sharp points. The bult is driven rapidly by the motor driving the lower pulley. When this positive charge reaches near the upper metal brush, the corona discharge takes place, and the positive charges are transferred to the metallic sphere through the metal brush. Thus, the positive charge from the electric source which supplies the charge continuously to the lower metal brush, is transferred to the outer surface of the larger sphere surrounding the upper pulley. The potential of the sphere, thus keeps on increasing till the dielectric breakdown of air surrounding the sphere.
The dielectric strength of air is 3 × 10 6 v / m 3 × 10 6 v / m 3xx10^(6)v//m3 \times 10^{6} \mathrm{v} / \mathrm{m}3×106v/m. Hence the sphere of radius 1 m 1 m 1m1 \mathrm{~m}1 m can have potential upto 3 × 10 6 V 3 × 10 6 V 3xx10^(6)V3 \times 10^{6} \mathrm{~V}3×106 V because V = E R V = E R V=ER\mathrm{V}=\mathrm{ER}V=ER. The potential can be raised by enclosing the sphere in a highly evacuated chamber.

47. COLOUMB'S LAW AND ELECTRIC FIELD

48. EXERCISE-1

Q.1 Two metallic spheres of same mass are given equal and opposite charges; then :
(1) the mass of positively charged sphere increases
(2) the mass of both remains the same
(3) the mass of negatively charged sphere increases
(4) the mass of both spheres increase
Q.2 Which one of the following statement regarding electrostatics is wrong ?
(1) Charge is quantized
(2) Charge is conserved
(3) There is an electric field near an isolated charge at rest
(4) A stationary charge produces both electric and magnetic fields
Q.3 Two identical metallic sphere are charged with 10 and -20 units of charge. If both the spheres are first brought into contact with each other and then are placed to their previous positions, then the ratio of the force in the two situations will be :-
(1) 8 : 1 8 : 1 8:18: 18:1
(2) 1 : 8 1 : 8 1:81: 81:8
(3) 2 : 1 2 : 1 2:12: 12:1
(4) 1 : 2 1 : 2 1:21: 21:2
Q.4 A stationary electric charge produces-
(1) Only electric fields
(2) Only magnetic field
(3) Both electric as magnetic field
(4) Neither electric Nor magnetic field
Q.5 It is observed that when a soap bubble is given some positive charge its radius increases. What will happen to the same soap bubble if it is given equal negative charge instead of positive charge?
(1) Its radius increases
(2) Its radius decreases
(3) Its radius remains same
(4) It gets burst
Q.6 Two charges placed in air repel each other by a force of 10 − 4 N 10 − 4 N 10^(-4)N10^{-4} \mathrm{~N}10−4 N. When oil is introduced between the charges, the force becomes 2.5 × 10 − 5 N 2.5 × 10 − 5 N 2.5 xx10^(-5)N2.5 \times 10^{-5} \mathrm{~N}2.5×10−5 N. The dielectric constant of oil is :
(1) 2.5
(2) 0.25
(3) 2.0
(4) 4.0
Q.7 The ratio of the forces between two small spheres with constant charges (a) in air and (b) in a medium of dielectric constant K K K\mathrm{K}K is respectively :
(1) 1 : K 1 : K 1:K1: \mathrm{K}1:K
(2) K : 1 K : 1 K:1\mathrm{K}: 1K:1
(3) 1 : K 2 1 : K 2 1:K^(2)1: \mathrm{K}^{2}1:K2
(4) K 2 : 1 K 2 : 1 K^(2):1\mathrm{K}^{2}: 1K2:1
Q.8 Two charges of + 1 μ C & + 5 μ C + 1 μ C & + 5 μ C +1muC&+5muC+1 \mu \mathrm{C} \&+5 \mu \mathrm{C}+1μC&+5μC are placed 4 c m 4 c m 4cm4 \mathrm{~cm}4 cm apart, the ratio of the force exerted by both charges on each other will be -
(1) 1 : 1 1 : 1 1:11: 11:1
(2) 1 : 5 1 : 5 1:51: 51:5
(3) 5 : 1 5 : 1 5:15: 15:1
(4) 25 : 1 25 : 1 25:125: 125:1
Q.9 The force between two point charges in vacuum is 15 N 15 N 15N15 \mathrm{~N}15 N, if a brass plate is introduced between the two charges, then force between them will-
(1) Becomes zero
(2) Remains the same
(3) Becomes 30 N 30 N 30N30 \mathrm{~N}30 N
(4) Becomes 60 N 60 N 60N60 \mathrm{~N}60 N

49. ELECTROSTATICS

Q.10 The force between two point charges placed in vacuum at distance 1 m m 1 m m 1mm1 \mathrm{~mm}1 mm is 18 N 18 N 18N18 \mathrm{~N}18 N. If a glass plate of thickness 1 m m 1 m m 1mm1 \mathrm{~mm}1 mm and dielectric constant 6 , be kept between the charges then new force between them would be-
(1) 18 N 18 N 18N18 \mathrm{~N}18 N
(2) 108 N 108 N 108N108 \mathrm{~N}108 N
(3) 3 N 3 N 3N3 \mathrm{~N}3 N
(4) 3 × 10 − 6 N 3 × 10 − 6 N 3xx10^(-6)N3 \times 10^{-6} \mathrm{~N}3×10−6 N
Q.11 Five point charges, each of value + q + q +q+\mathrm{q}+q Coulomb, are placed on five vertices of a regular hexagon of side L metre. The magnitude of the force on a point charge of value -q Coulomb placed at the centre of the hexagon is -
(1) k q 2 L 2 k q 2 L 2 (kq^(2))/(L^(2))\frac{\mathrm{kq}^{2}}{\mathrm{~L}^{2}}kq2 L2
(2) 5 k q 2 L 2 5 k q 2 L 2 sqrt5(kq^(2))/(L^(2))\sqrt{5} \frac{k q^{2}}{\mathrm{~L}^{2}}5kq2 L2
(3) 3 k q 2 L 2 3 k q 2 L 2 sqrt3(kq^(2))/(L^(2))\sqrt{3} \frac{k q^{2}}{\mathrm{~L}^{2}}3kq2 L2
(4) Zero
Q.12 A body has 80 microcoulomb of charge. Number of additional electrons on it will be :
(1) 8 × 10 − 5 8 × 10 − 5 8xx10^(-5)8 \times 10^{-5}8×10−5
(2) 80 × 10 15 80 × 10 15 80 xx10^(15)80 \times 10^{15}80×1015
(3) 5 × 10 14 5 × 10 14 5xx10^(14)5 \times 10^{14}5×1014
(4) 1.28 × 10 − 17 1.28 × 10 − 17 1.28 xx10^(-17)1.28 \times 10^{-17}1.28×10−17
Q.13 A pendulem bob of mass 80 m g 80 m g 80mg80 \mathrm{mg}80mg and carrying a charge of 2 × 10 − 8 2 × 10 − 8 2xx10^(-8)2 \times 10^{-8}2×10−8 Coulomb is at rest in a horizontal uniform electric field of 20 , 000 V m − 1 20 , 000 V m − 1 20,000Vm^(-1)20,000 \mathrm{~V} \mathrm{~m}^{-1}20,000 V m−1. Find the tension in the thread of pendulum -
(1) 8.8 × 10 − 2 N 8.8 × 10 − 2 N 8.8 xx10^(-2)N8.8 \times 10^{-2} \mathrm{~N}8.8×10−2 N
(2) 8.8 × 10 − 3 N 8.8 × 10 − 3 N 8.8 xx10^(-3)N8.8 \times 10^{-3} \mathrm{~N}8.8×10−3 N
(3) 8.8 × 10 − 4 N 8.8 × 10 − 4 N 8.8 xx10^(-4)N8.8 \times 10^{-4} \mathrm{~N}8.8×10−4 N
(4) 8.8 × 10 − 5 N 8.8 × 10 − 5 N 8.8 xx10^(-5)N8.8 \times 10^{-5} \mathrm{~N}8.8×10−5 N
Q.14 Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E E EEE, then the resultant electric field intensity at centre of square will be :
(1) Zero
(2) 4 E 4 E 4E4 \mathrm{E}4E
(3) E E E\mathrm{E}E
(4) 1 / 2 E 1 / 2 E 1//2E1 / 2 \mathrm{E}1/2E
Q.15 Three charge + 4 q , Q + 4 q , Q +4q,Q+4 \mathrm{q}, \mathrm{Q}+4q,Q and q q q\mathrm{q}q are placed in a straight line of length ℓ ℓ ℓ\ellℓ at points distance 0 , ℓ / 2 0 , ℓ / 2 0,ℓ//20, \ell / 20,ℓ/2 and ℓ ℓ ℓ\ellℓ respectively. What should be the value of Q Q Q\mathrm{Q}Q in order to make the net force on q q q\mathrm{q}q to be zero?
(1) − q − q -q-\mathrm{q}−q
(2) − 2 q − 2 q -2q-2 q−2q
(3) − q / 2 − q / 2 -q//2-\mathrm{q} / 2−q/2
(4) 4 q 4 q 4q4 \mathrm{q}4q
Q.16 Two point charges placed at a distance r r rrr in air exert a force F F FFF on each other. The value of distance R R R\mathrm{R}R at which they experience force 4 F 4 F 4F4 \mathrm{~F}4 F when placed in a medium of dielectric constant K = 16 K = 16 K=16\mathrm{K}=16K=16 is :
(1) r r r\mathrm{r}r
(2) r / 4 r / 4 r//4r / 4r/4
(3) r / 8 r / 8 r//8r / 8r/8
(4) 2 r 2 r 2r2 r2r
Q.17 Four charges + q , + q , − q + q , + q , − q +q,+q,-q+\mathrm{q},+\mathrm{q},-\mathrm{q}+q,+q,−q and − q − q -q-\mathrm{q}−q are placed respectively at the corners A , B , C A , B , C A,B,C\mathrm{A}, \mathrm{B}, \mathrm{C}A,B,C and D D D\mathrm{D}D of a square of side "a", arranged in the given order. Calculate the intensity at ( O ) ( O ) (O)(\mathrm{O})(O) the centre of the square
(1) 4 π ε 0 ⋅ a 2 4 2 q 4 π ε 0 ⋅ a 2 4 2 q (4piepsi_(0)*a^(2))/(4sqrt2q)\frac{4 \pi \varepsilon_{0} \cdot a^{2}}{4 \sqrt{2} q}4πε0⋅a242q
(2) 4 2 q 4 π ε 0 ⋅ a 2 4 2 q 4 π ε 0 ⋅ a 2 (4sqrt2q)/(4piepsi_(0)*a^(2))\frac{4 \sqrt{2} q}{4 \pi \varepsilon_{0} \cdot a^{2}}42q4πε0⋅a2
(3) π ε 0 ⋅ a 2 4 2 q π ε 0 ⋅ a 2 4 2 q (piepsi_(0)*a^(2))/(4sqrt2q)\frac{\pi \varepsilon_{0} \cdot a^{2}}{4 \sqrt{2} q}πε0⋅a242q
(4) 4 2 q π ε 0 ⋅ a 2 4 2 q π ε 0 ⋅ a 2 (4sqrt2q)/(piepsi_(0)*a^(2))\frac{4 \sqrt{2} q}{\pi \varepsilon_{0} \cdot a^{2}}42qπε0⋅a2
Q.18 Four charges are arranged at the corners of a square A B C D A B C D ABCD\mathrm{ABCD}ABCD, as shown. The force on a +ve charge kept at the centre of the square is
(1) zero
(2) along diagonal A C A C AC\mathrm{AC}AC
(3) along diagonal BD
(4) perpendicular to the side A B A B AB\mathrm{AB}AB

50. ELECTROSTATICS

Q.19 Three identical charges each of 1 μ C 1 μ C 1muC1 \mu \mathrm{C}1μC are kept on the circumference of a circle of radius 1 metre forming equilateral triangle. The electric intensity at the center of the circle in N / C N / C N//C\mathrm{N} / \mathrm{C}N/C is
(1) 9 × 10 3 9 × 10 3 9xx10^(3)9 \times 10^{3}9×103
(2) 13.5 × 10 3 13.5 × 10 3 13.5 xx10^(3)13.5 \times 10^{3}13.5×103
(3) 27 × 10 3 27 × 10 3 27 xx10^(3)27 \times 10^{3}27×103
(4) Zero
Q.20 The three charges each of 5 × 10 − 6 5 × 10 − 6 5xx10^(-6)5 \times 10^{-6}5×10−6 coulomb are placed at vertex of an equilateral triangle of side 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm. The force exerted on the charge of 1 μ C 1 μ C 1muC1 \mu \mathrm{C}1μC placed at centre of triangle in newton will be
(1) 13.5
(2) zero
(3) 4.5
(4) 6.75
Q.21 Four equal charges, each + q + q +q+\mathrm{q}+q are placed at the corners of a square of side a a aaa. Then the coulomb force experienced by one charge due to the rest of three is :
(1) ( 2 2 + 1 ) K q 2 / 2 a 2 ( 2 2 + 1 ) K q 2 / 2 a 2 (2sqrt2+1)Kq^(2)//2a^(2)(2 \sqrt{2}+1) \mathrm{Kq}^{2} / 2 \mathrm{a}^{2}(22+1)Kq2/2a2
(2) 3 K q 2 / a 2 3 K q 2 / a 2 3Kq^(2)//a^(2)3 \mathrm{Kq}^{2} / \mathrm{a}^{2}3Kq2/a2
(3) 2 2 K q 2 / a 2 2 2 K q 2 / a 2 2sqrt2Kq^(2)//a^(2)2 \sqrt{2} \mathrm{Kq}^{2} / \mathrm{a}^{2}22Kq2/a2
(4) zero
Q.22 Six charges + Q + Q +Q+Q+Q each are placed at the corners of a regular hexagon of side (a), the electric field at the centre of hexagon is-
(1) Zero
(2) 1 4 π ϵ 0 ⋅ 6 Q 2 a 2 1 4 π ϵ 0 ⋅ 6 Q 2 a 2 (1)/(4piepsilon_(0))*(6Q^(2))/(a^(2))\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{6 Q^{2}}{a^{2}}14πϵ0⋅6Q2a2
(3) 1 4 π ϵ 0 ⋅ Q 2 a 2 1 4 π ϵ 0 ⋅ Q 2 a 2 (1)/(4piepsilon_(0))*(Q^(2))/(a^(2))\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q^{2}}{a^{2}}14πϵ0⋅Q2a2
(4) 1 4 π ϵ 0 ⋅ 6 Q 2 a 2 1 4 π ϵ 0 ⋅ 6 Q 2 a 2 (1)/(4piepsilon_(0))*(6Q^(2))/(asqrt2)\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{6 Q^{2}}{a \sqrt{2}}14πϵ0⋅6Q2a2
Q.23 Two free positive charges 4 q 4 q 4q4 \mathrm{q}4q and q q q\mathrm{q}q are a distance l l lll apart. What charge Q Q Q\mathrm{Q}Q is needed to achieve equilibrium for the entire system and where should it be placed form charge q q qqq ?
(1) Q = 4 9 q Q = 4 9 q Q=(4)/(9)q\mathrm{Q}=\frac{4}{9} \mathrm{q}Q=49q (negative) at l 3 l 3 (l)/(3)\frac{l}{3}l3
(2) Q = 4 9 q ( Q = 4 9 q Q=(4)/(9)q(:}\mathrm{Q}=\frac{4}{9} \mathrm{q}\left(\right.Q=49q( positive) at l 3 l 3 (l)/(3)\frac{l}{3}l3
(3) Q = q ( Q = q Q=q(:}\mathrm{Q}=\mathrm{q}\left(\right.Q=q( positive) at l 3 l 3 (l)/(3)\frac{l}{3}l3
(4) Q = q Q = q Q=q\mathrm{Q}=\mathrm{q}Q=q (negative) at l 3 l 3 (l)/(3)\frac{l}{3}l3
Q.24 Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O O O\mathrm{O}O, force on it will be:
(1) Zero
(2) Along OF
(3) Along OC
(4) None of these

51. ELECTRIC FIELD

Q.25 Three charges + 3 q , + q + 3 q , + q +3q,+q+3 \mathrm{q},+\mathrm{q}+3q,+q and Q Q Q\mathrm{Q}Q are placed on a straight line with equal separation . In order to make the net force on q q qqq to be zero, the value of Q Q QQQ will be
(1) + 3 q + 3 q +3q+3 q+3q
( 2 ) + 2 q ( 2 ) + 2 q (2)+2q(2)+2 q(2)+2q
(3) − 3 q − 3 q -3q-3 q−3q
(4) − 4 q − 4 q -4q-4 \mathrm{q}−4q
Q.26 Two parallel large thin metal sheets have equal surface charge densities ( σ = 26.4 × 10 − 12 c / m 2 ) σ = 26.4 × 10 − 12 c / m 2 (sigma=26.4 xx10^(-12)c//m^(2))\left(\sigma=26.4 \times 10^{-12} \mathrm{c} / \mathrm{m}^{2}\right)(σ=26.4×10−12c/m2) of opposite signs. The electric field between these plates is :
(1) 1.5 N / C 1.5 N / C 1.5N//C1.5 \mathrm{~N} / \mathrm{C}1.5 N/C
(2) 1.5 × 10 − 10 N / C 1.5 × 10 − 10 N / C 1.5 xx10^(-10)N//C1.5 \times 10^{-10} \mathrm{~N} / \mathrm{C}1.5×10−10 N/C
(3) 3 N / C 3 N / C 3N//C3 \mathrm{~N} / \mathrm{C}3 N/C
(4) 3 × 10 − 10 N / C 3 × 10 − 10 N / C 3xx10^(-10)N//C3 \times 10^{-10} \mathrm{~N} / \mathrm{C}3×10−10 N/C
Q.27 If an electron is placed in a uniform electric field, then the electron will :
(1) experience no force.
(2) moving with constant velocity in the direction of the field.
(3) move with constant velocity in the direction opposite to the field.
(4) accelerate in direction opposite to field.

52. ELECTROSTATICS

Q.28 If Q = 2 Q = 2 Q=2\mathrm{Q}=2Q=2 coloumb and force on it is F = 100 F = 100 F=100\mathrm{F}=100F=100 newton, then the value of field intensity will be:
(1) 100 N / C 100 N / C 100N//C100 \mathrm{~N} / \mathrm{C}100 N/C
(2) 50 N / C 50 N / C 50N//C50 \mathrm{~N} / \mathrm{C}50 N/C
(3) 200 N / C 200 N / C 200N//C200 \mathrm{~N} / \mathrm{C}200 N/C
(4) 10 N / C 10 N / C 10N//C10 \mathrm{~N} / \mathrm{C}10 N/C
Q.29 The electric field intensity due to a uniformly charged sphere is zero :
(1) at the centre
(2) at infinity
(3) at the centre and at infinite distance
(4) on the surface
Q.30 Two spheres of radii 2 c m 2 c m 2cm2 \mathrm{~cm}2 cm and 4 c m 4 c m 4cm4 \mathrm{~cm}4 cm are charged equally, then the ratio of charge density on the surfaces of the spheres will be -
(1) 1 : 2 1 : 2 1:21: 21:2
(2) 4 : 1 4 : 1 4:14: 14:1
(3) 8 : 1 8 : 1 8:18: 18:1
(4) 1 : 4 1 : 4 1:41: 41:4
Q.31 A charged water drop of radius 0.1 μ m 0.1 μ m 0.1 mum0.1 \mu \mathrm{m}0.1μm is under equilibrium in some electric field. The charge on the drop is equivalent to electronic charge. The intensity of electric field is ( g = 10 m / s 2 ) g = 10 m / s 2 (g=10(m)//s^(2))\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)(g=10 m/s2) -
(1) 1.61 N C − 1 1.61 N C − 1 1.61NC^(-1)1.61 \mathrm{NC}^{-1}1.61NC−1
(2) 26.2 N C − 1 26.2 N C − 1 26.2NC^(-1)26.2 \mathrm{NC}^{-1}26.2NC−1
(3) 262 N C − 1 262 N C − 1 262NC^(-1)262 \mathrm{NC}^{-1}262NC−1
(4) 1610 N C − 1 1610 N C − 1 1610NC^(-1)1610 \mathrm{NC}^{-1}1610NC−1
Q.32 Two large sized charged plates have a charge density of + σ + σ +sigma+\sigma+σ and − σ − σ -sigma-\sigma−σ. The resultant force on the proton located midway between them will be -
(1) σ e / ∈ 0 σ e / ∈ 0 sigmae//in_(0)\sigma \mathrm{e} / \in_{0}σe/∈0
(2) σ e / 2 ϵ 0 σ e / 2 ϵ 0 sigma e//2epsilon_(0)\sigma e / 2 \epsilon_{0}σe/2ϵ0
(3) 2 σ e / ϵ 0 2 σ e / ϵ 0 2sigma e//epsilon_(0)2 \sigma e / \epsilon_{0}2σe/ϵ0
(4) zero
Q.33 There is a uniform electric field in x x x\mathrm{x}x-direction. If the work done by external agent in moving a charge of 0.2 C 0.2 C 0.2C0.2 \mathrm{C}0.2C through a distance of 2 metre slowly along the line making an angle of 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ with x x x\mathrm{x}x-direction is 4 joule, then the magnitude of E E E\mathrm{E}E is:
(1) 3 N / C 3 N / C sqrt3N//C\sqrt{3} \mathrm{~N} / \mathrm{C}3 N/C
(2) 4 N / C 4 N / C 4N//C4 \mathrm{~N} / \mathrm{C}4 N/C
(3) 5 N / C 5 N / C 5N//C5 \mathrm{~N} / \mathrm{C}5 N/C
(4) 20 N / C 20 N / C 20N//C20 \mathrm{~N} / \mathrm{C}20 N/C
Q.34 A simple pendulum has a length ℓ ℓ ℓ\ellℓ, mass of bob m m m\mathrm{m}m. The bob is given a charge q coulomb. The pendulum is suspended in a uniform horizontal electric field of strength E E EEE as shown in figure, then calculate the time period of oscillation when the bob is slightly displace from its mean position is :

(1) 2 π ℓ g 2 π ℓ g 2pisqrt((ℓ)/(g))2 \pi \sqrt{\frac{\ell}{g}}2πℓg
(2) 2 π { ℓ g + q E m } 2 π ℓ g + q E m 2pisqrt({(ℓ)/(g+(qE)/(m))})2 \pi \sqrt{\left\{\frac{\ell}{g+\frac{q E}{m}}\right\}}2π{ℓg+qEm}
(3) 2 π { ℓ g − q E m } 2 π ℓ g − q E m 2pisqrt({(ℓ)/(g-(qE)/(m))})2 \pi \sqrt{\left\{\frac{\ell}{g-\frac{q E}{m}}\right\}}2π{ℓg−qEm}
(4) ℓ g 2 + ( q E m ) 2 2 π ℓ g 2 + q E m 2 2 π root(2pi)((ℓ)/(sqrt(g^(2)+((qE)/(m))^(2))))\sqrt[2 \pi]{\frac{\ell}{\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}}}ℓg2+(qEm)22π

53. ELECTROSTATICS

Q.35 A charged particle of charge q q qqq and mass m m mmm is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time ' t t ttt ' seconds is
(1) E q m t E q m t (Eqm)/(t)\frac{E q m}{t}Eqmt
(2) E 2 q 2 t 2 2 m E 2 q 2 t 2 2 m (E^(2)q^(2)t^(2))/(2m)\frac{E^{2} q^{2} t^{2}}{2 m}E2q2t22m
(3) 2 E 2 t 2 m q 2 E 2 t 2 m q (2E^(2)t^(2))/(mq)\frac{2 E^{2} t^{2}}{m q}2E2t2mq
(4) E q 2 m 2 t 2 E q 2 m 2 t 2 (Eq^(2)(m))/(2t^(2))\frac{\mathrm{Eq}^{2} \mathrm{~m}}{2 \mathrm{t}^{2}}Eq2 m2t2
Q.36 Five balls, numbered 1 to 5 , are suspended using separate threads. Pairs ( 1 , 2 ) , ( 2 , 4 ) , ( 4 , 1 ) ( 1 , 2 ) , ( 2 , 4 ) , ( 4 , 1 ) (1,2),(2,4),(4,1)(1,2),(2,4),(4,1)(1,2),(2,4),(4,1) show electrostatic attraction, while pairs ( 2 , 3 ) ( 2 , 3 ) (2,3)(2,3)(2,3) and ( 4 , 5 ) ( 4 , 5 ) (4,5)(4,5)(4,5) show repulsion. Therefore ball 1 :
(1) Must be positively charged
(2) Must be negatively charged
(3) May be neutral
(4) Must be made of metal
Q.37 The electric field above a uniformly charged nonconducting sheet is E. If the nonconducting sheet is now replaced by a conducting sheet, with the charge same as before, the new electric field at the same point is :
(1) 2 E 2 E 2E2 \mathrm{E}2E
(2) E E E\mathrm{E}E
(3) E 2 E 2 (E)/(2)\frac{E}{2}E2
(4) None of these
Q.38 A positively charged pendulum is oscillating in a uniform electric field as shown in Figure. Its time period of SHM as compared to that when it was uncharged. ( m g > q E ) ( m g > q E ) (mg > qE)(\mathrm{mg}>\mathrm{qE})(mg>qE)
(1) Will increase
(2) Will decrease
(3) Will not change
(4) Will first increase then decrease
Q.39 Two small spherical balls each carrying a charge Q = 10 μ C ( 10 Q = 10 μ C ( 10 Q=10 muC(10\mathrm{Q}=10 \mu \mathrm{C}(10Q=10μC(10 micro-coulomb) are suspended by two insulating threads of equal lengths 1 each, from a point fixed in the ceiling. If is found that is equilibrium threads are separated by an angle 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ between them, as shown in the fig. What is the tension in the threads (Given 1 ( 4 π ε 0 ) = 9 × 10 9 N m / C 2 ) 1 4 π ε 0 = 9 × 10 9 N m / C 2 {sad1)/((4piepsi_(0)))=9xx10^(9)Nm//C^(2))\left.\frac{1}{\left(4 \pi \varepsilon_{0}\right)}=9 \times 10^{9} \mathrm{Nm} / \mathrm{C}^{2}\right)1(4πε0)=9×109Nm/C2) -

(1) 18 N 18 N 18N18 \mathrm{~N}18 N
(2) 1.8 N 1.8 N 1.8N1.8 \mathrm{~N}1.8 N
(3) 0.18 N 0.18 N 0.18N0.18 \mathrm{~N}0.18 N
(4) none of the above
Q.40 Two spheres of equal mass A and B are given + q + q +q+q+q and − q − q -q-q−q charge respectively then -
(1) mass of A increases
(2) mass of B increases
(3) mass of A remains constant
(4) mass of B decreases

54. ELECTROSTATICS

Q.41 The wrong statement about electric lines of force is -
(1) These originate from positive charge and end on negative charge
(2) they do no intersect each other at a point
(3) they have the same form for a point charge and a sphere
(4) they have physical existences
Q.42 Choose correct statement regarding electric lines of force :
(1) emerges from (-ve) charge and meet from (+ve) charge
(2) where the electric lines of force are close electic field in that region is strong
(3) just as it is shown for a point system in the same way it represent for a solid sphere
(4) has a physical nature
Q.43 The el ectric field required to keep a water drop of mass m m mmm and charge e just to remain suspended is :
(1) m g m g mg\mathrm{mg}mg
(2) emg
(3) m g e m g e (mg)/(e)\frac{m g}{e}mge
(4) e m g e m g (em)/(g)\frac{e m}{g}emg
Q.44 The separation between the two charges + q + q +q+q+q and − q − q -q-q−q becomes double. The value of force will be :
(1) two fold
(2) half
(3) four fold
(4) one fourth
Q.45 When a glass rod is rubbed with silk, the amount of positive charge acquired by glass rod in magnitude is:
(1) less than the charge on silk
(2) greater than the charge on slik
(3) equal to the charge on silk
(4) none of these
Q.46 The dielectric constant K K K\mathrm{K}K of an insulator can be :
(1) 5
(2) 0.5
(3) -1
(4) zero
Q.47 A cube has point charges of magnitude - q q q\mathrm{q}q at all its vertices. Electric field at the centre of the cube is :
(1) 1 4 π ε 0 6 q 3 a 2 1 4 π ε 0 6 q 3 a 2 (1)/(4piepsi_(0))(6q)/(3a^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{6 q}{3 a^{2}}14πε06q3a2
(2) 1 4 π ε 0 8 q a 2 1 4 π ε 0 8 q a 2 (1)/(4piepsi_(0))(8q)/(a^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{8 q}{a^{2}}14πε08qa2
(3) zero
(4) 1 4 π ε 0 − 8 q a 2 1 4 π ε 0 − 8 q a 2 (1)/(4piepsi_(0))(-8q)/(a^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{-8 q}{a^{2}}14πε0−8qa2
Q.48 A charged oil drop is suspended in uniform field of 3 × 10 4 V / m 3 × 10 4 V / m 3xx10^(4)V//m3 \times 10^{4} \mathrm{~V} / \mathrm{m}3×104 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the drop = 9.9 × 10 − 15 k g = 9.9 × 10 − 15 k g =9.9 xx10^(-15)kg=9.9 \times 10^{-15} \mathrm{~kg}=9.9×10−15 kg and g = 10 m / s 2 g = 10 m / s 2 g=10m//s^(2)\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}g=10 m/s2 )
(1) 3.3 × 10 − 18 C 3.3 × 10 − 18 C 3.3 xx10^(-18)C3.3 \times 10^{-18} \mathrm{C}3.3×10−18C
(2) 3.2 × 10 − 18 C 3.2 × 10 − 18 C 3.2 xx10^(-18)C3.2 \times 10^{-18} \mathrm{C}3.2×10−18C
(3) 1.6 × 10 − 18 C 1.6 × 10 − 18 C 1.6 xx10^(-18)C1.6 \times 10^{-18} \mathrm{C}1.6×10−18C
(4) 4.8 × 10 − 18 C 4.8 × 10 − 18 C 4.8 xx10^(-18)C4.8 \times 10^{-18} \mathrm{C}4.8×10−18C
Q.49 Two particle of equal mass m m mmm and charge q q qqq are placed at a distance of 16 c m 16 c m 16cm16 \mathrm{~cm}16 cm. Net force on each charge is zero then value of q m q m (q)/(m)\frac{\mathrm{q}}{\mathrm{m}}qm is
(1) 1
(2) π ε 0 G π ε 0 G sqrt((piepsi_(0))/(G))\sqrt{\frac{\pi \varepsilon_{0}}{\mathrm{G}}}πε0G
(3) G 4 π ε 0 G 4 π ε 0 sqrt((G)/(4piepsi_(0)))\sqrt{\frac{\mathrm{G}}{4 \pi \varepsilon_{0}}}G4πε0
(4) 4 π ε 0 G 4 π ε 0 G sqrt(4piepsi_(0)G)\sqrt{4 \pi \varepsilon_{0} \mathrm{G}}4πε0G
Q.50 An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r r r\mathrm{r}r. The Coulomb force F → F → vec(F)\overrightarrow{\mathrm{F}}F→ on the electron is :
(1) K e 2 r 2 r ^ K e 2 r 2 r ^ K(e^(2))/(r^(2)) hat(r)\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}Ke2r2r^
(2) − K e 2 r 3 r ^ − K e 2 r 3 r ^ -K(e^(2))/(r^(3)) hat(r)-\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{3}} \hat{\mathrm{r}}−Ke2r3r^
(3) K e 2 r 3 r → K e 2 r 3 r → K(e^(2))/(r^(3)) vec(r)\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{3}} \overrightarrow{\mathrm{r}}Ke2r3r→
(4) − K e 2 r 3 r → − K e 2 r 3 r → -K(e^(2))/(r^(3)) vec(r)-\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{3}} \overrightarrow{\mathrm{r}}−Ke2r3r→

55. ELECTROSTATICS

Q.51 Two positive charges of 1 μ C 1 μ C 1muC1 \mu \mathrm{C}1μC and 2 μ C 2 μ C 2muC2 \mu \mathrm{C}2μC are placed 1 metre apart. The value of electric field in N / N / N//\mathrm{N} /N/ C C C\mathrm{C}C at the middle point of the line joining the charges will be :-
(1) 10.8 × 10 4 10.8 × 10 4 10.8 xx10^(4)10.8 \times 10^{4}10.8×104
(2) 3.6 × 10 4 3.6 × 10 4 3.6 xx10^(4)3.6 \times 10^{4}3.6×104
(3) 1.8 × 10 4 1.8 × 10 4 1.8 xx10^(4)1.8 \times 10^{4}1.8×104
(4) 5.4 × 10 4 5.4 × 10 4 5.4 xx10^(4)5.4 \times 10^{4}5.4×104
Q.52 Which one of the following pattern of electric line of force can't possible :-
(1)
(2)
(3)
(4)

56. FLUX CALUCULATION AND GAUSS LAW :

Q.53 How much electric flux will come out through a surface of area vector S → = 10 j ^ S → = 10 j ^ vec(S)=10 hat(j)\overrightarrow{\mathrm{S}}=10 \hat{\mathrm{j}}S→=10j^ kept in an electrostatic field E = 2 i ^ + 4 j ^ + 3 k ^ E = 2 i ^ + 4 j ^ + 3 k ^ E=2 hat(i)+4 hat(j)+3 hat(k)\mathrm{E}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}E=2i^+4j^+3k^ ?
(1) 20 units
(2) 40 units
(3) 30 units
(4) 90 units
Q.54 Flux coming out from a unit positive charge placed in air and enclosed by a surface is :
(1) ε 0 ε 0 epsi_(0)\varepsilon_{0}ε0
(2) ε 0 − 1 ε 0 − 1 epsi_(0)^(-1)\varepsilon_{0}^{-1}ε0−1
(3) ( 4 π ε 0 ) − 1 4 π ε 0 − 1 (4piepsi_(0))^(-1)\left(4 \pi \varepsilon_{0}\right)^{-1}(4πε0)−1
(4) 4 π ε 0 4 π ε 0 4piepsi_(0)4 \pi \varepsilon_{0}4πε0
Q.55 The flux entering and leaving a closed surface are 5 × 10 5 5 × 10 5 5xx10^(5)5 \times 10^{5}5×105 and 4 × 10 5 4 × 10 5 4xx10^(5)4 \times 10^{5}4×105 MKS units respectively; then the charge inside the surface will be :
(1) − 8.85 × 10 − 7 C − 8.85 × 10 − 7 C -8.85 xx10^(-7)C-8.85 \times 10^{-7} \mathrm{C}−8.85×10−7C
(2) 8.85 × 10 − 7 C 8.85 × 10 − 7 C 8.85 xx10^(-7)C8.85 \times 10^{-7} \mathrm{C}8.85×10−7C
(3) 8.85 × 10 7 C 8.85 × 10 7 C 8.85 xx10^(7)C8.85 \times 10^{7} \mathrm{C}8.85×107C
(4) 6.85 × 10 − 7 C 6.85 × 10 − 7 C 6.85 xx10^(-7)C6.85 \times 10^{-7} \mathrm{C}6.85×10−7C
Q.56 In a region of space the electric field is given by E → = 8 i ^ + 4 j ^ + 3 k ^ E → = 8 i ^ + 4 j ^ + 3 k ^ vec(E)=8 hat(i)+4 hat(j)+3 hat(k)\vec{E}=8 \hat{i}+4 \hat{j}+3 \hat{k}E→=8i^+4j^+3k^. The electric flux through a surface of area of 100 units in x-y plane is :
(1) 800 units
(2) 300 units
(3) 400 units
(4) 1500 units
Q.57 Three charges q 1 = 1 μ c , q 2 = 2 μ c q 1 = 1 μ c , q 2 = 2 μ c q_(1)=1muc,q_(2)=2muc\mathrm{q}_{1}=1 \mu \mathrm{c}, \mathrm{q}_{2}=2 \mu \mathrm{c}q1=1μc,q2=2μc and q 3 = − 3 μ c q 3 = − 3 μ c q_(3)=-3muc\mathrm{q}_{3}=-3 \mu \mathrm{c}q3=−3μc and four surfaces S 1 , S 2 , S 3 S 1 , S 2 , S 3 S_(1),S_(2),S_(3)\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}S1, S2, S3 and S 4 S 4 S_(4)\mathrm{S}_{4}S4 are shown. The flux emerging through surface S 2 S 2 S_(2)\mathrm{S}_{2}S2 in N − m 2 / C N − m 2 / C N-m^(2)//C\mathrm{N}-\mathrm{m}^{2} / \mathrm{C}N−m2/C is -
(1) 36 π × 10 3 36 π × 10 3 36 pi xx10^(3)36 \pi \times 10^{3}36π×103
(2) − 36 π × 10 3 − 36 π × 10 3 -36 pi xx10^(3)-36 \pi \times 10^{3}−36π×103
(3) 36 π × 10 9 36 π × 10 9 36 pi xx10^(9)36 \pi \times 10^{9}36π×109
(4) − 36 π × 10 9 − 36 π × 10 9 -36 pi xx10^(9)-36 \pi \times 10^{9}−36π×109
Q.58 Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm surrounding the total charge is 25 V 25 V 25V25 \mathrm{~V}25 V-m. The flux over a concentricsphere of radius 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm will be :-
(1) 25 V − m 25 V − m 25V-m25 \mathrm{~V}-\mathrm{m}25 V−m
(2) 50 V − m 50 V − m 50V-m50 \mathrm{~V}-\mathrm{m}50 V−m
(3) 100 V − m 100 V − m 100V-m100 \mathrm{~V}-\mathrm{m}100 V−m
(4) 200 V − m 200 V − m 200V-m200 \mathrm{~V}-\mathrm{m}200 V−m
Q.59 A charge q q qqq is placed at the centre of the open end of a cylindrical vessel figure. The flux of the electric field through the surface of the vessel is :-
(1) zero
(2) q / ε 0 q / ε 0 q//epsi_(0)q / \varepsilon_{0}q/ε0
(3) q / 2 ε 0 q / 2 ε 0 q//2epsi_(0)q / 2 \varepsilon_{0}q/2ε0
(4) 2 q / ε 0 2 q / ε 0 2q//epsi_(0)2 q / \varepsilon_{0}2q/ε0

57. ELECTROSTATICS

Q.60 In a certain region of surface there exists a uniform electric field of 2 × 10 3 k ^ V / m 2 × 10 3 k ^ V / m 2xx10^(3) hat(k)V//m2 \times 10^{3} \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}2×103k^V/m. A rectangular coil of dimensions 10 c m × 20 c m 10 c m × 20 c m 10cmxx20cm10 \mathrm{~cm} \times 20 \mathrm{~cm}10 cm×20 cm is placed in x − y x − y x-yx-yx−y plane. The electric flux through the coil is -
(1) zero
(2) 4 × 10 − 3 V − m 4 × 10 − 3 V − m 4xx10^(-3)V-m4 \times 10^{-3} \mathrm{~V}-\mathrm{m}4×10−3 V−m
(3) 40 V − m 40 V − m 40V-m40 \mathrm{~V}-\mathrm{m}40 V−m
(4) 4 × 10 5 V − m 4 × 10 5 V − m 4xx10^(5)V-m4 \times 10^{5} \mathrm{~V}-\mathrm{m}4×105 V−m
Q.61 The electric flux from a cube of edge ℓ ℓ ℓ\ellℓ is ϕ ϕ phi\phiϕ. What will be its value if edge of cube is made 2 ℓ 2 ℓ 2ℓ2 \ell2ℓ and charge enclosed is halved -
(1) ϕ / 2 ϕ / 2 phi//2\phi / 2ϕ/2
(2) 2 ϕ 2 ϕ 2phi2 \phi2ϕ
(3) 4 ϕ 4 ϕ 4phi4 \phi4ϕ
(4) ϕ ϕ phi\phiϕ
Q.62 A cubical box of side 1 m 1 m 1m1 \mathrm{~m}1 m is immersed a uniform electric field of strength 10 4 N / C 10 4 N / C 10^(4)N//C10^{4} \mathrm{~N} / \mathrm{C}104 N/C. The flux through the cube is-
(1) 10 4 10 4 10^(4)10^{4}104
(2) 6 × 10 4 6 × 10 4 6xx10^(4)6 \times 10^{4}6×104
(3) 2 × 10 4 2 × 10 4 2xx10^(4)2 \times 10^{4}2×104
(4) Zero
Q.63 Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that

(1) E A > E B > E C E A > E B > E C E_(A) > E_(B) > E_(C)\mathrm{E}_{\mathrm{A}}>\mathrm{E}_{\mathrm{B}}>\mathrm{E}_{\mathrm{C}}EA>EB>EC
(2) E A = E B = E C E A = E B = E C E_(A)=E_(B)=E_(C)\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{B}}=\mathrm{E}_{\mathrm{C}}EA=EB=EC
(3) E A = E C > E B E A = E C > E B E_(A)=E_(C) > E_(B)\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{C}}>\mathrm{E}_{\mathrm{B}}EA=EC>EB
(4) E A = E C < E B E A = E C < E B E_(A)=E_(C) < E_(B)\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{C}}<\mathrm{E}_{\mathrm{B}}EA=EC<EB
Q.64 A charge of 1 coulomb is located at the centre of a sphere of radius 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm and a cube of side 20 c m c m cm\mathrm{cm}cm. The ratio of outgoing flux from the sphere and cube will be :
(1) More than one
(2) Less than one
(3) One
(4) Nothing certain can be said
Q.65 A cylinder of radius R R R\mathrm{R}R and length L L L\mathrm{L}L is placed in a uniform electric field E E E\mathrm{E}E parallel to the cylinder axis. The outward flux over the surface of the cylinder is given by :
(1) 2 π R 2 E 2 π R 2 E 2piR^(2)E2 \pi R^{2} E2πR2E
(2) zero
(3) 2 π R L E 2 π R L E 2piRLE2 \pi \mathrm{RLE}2πRLE
(4) π R 2 E π R 2 E piR^(2)E\pi R^{2} EπR2E
Q.66 A rectangular surface of sides 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm and 15 c m 15 c m 15cm15 \mathrm{~cm}15 cm is placed inside a uniform electric field of 25 V / 25 V / 25V//25 \mathrm{~V} /25 V/ m m m\mathrm{m}m, such that the surface makes an anagle of 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘ with the direction of electric field. Find the flux of the electric field throught he rectangular surface :
(1) 0.1675 N / m 2 C 0.1675 N / m 2 C 0.1675N//m^(2)C0.1675 \mathrm{~N} / \mathrm{m}^{2} \mathrm{C}0.1675 N/m2C
(2) 0.1875 N m 2 / C 0.1875 N m 2 / C 0.1875Nm^(2)//C0.1875 \mathrm{Nm}^{2} / \mathrm{C}0.1875Nm2/C
(3) Zero
(4) 0.1075 N m 2 / C 0.1075 N m 2 / C 0.1075Nm^(2)//C0.1075 \mathrm{Nm}^{2} / \mathrm{C}0.1075Nm2/C
Q.67 A charge Q Q QQQ is kept at the corner of a cube. Electric flux passing through one of those faces not touching that charge is :
(1) Q 24 ∈ 0 Q 24 ∈ 0 (Q)/(24in_(0))\frac{\mathrm{Q}}{24 \in_{0}}Q24∈0
(2) Q 3 ϵ 0 Q 3 ϵ 0 (Q)/(3epsilon_(0))\frac{Q}{3 \epsilon_{0}}Q3ϵ0
(3) Q 8 ϵ 0 Q 8 ϵ 0 (Q)/(8epsilon_(0))\frac{\mathrm{Q}}{8 \epsilon_{0}}Q8ϵ0
(4) Q 6 ϵ 0 Q 6 ϵ 0 (Q)/(6epsilon_(0))\frac{Q}{6 \epsilon_{0}}Q6ϵ0
Q.68 There is uniform electric field of 8 × 10 3 i ^ N / C 8 × 10 3 i ^ N / C 8xx10^(3) hat(i)N//C8 \times 10^{3} \hat{\mathrm{i}} \mathrm{N} / \mathrm{C}8×103i^N/C. What is the net flux (in SI units) of the uniform electric field through a cube of side 0.3 m 0.3 m 0.3m0.3 \mathrm{~m}0.3 m oriented so that its faces are parallel to the coordinates plane ?
(1) 2 × 8 × 10 3 2 × 8 × 10 3 2xx8xx10^(3)2 \times 8 \times 10^{3}2×8×103
(2) 0.3 × 8 × 10 3 0.3 × 8 × 10 3 0.3 xx8xx10^(3)0.3 \times 8 \times 10^{3}0.3×8×103
(3) Zero
(4) 8 × 10 6 × 6 8 × 10 6 × 6 8xx10^(6)xx68 \times 10^{6} \times 68×106×6

58. ELECTROSTATICS

Q.69 If electric lines of force in are represented as shown in the figure, then one can conclude that, electric field is :

(1) Non-uniform
(2) Uniform
(3) Both uniform and non-uniform
(4) Zero everywhere
Q.70 Total flux coming out of some closed surface is :
(1) q / ε 0 q / ε 0 q//epsi_(0)q / \varepsilon_{0}q/ε0
(2) ε 0 / q ε 0 / q epsi_(0)//q\varepsilon_{0} / \mathrm{q}ε0/q
(3) q ε 0 q ε 0 qepsi_(0)\mathrm{q} \varepsilon_{0}qε0
(4) q / ε 0 q / ε 0 sqrt(q//epsi_(0))\sqrt{\mathrm{q} / \varepsilon_{0}}q/ε0
Q.71 Eight charges, 1 μ C , . − 7 μ C , − 4 μ C , 10 μ C , 2 μ C , − 5 μ C , − 3 μ C 1 μ C , . − 7 μ C , − 4 μ C , 10 μ C , 2 μ C , − 5 μ C , − 3 μ C 1muC,.-7muC,-4muC,10 muC,2muC,-5muC,-3muC1 \mu \mathrm{C}, .-7 \mu \mathrm{C},-4 \mu \mathrm{C}, 10 \mu \mathrm{C}, 2 \mu \mathrm{C},-5 \mu \mathrm{C},-3 \mu \mathrm{C}1μC,.−7μC,−4μC,10μC,2μC,−5μC,−3μC and 6 μ C 6 μ C 6muC6 \mu \mathrm{C}6μC are situated at the eight corners of a cube of side 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm. A spherical surface of radius 80 c m 80 c m 80cm80 \mathrm{~cm}80 cm encloses this cube. The centre of the sphere coincides with the centre of the cube. Then the total outgoing flux from the spherical surface (in unit of volt meter) is-
(1) 36 π × 10 3 36 π × 10 3 36 pi xx10^(3)36 \pi \times 10^{3}36π×103
(2) 684 π × 10 3 684 π × 10 3 684 pi xx10^(3)684 \pi \times 10^{3}684π×103
(3) zero
(4) none of the above
Q.72 Gauss law is given by ∈ 0 ∮ s E → ⋅ d → s = q ∈ 0 ∮ s E → ⋅ d → s = q in_(0)oint_(s) vec(E)* vec(d)s=q\in_{0} \oint_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d}} \mathrm{s}=\mathrm{q}∈0∮sE→⋅d→s=q, if net charge enclosed in gaussian surface is zero then :-
(1) E on surface must be zero
(2) incoming and outgoing electric lines are equal
(3) there is a net incoming electric lines
(4) none
Q.73 If the electric flux entering and leaving an closed surface respectively is ϕ 1 ϕ 1 phi_(1)\phi_{1}ϕ1 and ϕ 2 ϕ 2 phi_(2)\phi_{2}ϕ2 the electric charge inside the surface will be
(1) ( ϕ 1 + ϕ 2 ) ε 0 ϕ 1 + ϕ 2 ε 0 (phi_(1)+phi_(2))epsi_(0)\left(\phi_{1}+\phi_{2}\right) \varepsilon_{0}(ϕ1+ϕ2)ε0
(2) ( ϕ 2 − ϕ 1 ) ε 0 ϕ 2 − ϕ 1 ε 0 (phi_(2)-phi_(1))epsi_(0)\left(\phi_{2}-\phi_{1}\right) \varepsilon_{0}(ϕ2−ϕ1)ε0
(3) ϕ 1 + ϕ 2 ε 0 ϕ 1 + ϕ 2 ε 0 (phi_(1)+phi_(2))/(epsi_(0))\frac{\phi_{1}+\phi_{2}}{\varepsilon_{0}}ϕ1+ϕ2ε0
(4) ϕ 2 − ϕ 1 ε 0 ϕ 2 − ϕ 1 ε 0 (phi_(2)-phi_(1))/(epsi_(0))\frac{\phi_{2}-\phi_{1}}{\varepsilon_{0}}ϕ2−ϕ1ε0
Q.74 A nonconducting solid sphere of radius R R R\mathrm{R}R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r r rrr from its centre -
(a) increases as r r rrr increases, for r < R r < R r < Rr<Rr<R
(b) decreases as r r r\mathrm{r}r increases, for 0 < r < ∞ 0 < r < ∞ 0 < r < oo0<\mathrm{r}<\infty0<r<∞
(c) decreases as r r r\mathrm{r}r increases, for R < r < ∞ R < r < ∞ R < r < oo\mathrm{R}<\mathrm{r}<\inftyR<r<∞
(d) is discontinuous at r = R r = R r=R\mathrm{r}=\mathrm{R}r=R
(1) a, c
(2) c , d c , d c,dc, dc,d
(3) a , b a , b a,ba, ba,b
(4) b, d
Q.75 A sphere of radius R R RRR and charge Q Q QQQ is placed inside an imaginary sphere of radius 2 R 2 R 2R2 R2R whose centre coincides with the given sphere. the flux related to imaginary sphere is :-
(1) Q ϵ 0 Q ϵ 0 (Q)/(epsilon_(0))\frac{\mathrm{Q}}{\epsilon_{0}}Qϵ0
(2) Q 2 ϵ 0 Q 2 ϵ 0 (Q)/(2epsilon_(0))\frac{\mathrm{Q}}{2 \epsilon_{0}}Q2ϵ0
(3) 4 Q ϵ 0 4 Q ϵ 0 (4Q)/(epsilon_(0))\frac{4 \mathrm{Q}}{\epsilon_{0}}4Qϵ0
(4) 2 Q ϵ 0 2 Q ϵ 0 (2Q)/(epsilon_(0))\frac{2 Q}{\epsilon_{0}}2Qϵ0
Q.76 20 μ C 20 μ C 20 muC20 \mu \mathrm{C}20μC charge is placed inside a closed surface then flux related to surface is ϕ ϕ phi\phiϕ. If 80 μ C 80 μ C 80 muC80 \mu \mathrm{C}80μC charge is added inside the surface then change in flux is :-
(1) 4 ϕ 4 ϕ 4phi4 \phi4ϕ
(2) 5 ϕ 5 ϕ 5phi5 \phi5ϕ
(3) ϕ ϕ phi\phiϕ
(4) 8 ϕ 8 ϕ 8phi8 \phi8ϕ

59. ELECTROSTATICS

Q.77 A point charge is placed at a distance a 2 a 2 (a)/(2)\frac{\mathrm{a}}{2}a2 perpendicular to the plane and above the centre of a square of side a. The electric flux through the square is :-
(1) q ϵ 0 q ϵ 0 (q)/(epsilon_(0))\frac{\mathrm{q}}{\epsilon_{0}}qϵ0
(2) q π ∈ 0 q π ∈ 0 (q)/(piin_(0))\frac{\mathrm{q}}{\pi \in_{0}}qπ∈0
(3) q 4 ∈ 0 q 4 ∈ 0 (q)/(4in_(0))\frac{q}{4 \in_{0}}q4∈0
(4) q 6 ϵ 0 q 6 ϵ 0 (q)/(6epsilon_(0))\frac{\mathrm{q}}{6 \epsilon_{0}}q6ϵ0

60. ELECTRIC POTENTIAL AND CONDUCTORS

Q.78 In electric field, a 6.75 μ C 6.75 μ C 6.75 muC6.75 \mu \mathrm{C}6.75μC charge experiences 2.5 N 2.5 N 2.5N2.5 \mathrm{~N}2.5 N force, when placed at distance of 5 m 5 m 5m5 \mathrm{~m}5 m from the origin. Then potential gradient at this point will be- (in M.K.S.)
(1) 5.71 × 10 5 5.71 × 10 5 5.71 xx10^(5)5.71 \times 10^{5}5.71×105
(2) 3.71 × 10 5 3.71 × 10 5 3.71 xx10^(5)3.71 \times 10^{5}3.71×105
(3) 18.81 × 10 5 18.81 × 10 5 18.81 xx10^(5)18.81 \times 10^{5}18.81×105
(4) 1.881 × 10 5 1.881 × 10 5 1.881 xx10^(5)1.881 \times 10^{5}1.881×105
Q.79 When charge of 3 coulomb is placed in a uniform electric field, it experiences a force of 3000 newton, within this field, potential difference between two points separated by a distance of 1 c m 1 c m 1cm1 \mathrm{~cm}1 cm is-
(1) 10 Volt
(2) 90 Volt
(3) 1000 Volt
(4) 3000 Volt.
Q.80 A uniform electric field having a magnitude E 0 E 0 E_(0)\mathrm{E}_{0}E0 and direction along positive x x x\mathrm{x}x-axis exists.If the electric potential ( V ) ( V ) (V)(\mathrm{V})(V) is zero at x = 0 x = 0 x=0\mathrm{x}=0x=0, then its value at x = + x x = + x x=+x\mathrm{x}=+\mathrm{x}x=+x will be-
(1) V x = x E 0 V x = x E 0 V_(x)=xE_(0)\mathrm{V}_{\mathrm{x}}=\mathrm{x} \mathrm{E}_{0}Vx=xE0
(2) V x = − x ⋅ E 0 V x = − x ⋅ E 0 V_(x)=-x*E_(0)\mathrm{V}_{\mathrm{x}}=-\mathrm{x} \cdot \mathrm{E}_{0}Vx=−x⋅E0
(3) V x = x 2 E 0 V x = x 2 E 0 V_(x)=x^(2)E_(0)\mathrm{V}_{\mathrm{x}}=\mathrm{x}^{2} \mathrm{E}_{0}Vx=x2E0
(4) V x = x 2 E 0 V x = x 2 E 0 V_(x)=x^(2)E_(0)\mathrm{V}_{\mathrm{x}}=\mathrm{x}^{2} \mathrm{E}_{0}Vx=x2E0
Q.81 The electric potential V V V\mathrm{V}V at any point ( x , y , z ) ( x , y , z ) (x,y,z)(\mathrm{x}, \mathrm{y}, \mathrm{z})(x,y,z) in space is given by V = 4 x 2 V = 4 x 2 V=4x^(2)\mathrm{V}=4 \mathrm{x}^{2}V=4x2 volt. The electric field E E E\mathrm{E}E in V / m V / m V//m\mathrm{V} / \mathrm{m}V/m at the point ( 1 , 0 , 2 ) ( 1 , 0 , 2 ) (1,0,2)(1,0,2)(1,0,2) is -
(1) +8 in x x xxx direction
(2) 8 in − x − x -x-\mathrm{x}−x direction
(3) 16 in + x + x +x+x+x direction
(4) 16 in − x − x -x-\mathrm{x}−x direction
Q.82 Charges of + ( 10 3 ) × 10 − 9 + 10 3 × 10 − 9 +((10)/(3))xx10^(-9)+\left(\frac{10}{3}\right) \times 10^{-9}+(103)×10−9 are placed at each of the four corners of a square of side 8 c m 8 c m 8cm8 \mathrm{~cm}8 cm. The potential at the intersection of the diagonals is
(1) 150 2 150 2 150sqrt2150 \sqrt{2}1502 Volt
(2) 1500 2 1500 2 1500sqrt21500 \sqrt{2}15002 Volt
(3) 900 2 900 2 900sqrt2900 \sqrt{2}9002 Volt
(4) 900 Volt
Q.83 The electric potential (V) as a function of distance (x) [in meters] is given by V = ( 5 x 2 + 10 x − 9 ) V = 5 x 2 + 10 x − 9 V=(5x^(2)+10x-9)\mathrm{V}=\left(5 \mathrm{x}^{2}+10 \mathrm{x}-9\right)V=(5x2+10x−9) Volt. The value of electric field at x = 1 m x = 1 m x=1m\mathrm{x}=1 \mathrm{~m}x=1 m would be-
(1) 20 Volt / m / m //m/ m/m (2) 6 V o l t / m 6 V o l t / m 6Volt//m6 \mathrm{Volt} / \mathrm{m}6Volt/m
(3) 11 V o l t / m 11 V o l t / m 11Volt//m11 \mathrm{Volt} / \mathrm{m}11Volt/m
(4) -23 Volt / m / m //m/ \mathrm{m}/m
Q.84 A positive point charge q q q\mathrm{q}q is carried from a point B B B\mathrm{B}B to a point A A A\mathrm{A}A in the electric field of a point charge + Q + Q +Q+Q+Q at O O OOO. If the permittivity of free space is ε 0 ε 0 epsi_(0)\varepsilon_{0}ε0, the work done in the process is given by ( ( ((( where a = O A a = O A a=OA\mathrm{a}=\mathrm{OA}a=OA and b = O B ) b = O B ) b=OB)\mathrm{b}=\mathrm{OB})b=OB) -
(1) q Q 4 π ε 0 ( 1 a + 1 b ) q Q 4 π ε 0 1 a + 1 b (qQ)/(4piepsi_(0))((1)/(a)+(1)/(b))\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a}+\frac{1}{b}\right)qQ4πε0(1a+1b)
(2) q Q 4 π ε 0 ( 1 a − 1 b ) q Q 4 π ε 0 1 a − 1 b (qQ)/(4piepsi_(0))((1)/(a)-(1)/(b))\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a}-\frac{1}{b}\right)qQ4πε0(1a−1b)
(3) q Q 4 π ε 0 ( 1 a 2 − 1 b 2 ) q Q 4 π ε 0 1 a 2 − 1 b 2 (qQ)/(4piepsi_(0))((1)/(a^(2))-(1)/(b^(2)))\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)qQ4πε0(1a2−1b2)
(4) q Q 4 π ε 0 ( 1 a 2 + 1 b 2 ) q Q 4 π ε 0 1 a 2 + 1 b 2 (qQ)/(4piepsi_(0))((1)/(a^(2))+(1)/(b^(2)))\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)qQ4πε0(1a2+1b2)

61. ELECTROSTATICS

Q.85 Some equipotential lines are as shown is fig. E 1 , E 2 E 1 , E 2 E_(1),E_(2)\mathrm{E}_{1}, \mathrm{E}_{2}E1,E2 and E 3 E 3 E_(3)\mathrm{E}_{3}E3 are the electric fields at points 1,2 and 3 then -
(1) E 1 = E 2 = E 3 E 1 = E 2 = E 3 E_(1)=E_(2)=E_(3)\mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}E1=E2=E3
(2) E 1 > E 2 > E 3 E 1 > E 2 > E 3 E_(1) > E_(2) > E_(3)\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}E1>E2>E3
(3) E 1 > E 2 , E 2 < E 3 E 1 > E 2 , E 2 < E 3 E_(1) > E_(2),E_(2) < E_(3)\mathrm{E}_{1}>\mathrm{E}_{2}, \mathrm{E}_{2}<\mathrm{E}_{3}E1>E2,E2<E3
(4) E 1 < E 2 < E 3 E 1 < E 2 < E 3 E_(1) < E_(2) < E_(3)\mathrm{E}_{1}<\mathrm{E}_{2}<\mathrm{E}_{3}E1<E2<E3
Q.86 A α α alpha\alphaα-particle moves towards a rest nucleus, if kinetic energy of α α alpha\alphaα-particle is 10 M e V 10 M e V 10MeV10 \mathrm{MeV}10MeV and atomic number of nucleus is 50 . The closest approach will be -
(1) 1.44 × 10 − 14 m 1.44 × 10 − 14 m 1.44 xx10^(-14)m1.44 \times 10^{-14} \mathrm{~m}1.44×10−14 m
(2) 2.88 × 10 − 14 m 2.88 × 10 − 14 m 2.88 xx10^(-14)m2.88 \times 10^{-14} \mathrm{~m}2.88×10−14 m
(3) 1.44 × 10 − 10 m 1.44 × 10 − 10 m 1.44 xx10^(-10)m1.44 \times 10^{-10} \mathrm{~m}1.44×10−10 m
(4) 2.88 × 10 − 10 m 2.88 × 10 − 10 m 2.88 xx10^(-10)m2.88 \times 10^{-10} \mathrm{~m}2.88×10−10 m
Q.87 A , B , C A , B , C A,B,CA, B, CA,B,C and D D DDD are four points on an imaginary circle in region containing uniform electric field as shown in figure. Select the incorrect option
(1) V B > V A V B > V A V_(B) > V_(A)V_{B}>V_{A}VB>VA
(2) V B > V C V B > V C V_(B) > V_(C)V_{B}>V_{C}VB>VC
(3) V B < V D V B < V D V_(B) < V_(D)V_{B}<V_{D}VB<VD
(4) V A > V D V A > V D V_(A) > V_(D)V_{A}>V_{D}VA>VD
Q.88 When a negative charge is released and moves in electric field, it moves toward a position of
(1) lower electric potential and lower potential energy
(2) lower electric potential and higher potential energy
(3) higher electric potential and lower potential energy
(4) higher electric potential and higher potential energy
Q.89 Four equal charges of charge q are placed at corner of a square of side a. Potential energy of the whole system is-
(1) 4 k q 2 a 4 k q 2 a (4kq^(2))/(a)\frac{4 k q^{2}}{a}4kq2a
(2) 4 k q a ( 1 + 1 2 2 ) 4 k q a 1 + 1 2 2 (4kq)/(a)(1+(1)/(2sqrt2))\frac{4 \mathrm{kq}}{\mathrm{a}}\left(1+\frac{1}{2 \sqrt{2}}\right)4kqa(1+122)
(3) 1 2 2 k q 2 a 1 2 2 k q 2 a (1)/(2sqrt2)(kq^(2))/(a)\frac{1}{2 \sqrt{2}} \frac{k q^{2}}{a}122kq2a
(4) k q 2 a ( 4 + 1 2 2 ) k q 2 a 4 + 1 2 2 (kq^(2))/(a)(4+(1)/(2sqrt2))\frac{k q^{2}}{a}\left(4+\frac{1}{2 \sqrt{2}}\right)kq2a(4+122)
Q.90 Two points P P P\mathrm{P}P and Q Q Q\mathrm{Q}Q are maintained at the potentials of 10 V 10 V 10V10 \mathrm{~V}10 V and − 4 V − 4 V -4V-4 \mathrm{~V}−4 V, respectively. The work done in moving 100 electrons from P P P\mathrm{P}P to Q Q Q\mathrm{Q}Q is -
(1) − 9.60 × 10 − 17 J − 9.60 × 10 − 17 J -9.60 xx10^(-17)J-9.60 \times 10^{-17} \mathrm{~J}−9.60×10−17 J
(2) 9.60 × 10 − 17 J 9.60 × 10 − 17 J 9.60 xx10^(-17)J9.60 \times 10^{-17} \mathrm{~J}9.60×10−17 J
(3) − 2.24 × 10 − 16 J − 2.24 × 10 − 16 J -2.24 xx10^(-16)J-2.24 \times 10^{-16} \mathrm{~J}−2.24×10−16 J
(4) 2.24 × 10 − 16 J 2.24 × 10 − 16 J 2.24 xx10^(-16)J2.24 \times 10^{-16} \mathrm{~J}2.24×10−16 J
Q.91 The work done required to put the four charges together at the corners of a square of side a, as shown in the figure is :

(1) 1 4 π ε 0 q 2 a 1 4 π ε 0 q 2 a (1)/(4piepsi_(0))(q^(2))/(a)\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a}14πε0q2a
(2) − 2.6 4 π ε 0 q 2 a − 2.6 4 π ε 0 q 2 a -(2.6)/(4piepsi_(0))(q^(2))/(a)-\frac{2.6}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a}−2.64πε0q2a
(3) + 2.6 4 π ε 0 q 2 a + 2.6 4 π ε 0 q 2 a +(2.6)/(4piepsi_(0))(q^(2))/(a)+\frac{2.6}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a}+2.64πε0q2a
(4) None of these

62. ELECTROSTATICS

Q.92 Three charges − q , + Q − q , + Q -q,+Q-\mathrm{q},+\mathrm{Q}−q,+Q and − q − q -q-\mathrm{q}−q are placed in a straight line as shown.
If the total potential energy of the system is zero, then the ratio q Q q Q (q)/(Q)\frac{\mathrm{q}}{\mathrm{Q}}qQ is
(1) 2
(2) 5.5
(3) 4
(4) 1.5
Q.93 Electric field at a distance x x x\mathrm{x}x from origin is given as E = 100 x 2 E = 100 x 2 E=(100)/(x^(2))\mathrm{E}=\frac{100}{\mathrm{x}^{2}}E=100x2, then potential between points situated at x = 10 m x = 10 m x=10m\mathrm{x}=10 \mathrm{~m}x=10 m and x = 20 m x = 20 m x=20m\mathrm{x}=20 \mathrm{~m}x=20 m.
(1) 5 V 5 V 5V5 \mathrm{~V}5 V
(2) 10 V 10 V 10V10 \mathrm{~V}10 V
(3) 15 V 15 V 15V15 \mathrm{~V}15 V
(4) 4 V 4 V 4V4 \mathrm{~V}4 V
Q.94 Figure shows a set of equipotential surfaces. The magnitude and direction of electric field that exists in the region is-
(1) 10 2 V / m 10 2 V / m 10sqrt2V//m10 \sqrt{2} \mathrm{~V} / \mathrm{m}102 V/m at 45 ∘ 45 ∘ 45^(@)45^{\circ}45∘ with x x x\mathrm{x}x-axis
(2) 10 2 V / m 10 2 V / m 10sqrt2V//m10 \sqrt{2} \mathrm{~V} / \mathrm{m}102 V/m at − 45 ∘ − 45 ∘ -45^(@)-45^{\circ}−45∘ with x x x\mathrm{x}x-axis
(3) 5 2 V / m 5 2 V / m 5sqrt2V//m5 \sqrt{2} \mathrm{~V} / \mathrm{m}52 V/m at 45 ∘ 45 ∘ 45^(@)45^{\circ}45∘ with x x xxx-axis
(4) 5 2 V / m 5 2 V / m 5sqrt2V//m5 \sqrt{2} \mathrm{~V} / \mathrm{m}52 V/m at − 45 ∘ − 45 ∘ -45^(@)-45^{\circ}−45∘ with x x x\mathrm{x}x-axis
Q.95 The variation of potential with distance R R R\mathrm{R}R from a fixed point is as shown in the figure. The electric field at R = 5 m R = 5 m R=5mR=5 \mathrm{~m}R=5 m is :
(1) 2.5 V / m 2.5 V / m 2.5V//m2.5 \mathrm{~V} / \mathrm{m}2.5 V/m
( 2 ) − 2.5 V / m ( 2 ) − 2.5 V / m (2)-2.5V//m(2)-2.5 \mathrm{~V} / \mathrm{m}(2)−2.5 V/m
(3) ( 2 / 5 ) V / m ( 2 / 5 ) V / m (2//5)V//m(2 / 5) \mathrm{V} / \mathrm{m}(2/5)V/m
(4) − ( 2 / 5 ) V / m − ( 2 / 5 ) V / m -(2//5)V//m-(2 / 5) \mathrm{V} / \mathrm{m}−(2/5)V/m
Q.96 A hollow conducting sphere of radius R R R\mathrm{R}R has charge ( + Q ) ( + Q ) (+Q)(+\mathrm{Q})(+Q) on its surface. The electric potential within the sphere at a distance r = R 3 r = R 3 r=(R)/(3)r=\frac{R}{3}r=R3 from the centre is -
(1) Zero
(2) 1 4 π ε 0 Q r 1 4 π ε 0 Q r (1)/(4piepsi_(0))(Q)/(r)\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}14πε0Qr
(3) 1 4 π ε 0 Q R 1 4 π ε 0 Q R (1)/(4piepsi_(0))(Q)/(R)\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}14πε0QR
(4) 1 4 π ε 0 Q r 2 1 4 π ε 0 Q r 2 (1)/(4piepsi_(0))(Q)/(r^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}14πε0Qr2
Q.97 A large isolated metal sphere of radius (R) carries a fixed charge. A small charge is placed at a distance (r) from its surface experiences a force which is -
(1) Proportional to R R R\mathrm{R}R
(2) Independent of R R R\mathrm{R}R and
(3) Inversely proportional to ( R + r ) 2 ( R + r ) 2 (R+r)^(2)(\mathrm{R}+\mathrm{r})^{2}(R+r)2
(4) inversely proportional to r 2 r 2 r^(2)r^{2}r2
Q.98 A uniform hollow sphere of charge does not produce an electric field at any-
(1) Interior point
(2) Outer point
(3) Surface point
(4) None of the above

63. ELECTROSTATICS

Q.99 Which of the following graphs show how the electric field strength E varies with distance r r rrr from the centre of the some conducting sphere ?
Q.100 The electric field intensity at P P P\mathrm{P}P and Q Q Q\mathrm{Q}Q, in the shown arrangement, are in the ratio :

(1) 1 : 2 1 : 2 1:21: 21:2
(2) 2 : 1 2 : 1 2:12: 12:1
(3) 1 : 1 1 : 1 1:11: 11:1
(4) 4 : 1 4 : 1 4:14: 14:1
Q.101 Two isolated metallic spheres of radii 2 c m 2 c m 2cm2 \mathrm{~cm}2 cm and 4 c m 4 c m 4cm4 \mathrm{~cm}4 cm are given equal charge, then the ratio of charge density on the surfaces of the spheres will be ? :
(1) 1 : 2 1 : 2 1:21: 21:2
(2) 4 : 1 4 : 1 4:14: 14:1
(3) 8 : 1 8 : 1 8:18: 18:1
(4) 1 : 4 1 : 4 1:41: 41:4
Q.102 Two concentric hollow conducting spheres of radius r r r\mathrm{r}r and R R R\mathrm{R}R are shown. The charge on outer shell is Q Q Q\mathrm{Q}Q. What charge should be given to inner sphere so that the potential at any point P P P\mathrm{P}P outside the outer sphere is zero ?

(1) − Q r R − Q r R -(Qr)/(R)-\frac{\mathrm{Qr}}{\mathrm{R}}−QrR
(2) − Q R r − Q R r -(QR)/(r)-\frac{\mathrm{QR}}{\mathrm{r}}−QRr
( 3 ) − Q ( 3 ) − Q (3)-Q(3)-Q(3)−Q
(4) − 2 Q R r − 2 Q R r -(2QR)/(r)-\frac{2 Q R}{r}−2QRr

64. ELECTRIC DIPOLE

Q.103 An electric dipole with dipole moment P → = ( 2 i ^ + 3 j ^ ) c m P → = ( 2 i ^ + 3 j ^ ) c m vec(P)=(2 hat(i)+3 hat(j))cm\vec{P}=(2 \hat{i}+3 \hat{j}) \mathrm{cm}P→=(2i^+3j^)cm is kept in electric field E → = 4 i ^ N / C E → = 4 i ^ N / C vec(E)=4 hat(i)N//C\vec{E}=4 \hat{i} \mathrm{~N} / \mathrm{C}E→=4i^ N/C. The torque acting on it is
(1) − 12 k ^ ( N m ) − 12 k ^ ( N m ) -12 hat(k)(Nm)-12 \hat{k}(\mathrm{Nm})−12k^(Nm)
(2) 8 k ^ 8 k ^ 8 hat(k)8 \hat{k}8k^
(3) 12 k ^ ( N m ) 12 k ^ ( N m ) 12 hat(k)(Nm)12 \hat{k}(\mathrm{Nm})12k^(Nm)
(4) − 8 k ^ − 8 k ^ -8 hat(k)-8 \hat{k}−8k^
Q.104 An electric dipole when placed in a uniform electric field E → E → vec(E)\vec{E}E→ will have maximum potential energy if the dipole moment makes the following angle with E → E → vec(E)\vec{E}E→
( 1 ) π ( 1 ) π (1)pi(1) \pi(1)π
(2) 3 π 2 3 π 2 (3pi)/(2)\frac{3 \pi}{2}3π2
(3) Zero
(4) π 2 π 2 (pi)/(2)\frac{\pi}{2}π2

65. ELECTROSTATICS

Q.105 An electric dipole has the magnitude of its charge as q q q\mathrm{q}q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively
(1) 2 q.E and minimum
(2) q.E and p.E
(3) Zero and minimum
(4) Zero and maximum
Q.106 An electric dipole is placed in non-uniform electric field. It may experience :
(1) Resultant force and couple
(2) Only resultant force
(3) Only couple
(4) All of these
Q.107 An electron released on the axis of a positively charged ring at a large distance from the centre will :
(1) Not move
(2) Do oscillatory motion
(3) Do SHM
(4) Do non periodic motion
Q.108 Two charges of + 25 × 10 − 9 + 25 × 10 − 9 +25 xx10^(-9)+25 \times 10^{-9}+25×10−9 coulomb and − 25 × 10 − 9 − 25 × 10 − 9 -25 xx10^(-9)-25 \times 10^{-9}−25×10−9 coulomb are placed 6 m 6 m 6m6 \mathrm{~m}6 m apart. Find the electric field intensity ratio at points 4 m 4 m 4m4 \mathrm{~m}4 m from the centre of the electric dipole (i) on axial line (ii) on equatorial line:
(1) 1000 49 1000 49 (1000)/(49)\frac{1000}{49}100049
(2) 49 1000 49 1000 (49)/(1000)\frac{49}{1000}491000
(3) 500 49 500 49 (500)/(49)\frac{500}{49}50049
(4) 49 500 49 500 (49)/(500)\frac{49}{500}49500
Q.109 The electric force on a point charge situated on the axis of a short dipole is F. If the charge is shifted along the axis to do+uble the distance, the electric force acting will be :
(1) 4 F 4 F 4F4 \mathrm{~F}4 F
(2) F 2 F 2 (F)/(2)\frac{F}{2}F2
(3) F 4 F 4 (F)/(4)\frac{F}{4}F4
(4) F 8 F 8 (F)/(8)\frac{F}{8}F8
Q.110 An electric dipole when placed in a uniform electric field E will have minimum potential energy, when the angle made by dipole moment with field E E E\mathrm{E}E is :
(1) π π pi\piπ
(2) 3 π 2 3 π 2 (3pi)/(2)\frac{3 \pi}{2}3π2
(3) Zero
(4) π 2 π 2 (pi)/(2)\frac{\pi}{2}π2
Q.111 The electric potential in volts due to an electric dipole of dipole moment 2 × 10 − 8 2 × 10 − 8 2xx10^(-8)2 \times 10^{-8}2×10−8 coulomb-metre at a distance of 3 m 3 m 3m3 \mathrm{~m}3 m on a line making an angle of 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ with the axis of the dipole is :
(1) 0
(2) 10
(3) 20
(4) 40
Q.112 An electric dipole of length 2 c m 2 c m 2cm2 \mathrm{~cm}2 cm is placed with its axis making an angle of 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘ to a uniform electric field 10 5 N / C 10 5 N / C 10^(5)N//C10^{5} \mathrm{~N} / \mathrm{C}105 N/C. If it experiences a torque of 10 3 N m 10 3 N m 10sqrt3Nm10 \sqrt{3} \mathrm{Nm}103Nm, then potential energy of the dipole
( 1 ) − 10 J ( 1 ) − 10 J (1)-10J(1)-10 \mathrm{~J}(1)−10 J
( 2 ) − 20 J ( 2 ) − 20 J (2)-20J(2)-20 \mathrm{~J}(2)−20 J
( 3 ) − 30 J ( 3 ) − 30 J (3)-30J(3)-30 \mathrm{~J}(3)−30 J
( 4 ) − 40 J ( 4 ) − 40 J (4)-40J(4)-40 \mathrm{~J}(4)−40 J
Q.113 If an electric dipole is kept in a non-uniform electric field, then it will experience -
(1) only torque
(2) no torque
(3) a resultant force and a torque
(4) only a force

66. ELECTROSTATICS

Q.114 A dipole of dipole moment p p p\mathrm{p}p, is placed in an electric field E → E → vec(E)\overrightarrow{\mathrm{E}}E→ and is in stable equilibrium. The torque required to rotate the dipole from this position by angle θ θ theta\thetaθ will be -
(1) p E cos ⁡ θ p E cos ⁡ θ pEcos theta\mathrm{pE} \cos \thetapEcos⁡θ
(2) p E sin ⁡ θ p E sin ⁡ θ pEsin theta\mathrm{pE} \sin \thetapEsin⁡θ
(3) p E tan ⁡ θ p E tan ⁡ θ pEtan theta\mathrm{pE} \tan \thetapEtan⁡θ
(4) − p E cos ⁡ θ − p E cos ⁡ θ -pEcos theta-\mathrm{pE} \cos \theta−pEcos⁡θ
Q.115 The electric potential at a point due to an electric dipole will be -
(1) k ( p → ⋅ r → ) r 3 k ( p → ⋅ r → ) r 3 (k(( vec(p))*( vec(r))))/(r^(3))\frac{k(\vec{p} \cdot \vec{r})}{r^{3}}k(p→⋅r→)r3
(2) k ( p → ⋅ r → ) r 2 k ( p → ⋅ r → ) r 2 (k(( vec(p))*( vec(r))))/(r^(2))\frac{k(\vec{p} \cdot \vec{r})}{r^{2}}k(p→⋅r→)r2
(3) k ( p → × r → ) r k ( p → × r → ) r (k(( vec(p))xx( vec(r))))/(r)\frac{k(\vec{p} \times \vec{r})}{r}k(p→×r→)r
(4) k ( p → × r → ) r 2 k ( p → × r → ) r 2 (k(( vec(p))xx( vec(r))))/(r^(2))\frac{k(\vec{p} \times \vec{r})}{r^{2}}k(p→×r→)r2
Q.116 An electric dipole is made up of two equal and opposite charges of 2 × 10 − 6 2 × 10 − 6 2xx10^(-6)2 \times 10^{-6}2×10−6 coulomb at a distance of 3 c m 3 c m 3cm3 \mathrm{~cm}3 cm. This is kept in an electric field of 2 × 10 5 N / C 2 × 10 5 N / C 2xx10^(5)N//C2 \times 10^{5} \mathrm{~N} / \mathrm{C}2×105 N/C, then the maximum torque acting on the dipole -
(1) 12 × 10 − 1 N m 12 × 10 − 1 N m 12 xx10^(-1)Nm12 \times 10^{-1} \mathrm{Nm}12×10−1Nm
(2) 12 × 10 − 3 N m 12 × 10 − 3 N m 12 xx10^(-3)Nm12 \times 10^{-3} \mathrm{Nm}12×10−3Nm
(3) 24 × 10 − 3 N m 24 × 10 − 3 N m 24 xx10^(-3)Nm24 \times 10^{-3} \mathrm{Nm}24×10−3Nm
(4) 24 × 10 − 1 N m 24 × 10 − 1 N m 24 xx10^(-1)Nm24 \times 10^{-1} \mathrm{Nm}24×10−1Nm
Q.117 The electric potential in volt at a distance of 0.01 m 0.01 m 0.01m0.01 \mathrm{~m}0.01 m on the equatorial line of an electric dipole of dipole moment p p p\mathrm{p}p is -
(1) p / 4 π ϵ 0 × 10 − 4 p / 4 π ϵ 0 × 10 − 4 p//4piepsilon_(0)xx10^(-4)p / 4 \pi \epsilon_{0} \times 10^{-4}p/4πϵ0×10−4
(2) zero
(3) 4 π ϵ 0 p × 10 − 4 4 π ϵ 0 p × 10 − 4 4piepsilon_(0)p xx10^(-4)4 \pi \epsilon_{0} p \times 10^{-4}4πϵ0p×10−4
(4) 4 π ∈ 0 / p × 10 − 4 4 π ∈ 0 / p × 10 − 4 4piin_(0)//p xx10^(-4)4 \pi \in_{0} / p \times 10^{-4}4π∈0/p×10−4
Q.118 When an electric dipole p → p → vec(p)\overrightarrow{\mathrm{p}}p→ is kept in a uniform electric field E → E → vec(E)\overrightarrow{\mathrm{E}}E→ then for what value of the angle between p → p → vec(p)\overrightarrow{\mathrm{p}}p→ and E → E → vec(E)\overrightarrow{\mathrm{E}}E→, torque will be maximum :-
(1) 90 ∘ 90 ∘ 90^(@)90^{\circ}90∘
(2) 0 ∘ 0 ∘ 0^(@)0^{\circ}0∘
(3) 180 ∘ 180 ∘ 180^(@)180^{\circ}180∘
(4) 45 ∘ 45 ∘ 45^(@)45^{\circ}45∘
Q.119 What will be the ratio of electric field at the axis and at equatorial line of a dipole :-
(1) 1 : 2 1 : 2 1:21: 21:2
(2) 2 : 1 2 : 1 2:12: 12:1
(3) 4 : 1 4 : 1 4:14: 14:1
(4) 1 : 4 1 : 4 1:41: 41:4
Q.120 For a dipole q = 2 × 10 − 6 C ; d = 0.01 m q = 2 × 10 − 6 C ; d = 0.01 m q=2xx10^(-6)C;d=0.01m\mathrm{q}=2 \times 10^{-6} \mathrm{C} ; \mathrm{d}=0.01 \mathrm{~m}q=2×10−6C;d=0.01 m find the maximum torque on the dipole if E = 5 × 10 5 N / C : − E = 5 × 10 5 N / C : − E=5xx10^(5)N//C:-\mathrm{E}=5 \times 10^{5} \mathrm{~N} / \mathrm{C}:-E=5×105 N/C:−
(1) 1 × 10 − 3 N m − 1 1 × 10 − 3 N m − 1 1xx10^(-3)Nm^(-1)1 \times 10^{-3} \mathrm{Nm}^{-1}1×10−3Nm−1
(2) 10 × 10 − 3 N m − 1 10 × 10 − 3 N m − 1 10 xx10^(-3)Nm^(-1)10 \times 10^{-3} \mathrm{Nm}^{-1}10×10−3Nm−1
(3) 10 × 10 − 3 N m 10 × 10 − 3 N m 10 xx10^(-3)Nm10 \times 10^{-3} \mathrm{Nm}10×10−3Nm
(4) 1 × 10 2 N m 2 1 × 10 2 N m 2 1xx10^(2)Nm^(2)1 \times 10^{2} \mathrm{Nm}^{2}1×102Nm2
Q.121 In an electric field electric dipole is rotated though an angle θ θ theta\thetaθ, then work done will be
(1) p E ( 1 − cos ⁡ θ ) p E ( 1 − cos ⁡ θ ) pE(1-cos theta)\mathrm{pE}(1-\cos \theta)pE(1−cos⁡θ)
(2) p E sin ⁡ θ p E sin ⁡ θ pEsin theta\mathrm{pE} \sin \thetapEsin⁡θ
(3) zero
(4) − p E cos ⁡ θ − p E cos ⁡ θ -pEcos theta-\mathrm{pE} \cos \theta−pEcos⁡θ

67. EXERCISE-2

ELECTROSTATICS
Q.1 A charged ball B hangs from a silk thread S, which makes an angle θ θ theta\thetaθ with a large charged conducting sheet P P PPP, as shown in the figure. The surface charge density σ σ sigma\sigmaσ of the sheet is proportional to

(1) cot ⁡ θ cot ⁡ θ cot theta\cot \thetacot⁡θ
( 2 ) cos ⁡ θ ( 2 ) cos ⁡ θ (2)cos theta(2) \cos \theta(2)cos⁡θ
(3) tan ⁡ θ tan ⁡ θ tan theta\tan \thetatan⁡θ
(4) sin ⁡ θ sin ⁡ θ sin theta\sin \thetasin⁡θ
Q.2 V = V = V=\mathrm{V}=V= axy, then electric field at a point will be proportional to :
(1) r r rrr
( 2 ) r − 1 ( 2 ) r − 1 (2)r^(-1)(2) \mathrm{r}^{-1}(2)r−1
(3) r − 2 r − 2 r^(-2)\mathrm{r}^{-2}r−2
(4) r 2 r 2 r^(2)r^{2}r2
Q.3 The insulation property of air breaks down at intensity as 3 × 10 6 V / m 3 × 10 6 V / m 3xx10^(6)V//m3 \times 10^{6} \mathrm{~V} / \mathrm{m}3×106 V/m. The maximum charge that can be given to a sphere of diameter 5 m 5 m 5m5 \mathrm{~m}5 m is :
(1) 2 × 10 − 2 C 2 × 10 − 2 C 2xx10^(-2)C2 \times 10^{-2} \mathrm{C}2×10−2C
(2) 2 × 10 − 3 C 2 × 10 − 3 C 2xx10^(-3)C2 \times 10^{-3} \mathrm{C}2×10−3C
(3) 2 × 10 − 4 C 2 × 10 − 4 C 2xx10^(-4)C2 \times 10^{-4} \mathrm{C}2×10−4C
(4) 0
Q.4 If a square coil is making an angle 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ with electric field E E E\mathrm{E}E according to figure, the electric flux passing through the square coil is (the side of square is 4 c m 4 c m 4cm4 \mathrm{~cm}4 cm ) :

(1) 853 E 853 E 853E853 \mathrm{E}853E
(2) 8 E 8 E 8E8 \mathrm{E}8E
(3) 16 E 16 E 16E16 \mathrm{E}16E
(4) none of these
Q.5 If n n n\mathrm{n}n drops of potential V V V\mathrm{V}V merge, find new potential on the big drop :
(1) n 2 / 3 V n 2 / 3 V n^(2//3)Vn^{2 / 3} \mathrm{~V}n2/3 V
( 2 ) n 1 / 3 V ( 2 ) n 1 / 3 V (2)n^(1//3)V(2) n^{1 / 3} \mathrm{~V}(2)n1/3 V
(3) n V n V nV\mathrm{nV}nV
(4) V n / 3 V n / 3 V^(n//3)\mathrm{V}^{\mathrm{n} / 3}Vn/3
Q.6 Two conducting spheres of radii R 1 R 1 R_(1)\mathrm{R}_{1}R1 and R 2 R 2 R_(2)\mathrm{R}_{2}R2 respectively are charged and joined by a wire. The ratio of electric fields of spheres is
(1) R 2 2 R 1 2 R 2 2 R 1 2 (R_(2)^(2))/(R_(1)^(2))\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}^{2}}R22R12
(2) R 1 2 R 2 2 R 1 2 R 2 2 (R_(1)^(2))/(R_(2)^(2))\frac{R_{1}^{2}}{R_{2}^{2}}R12R22
(3) R 2 R 1 R 2 R 1 (R_(2))/(R_(1))\frac{R_{2}}{R_{1}}R2R1
(4) R 1 R 2 R 1 R 2 (R_(1))/(R_(2))\frac{R_{1}}{R_{2}}R1R2
Q.7 Two equal negative charges − q − q -q-\mathrm{q}−q, are placed at points ( 0 , a ) ( 0 , a ) (0,a)(0, a)(0,a) and ( 0 , − a ) ( 0 , − a ) (0,-a)(0,-\mathrm{a})(0,−a) on y axis, one positive charge q q qqq at rest is left to move from point ( 2 a , 0 ) ( 2 a , 0 ) (2a,0)(2 \mathrm{a}, 0)(2a,0). This charge will be
(1) execute S.H.M. about the origin.
(2) oscillate but not execute S. H. M.
(3) move towards origin and will become stationary.
(4) S. H. M. along x x xxx axis.

68. ELECTROSTATICS

Q.8 Two point charges placed at a distance ' r r rrr ' in air exert a force ' F F FFF '. The value of distance at which they exerts same force when placed in medium (dielectric constant K K K\mathrm{K}K ) is :-
(1) r K r K rK\mathrm{rK}rK
(2) r / K r / K r//K\mathrm{r} / \mathrm{K}r/K
(3) r / K r / K r//sqrtK\mathrm{r} / \sqrt{\mathrm{K}}r/K
(4) r K r K rsqrtKr \sqrt{\mathrm{K}}rK
Q.9 A charge q q q\mathrm{q}q is placed in the middle of a line joining the two equal and like point charges Q. This system will remain in equilibrium for which the value of q q qqq is -
(1) − Q 3 − Q 3 -(Q)/(3)-\frac{\mathrm{Q}}{3}−Q3
( 2 ) − Q 4 ( 2 ) − Q 4 (2)-(Q)/(4)(2)-\frac{Q}{4}(2)−Q4
(3) Q 2 Q 2 (Q)/(2)\frac{Q}{2}Q2
(4) − Q 2 − Q 2 -(Q)/(2)-\frac{Q}{2}−Q2
Q.10 There is a uniform electric field of strength 10 3 V / m 10 3 V / m 10^(3)V//m10^{3} \mathrm{~V} / \mathrm{m}103 V/m along y y y\mathrm{y}y-axis. A body of mass 1 g 1 g 1g1 \mathrm{~g}1 g and charge 10 − 6 C 10 − 6 C 10^(-6)C10^{-6} \mathrm{C}10−6C is projected into the field from origin along the positive x x x\mathrm{x}x-axis with a velocity 10 m / s 10 m / s 10m//s10 \mathrm{~m} / \mathrm{s}10 m/s. Its speed in m / s m / s m//s\mathrm{m} / \mathrm{s}m/s after 10 s 10 s 10s10 \mathrm{~s}10 s is (Neglect gravitation)
(1) 10
(2) 5 2 5 2 5sqrt25 \sqrt{2}52
(3) 10 2 10 2 10sqrt210 \sqrt{2}102
(4) 20
Q.11 Three point charges are placed at the corners of an equilateral triangle. Assuming only electrostatics forces are acting-
(1) if the charges have different magnitudes and different signs, the system will be in equilibrium.
(2) the system will be in equilibrium if the charges have the same magnitudes but different signs.
(3) the system can never be in equillibrium.
(4) the system will be in equillibrium if the charges rotate about the centre of the triangle.
Q.12 An electron and a proton are set free in a uniform electric field. The ratio of their acceleration is
(1) unity
(2) zero
(3) m p m e m p m e (m_(p))/(m_(e))\frac{m_{p}}{m_{e}}mpme
(4) m e m p m e m p (m_(e))/(m_(p))\frac{m_{e}}{m_{p}}memp
Q.13 Two point charges + 9 e + 9 e +9e+9 \mathrm{e}+9e and + e + e +e+\mathrm{e}+e are kept 16 c m 16 c m 16cm16 \mathrm{~cm}16 cm. apart to each other. Where should a third charge q q q\mathrm{q}q be placed between them so that the system remains in the equilibrium state :-
(1) 24 c m f r o m + 9 e 24 c m f r o m + 9 e 24cmfrom+9e24 \mathrm{~cm} \mathrm{from}+9 \mathrm{e}24 cmfrom+9e
(2) 12 c m 12 c m 12cm12 \mathrm{~cm}12 cm from + 9 e + 9 e +9e+9 \mathrm{e}+9e
(3) 24 c m 24 c m 24cm24 \mathrm{~cm}24 cm from + e + e +e+\mathrm{e}+e
(4) 12 c m 12 c m 12cm12 \mathrm{~cm}12 cm from + e + e +e+\mathrm{e}+e
Q.14 Conservation of charge is a consequence of
(1) Columb law
(2) Gauss law
(3) continuity equation
(4) Hygen's wave equation
Q.15 A sphere of radius R R R\mathrm{R}R, is charged uniformly with total charge Q Q Q\mathrm{Q}Q. Then correct statement for electric field is ( r = r = r=\mathrm{r}=r= distance from centre ) ) ))) :-
(1) K Q r R 3 K Q r R 3 (KQr)/(R^(3))\frac{\mathrm{KQr}}{\mathrm{R}^{3}}KQrR3, where r < R r < R r < R\mathrm{r}<\mathrm{R}r<R
(2) K Q r 2 K Q r 2 (KQ)/(r^(2))\frac{\mathrm{KQ}}{\mathrm{r}^{2}}KQr2, where r ≥ R r ≥ R r >= Rr \geq Rr≥R
(3) it is zero, at all points
(4) (1) and (2) both
Q.16 A ring of radius R R R\mathrm{R}R is charged uniformly with a charge + Q + Q +Q+\mathrm{Q}+Q. The electric field at any point on its axis at a distance r r rrr from the circumference of the ring will be:-
(1) K Q r K Q r (KQ)/(r)\frac{\mathrm{KQ}}{\mathrm{r}}KQr
(2) K Q r 2 K Q r 2 (KQ)/(r^(2))\frac{K Q}{r^{2}}KQr2
(3) K Q r 3 ( r 2 − R 2 ) 1 / 2 K Q r 3 r 2 − R 2 1 / 2 (KQ)/(r^(3))(r^(2)-R^(2))^(1//2)\frac{\mathrm{KQ}}{\mathrm{r}^{3}}\left(\mathrm{r}^{2}-\mathrm{R}^{2}\right)^{1 / 2}KQr3(r2−R2)1/2
(4) K Q r R 3 K Q r R 3 (KQr)/(R^(3))\frac{\mathrm{KQr}}{\mathrm{R}^{3}}KQrR3 ELECTROSTATICS
Q.17 If in Millikan's oil drop experiment charges on drops are found to be 8 μ C , 12 μ C , 20 μ C 8 μ C , 12 μ C , 20 μ C 8muC,12 muC,20 muC8 \mu \mathrm{C}, 12 \mu \mathrm{C}, 20 \mu \mathrm{C}8μC,12μC,20μC, then quanta of charge is :-
(1) 8 μ C 8 μ C 8muC8 \mu \mathrm{C}8μC
(2) 4 μ C 4 μ C 4muC4 \mu \mathrm{C}4μC
(3) 20 μ C 20 μ C 20 muC20 \mu \mathrm{C}20μC
(4) 12 μ C 12 μ C 12 muC12 \mu \mathrm{C}12μC
Q.18 Force between two identical spheres charged with same charge is F F F\mathrm{F}F. If 50 % 50 % 50%50 \%50% charge of one sphere is transferred to second sphere then new force will be :-
(1) 3 4 F 3 4 F (3)/(4)F\frac{3}{4} \mathrm{~F}34 F
(2) 3 8 F 3 8 F (3)/(8)F\frac{3}{8} \mathrm{~F}38 F
(3) 3 2 F 3 2 F (3)/(2)F\frac{3}{2} \mathrm{~F}32 F
(4) none of these
Q.19 Two balls carrying charges + 7 μ C + 7 μ C +7muC+7 \mu \mathrm{C}+7μC and − 5 μ C − 5 μ C -5muC-5 \mu \mathrm{C}−5μC attract each other with a force F F F\mathrm{F}F. If a charge − 2 μ C − 2 μ C -2muC-2 \mu \mathrm{C}−2μC is added to both, the force between them will be :-
(1) F F F\mathrm{F}F
(2) F 2 F 2 (F)/(2)\frac{F}{2}F2
(3) 2 F 2 F 2F2 \mathrm{~F}2 F
(4) zero
Q.20 Two charges of equal magnitude q q q\mathrm{q}q are placed at a distance 2 a 2 a 2a2 \mathrm{a}2a. Another charge q q q\mathrm{q}q of mass m m m\mathrm{m}m, is placed midway between the two charges on X X X\mathrm{X}X-axis. If this charges is displaced from equilibrium state to a distance x ( x << a ) x ( x << a ) x(x<<a)\mathrm{x}(\mathrm{x}<<\mathrm{a})x(x<<a), then the particle :-
(1) will execute simple harmonic motion about equilibrium position
(2) will be oscillating about equilibrium position but will not execute simple harmonic motion
(3) will not return back to the equilibrium position
(4) will stop at equilibrium position
Q.21 In 1 g 1 g 1g1 \mathrm{~g}1 g of a solid, there are 5 × 10 21 5 × 10 21 5xx10^(21)5 \times 10^{21}5×1021 atoms. If one electron is removed from everyone of 0.01 % 0.01 % 0.01%0.01 \%0.01% atoms of the solid, the charge gained by the solid is :- (electronic charge is 1.6 × 10 − 19 C 1.6 × 10 − 19 C 1.6 xx10^(-19)C1.6 \times 10^{-19} \mathrm{C}1.6×10−19C )
(1) + 0.08 C + 0.08 C +0.08C+0.08 \mathrm{C}+0.08C
( 2 ) + 0.8 C ( 2 ) + 0.8 C (2)+0.8C(2)+0.8 \mathrm{C}(2)+0.8C
(3) − 0.08 C − 0.08 C -0.08C-0.08 \mathrm{C}−0.08C
(4) − 0.8 C − 0.8 C -0.8C-0.8 \mathrm{C}−0.8C
Q.22 A point charge q q qqq of mass m m mmm is located at the centre of a ring having radius R R R\mathrm{R}R and charge Q Q Q\mathrm{Q}Q. When it is displaced slightly, the point charge accelerates along the x x xxx-axis to infinity, the ultimate speed of the point charge :-
(1) 2 k Q q m R 2 k Q q m R sqrt((2kQq)/(mR))\sqrt{\frac{2 \mathrm{kQq}}{\mathrm{mR}}}2kQqmR
(2) k Q q m R k Q q m R sqrt((kQq)/(mR))\sqrt{\frac{\mathrm{kQq}}{\mathrm{mR}}}kQqmR
(3) k Q q 2 m R k Q q 2 m R sqrt((kQq)/(2mR))\sqrt{\frac{\mathrm{kQq}}{2 \mathrm{mR}}}kQq2mR
(4) zero
Q.23 A point particle of mass M M M\mathrm{M}M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of same mass is attached to the other end of the rod. The two particles carry charges + q + q +q+\mathrm{q}+q and − q − q -q-\mathrm{q}−q repectively. This arrangement is held in a region of uniform electric field E such that the rod makes a small angle θ ( < 5 0 ) θ < 5 0 theta( < 5^(0))\theta\left(<5^{0}\right)θ(<50) with the field direction. The minimum time needed for the rod to become parallel to the field after it is set free. (rod rotates about centre of mass)
(1) 2 π M L 2 q E 2 π M L 2 q E 2pisqrt((ML)/(2qE))2 \pi \sqrt{\frac{M L}{2 q E}}2πML2qE
(2) π M L 2 q E π M L 2 q E pisqrt((ML)/(2qE))\pi \sqrt{\frac{M L}{2 q E}}πML2qE
(3) π 2 M L 2 q E π 2 M L 2 q E (pi)/(2)sqrt((ML)/(2qE))\frac{\pi}{2} \sqrt{\frac{M L}{2 q E}}π2ML2qE
(4) 4 π M L 2 q E 4 π M L 2 q E 4pisqrt((ML)/(2qE))4 \pi \sqrt{\frac{M L}{2 q E}}4πML2qE

69. ELECTROSTATICS

Q.24 Two charges 9 e 9 e 9e9 \mathrm{e}9e and 3 e 3 e 3e3 \mathrm{e}3e are placed at a distance r r r\mathrm{r}r. The distance of the point where the electric field intensity will be zero, is :-
(1) r ( 1 + 3 ) r ( 1 + 3 ) (r)/((1+sqrt3))\frac{r}{(1+\sqrt{3})}r(1+3) from 9e charge
(2) r ( 1 + 1 3 ) r 1 + 1 3 (r)/((1+sqrt((1)/(3))))\frac{\mathrm{r}}{\left(1+\sqrt{\frac{1}{3}}\right)}r(1+13) from 9e charge
(3) r ( 1 − 3 ) r ( 1 − 3 ) (r)/((1-sqrt3))\frac{r}{(1-\sqrt{3})}r(1−3) from 3e charge
(4) r ( 1 + 1 3 ) r 1 + 1 3 (r)/((1+sqrt((1)/(3))))\frac{r}{\left(1+\sqrt{\frac{1}{3}}\right)}r(1+13) from 3e charge
Q.25 Two infinitely long parallel wires having linear charge densities λ 1 λ 1 lambda_(1)\lambda_{1}λ1 and λ 2 λ 2 lambda_(2)\lambda_{2}λ2 respectively are placed at a distance of R R R\mathrm{R}R. The force per unit length of an either wire will be :-
(1) k 2 λ 1 λ 2 R 2 k 2 λ 1 λ 2 R 2 (k2lambda_(1)lambda_(2))/(R^(2))\frac{\mathrm{k} 2 \lambda_{1} \lambda_{2}}{\mathrm{R}^{2}}k2λ1λ2R2
(2) k 2 λ 1 λ 2 R k 2 λ 1 λ 2 R (k2lambda_(1)lambda_(2))/(R)\frac{\mathrm{k} 2 \lambda_{1} \lambda_{2}}{\mathrm{R}}k2λ1λ2R
(3) k λ 1 λ 2 R 2 k λ 1 λ 2 R 2 (klambda_(1)lambda_(2))/(R^(2))\frac{k \lambda_{1} \lambda_{2}}{R^{2}}kλ1λ2R2
(4) λ 1 λ 2 R λ 1 λ 2 R (lambda_(1)lambda_(2))/(R)\frac{\lambda_{1} \lambda_{2}}{R}λ1λ2R
Q.26 An electric dipole is placed in an electric field generated by a point charge -
(1) the net electric force on the dipole must be zero
(2) the net electric force on the dipole may be zero
(3) the torque on the dipole due to the field must be zero
(4) the torque on the dipole due to the field may be zero
Q.27 A conducting sphere of radius 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm is charged with 10 μ C 10 μ C 10 muC10 \mu \mathrm{C}10μC. another uncharged sphere of radius 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm is allowed to touch it for some time after two spheres are separated, then surface density of the charge on the sphere will be in the ratio of :
(1) 2 : 1 2 : 1 2:12: 12:1
(2) 1 : 1 1 : 1 1:11: 11:1
(3) 3 : 1 3 : 1 3:13: 13:1
(4) 4 : 1 4 : 1 4:14: 14:1
Q.28 A small electric dipole is of dipole moment p p p\mathrm{p}p. The electric potential at a distance ' r r rrr ' from its centre and making an angle θ θ theta\thetaθ from the axis of dipole will be :-
(1) k p sin ⁡ θ r 2 k p sin ⁡ θ r 2 (kpsin theta)/(r^(2))\frac{\mathrm{kp} \sin \theta}{\mathrm{r}^{2}}kpsin⁡θr2
(2) k p cos ⁡ θ r 2 k p cos ⁡ θ r 2 (kpcos theta)/(r^(2))\frac{\mathrm{kp} \cos \theta}{\mathrm{r}^{2}}kpcos⁡θr2
(3) k p r 3 1 + 3 cos 2 ⁡ θ k p r 3 1 + 3 cos 2 ⁡ θ (kp)/(r^(3))sqrt(1+3cos^(2)theta)\frac{\mathrm{kp}}{\mathrm{r}^{3}} \sqrt{1+3 \cos ^{2} \theta}kpr31+3cos2⁡θ
(4) k p r 3 1 + 3 sin 2 ⁡ θ k p r 3 1 + 3 sin 2 ⁡ θ (kp)/(r^(3))sqrt(1+3sin^(2)theta)\frac{\mathrm{kp}}{\mathrm{r}^{3}} \sqrt{1+3 \sin ^{2} \theta}kpr31+3sin2⁡θ
Q.29 The electric dipole is placed along the x x x\mathrm{x}x - axis at the origin O O O\mathrm{O}O. A point P P P\mathrm{P}P is at a distance of 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm from this origin such that OP makes an angle π 3 π 3 (pi)/(3)\frac{\pi}{3}π3, with the x x xxx - axis. If the electric field at P makes an angle θ θ theta\thetaθ with the x x x\mathrm{x}x - axis, the value of θ θ theta\thetaθ would be :-
(1) π 3 π 3 (pi)/(3)\frac{\pi}{3}π3
(2) π 3 + tan − 1 ⁡ 3 2 π 3 + tan − 1 ⁡ 3 2 (pi)/(3)+tan^(-1)(sqrt3)/(2)\frac{\pi}{3}+\tan ^{-1} \frac{\sqrt{3}}{2}π3+tan−1⁡32
(3) 2 π / 3 2 π / 3 2pi//32 \pi / 32π/3
(4) tan − 1 ⁡ 3 2 tan − 1 ⁡ 3 2 tan^(-1)(sqrt3)/(2)\tan ^{-1} \frac{\sqrt{3}}{2}tan−1⁡32
Q.30 The given figure gives electric lines of force due to two charges q 1 q 1 q_(1)\mathrm{q}_{1}q1 and q 2 q 2 q_(2)\mathrm{q}_{2}q2. What are the signs of the two charges?
(1) Both are negative
(2) Both are positive
(3) q 1 q 1 q_(1)q_{1}q1 is positive but q 2 q 2 q_(2)q_{2}q2 is negative
(4) q 1 q 1 q_(1)q_{1}q1 is negative but q 2 q 2 q_(2)q_{2}q2 is positive

70. ELECTROSTATICS

Q.31 A charge Q Q QQQ is kept at the centre of a conducting sphere of inner radius R 1 R 1 R_(1)R_{1}R1 and outer radius R 2 R 2 R_(2)R_{2}R2. A point charge q q qqq is kept at a distance r ( > R 2 ) r > R 2 r( > R_(2))r\left(>R_{2}\right)r(>R2) from the centre. If q q qqq experiences an electrostatic force 10 N 10 N 10N10 \mathrm{~N}10 N then assuming that no other charges are present, electrostatic force experienced by Q Q QQQ will be:
(1) − 10 N − 10 N -10N-10 \mathrm{~N}−10 N
(2) 0
(3) 20 N 20 N 20N20 \mathrm{~N}20 N
(4) none of these
Q.32 A charge ' q q qqq ' is placed at the centre of a conducting spherical shell of radius R R RRR, which is given a charge Q. An external charge Q ′ Q ′ Q^(')Q^{\prime}Q′ is also present at distance R ′ ( R ′ > R ) R ′ R ′ > R R^(')(R^(') > R)R^{\prime}\left(R^{\prime}>R\right)R′(R′>R) from ' q q qqq '. Then the resultant field will be best represented for region r < R r < R r < R\mathrm{r}<\mathrm{R}r<R by:
[ where r r r\mathrm{r}r is the distance of the point from q q q\mathrm{q}q ]
(1)
(2)
(3)
( 4 ) ( 4 ) (4)(4)(4)
Q.33 The linear charge density on upper half of a segment of ring is λ λ lambda\lambdaλ and at lower half, it is − λ − λ -lambda-\lambda−λ. The direction of electric field at centre O O O\mathrm{O}O of ring is :

(1) along O A O A OA\mathrm{OA}OA
(2) along O B O B OB\mathrm{OB}OB
(3) along O C O C OC\mathrm{OC}OC
(4) along OD
Q.34 A charged particle ' q q qqq ' is shot from a large distance with speed v v vvv towards a fixed charged particle Q Q QQQ. It approaches Q Q QQQ upto a closest distance r r r\mathrm{r}r and then returns. If q q qqq were given a speed ' 2 v 2 v 2v2 \mathrm{v}2v ', the closest distance of approach would be :
q v ⟶ − → − → − Q q v ⟶ − → − → − Q q^(v)longrightarrow-rarr-rarr-Q\stackrel{v}{\mathrm{q}} \longrightarrow-\rightarrow-\rightarrow-Qqv⟶−→−→−Q
(1) r r r\mathrm{r}r
(2) 2 r 2 r 2r2 \mathrm{r}2r
(3) r 2 r 2 (r)/(2)\frac{r}{2}r2
(4) r 4 r 4 (r)/(4)\frac{r}{4}r4
Q.35 In an electron gun, electrons are accelerated through a potential difference of V V V\mathrm{V}V volt. Taking electronic charge and mass to be respectively e and m m mmm, the maximum velocity attained by them is :
(1) 2 e V m 2 e V m (2eV)/(m)\frac{2 \mathrm{eV}}{\mathrm{m}}2eVm
(2) 2 e V m 2 e V m sqrt((2eV)/(m))\sqrt{\frac{2 e V}{m}}2eVm
(3) 2 m / e V 2 m / e V 2m//eV2 \mathrm{~m} / \mathrm{eV}2 m/eV
(4) ( V 2 / 2 e m ) V 2 / 2 e m (V^(2)//2em)\left(\mathrm{V}^{2} / 2 \mathrm{em}\right)(V2/2em)
Q.36 A dipole having dipole moment p p p\mathrm{p}p is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point A lying on the surface of the sphere is :
(1) k p cos ⁡ ϕ r 2 k p cos ⁡ ϕ r 2 (kp cos phi)/(r^(2))\frac{k p \cos \phi}{r^{2}}kpcos⁡ϕr2
(2) k p cos 2 ⁡ ϕ r 2 k p cos 2 ⁡ ϕ r 2 (kpcos^(2)phi)/(r^(2))\frac{k p \cos ^{2} \phi}{r^{2}}kpcos2⁡ϕr2
(3) zero
(4) 2 k p cos 2 ⁡ ϕ r 2 2 k p cos 2 ⁡ ϕ r 2 (2kpcos^(2)phi)/(r^(2))\frac{2 \mathrm{kp} \cos ^{2} \phi}{\mathrm{r}^{2}}2kpcos2⁡ϕr2

71. ELECTROSTATICS

Q.37 A charge q q q\mathrm{q}q is placed at the centre of the cubical vessel (with one face open) as shown in figure. The flux of the electric field through the surface of the vessel is

(1) zero
( 2 ) q / ε 0 ( 2 ) q / ε 0 (2)q//epsi_(0)(2) q / \varepsilon_{0}(2)q/ε0
(3) q 4 ε 0 q 4 ε 0 (q)/(4epsi_(0))\frac{q}{4 \varepsilon_{0}}q4ε0
(4) 5 q / 6 ε 0 5 q / 6 ε 0 5q//6epsi_(0)5 \mathrm{q} / 6 \varepsilon_{0}5q/6ε0
Q.38 A hollow conducting sphere of radius R R R\mathrm{R}R has a charge ( + Q ) ( + Q ) (+Q)(+\mathrm{Q})(+Q) on its surface. What it the electric potential within the sphere at a distance r = R 3 r = R 3 r=(R)/(3)r=\frac{R}{3}r=R3 from its centre -
(1) zero
(2) 1 4 π ε 0 Q r 1 4 π ε 0 Q r (1)/(4piepsi_(0))(Q)/(r)\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}14πε0Qr
(3) 1 4 π ε 0 Q R 1 4 π ε 0 Q R (1)/(4piepsi_(0))(Q)/(R)\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}14πε0QR
(4) 1 4 π ε 0 Q r 2 1 4 π ε 0 Q r 2 (1)/(4piepsi_(0))(Q)/(r^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}14πε0Qr2
Q.39 Electric charge is uniformly distributed along a long straight wire of radius 1 m m 1 m m 1mm1 \mathrm{~mm}1 mm. The charge per c m c m cm\mathrm{cm}cm length of the wire is Q Q QQQ coulomb. Another cylindrical surface of radius 50 c m 50 c m 50cm50 \mathrm{~cm}50 cm and length 1 m 1 m 1m1 \mathrm{~m}1 m symmetrically encloses the wire as shown in fig. The total electric flux passing through the cylindrical surface is -

(1) Q ε 0 Q ε 0 (Q)/(epsi_(0))\frac{\mathrm{Q}}{\varepsilon_{0}}Qε0
(2) 100 Q ε 0 100 Q ε 0 (100Q)/(epsi_(0))\frac{100 \mathrm{Q}}{\varepsilon_{0}}100Qε0
(3) 10 Q ( π ε 0 ) 10 Q π ε 0 (10 Q)/((piepsi_(0)))\frac{10 Q}{\left(\pi \varepsilon_{0}\right)}10Q(πε0)
(4) 100 Q ( π ε 0 ) 100 Q π ε 0 (100 Q)/((piepsi_(0)))\frac{100 Q}{\left(\pi \varepsilon_{0}\right)}100Q(πε0)
Q.40 The work done in placing four charges at the corners of a square as shown in the figure, will be -

(1) ( 4 − 2 ) K q 2 a ( 4 − 2 ) K q 2 a (4-sqrt2)(Kq^(2))/(a)(4-\sqrt{2}) \frac{\mathrm{Kq}^{2}}{\mathrm{a}}(4−2)Kq2a
(2) ( 4 + 2 ) K q 2 a ( 4 + 2 ) K q 2 a (4+sqrt2)(Kq^(2))/(a)(4+\sqrt{2}) \frac{\mathrm{Kq}^{2}}{\mathrm{a}}(4+2)Kq2a
(3) ( 4 − 2 ) K q 2 a 2 ( 4 − 2 ) K q 2 a 2 (4-sqrt2)(Kq^(2))/(a^(2))(4-\sqrt{2}) \frac{K q^{2}}{a^{2}}(4−2)Kq2a2
(4) ( 4 + 2 ) K 2 a 2 ( 4 + 2 ) K 2 a 2 (4+sqrt2)(K^(2))/(a^(2))(4+\sqrt{2}) \frac{K^{2}}{\mathrm{a}^{2}}(4+2)K2a2
Q.41 Point charge (q) moves from point (P) to point (S) along the path PQRS as shown in fig. in a uniform electric field E pointing coparallel to the positive direction of the x x x\mathrm{x}x-axis. The coordinates of the points P , Q , R P , Q , R P,Q,RP, Q, RP,Q,R and S S SSS are ( a , b , 0 ) , ( 2 a , 0 , 0 ) , ( a , − b , 0 ) ( a , b , 0 ) , ( 2 a , 0 , 0 ) , ( a , − b , 0 ) (a,b,0),(2a,0,0),(a,-b,0)(a, b, 0),(2 a, 0,0),(a,-b, 0)(a,b,0),(2a,0,0),(a,−b,0) and ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) respectively. The work done by the field in the above process is given by the expression
(1) q E q
(2) − q E − q E -qE-\mathrm{q} \mathrm{E}−qE a
(3) q E E EEE a 2 2 sqrt2\sqrt{2}2
(4) q E [ ( 2 a ) 2 + b 2 ] q E ( 2 a ) 2 + b 2 qEsqrt([(2a)^(2)+b^(2)])\mathrm{qE} \sqrt{\left[(2 a)^{2}+b^{2}\right]}qE[(2a)2+b2]

72. ELECTROSTATICS

Q.42 If three electric di-poles are placed in some closed surface, then the electric flux emitting from the surface will be-
(1) zero
(2) positive
(3) negative
(4) None
Q.43 In a regular polygon of n n n\mathrm{n}n sides, each corner is at a distance r r r\mathrm{r}r from the centre. Identical charges are placed at ( n − 1 ) ( n − 1 ) (n-1)(\mathrm{n}-1)(n−1) corners. At the centre, the intensity is E E E\mathrm{E}E and the potential is V V V\mathrm{V}V. The ratio V / E V / E V//E\mathrm{V} / \mathrm{E}V/E has magnitude.
(1) r n r n rnr \mathrm{n}rn
(2) r ( n − 1 ) r ( n − 1 ) r(n-1)\mathrm{r}(\mathrm{n}-1)r(n−1)
(3) ( n − 1 ) / r ( n − 1 ) / r (n-1)//r(\mathrm{n}-1) / \mathrm{r}(n−1)/r
(4) r ( n − 1 ) / n r ( n − 1 ) / n r(n-1)//nr(n-1) / nr(n−1)/n
Q.44 At any point ( x , 0 , 0 ) ( x , 0 , 0 ) (x,0,0)(x, 0,0)(x,0,0) the electric potential V V VVV is ( 1000 x + 1500 x 2 + 500 x 3 ) 1000 x + 1500 x 2 + 500 x 3 ((1000 )/(x)+(1500)/(x^(2))+(500)/(x^(3)))\left(\frac{1000}{x}+\frac{1500}{x^{2}}+\frac{500}{x^{3}}\right)(1000x+1500x2+500x3) volt, then electric field at x = 1 m x = 1 m x=1m\mathrm{x}=1 \mathrm{~m}x=1 m -
(1) 5500 ( j ^ + k ^ ) V / m 5500 ( j ^ + k ^ ) V / m 5500( hat(j)+ hat(k))V//m5500(\hat{j}+\hat{k}) V / m5500(j^+k^)V/m
(2) 5500 i ^ V / m 5500 i ^ V / m 5500 hat(i)V//m5500 \hat{i} V / m5500i^V/m
(3) 5500 2 ( j ^ + k ^ ) V / m 5500 2 ( j ^ + k ^ ) V / m (5500)/(sqrt2)( hat(j)+ hat(k))V//m\frac{5500}{\sqrt{2}}(\hat{j}+\hat{k}) V / m55002(j^+k^)V/m
(4) 5500 2 ( i ^ + k ^ ) V / m 5500 2 ( i ^ + k ^ ) V / m (5500)/(sqrt2)( hat(i)+ hat(k))V//m\frac{5500}{\sqrt{2}}(\hat{i}+\hat{k}) V / m55002(i^+k^)V/m
Q.45 There is an electric field E E E\mathrm{E}E in X X X\mathrm{X}X-direction. If the work done on moving a charge 0.2 C 0.2 C 0.2C0.2 \mathrm{C}0.2C through a distance of 2 m 2 m 2m2 \mathrm{~m}2 m along a line making an angle 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ with the X X XXX-axis is 4.0 , what is the value of E E E\mathrm{E}E ?
(1) 3 N / C 3 N / C sqrt3N//C\sqrt{3} \mathrm{~N} / \mathrm{C}3 N/C
(2) 4 N / C 4 N / C 4N//C4 \mathrm{~N} / \mathrm{C}4 N/C
(3) 5 N / C 5 N / C 5N//C5 \mathrm{~N} / \mathrm{C}5 N/C
(4) None of these
Q.46 Two similar rings P P PPP and Q Q QQQ ( radius = 0.1 m t = 0.1 m t =0.1mt=0.1 \mathrm{mt}=0.1mt ) are placed co-axially at a distance 0.5 m t 0.5 m t 0.5mt0.5 \mathrm{mt}0.5mt.apart .The charge on P P P\mathrm{P}P and Q Q Q\mathrm{Q}Q is 2 μ C 2 μ C 2muC2 \mu \mathrm{C}2μC and 4 μ C 4 μ C 4muC4 \mu \mathrm{C}4μC respectively. Work done to move a 5 μ C 5 μ C 5muC5 \mu \mathrm{C}5μC charge from centre of P P PPP to the center of Q Q QQQ is-
(1) 1.28 J 1.28 J 1.28J1.28 \mathrm{~J}1.28 J
(2) 0.72 J 0.72 J 0.72J0.72 \mathrm{~J}0.72 J
(3) 0.144 J 0.144 J 0.144J0.144 \mathrm{~J}0.144 J
(4) 1.44 J 1.44 J 1.44J1.44 \mathrm{~J}1.44 J
Q.47 Potential and field strength at a certain distance from a point charge are 600 V 600 V 600V600 \mathrm{~V}600 V and 200 N / C 200 N / C 200N//C200 \mathrm{~N} / \mathrm{C}200 N/C. Distance of the point from the charge is :
(1) 2 m 2 m 2m2 \mathrm{~m}2 m
(2) 4 m 4 m 4m4 \mathrm{~m}4 m
(3) 8 m 8 m 8m8 \mathrm{~m}8 m
(4) 3 m 3 m 3m3 \mathrm{~m}3 m
Q.48 The work done required to change the structure (1) into structure (2)?
structure ( 2 ) ( 2 ) (2)(2)(2)
( 1 ) − 1 4 π ε 0 q 2 ℓ ( 1 ) − 1 4 π ε 0 q 2 ℓ (1)-(1)/(4piepsi_(0))(q^(2))/(ℓ)(1)-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{\ell}(1)−14πε0q2ℓ
( 2 ) + 1 4 π ε 0 q 2 2 ℓ ( 2 ) + 1 4 π ε 0 q 2 2 ℓ (2)+(1)/(4piepsi_(0))(q^(2))/(2ℓ)(2)+\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{2 \ell}(2)+14πε0q22ℓ
(3) − 1 4 π ε 0 q 2 2 ℓ − 1 4 π ε 0 q 2 2 ℓ -(1)/(4piepsi_(0))(q^(2))/(2ℓ)-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{2 \ell}−14πε0q22ℓ
( 4 ) + 1 4 π ε 0 q 2 ℓ ( 4 ) + 1 4 π ε 0 q 2 ℓ (4)+(1)/(4piepsi_(0))(q^(2))/(ℓ)(4)+\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{\ell}(4)+14πε0q2ℓ
Q.49 Charges are placed on the vertices of a square as shown. Let E → E → vec(E)\overrightarrow{\mathrm{E}}E→ be the electric field and V the potential at the centre. If the charges on A A AAA and B B BBB are interchanged with those on D D DDD and C C CCC respectively, then
(1) E → E → vec(E)\overrightarrow{\mathrm{E}}E→ remains unchanged, V changes
(2) Both E → E → vec(E)\vec{E}E→ and V change
(3) E → E → vec(E)\vec{E}E→ and V V VVV remain unchanged
(4) E → E → vec(E)\overrightarrow{\mathrm{E}}E→ changes, V V V\mathrm{V}V remains unchanged

73. ELECTROSTATICS

Q.50 A speed v v vvv is given to a particle A A AAA of charge q q qqq and mass m m mmm so that it moves towards another particle B B BBB of charge 4 q 4 q 4q4 q4q and mass 4 m 4 m 4m4 m4m placed at rest initially. Particle A A AAA approaches B B BBB upto a closest distance. The speed of particle B B BBB at this moment is
(1) v v vvv
(2) v 2 v 2 (v)/(2)\frac{v}{2}v2
(3) 4 v 4 v 4v4 v4v
(4) v 5 v 5 (v)/(5)\frac{v}{5}v5
Q.51 If an electric field is given by 10 i ^ + 3 j ^ + 4 k ^ 10 i ^ + 3 j ^ + 4 k ^ 10 hat(i)+3 hat(j)+4 hat(k)10 \hat{i}+3 \hat{j}+4 \hat{k}10i^+3j^+4k^. Calculate the electric flux through a surface of area 1 unit lying in y z y z yzy zyz plane
(1) 10 units
(2) 17 units
(3) 30 units
(4) 40 units
Q.52 Two parallel plates of infinite dimensions are uniformly charged. The surface charge density on one is σ A σ A sigma_(A)\sigma_{\mathrm{A}}σA and on the other is σ B σ B sigma_(B)\sigma_{\mathrm{B}}σB, field intensity at point C C C\mathrm{C}C will be- ⋅ D ⋅ D *D\cdot \mathrm{D}⋅D
(1) Proportional to ( σ A − σ B ) σ A − σ B (sigma_(A)-sigma_(B))\left(\sigma_{\mathrm{A}}-\sigma_{\mathrm{B}}\right)(σA−σB)
(2) Proportional to ( σ A + σ B ) σ A + σ B (sigma_(A)+sigma_(B))\left(\sigma_{\mathrm{A}}+\sigma_{\mathrm{B}}\right)(σA+σB)
(3) Zero
(4) 2 σ A 2 σ A 2sigma_(A)2 \sigma_{\mathrm{A}}2σA
Q.53 The electric potential at a point ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z) is given by
V = − x 2 y − x z 3 + 4 V = − x 2 y − x z 3 + 4 V=-x^(2)y-xz^(3)+4V=-\mathrm{x}^{2} \mathrm{y}-\mathrm{xz}^{3}+4V=−x2y−xz3+4
The electric field E E E\mathrm{E}E at that point is
(1) E = i ( 2 x y + z 3 ) + j x 2 + k 3 x z 2 E = i 2 x y + z 3 + j x 2 + k 3 x z 2 E=i(2xy+z^(3))+jx^(2)+k3xz^(2)E=i\left(2 x y+z^{3}\right)+j x^{2}+k 3 x z^{2}E=i(2xy+z3)+jx2+k3xz2
(2) E = i 2 x y + j ( x 2 + y 2 ) + k ( 3 x z − y 2 ) E = i 2 x y + j x 2 + y 2 + k 3 x z − y 2 E=i2xy+j(x^(2)+y^(2))+k(3xz-y^(2))E=i 2 x y+j\left(x^{2}+y^{2}\right)+k\left(3 x z-y^{2}\right)E=i2xy+j(x2+y2)+k(3xz−y2)
(3) E = i z 3 + j x y z + k z 2 E = i z 3 + j x y z + k z 2 E=iz^(3)+jxyz+kz^(2)E=i z^{3}+j x y z+k z^{2}E=iz3+jxyz+kz2
(4) E = i ( 2 x y − z 3 ) + j x y 2 + k 3 z 2 x E = i 2 x y − z 3 + j x y 2 + k 3 z 2 x E=i(2xy-z^(3))+jxy^(2)+k3z^(2)xE=i\left(2 x y-z^{3}\right)+j x y^{2}+k 3 z^{2} xE=i(2xy−z3)+jxy2+k3z2x
Q.54 A particle has a mass 400 times than that of the electron and charge is double than that of a electron. It is accelerated by 5 V 5 V 5V5 \mathrm{~V}5 V of potential difference. Initially the particle was at rest. Then its final kinetic energy will be -
(1) 5 e V 5 e V 5eV5 \mathrm{eV}5eV
(2) 10 e V 10 e V 10eV10 \mathrm{eV}10eV
(3) 100 e V 100 e V 100eV100 \mathrm{eV}100eV
(4) 2000 e V 2000 e V 2000eV2000 \mathrm{eV}2000eV
Q.55 Figure shows a closed surface which intersects a conducting sphere. If a positive charged is placed at the point P P PPP, the flux of the electric field through the closed surface :-
(1) will remain zero
(2) will become positive
(3) will become negative
(4) will become undefined
Q.56 Consider the situation of figure. The work done in taking a point charge from P P P\mathrm{P}P to A A A\mathrm{A}A is W A W A W_(A)\mathrm{W}_{\mathrm{A}}WA, from P P P\mathrm{P}P to B B B\mathrm{B}B is W B W B W_(B)\mathrm{W}_{\mathrm{B}}WB and from P P P\mathrm{P}P to C C C\mathrm{C}C is W C W C W_(C)\mathrm{W}_{\mathrm{C}}WC :-
(1) W A < W B < W C W A < W B < W C W_(A) < W_(B) < W_(C)\mathrm{W}_{\mathrm{A}}<\mathrm{W}_{\mathrm{B}}<\mathrm{W}_{\mathrm{C}}WA<WB<WC
(2) W A > W B > W C W A > W B > W C W_(A) > W_(B) > W_(C)\mathrm{W}_{\mathrm{A}}>\mathrm{W}_{\mathrm{B}}>\mathrm{W}_{\mathrm{C}}WA>WB>WC
(3) W A = W B = W C W A = W B = W C W_(A)=W_(B)=W_(C)\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}=\mathrm{W}_{\mathrm{C}}WA=WB=WC
(4) None of these

74. ELECTROSTATICS

Q.57 The figure shows the electric field lines in the vicinity of two point charges. Which one of the following statements concerning this situation is true?
(1) q 1 q 1 q_(1)q_{1}q1 is negative and q 2 q 2 q_(2)q_{2}q2 is positive
(2) The magnitude of the ratio ( q 2 / q 1 ) q 2 / q 1 (q_(2)//q_(1))\left(q_{2} / q_{1}\right)(q2/q1) is less than one
(3) Both q 1 q 1 q_(1)q_{1}q1 and q 2 q 2 q_(2)q_{2}q2 have the same sign of charge
(4) The electric field is strongest midway between the charges.
Q.58 Two identical particles of mass m m mmm carry a charge Q Q QQQ each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v v vvv. The closest distance of approach is
(1A) 1 4 π ε 0 Q 2 m v 1 4 π ε 0 Q 2 m v (1)/(4piepsi_(0))(Q^(2))/(mv)\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{\mathrm{mv}}14πε0Q2mv
(2) 1 4 π ε 0 4 Q 2 m v 2 1 4 π ε 0 4 Q 2 m v 2 (1)/(4piepsi_(0))(4Q^(2))/(mv^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{4 Q^{2}}{m v^{2}}14πε04Q2mv2
(3) 1 4 π ε 0 2 Q 2 m v 2 1 4 π ε 0 2 Q 2 m v 2 (1)/(4piepsi_(0))(2Q^(2))/(mv^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Q^{2}}{m v^{2}}14πε02Q2mv2
(4) 1 4 π ε 0 3 Q 2 m v 2 1 4 π ε 0 3 Q 2 m v 2 (1)/(4piepsi_(0))(3Q^(2))/(mv^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q^{2}}{m v^{2}}14πε03Q2mv2
Q.59 An uncharged conductor A is brought close to another positive charged conductor B B B\mathrm{B}B, then the charge on B B B\mathrm{B}B -
(1) will increase but potential will be constant.
(2) will be constant but potential will increase
(3) will be constant but potential decreases.
(4) the potential and charge on both are constant.
Q.60 The fig. shows lines of constant potential in a region in which an electric field is present. The value of the potential are written in brackets of the points A , B A , B A,B\mathrm{A}, \mathrm{B}A,B and C C C\mathrm{C}C, the magnitude of the electric field is greatest at the point -

(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) A & C
Q.61 The electric charge in uniform motion produces -
(1) an electric field only
(2) a magnetic field only
(3) both electric and magnetic fields
(4) neither electric nor magnetic fields
Q.62 A neutral metallic object is placed near a finite metal plate carrying a positive charge. The electric force on the object will be :
(1) towards the plate
(2) away from the plate
(3) parallel to the plate
(4) zero

75. ELECTROSTATICS

Q.63 Figure shows a thick metallic sphere. If it is given a charge + Q + Q +Q+Q+Q, then electric field will be present in the region

(1) r < R 1 r < R 1 r < R_(1)\mathrm{r}<\mathrm{R}_{1}r<R1 only
(2) r > R 1 r > R 1 r > R_(1)r>R_{1}r>R1 and R 1 < r < R 2 R 1 < r < R 2 R_(1) < r < R_(2)R_{1}<r<R_{2}R1<r<R2
(3) r ≥ R 2 r ≥ R 2 r >= R_(2)\mathrm{r} \geq \mathrm{R}_{2}r≥R2 only
(4) r ≤ R 2 r ≤ R 2 r <= R_(2)\mathrm{r} \leq \mathrm{R}_{2}r≤R2 only
Q.64 An uncharged sphere of metal is placed in a uniform electric field produced by two large conducting parallel plates having equal and opposite charges, then lines of force look like
(3)
(4)
Q.65 A sphere of radius 1 c m 1 c m 1cm1 \mathrm{~cm}1 cm has potential of 8000 V 8000 V 8000V8000 \mathrm{~V}8000 V. The energy density near the surface of sphere will be:
(1) 64 × 10 5 J / m 3 64 × 10 5 J / m 3 64 xx10^(5)J//m^(3)64 \times 10^{5} \mathrm{~J} / \mathrm{m}^{3}64×105 J/m3
(2) 8 × 10 3 J / m 3 8 × 10 3 J / m 3 8xx10^(3)J//m^(3)8 \times 10^{3} \mathrm{~J} / \mathrm{m}^{3}8×103 J/m3
(3) 32 J / m 3 32 J / m 3 32J//m^(3)32 \mathrm{~J} / \mathrm{m}^{3}32 J/m3
(4) 2.83 J / m 3 2.83 J / m 3 2.83J//m^(3)2.83 \mathrm{~J} / \mathrm{m}^{3}2.83 J/m3
Q.66 If ' n n nnn ' identical water drops assumed spherical each charged to a potential energy U coalesce to a single drop, the potential energy of the single drop is(Assume that drops are uniformly charged):
(1) n 1 / 3 U n 1 / 3 U n^(1//3)U\mathrm{n}^{1 / 3} \mathrm{U}n1/3U
(2) n 2 / 3 U n 2 / 3 U n^(2//3)U\mathrm{n}^{2 / 3} \mathrm{U}n2/3U
(3) n 4 / 3 U n 4 / 3 U n^(4//3)U\mathrm{n}^{4 / 3} \mathrm{U}n4/3U
(4) n 5 / 3 U n 5 / 3 U n^(5//3)Un^{5 / 3} Un5/3U

76. Direction for Assertion & Reason Questions

77. EXERCISE-3

ELECTROSTATICS
A. If both Assertion & Reason are True and the Reason is a correct explanation of the Assertion.
B. If both Assertion and Reason are True but Reason is not a correct explanation of the Assertion
C. If Assertion is True but the Reason is False.
D. If both Assertion and Reason are False.
Q.1 Assertion: If there exists coulomb attraction between two bodies, both of them may not be charged. Reason : In coulomb attraction two bodies are oppositely charged.
(1) A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.2 Assertion: If a conducting medium is placed between two charges, then electric force between them becomes zero.
Reason Reduction in a force due to introduced conductor is proportional to its dielectric constant.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.3 Assertion Force between two charges decreases when air separating the charges is replaced by water. Reason: Medium intervening the charges has no effect on force.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D
Q.4 Assertion Three equal charges are situated on a circle of radius r r rrr such that they form on equilateral triangle, then the electric field intensity at the centre is zero.
Reason: The force on unit positive charge at the centre, due to the three equal charges are represented by the three sides of a triangle taken in the same order.Therefore, electric field intensity at the centre is zero.
(1) A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D
Q.5 Assertion A charged particle free to move in an electric field always moves along an electric line of force.
Reason: The electric lines force diverge from a positive charge and converge at a negative charge.
(1) A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.6 Assertion If a point charge q q q\mathrm{q}q is placed in front of an infinite grounded conducting plane surface, the point charge will experience a force.
Reason: The force is due to the induced charge on the conducting surface which is at zero potential.
(1) A
(2) B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.7 Assertion Positive charge always moves from a higher potential point to a lower potential point. Reason: Electric potential is a vector quantity.
(1) A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.8 Assertion : Electric potential is scalar and electric intensity is vector quantity.
Reason : The value of electric potential and electric intensity at the middle point of the line joining an electron and a proton is zero.
(1) A
( 2 ) B ( 2 ) B (2)B(2) \mathrm{B}(2)B
(3) C C C\mathrm{C}C
(4) D
Q.9 Assertion: Electrons move away from a region of lower potential to a region of higher potential. Reason: Since an e − e − e^(-)\mathrm{e}^{-}e−has negative charge.
(1) A A A\mathrm{A}A
(2) B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.10 Assertion: No work is done in taking a positive charge from one point to other inside a positively charged metallic sphere while outside the sphere work is done in taking the charge towards the sphere. Reason: Inside the sphere electric potential is same at each point, but outside it is different on different point.
(1) A A A\mathrm{A}A
(2) B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.11 Assertion: A point charge is lying at the centre of a cube of each side ℓ ℓ ℓ\ellℓ. The electric flux emanating from each surface of the cube is ( 1 / 6 ) ( 1 / 6 ) (1//6)(1 / 6)(1/6) th of total flux.
Reason: According to Gauss' theorem, total electric flux through a closed surface enclosing a charge is equal to ( 1 / ε 0 ) 1 / ε 0 (1//epsi_(0))\left(1 / \varepsilon_{0}\right)(1/ε0) times the magnitude of the charge enclosed.
(1) A A A\mathrm{A}A
(2) B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D

78. ELECTROSTATICS

Q.12 Assertion: Increasing the charge on the plates of a capacitor means increasing the capacitance. Reason : Because Q = C V ⇒ Q ∝ C Q = C V ⇒ Q ∝ C Q=CV=>QpropC\mathrm{Q}=\mathrm{CV} \Rightarrow \mathrm{Q} \propto \mathrm{C}Q=CV⇒Q∝C.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.13 Assertion: The capacitance of a capacitor depends on the shape, size and geometrical placing of the conductors and its medium between them.
Reason : when a charge q passes through a battery of emf E E EEE from the negative terminal to an positive terminal, an amount q E q E qE\mathrm{qE}qE of work is done by the battery.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.14 Assertion: a dielectric slab is inserted between the plates of an isolated charged capacitor. The charge on the capacitor will remains the same.
Reason : Charge on a isolated system is conserved.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.15 Assertion : A point charge q q q\mathrm{q}q is placed at centre of spherical cavity inside a spherical conductor as shown. Another point charge Q Q QQQ is placed outside the conductor as shown. Now as the point charge Q Q QQQ is pushed away from conductor, the potential difference ( V A − V B ) V A − V B (V_(A)-V_(B))\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right)(VA−VB) between two points A A A\mathrm{A}A and B B B\mathrm{B}B within the cavity of sphere remains constant.
Reason : The electric field due to charges on outer surface of conductor and outside the conductor is zero at all points inside the conductor.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.16 Assertion : A solid uncharged conducting cylinder moves with acceleration a (w.r.t ground). As a result of acceleration of cylinder, an electric field is produced within cylinder.
solid conducting cylinder
Reason : When a solid conductor moves with acceleration a, then from frame of conductor a pseudoforce (of magnitude ma; where m m m\mathrm{m}m is mass of electron) will act on free electrons in the conductor. As a result some portion of the surface of conductor acquires negative charge and remaining portion of surface of conductor acquires positive charge.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.17 Assertion : The electric potential and the electric field intensity at the centre of a square having four point charges at their vertices (as shown) are zero.
Reason : Electric field is negative derivative of the potential.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.18 Assertion : An applied electric field will polarize the polar dielectric material.
Reason : In polar dielectrics, each molecule has a permanent dipole moment but these are randomly oriented in the absence of an externally applied electric field.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.19 Assertion : The surface of a conductor is an equipotential surface.
Reason : Conductor allows the flow of charge.
(1) A A A\mathrm{A}A
(2) B B BBB
(3) C C C\mathrm{C}C
(4) D

79. ELECTROSTATICS

80. EXERCISE-4 (A)
PREVIOUS YEAR'S QUESTIONS (AIPMT)

Q.1 Electric field strength due to a point charge of 5 μ C 5 μ C 5muC5 \mu \mathrm{C}5μC at a distance of 80 c m 80 c m 80cm80 \mathrm{~cm}80 cm from the charge is :
(1) 8 × 10 4 N / C 8 × 10 4 N / C 8xx10^(4)N//C8 \times 10^{4} \mathrm{~N} / \mathrm{C}8×104 N/C
(2) 7 × 10 4 N / C 7 × 10 4 N / C 7xx10^(4)N//C7 \times 10^{4} \mathrm{~N} / \mathrm{C}7×104 N/C
(3) 5 × 10 4 N / C 5 × 10 4 N / C 5xx10^(4)N//C5 \times 10^{4} \mathrm{~N} / \mathrm{C}5×104 N/C
(4) 4 × 10 4 N / C 4 × 10 4 N / C 4xx10^(4)N//C4 \times 10^{4} \mathrm{~N} / \mathrm{C}4×104 N/C
[CBSE PMT 2000]
Q.2 The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 c m c m cm\mathrm{cm}cm is 2 μ F 2 μ F 2muF2 \mu \mathrm{F}2μF. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is :
(1) 11.2 μ F 11.2 μ F 11.2 muF11.2 \mu \mathrm{F}11.2μF
(2) 15.6 μ F 15.6 μ F 15.6 muF15.6 \mu \mathrm{F}15.6μF
(3) 19.2 μ F 19.2 μ F 19.2 muF19.2 \mu \mathrm{F}19.2μF
(4) 22.4 μ F 22.4 μ F 22.4 muF22.4 \mu \mathrm{F}22.4μF
[CBSE PMT 2000]
Q.3 A capacitor is charged by connecting a battery across its plates. It stores energy U. Now the battery is disconnected and another identical capacitor is connected across it, then the energy stored by both capacitors of the system will be
[AIPMT 2000]
(1) U U UUU
(2) U 2 U 2 (U)/(2)\frac{U}{2}U2
(3) 2 U 2 U 2U2 \mathrm{U}2U
(4) 3 2 U 3 2 U (3)/(2)U\frac{3}{2} \mathrm{U}32U
Q.4 The capacity of a parallel plate condenser is 15 μ F 15 μ F 15 muF15 \mu \mathrm{F}15μF, when the distance between its plates is 6 c m 6 c m 6cm6 \mathrm{~cm}6 cm. If the distance between the plates is reduced to 2 c m 2 c m 2cm2 \mathrm{~cm}2 cm, then the capacity of this parallel plate condenser will be :
[CBSE PMT 2001]
(1) 15 μ F 15 μ F 15 muF15 \mu \mathrm{F}15μF
(2) 30 μ F 30 μ F 30 muF30 \mu \mathrm{F}30μF
(3) 45 μ F 45 μ F 45 muF45 \mu \mathrm{F}45μF
(4) 60 μ F 60 μ F 60 muF60 \mu \mathrm{F}60μF
Q.5 A simple pendulum of period T T T\mathrm{T}T has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will :
(1) Remains equal to T T TTT
(2) Less than T
(2) Greater than T
(4) Infinite
[CBSE PMT 2001]
Q.6 The electric intensity due to a dipole of length 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm and having a charge of 500 μ C 500 μ C 500 muC500 \mu \mathrm{C}500μC, at a point on the axis at a distance 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm from one of the charges in air, is :
[CBSE PMT - 2001]
(1) 6.25 × 10 7 N / C 6.25 × 10 7 N / C 6.25 xx10^(7)N//C6.25 \times 10^{7} \mathrm{~N} / \mathrm{C}6.25×107 N/C
(2) 9.28 × 10 7 N / C 9.28 × 10 7 N / C 9.28 xx10^(7)N//C9.28 \times 10^{7} \mathrm{~N} / \mathrm{C}9.28×107 N/C
(3) 13.1 × 10 11 N / C 13.1 × 10 11 N / C 13.1 xx10^(11)N//C13.1 \times 10^{11} \mathrm{~N} / \mathrm{C}13.1×1011 N/C
(4) 20.5 × 10 7 N / C 20.5 × 10 7 N / C 20.5 xx10^(7)N//C20.5 \times 10^{7} \mathrm{~N} / \mathrm{C}20.5×107 N/C
Q.7 In a parallel plate capacitor, the distance between the plates is d and potential difference across plates is V. Energy stored per unit volume between the plates of capacitor is [AIPMT 2001]
(1) Q 2 2 V 2 Q 2 2 V 2 (Q^(2))/(2V^(2))\frac{Q^{2}}{2 \mathrm{~V}^{2}}Q22 V2
(2) 1 2 ε 0 V 2 d 2 1 2 ε 0 V 2 d 2 (1)/(2)(epsi_(0)V^(2))/(d^(2))\frac{1}{2} \frac{\varepsilon_{0} V^{2}}{d^{2}}12ε0V2d2
(3) 1 2 V 2 ε 0 d 2 1 2 V 2 ε 0 d 2 (1)/(2)(V^(2))/(epsi_(0)d^(2))\frac{1}{2} \frac{\mathrm{V}^{2}}{\varepsilon_{0} \mathrm{~d}^{2}}12V2ε0 d2
(4) 1 2 ε 0 V 2 d 2 1 2 ε 0 V 2 d 2 (1)/(2)epsi_(0)(V^(2))/(d^(2))\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}12ε0V2d2
Q.8 If identical charges ( − q ) ( − q ) (-q)(-\mathrm{q})(−q) are placed at each corner of a cube of side b b bbb, then electric potential energy of charge ( + q ) ( + q ) (+q)(+q)(+q) which is placed at centre of the cube will be :
[CBSE PMT 2002]
(1) 8 2 q 2 4 π ε 0 b 8 2 q 2 4 π ε 0 b (8sqrt2q^(2))/(4piepsi_(0)(b))\frac{8 \sqrt{2} q^{2}}{4 \pi \varepsilon_{0} \mathrm{~b}}82q24πε0 b
(2) − 8 2 q 2 π ε 0 b − 8 2 q 2 π ε 0 b (-8sqrt2q^(2))/(piepsi_(0)b)\frac{-8 \sqrt{2} q^{2}}{\pi \varepsilon_{0} b}−82q2πε0b
(3) − 4 2 q 2 π ε 0 b − 4 2 q 2 π ε 0 b (-4sqrt2q^(2))/(piepsi_(0)b)\frac{-4 \sqrt{2} q^{2}}{\pi \varepsilon_{0} b}−42q2πε0b
(4) − 4 q 2 3 π ε 0 b − 4 q 2 3 π ε 0 b (-4q^(2))/(sqrt3piepsi_(0)b)\frac{-4 q^{2}}{\sqrt{3} \pi \varepsilon_{0} b}−4q23πε0b
Q.9 A capacitor of capacity C 1 C 1 C_(1)\mathrm{C}_{1}C1 is charged upto potential V V V\mathrm{V}V volt and then connected in parallel to an uncharged capacitor of capacity C 2 C 2 C_(2)\mathrm{C}_{2}C2. The final potential difference across each capacitor will be :
[AIPMT-2002]
(1) C 2 V ( C 1 + C 2 ) C 2 V C 1 + C 2 (C_(2)(V))/((C_(1)+C_(2)))\frac{\mathrm{C}_{2} \mathrm{~V}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}C2 V(C1+C2)
(2) C 1 V C 1 + C 2 C 1 V C 1 + C 2 (C_(1)(V))/(C_(1)+C_(2))\frac{\mathrm{C}_{1} \mathrm{~V}}{\mathrm{C}_{1}+\mathrm{C}_{2}}C1 VC1+C2
(3) ( 1 + C 2 C 1 ) V 1 + C 2 C 1 V (1+(C_(2))/(C_(1)))V\left(1+\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\right) V(1+C2C1)V
(4) ( 1 − C 2 C 1 ) V 1 − C 2 C 1 V (1-(C_(2))/(C_(1)))V\left(1-\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\right) \mathrm{V}(1−C2C1)V

81. ELECTROSTATICS

Q.10 An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius r r r\mathrm{r}r. The coulomb force F → F → vec(F)\overrightarrow{\mathrm{F}}F→ between the two is (Where K = 1 4 π ε 0 K = 1 4 π ε 0 K=(1)/(4piepsi_(0))\mathrm{K}=\frac{1}{4 \pi \varepsilon_{0}}K=14πε0 ) :
[CBSE PMT 2003]
(1) − K e 2 r 3 r ^ − K e 2 r 3 r ^ -K(e^(2))/(r^(3)) hat(r)-\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{3}} \hat{\mathrm{r}}−Ke2r3r^
(2) K e 2 r 3 r → K e 2 r 3 r → K(e^(2))/(r^(3)) vec(r)\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{3}} \overrightarrow{\mathrm{r}}Ke2r3r→
(3) − K e 2 r 3 r → − K e 2 r 3 r → -K(e^(2))/(r^(3)) vec(r)-K \frac{e^{2}}{r^{3}} \vec{r}−Ke2r3r→
(4) K e 2 r 2 r ^ K e 2 r 2 r ^ K(e^(2))/(r^(2)) hat(r)\mathrm{K} \frac{\mathrm{e}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}Ke2r2r^
Q.11 Three capacitors each of capacity 4 μ F 4 μ F 4muF4 \mu \mathrm{F}4μF are to be connected in such a way that the effective capacitance is 6 μ F 6 μ F 6muF6 \mu \mathrm{F}6μF. This can be done by [2003] :
[AIPMT-2003]
(1) Connecting two in series and one in parallel
(2) Connecting two in parallel and one in series
(3) Connecting all of them is series
(4) Connecting all of them in parallel
Q.12 A charge q q qqq is located at the centre of a cube. The electric flux through any face is :
[AIPMT 2003]
(1) 2 π q 6 ( 4 π ε 0 ) 2 π q 6 4 π ε 0 (2pi q)/(6(4piepsi_(0)))\frac{2 \pi q}{6\left(4 \pi \varepsilon_{0}\right)}2πq6(4πε0)
(2) 4 π q 6 ( 4 π ε 0 ) 4 π q 6 4 π ε 0 (4pi q)/(6(4piepsi_(0)))\frac{4 \pi q}{6\left(4 \pi \varepsilon_{0}\right)}4πq6(4πε0)
(3) π q 6 ( 4 π ε 0 ) π q 6 4 π ε 0 (pi q)/(6(4piepsi_(0)))\frac{\pi q}{6\left(4 \pi \varepsilon_{0}\right)}πq6(4πε0)
(4) q 6 ( 4 π ε 0 ) q 6 4 π ε 0 (q)/(6(4piepsi_(0)))\frac{q}{6\left(4 \pi \varepsilon_{0}\right)}q6(4πε0)
Q.13 An electric dipole has the magnitude of its charge as q q qqq and its dipole moment is p p p\mathrm{p}p. It is placed in a uniform electric field E E EEE. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively :
(1) 2q. E and minimum
(2) q.E and p.E
(3) zero and minimum
(4) q − E q − E q-Eq-Eq−E and maximum
[AIPMT 2004]
Q.14 A bullet of mass 2 g 2 g 2g2 \mathrm{~g}2 g is having a charge of 2 μ C 2 μ C 2muC2 \mu \mathrm{C}2μC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m / s 10 m / s 10m//s10 \mathrm{~m} / \mathrm{s}10 m/s ?
(1) 5 k V 5 k V 5kV5 \mathrm{kV}5kV
(2) 50 k V 50 k V 50kV50 \mathrm{kV}50kV
(3) 5 V 5 V 5V5 \mathrm{~V}5 V
(4) 50 V 50 V 50V50 \mathrm{~V}50 V
[AIPMT 2004]
Q.15 As per the diagram a point charge + q + q +q+\mathrm{q}+q is placed at the origin O O O\mathrm{O}O. Work done in taking another point charge − Q − Q -Q-\mathrm{Q}−Q from the point A A A\mathrm{A}A [coordinates ( 0 , a ) ] ( 0 , a ) ] (0,a)](0, \mathrm{a})](0,a)] to another point B B B\mathrm{B}B [coordinates ( a , 0 ) ] ( a , 0 ) ] (a,0)](\mathrm{a}, 0)](a,0)] along the straight path A B A B AB\mathrm{AB}AB is :
[AIPMT 2005]
(1) zero
(2) ( q Q 4 π ε 0 1 a 2 ) ⋅ 2 a q Q 4 π ε 0 1 a 2 ⋅ 2 a ((qQ)/(4piepsi_(0))(1)/(a^(2)))*sqrt2a\left(\frac{\mathrm{qQ}}{4 \pi \varepsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right) \cdot \sqrt{2} \mathrm{a}(qQ4πε01a2)⋅2a
(3) ( − q Q 4 π ε 0 1 a 2 ) ⋅ 2 a − q Q 4 π ε 0 1 a 2 ⋅ 2 a ((-qQ)/(4piepsi_(0))(1)/(a^(2)))*sqrt2a\left(\frac{-q Q}{4 \pi \varepsilon_{0}} \frac{1}{a^{2}}\right) \cdot \sqrt{2} a(−qQ4πε01a2)⋅2a
(4) ( q Q 4 π ε 0 1 a 2 ) ⋅ a 2 q Q 4 π ε 0 1 a 2 ⋅ a 2 ((qQ)/(4piepsi_(0))(1)/(a^(2)))*(a)/(sqrt2)\left(\frac{\mathrm{qQ}}{4 \pi \varepsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right) \cdot \frac{\mathrm{a}}{\sqrt{2}}(qQ4πε01a2)⋅a2
Q.16 Two charges q 1 q 1 q_(1)q_{1}q1 and q 2 q 2 q_(2)q_{2}q2 are placed 30 c m 30 c m 30cm30 \mathrm{~cm}30 cm apart as shown in the figure. A third charge q 3 q 3 q_(3)q_{3}q3 is moved along the arc of a circle of radius 40 c m 40 c m 40cm40 \mathrm{~cm}40 cm from C C CCC to D D DDD. The change in the potential energy of the system is q 3 4 π ε 0 k q 3 4 π ε 0 k (q_(3))/(4piepsi_(0))k\frac{\mathrm{q}_{3}}{4 \pi \varepsilon_{0}} \mathrm{k}q34πε0k, where k k k\mathrm{k}k is :
(1) 8 q 1 8 q 1 8q_(1)8 \mathrm{q}_{1}8q1
(2) 6 q 1 6 q 1 6q_(1)6 q_{1}6q1
(3) 8 q 2 8 q 2 8q_(2)8 \mathrm{q}_{2}8q2
(4) 6 q 2 6 q 2 6q_(2)6 q_{2}6q2
[AIPMT 2005]

82. ELECTROSTATICS

Q.17 A network of four capacitors of capacity equal to C 1 = C C 1 = C C_(1)=C\mathrm{C}_{1}=\mathrm{C}C1=C, C 2 = 2 C , C 3 = 3 C C 2 = 2 C , C 3 = 3 C C_(2)=2C,C_(3)=3C\mathrm{C}_{2}=2 \mathrm{C}, \mathrm{C}_{3}=3 \mathrm{C}C2=2C,C3=3C and C 4 = 4 C C 4 = 4 C C_(4)=4C\mathrm{C}_{4}=4 \mathrm{C}C4=4C are connected to a battery as shown in the figure. The ratio of the charges on C 2 C 2 C_(2)\mathrm{C}_{2}C2 and C 4 C 4 C_(4)\mathrm{C}_{4}C4 is :
[AIPMT-2005]
(1) 22 3 22 3 (22)/(3)\frac{22}{3}223
(2) 3 22 3 22 (3)/(22)\frac{3}{22}322
(3) 7 4 7 4 (7)/(4)\frac{7}{4}74
(4) 4 7 4 7 (4)/(7)\frac{4}{7}47
Q.18 A parallel plate air capacitor is charged to a potential difference of V V V\mathrm{V}V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates : [AIPMT-2006]
(1) decreases
(2) does not change
(3) becomes zero
(4) increases
Q.19 An electric dipole of moment p → p → vec(p)\vec{p}p→ is lying along a uniform electric field E → E → vec(E)\overrightarrow{\mathrm{E}}E→. The work done in rotating the dipole by 90 ∘ 90 ∘ 90^(@)90^{\circ}90∘ is :
[AIPMT 2006]
(1) p E p E pE\mathrm{pE}pE
(2) 2 p E 2 p E sqrt2pE\sqrt{2} \mathrm{pE}2pE
(3) p E / 2 p E / 2 pE//2\mathrm{pE} / 2pE/2
(4) 2 p E 2 p E 2pE2 \mathrm{pE}2pE
Q.20 A square surface of side L metres is in the plane of the paper. A uniform electric field E → E → vec(E)\overrightarrow{\mathrm{E}}E→ (volt / m / m //m/ \mathrm{m}/m ), also in the plane of the paper, is limited only to the lower half of the square surface (see figure). The electric flux in SI units associated with the surface is :
[AIPMT 2006]

(1) E L 2 E L 2 EL^(2)\mathrm{EL}^{2}EL2
(2) E L 2 / 2 ε 0 E L 2 / 2 ε 0 EL^(2)//2epsi_(0)\mathrm{EL}^{2} / 2 \varepsilon_{0}EL2/2ε0
(3) E L 2 / 2 E L 2 / 2 EL^(2)//2\mathrm{EL}^{2} / 2EL2/2
(4) zero
Q.21 A hollow cylinder has a charge q q q\mathrm{q}q coulomb within it. If ϕ ϕ phi\phiϕ is the electric flux in units of voltmeter associated with the curved surface B B BBB, the flux linked with the plane surface A A AAA in units of V-m will be :
[AIPMT 2007]
(1) q 2 ε 0 q 2 ε 0 (q)/(2epsi_(0))\frac{q}{2 \varepsilon_{0}}q2ε0
(2) ϕ 3 ϕ 3 (phi)/(3)\frac{\phi}{3}ϕ3
(3) q ε 0 − ϕ q ε 0 − ϕ (q)/(epsi_(0))-phi\frac{\mathrm{q}}{\varepsilon_{0}}-\phiqε0−ϕ
(4) 1 2 ( q ε 0 − ϕ ) 1 2 q ε 0 − ϕ (1)/(2)((q)/(epsi_(0))-phi)\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)12(qε0−ϕ)
Q.22 Charges + q + q +q+\mathrm{q}+q and − q − q -q-\mathrm{q}−q are placed at point A A A\mathrm{A}A and B B B\mathrm{B}B respetively which are a distance 2 L 2 L 2L2 \mathrm{~L}2 L apart, C C C\mathrm{C}C is the midpoint between A A A\mathrm{A}A and B B B\mathrm{B}B. The work done in moving charge + Q + Q +Q+\mathrm{Q}+Q along the semicircle C R D C R D CRD\mathrm{CRD}CRD is :
[AIPMT 2007]
(1) q Q 2 π ε 0 L q Q 2 π ε 0 L (qQ)/(2piepsi_(0)(L))\frac{\mathrm{qQ}}{2 \pi \varepsilon_{0} \mathrm{~L}}qQ2πε0 L
(2) q Q 6 π ε 0 L q Q 6 π ε 0 L (qQ)/(6piepsi_(0)(L))\frac{\mathrm{qQ}}{6 \pi \varepsilon_{0} \mathrm{~L}}qQ6πε0 L
(3) − q Q 6 π ε 0 L − q Q 6 π ε 0 L -(qQ)/(6piepsi_(0)(L))-\frac{\mathrm{qQ}}{6 \pi \varepsilon_{0} \mathrm{~L}}−qQ6πε0 L
(4) q Q 4 π ε 0 L q Q 4 π ε 0 L (qQ)/(4piepsi_(0)(L))\frac{\mathrm{qQ}}{4 \pi \varepsilon_{0} \mathrm{~L}}qQ4πε0 L
Q.23 Two condensers, one of capacity C C C\mathrm{C}C and the other of capacity C 2 C 2 (C)/(2)\frac{\mathrm{C}}{2}C2, are connected to a V V V\mathrm{V}V battery, as shown. The work done in charging fully both the condensers is:[AIPMT-2007]

(1) 2 C V 2 2 C V 2 2CV^(2)2 \mathrm{CV}^{2}2CV2
(2) 1 4 C V 2 1 4 C V 2 (1)/(4)CV^(2)\frac{1}{4} \mathrm{CV}^{2}14CV2
(3) 3 4 C V 2 3 4 C V 2 (3)/(4)CV^(2)\frac{3}{4} \mathrm{CV}^{2}34CV2
(4) 1 2 C V 2 1 2 C V 2 (1)/(2)CV^(2)\frac{1}{2} \mathrm{CV}^{2}12CV2
Q.24 Three point charges + q , − 2 q + q , − 2 q +q,-2q+\mathrm{q},-2 \mathrm{q}+q,−2q and + q + q +q+\mathrm{q}+q are placed at points ( x = 0 , y = a , z = 0 ) , ( x = 0 , y = 0 , z = 0 ) ( x = 0 , y = a , z = 0 ) , ( x = 0 , y = 0 , z = 0 ) (x=0,y=a,z=0),(x=0,y=0,z=0)(x=0, y=a, z=0),(x=0, y=0, z=0)(x=0,y=a,z=0),(x=0,y=0,z=0) and ( x = a , y = 0 , z = 0 ) ( x = a , y = 0 , z = 0 ) (x=a,y=0,z=0)(\mathrm{x}=\mathrm{a}, \mathrm{y}=0, \mathrm{z}=0)(x=a,y=0,z=0) respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are :
[AIPMT 2007]
(1) 2 2 sqrt2\sqrt{2}2 qa along the line joining points ( x = 0 , y = 0 , z = 0 ) ( x = 0 , y = 0 , z = 0 ) (x=0,y=0,z=0)(x=0, y=0, z=0)(x=0,y=0,z=0) and ( x = a , y = a , z = 0 ) ( x = a , y = a , z = 0 ) (x=a,y=a,z=0)(x=a, y=a, z=0)(x=a,y=a,z=0)
(2) qa along the line joining points ( x = 0 , y = 0 , z = 0 ) ( x = 0 , y = 0 , z = 0 ) (x=0,y=0,z=0)(x=0, y=0, z=0)(x=0,y=0,z=0) and ( x = a , y = a , z = 0 ) ( x = a , y = a , z = 0 ) (x=a,y=a,z=0)(x=a, y=a, z=0)(x=a,y=a,z=0)
(3) 2 2 sqrt2\sqrt{2}2 qa along + x + x +x+x+x direction
(4) 2 2 sqrt2\sqrt{2}2 qa along + y + y +y+\mathrm{y}+y direction
Q.25 Three capacitors each of capacitance C C C\mathrm{C}C and of breakdown voltage V V V\mathrm{V}V are joined in series. The

(1) C 3 , V 3 C 3 , V 3 (C)/(3),(V)/(3)\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}C3,V3
(2) 3 C , V 3 3 C , V 3 3C,(V)/(3)3 \mathrm{C}, \frac{\mathrm{V}}{3}3C,V3
(3) C 3 , 3 V C 3 , 3 V (C)/(3),3V\frac{\mathrm{C}}{3}, 3 \mathrm{~V}C3,3 V
(4) 3 C , 3 V 3 C , 3 V 3C,3V3 \mathrm{C}, 3 \mathrm{~V}3C,3 V
Q.26 A series combination of n 1 n 1 n_(1)\mathrm{n}_{1}n1 capacitors, each of value C 1 C 1 C_(1)\mathrm{C}_{1}C1, is charged by a source of potential difference 4 V 4 V 4V4 \mathrm{~V}4 V. when another parallel combination of n 2 n 2 n_(2)\mathrm{n}_{2}n2 capacitors, each of value C 2 C 2 C_(2)\mathrm{C}_{2}C2, is charged by a source of potential difference V V V\mathrm{V}V, it has the same (total) energy stored in it, as the first combination has. The value of C 2 C 2 C_(2)\mathrm{C}_{2}C2, in terms of C 1 C 1 C_(1)\mathrm{C}_{1}C1, is then
[AIPMT-2010]
(1) 2 C 1 n 1 n 2 2 C 1 n 1 n 2 (2C_(1))/(n_(1)n_(2))\frac{2 \mathrm{C}_{1}}{\mathrm{n}_{1} \mathrm{n}_{2}}2C1n1n2
(2) 16 n 2 n 1 C 1 16 n 2 n 1 C 1 16(n_(2))/(n_(1))C_(1)16 \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \mathrm{C}_{1}16n2n1C1
(3) 2 n 2 n 1 C 1 2 n 2 n 1 C 1 2(n_(2))/(n_(1))C_(1)2 \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \mathrm{C}_{1}2n2n1C1
(4) 16 c 1 n 1 n 2 16 c 1 n 1 n 2 (16c_(1))/(n_(1)n_(2))\frac{16 c_{1}}{n_{1} n_{2}}16c1n1n2
Q.27 Two parallel metal plates having charges + Q + Q +Q+\mathrm{Q}+Q and − Q − Q -Q-\mathrm{Q}−Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will :
[AIPMT 2010]
(1) become zero
(2) increase
(3) decrease
(4) remain same
Q.28 The electric field at a distance 3 R 2 3 R 2 (3R)/(2)\frac{3 R}{2}3R2 from the centre of a charged conducting spherical shell of radius R R R\mathrm{R}R is E. The electric field at a distance R 2 R 2 (R)/(2)\frac{\mathrm{R}}{2}R2 from the centre of the sphere is : [AIPMT 2010]
(1) zero
(2) E E EEE
(3) E 2 E 2 (E)/(2)\frac{E}{2}E2
(4) E 3 E 3 (E)/(3)\frac{E}{3}E3 ELECTROSTATICS
Q.29 A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :
(1) increase four time
(2) be reduced to half
(3) remains the same
(4) be doubled
[AIPMT 2011]
Q.30 Four electric charges + q , + q , − q + q , + q , − q +q,+q,-q+\mathrm{q},+\mathrm{q},-\mathrm{q}+q,+q,−q and − q − q -q-\mathrm{q}−q are placed at the corners of a square of side 2 L 2 L 2L2 \mathrm{~L}2 L (see figure). The electric potential at point A A A\mathrm{A}A, midway between the two charges + q + q +q+\mathrm{q}+q and + q + q +q+\mathrm{q}+q, is :
[AIPMT 2011]

(1) 1 4 π ε 0 2 q L ( 1 + 5 ) 1 4 π ε 0 2 q L ( 1 + 5 ) (1)/(4piepsi_(0))(2q)/(L)(1+sqrt5)\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{q}}{\mathrm{L}}(1+\sqrt{5})14πε02qL(1+5)
(2) 1 4 π ε 0 2 q L ( 1 + 1 5 ) 1 4 π ε 0 2 q L 1 + 1 5 (1)/(4piepsi_(0))(2q)/(L)(1+(1)/(sqrt5))\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{q}}{\mathrm{L}}\left(1+\frac{1}{\sqrt{5}}\right)14πε02qL(1+15)
(3) 1 4 π ε 0 2 q L ( 1 − 1 5 ) 1 4 π ε 0 2 q L 1 − 1 5 (1)/(4piepsi_(0))(2q)/(L)(1-(1)/(sqrt5))\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right)14πε02qL(1−15)
(4) zero
Q.31 What is the flux through a cube of side ' a a aaa ' if a point charge of q q qqq is at one of its corner :
[AIPMT 2012]
(1) q ϵ 0 q ϵ 0 (q)/(epsilon_(0))\frac{q}{\epsilon_{0}}qϵ0
(2) q 2 ϵ 0 6 a 2 q 2 ϵ 0 6 a 2 (q)/(2epsilon_(0))6a^(2)\frac{\mathrm{q}}{2 \epsilon_{0}} 6 \mathrm{a}^{2}q2ϵ06a2
(3) 2 q ϵ 0 2 q ϵ 0 (2q)/(epsilon_(0))\frac{2 q}{\epsilon_{0}}2qϵ0
(4) q 8 ∈ 0 q 8 ∈ 0 (q)/(8in_(0))\frac{q}{8 \in_{0}}q8∈0
Q.32 An electric dipole of moment ' p p ppp ' is placed in an electric field of intensity 'E'. The dipole acquires a position such that the axis of the dipole makes an angle θ θ theta\thetaθ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ = 90 ∘ θ = 90 ∘ theta=90^(@)\theta=90^{\circ}θ=90∘. the torque and the potential energy of the dipole will respectively be
(1) p E sin ⁡ θ , 2 p E cos ⁡ θ p E sin ⁡ θ , 2 p E cos ⁡ θ pEsin theta,2pEcos theta\mathrm{pE} \sin \theta, 2 \mathrm{pE} \cos \thetapEsin⁡θ,2pEcos⁡θ
(2) p E cos ⁡ θ , − p E sin ⁡ θ p E cos ⁡ θ , − p E sin ⁡ θ pEcos theta,-pEsin theta\mathrm{pE} \cos \theta,-\mathrm{pE} \sin \thetapEcos⁡θ,−pEsin⁡θ
(3) p E sin ⁡ θ , − p E cos ⁡ θ p E sin ⁡ θ , − p E cos ⁡ θ pEsin theta,-pEcos theta\mathrm{pE} \sin \theta,-\mathrm{pE} \cos \thetapEsin⁡θ,−pEcos⁡θ
(4) p E sin ⁡ θ , − 2 p E cos ⁡ θ p E sin ⁡ θ , − 2 p E cos ⁡ θ pEsin theta,-2pEcos theta\mathrm{pE} \sin \theta,-2 \mathrm{pE} \cos \thetapEsin⁡θ,−2pEcos⁡θ
[AIPMT 2012]
Q.33 Four point charges − Q , − q , 2 q − Q , − q , 2 q -Q,-q,2q-\mathrm{Q},-\mathrm{q}, 2 \mathrm{q}−Q,−q,2q and 2 Q 2 Q 2Q2 \mathrm{Q}2Q are placed one at each corner of the square. The relation between Q Q Q\mathrm{Q}Q and q q q\mathrm{q}q for which the potential at the centre of the square is zero is [AIPMT 2012]
(1) Q = q Q = q Q=qQ=qQ=q
(2) Q = 1 q Q = 1 q Q=(1)/(q)\mathrm{Q}=\frac{1}{\mathrm{q}}Q=1q
(3) Q = − q Q = − q Q=-qQ=-qQ=−q
(4) Q = − 1 q Q = − 1 q Q=-(1)/(q)Q=-\frac{1}{q}Q=−1q
Q.34 A parallel plate condenser has a uniform electric field E ( V / m ) E ( V / m ) E(V//m)\mathrm{E}(\mathrm{V} / \mathrm{m})E(V/m) in the space between the plates. If the distance between the plates is d ( m ) d ( m ) d(m)d(m)d(m) and area of each plate is A ( m 2 ) A m 2 A(m^(2))A\left(m^{2}\right)A(m2) the energy (joule) stored in the condenser is :
[AIPMT-2012]
(1) 1 2 ε 0 E 2 1 2 ε 0 E 2 (1)/(2)epsi_(0)E^(2)\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}12ε0E2
(2) ε 0 E A d ε 0 E A d epsi_(0)EAd\varepsilon_{0} \mathrm{EAd}ε0EAd
(3) 1 2 ε 0 E 2 A d 1 2 ε 0 E 2 A d (1)/(2)epsi_(0)E^(2)Ad\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}12ε0E2Ad
(4) E 2 A d / ε 0 E 2 A d / ε 0 E^(2)Ad//epsi_(0)\mathrm{E}^{2} \mathrm{Ad} / \varepsilon_{0}E2Ad/ε0 ELECTROSTATICS
Q.35 A, B, and C C C\mathrm{C}C are three points in a uniform electric field. The electric potential is : [AIPMT 2013]
(1) maximum at B B BBB
(2) maximum at C C C\mathrm{C}C
(3) same at all the three points A, B and C
(4) maximum at A A A\mathrm{A}A
Q.36 Two pith balls carrying equal charges are suspended from a common point by strings on equal length, the equilibrium separation between them is r r r\mathrm{r}r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes :
[AIPMT 2013]

(1) ( r 2 3 ) r 2 3 ((r)/(root(3)(2)))\left(\frac{r}{\sqrt[3]{2}}\right)(r23)
( 2 ) ( 2 r 3 ) ( 2 ) 2 r 3 (2)((2r)/(sqrt3))(2)\left(\frac{2 r}{\sqrt{3}}\right)(2)(2r3)
(3) ( 2 r 3 ) 2 r 3 ((2r)/(3))\left(\frac{2 r}{3}\right)(2r3)
(4) ( 1 2 ) 2 1 2 2 ((1)/(sqrt2))^(2)\left(\frac{1}{\sqrt{2}}\right)^{2}(12)2
Q.37 In a region, the potential is represented by V ( x , y , z ) = 6 x − 8 x y − 8 y + 6 y z V ( x , y , z ) = 6 x − 8 x y − 8 y + 6 y z V(x,y,z)=6x-8xy-8y+6yzV(x, y, z)=6 x-8 x y-8 y+6 y zV(x,y,z)=6x−8xy−8y+6yz, where V V VVV is in volts and x , y , z x , y , z x,y,z\mathrm{x}, \mathrm{y}, \mathrm{z}x,y,z are in meters. The electric force experienced by a charge of 2 coulomb situated at point ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) is
[AIPMT-2014]
(1) 24 N 24 N 24N24 \mathrm{~N}24 N
(2) 4 35 N 4 35 N 4sqrt35N4 \sqrt{35} \mathrm{~N}435 N
(3) 6 5 N 6 5 N 6sqrt5N6 \sqrt{5} \mathrm{~N}65 N
(4) 30 N 30 N 30N30 \mathrm{~N}30 N
Q.38 Two thin dielectric slabs of dielectric constants K 1 K 1 K_(1)\mathrm{K}_{1}K1 and K 2 ( K 1 < K 2 ) K 2 K 1 < K 2 K_(2)(K_(1) < K_(2))\mathrm{K}_{2}\left(\mathrm{~K}_{1}<\mathrm{K}_{2}\right)K2( K1<K2) are inserted between plates of a parallel plate capacitor, as shown in figure. The variation of electric field 'E' between the plates with distance ' d d d\mathrm{d}d ' as measured from plate P P PPP is correctly shown by:
[AIPMT-2014]
( 2 ) ( 2 ) (2)(2)(2)
(3)

83. ELECTROSTATICS

Q.39 A conducting sphere of radius R R R\mathrm{R}R is given a charge Q Q Q\mathrm{Q}Q. The electric potential and the electric field at the centre of the sphere respectively are :
[AIPMT-2014]
(1) Q 4 π ∈ 0 R Q 4 π ∈ 0 R (Q)/(4piin_(0)R)\frac{\mathrm{Q}}{4 \pi \in_{0} R}Q4π∈0R and Q 4 π ∈ 0 R 2 Q 4 π ∈ 0 R 2 (Q)/(4piin_(0)R^(2))\frac{\mathrm{Q}}{4 \pi \in_{0} \mathrm{R}^{2}}Q4π∈0R2
(2) both are zero
(3) zero and Q 4 π ∈ 0 R 2 Q 4 π ∈ 0 R 2 (Q)/(4piin_(0)R^(2))\frac{Q}{4 \pi \in_{0} R^{2}}Q4π∈0R2
(4) Q 4 π ϵ 0 R Q 4 π ϵ 0 R (Q)/(4piepsilon_(0)R)\frac{Q}{4 \pi \epsilon_{0} R}Q4πϵ0R and zero
Q.40 A parallel plate air capacitor of capacitance C C C\mathrm{C}C is connected to a cell of emf V V V\mathrm{V}V and then disconnected from it. A dielectric slab of dielectric constant K K K\mathrm{K}K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?
[AIPMT NEET 2015]
(1) The energy stored in the capacitor decreases K K K\mathrm{K}K times
(2) The change in energy stored is 1 2 C V 2 ( 1 K − 1 ) 1 2 C V 2 1 K − 1 (1)/(2)CV^(2)((1)/((K))-1)\frac{1}{2} \mathrm{CV}^{2}\left(\frac{1}{\mathrm{~K}}-1\right)12CV2(1 K−1)
(3) The charge on the capacitor is not conserved
(4) The potential difference between the plates decreases K K K\mathrm{K}K times.
Q.41 The electric field in a certain region is acting radially outward and is given by E = E = E=\mathrm{E}=E= Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by
[AIPMT NEET 2015]
(1) A ∈ 0 a 2 A ∈ 0 a 2 Ain_(0)a^(2)\mathrm{A} \in_{0} \mathrm{a}^{2}A∈0a2
(2) 4 π ∈ 0 A a 3 4 π ∈ 0 A a 3 4piin_(0)Aa^(3)4 \pi \in_{0} \mathrm{Aa}^{3}4π∈0Aa3
(3) ∈ 0 A a 3 ∈ 0 A a 3 in_(0)Aa^(3)\in_{0} \mathrm{Aa}^{3}∈0Aa3
(4) 4 π ∈ 0 A a 2 4 π ∈ 0 A a 2 4pi in_(0)Aa^(2)4 \pi \in{ }_{0} \mathrm{Aa}^{2}4π∈0Aa2
Q.42 An electric dipole is placed at an angle of 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘ with an electric field intensity 2 × 10 5 N / C 2 × 10 5 N / C 2xx10^(5)N//C2 \times 10^{5} \mathrm{~N} / \mathrm{C}2×105 N/C. It experiences a torque equal to 4 N m 4 N m 4Nm4 \mathrm{~N} \mathrm{~m}4 N m. The charge on the dipole, if the dipole length is 2 c m 2 c m 2cm2 \mathrm{~cm}2 cm, is
[NEET 2016]
(1) 7 μ C 7 μ C 7muC7 \mu \mathrm{C}7μC
(2) 8 m C 8 m C 8mC8 \mathrm{mC}8mC
(3) 2 m C 2 m C 2mC2 \mathrm{mC}2mC
(4) 5 m C 5 m C 5mC5 \mathrm{mC}5mC
Q.43 Two identical charged spheres suspended from a common point by two massless strings of lengths ℓ ℓ ℓ\ellℓ are initially at a distance d ( d << ℓ ) d ( d << ℓ ) d(d<<ℓ)\mathrm{d}(\mathrm{d}<<\ell)d(d<<ℓ) a part because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v v vvv. Then v v vvv varies as a function of the distance x x xxx between the spheres, as :
[NEET 2016]
( 1 ) v ∝ x − 1 ( 1 ) v ∝ x − 1 (1)vpropx^(-1)(1) \mathrm{v} \propto \mathrm{x}^{-1}(1)v∝x−1
( 2 ) v ∝ x 1 / 2 ( 2 ) v ∝ x 1 / 2 (2)v propx^(1//2)(2) v \propto x^{1 / 2}(2)v∝x1/2
(3) v ∝ x v ∝ x v prop xv \propto xv∝x
(4) v ∝ x − 1 / 2 v ∝ x − 1 / 2 v propx^(-1//2)v \propto x^{-1 / 2}v∝x−1/2
Q.44 Three identical charges are placed at the vertices of an equilateral triangle. The force experienced by each charge, (if k = 1 / 4 π ε 0 k = 1 / 4 π ε 0 k=1//4piepsi_(0)\mathrm{k}=1 / 4 \pi \varepsilon_{0}k=1/4πε0 ) is
[NEET-2016]
(1) 2 k q 2 r 2 2 k q 2 r 2 2k(q^(2))/(r^(2))2 \mathrm{k} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}}2kq2r2
(2) k q 2 2 r 2 k q 2 2 r 2 (kq^(2))/(2r^(2))\frac{\mathrm{kq}^{2}}{2 \mathrm{r}^{2}}kq22r2
(3) 3 k q 2 r 2 3 k q 2 r 2 sqrt3k(q^(2))/(r^(2))\sqrt{3} \mathrm{k} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}}3kq2r2
(4) k q 2 2 r 2 k q 2 2 r 2 (kq^(2))/(sqrt2r^(2))\frac{\mathrm{kq}^{2}}{\sqrt{2} r^{2}}kq22r2

84. ELECTROSTATICS

Q.45 Suppose the charge of a proton and an electron differ slightly. One of them is − e − e -e-\mathrm{e}−e, the other is ( e + e + e+\mathrm{e}+e+ Δ e ) Δ e ) Deltae)\Delta \mathrm{e})Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d d d\mathrm{d}d (much greater than atomic size) apart is zero, then Δ e Δ e Deltae\Delta \mathrm{e}Δe is of the order of [Given mass of hydrogen m h = 1.67 × 10 − 27 k g m h = 1.67 × 10 − 27 k g m_heart=1.67 xx10^(-27)kg\mathrm{m}_{\mathrm{h}}=1.67 \times 10^{-27} \mathrm{~kg}mh=1.67×10−27 kg ]
(1) 10 − 20 C 10 − 20 C 10^(-20)C10^{-20} \mathrm{C}10−20C
(2) 10 − 23 C 10 − 23 C 10^(-23)C10^{-23} \mathrm{C}10−23C
(3) 10 − 37 C 10 − 37 C 10^(-37)C10^{-37} \mathrm{C}10−37C
(4) 10 − 47 C 10 − 47 C 10^(-47)C10^{-47} \mathrm{C}10−47C
[NEET2017]
Q.46 The diagrams below show regions of equipotentials.
[NEET2017]
(a)
(b)
(c)
(d)
A positive charge is moved from A A A\mathrm{A}A to B B B\mathrm{B}B in each diagram
(1) Maximum work is required to move q q qqq in figure (c).
(2) In all the four cases the work done is the same.
(3) Minimum work is required to move q q qqq in figure (a)
(4) Maximum work is required to move q q qqq in figure (b).

85. ELECTROSTATICS

86. EXERCISE-4 (B)
PREVIOUS YEAR'S QUESTIONS (AIIMS)

Q.1 Three capacitors of capacitance 3 μ F , 10 μ F 3 μ F , 10 μ F 3muF,10 muF3 \mu \mathrm{F}, 10 \mu \mathrm{F}3μF,10μF and 15 μ F 15 μ F 15 muF15 \mu \mathrm{F}15μF are connected in series to a voltage source of 100 V 100 V 100V100 \mathrm{~V}100 V. The charge of 15 μ F 15 μ F 15 muF15 \mu \mathrm{F}15μF is :
(1) 50 μ C 50 μ C 50 muC50 \mu \mathrm{C}50μC
(2) 160 μ C 160 μ C 160 muC160 \mu \mathrm{C}160μC
(3) 200 μ C 200 μ C 200 muC200 \mu \mathrm{C}200μC
(4) 280 μ C 280 μ C 280 muC280 \mu \mathrm{C}280μC
[AIIMS 2000]
Q.2 Minimum number of 8 μ F 8 μ F 8muF8 \mu \mathrm{F}8μF and 250 V 250 V 250V250 \mathrm{~V}250 V capacitors used to make a combination of 16 μ F 16 μ F 16 muF16 \mu \mathrm{F}16μF and 1000 V V V\mathrm{V}V are:
(1) 32
(2) 16
(3) 8
(4) 4
[AIIMS 2000]
Q.3 Assertion : A metallic shield in form of a hollow shell may be built to block an electric field. Reason : In a hollow spherical shield, the electric field inside it is zero at every point.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
[AIIMS-2001]
Q.4 Two materials of dielectric constant k 1 k 1 k_(1)\mathrm{k}_{1}k1 and k 2 k 2 k_(2)\mathrm{k}_{2}k2 are filled between two parallel plates of a capacitor is :
[AIIMS 2001]

(1) A ∈ 0 ( k 1 + k 2 ) 2 d A ∈ 0 k 1 + k 2 2 d (Ain_(0)(k_(1)+k_(2)))/(2(d))\frac{\mathrm{A} \in_{0}\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right)}{2 \mathrm{~d}}A∈0(k1+k2)2 d
(2) 2 A ∈ 0 d ( k 1 k 2 k 1 + k 2 ) 2 A ∈ 0 d k 1 k 2 k 1 + k 2 (2(A)in_(0))/((d))((k_(1)k_(2))/(k_(1)+k_(2)))\frac{2 \mathrm{~A} \in_{0}}{\mathrm{~d}}\left(\frac{\mathrm{k}_{1} \mathrm{k}_{2}}{\mathrm{k}_{1}+\mathrm{k}_{2}}\right)2 A∈0 d(k1k2k1+k2)
(3) A ∈ 0 d ( k 1 k 2 k 1 + k 2 ) A ∈ 0 d k 1 k 2 k 1 + k 2 (Ain_(0))/(d)((k_(1)k_(2))/(k_(1)+k_(2)))\frac{A \in_{0}}{d}\left(\frac{k_{1} k_{2}}{k_{1}+k_{2}}\right)A∈0d(k1k2k1+k2)
(4) A ∈ 0 2 d ( k 1 + k 2 k 1 k 2 ) A ∈ 0 2 d k 1 + k 2 k 1 k 2 (Ain_(0))/(2(d))((k_(1)+k_(2))/(k_(1)k_(2)))\frac{\mathrm{A} \in_{0}}{2 \mathrm{~d}}\left(\frac{\mathrm{k}_{1}+\mathrm{k}_{2}}{\mathrm{k}_{1} \mathrm{k}_{2}}\right)A∈02 d(k1+k2k1k2)
Q.5 Assertion : If three capacitors of capacitance C 1 < C 2 < C 3 C 1 < C 2 < C 3 C_(1) < C_(2) < C_(3)\mathrm{C}_{1}<\mathrm{C}_{2}<\mathrm{C}_{3}C1<C2<C3 are connected in parallel then their equivalent capacitance C p > C 3 C p > C 3 C_(p) > C_(3)\mathrm{C}_{\mathrm{p}}>\mathrm{C}_{3}Cp>C3.
Reason : 1 C p = 1 C 1 + 1 C 2 + 1 C 3 1 C p = 1 C 1 + 1 C 2 + 1 C 3 (1)/(C_(p))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))\frac{1}{\mathrm{C}_{\mathrm{p}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}1Cp=1C1+1C2+1C3
[AIIMS - 2002]
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.6 Assertion : Electric lines of force never cross each other.
[AIIMS 2002]
Reason : Electric field at a point superimpose to give one resultant electric field.
(1) A A A\mathrm{A}A
(2) B B B\mathrm{B}B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
Q.7 An electron having charge ' e e e\mathrm{e}e ' and mass ' m m m\mathrm{m}m ' is moving in a uniform electric field E. Its acceleration will be :
[AIIMS-2002]
(1) e 2 m e 2 m (e^(2))/((m))\frac{\mathrm{e}^{2}}{\mathrm{~m}}e2 m
(2) E 2 e m E 2 e m (E^(2)e)/(m)\frac{E^{2} e}{m}E2em
(3) e E m e E m (eE)/(m)\frac{\mathrm{eE}}{\mathrm{m}}eEm
(4) m E e m E e (mE)/(e)\frac{m E}{e}mEe
Q.8 A conducting sphere of radius 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm is charged 10 μ C 10 μ C 10 muC10 \mu \mathrm{C}10μC. Another uncharged sphere of radius 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm is allowed to touch it for some time. After that if the sphere are separated, then surface density of charges, on the spheres will be in the ratio of :
[AIIMS 2002]
(1) 1 : 4 1 : 4 1:41: 41:4
(2) 1 : 3 1 : 3 1:31: 31:3
(3) 2 : 1 2 : 1 2:12: 12:1
(4) 1 : 1 1 : 1 1:11: 11:1

87. ELECTROSTATICS

Q.9 A conducting sphere of radius 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm is charged with 10 μ C 10 μ C 10 muC10 \mu \mathrm{C}10μC. Another uncharged sphere of radius 20 c m 20 c m 20cm20 \mathrm{~cm}20 cm is allowed to touch it for some time. After that if the sphere are separated, then surface density of charged on the spheres will be in ratio of :
[AIIMS 2002]
(1) 1 : 4 1 : 4 1:41: 41:4
(2) 1 : 3 1 : 3 1:31: 31:3
(3) 2 : 1 2 : 1 2:12: 12:1
(4) 1 : 1 1 : 1 1:11: 11:1
Q.10 In the given figure, the capacitor C 1 , C 3 , C 4 , C 5 C 1 , C 3 , C 4 , C 5 C_(1),C_(3),C_(4),C_(5)\mathrm{C}_{1}, \mathrm{C}_{3}, \mathrm{C}_{4}, \mathrm{C}_{5}C1,C3,C4,C5 have a capacitor 4 μ F 4 μ F 4muF4 \mu \mathrm{F}4μF each. If the capacitor C 2 C 2 C_(2)\mathrm{C}_{2}C2 has a capacitance 10 μ F 10 μ F 10 muF10 \mu \mathrm{F}10μF, then effective capacitance between A and B will be : [AIIMS 2002]

(1) 2 μ F 2 μ F 2muF2 \mu \mathrm{F}2μF
(2) 4 μ F 4 μ F 4muF4 \mu \mathrm{F}4μF
(3) 6 μ F 6 μ F 6muF6 \mu \mathrm{F}6μF
(4) 8 μ F 8 μ F 8muF8 \mu \mathrm{F}8μF
Q.11 Assertion : The coulomb force is the dominating force in the universe.
Reason : The coulomb force is weaker than the gravitational force.
(1) A A A\mathrm{A}A
(2) B
(3) C C C\mathrm{C}C
(4) D D D\mathrm{D}D
[AIIMS-2003]
Q.12 A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 k V 1 k V 1kV1 \mathrm{kV}1kV, its kinetic energy will be :
(1) 1840 k e V 1840 k e V 1840keV1840 \mathrm{keV}1840keV
(2) 1 / 1840 k e V 1 / 1840 k e V 1//1840keV1 / 1840 \mathrm{keV}1/1840keV
(3) 1 k e V 1 k e V 1keV1 \mathrm{keV}1keV
(4) 920 k e V 920 k e V 920keV920 \mathrm{keV}920keV
[AIIMS 2003]
Q.13 A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 k e V 1 k e V 1keV1 \mathrm{keV}1keV, its kinetic energy will be :
[AIIMS 2003]
(1) 1840 k e V 1840 k e V 1840keV1840 \mathrm{keV}1840keV
(2) 1 / 1840 k e V 1 / 1840 k e V 1//1840keV1 / 1840 \mathrm{keV}1/1840keV
(3) 1 k e V 1 k e V 1keV1 \mathrm{keV}1keV
(4) 920 k e V 920 k e V 920keV920 \mathrm{keV}920keV
Q.14 An electric dipole placed in a non-uniform electric field experiences :
[AIIMS 2003]
(1) both, a torque and a net force
(2) only a force but no torque
(3) only a torque but no net force
(4) no torque and no net force
Q.15 Three charges are placed at the vertices of an equilateral triangle of side a as shown in the following figure. The force experienced by the charge placed at the vertex A A A\mathrm{A}A in a direction normal to B C B C BC\mathrm{BC}BC is :
[AIIMS 2003]
(1) Q 2 ( 4 π ε 0 a 2 ) Q 2 4 π ε 0 a 2 Q^(2)(4piepsi_(0)a^(2))\mathrm{Q}^{2}\left(4 \pi \varepsilon_{0} \mathrm{a}^{2}\right)Q2(4πε0a2)
( 2 ) − Q 2 ( 4 π ε 0 a 2 ) ( 2 ) − Q 2 4 π ε 0 a 2 (2)-Q^(2)(4piepsi_(0)a^(2))(2)-Q^{2}\left(4 \pi \varepsilon_{0} a^{2}\right)(2)−Q2(4πε0a2)
(3) zero
(4) Q 2 / ( 2 π ε 0 a 2 ) Q 2 / 2 π ε 0 a 2 Q^(2)//(2piepsi_(0)a^(2))\mathrm{Q}^{2} /\left(2 \pi \varepsilon_{0} \mathrm{a}^{2}\right)Q2/(2πε0a2)

88. ELECTROSTATICS

Q.16 The electric field due to a uniformly charged sphere of radius R R R\mathrm{R}R as a function of the distance from its centre is represented graphically by :
[AIIMS - 2004]
( 1 ) ( 1 ) (1)(1)(1)
( 2 ) ( 2 ) (2)(2)(2)
(3)
( 4 ) ( 4 ) (4)(4)(4)
Q.17 A 40 μ F 40 μ F 40 muF40 \mu \mathrm{F}40μF capacitor in a defibrillator is charged to 3000 V 3000 V 3000V3000 \mathrm{~V}3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 m s 2 m s 2ms2 \mathrm{~ms}2 ms. The power delivered to the patient is :
(1) 45 k W 45 k W 45kW45 \mathrm{~kW}45 kW
(2) 90 k W 90 k W 90kW90 \mathrm{~kW}90 kW
(3) 180 k W 180 k W 180kW180 \mathrm{~kW}180 kW
(4) 360 k W 360 k W 360kW360 \mathrm{~kW}360 kW
[AIIMS 2004]
Q.18 Equipotential surfaces associated with an electric field which is increasing in magnitude along the x x x\mathrm{x}x-direction are :
[AIIMS-2004]
(1) Planes parallel to yz-plane
(2) Planes parallel to xy-plane
(3) Planes parallel to xz-plane
(4) Coaxial cylinders of increasing radii around the x x xxx-axis
Q.19 In the basic C s C l C s C l CsCl\mathrm{CsCl}CsCl crystal structure, C s + C s + Cs^(+)\mathrm{Cs}^{+}Cs+and C l − C l − Cl^(-)\mathrm{Cl}^{-}Cl−ions are arranged in a bec configuration as shown in the figure. The net electrostatic force exerted by the eight C s + C s + Cs^(+)\mathrm{Cs}^{+}Cs+ions on the C l − C l − Cl^(-)\mathrm{Cl}^{-}Cl−ion is :
[AIIMS 2004]

(1) 1 4 π ε 0 4 e 2 3 a 2 1 4 π ε 0 4 e 2 3 a 2 (1)/(4piepsi_(0))(4e^(2))/(3a^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \mathrm{e}^{2}}{3 \mathrm{a}^{2}}14πε04e23a2
(2) 1 4 π ε 0 16 e 2 3 a 2 1 4 π ε 0 16 e 2 3 a 2 (1)/(4piepsi_(0))(16e^(2))/(3a^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{16 \mathrm{e}^{2}}{3 \mathrm{a}^{2}}14πε016e23a2
(3) 1 4 π ε 0 32 e 2 3 a 2 1 4 π ε 0 32 e 2 3 a 2 (1)/(4piepsi_(0))(32e^(2))/(3a^(2))\frac{1}{4 \pi \varepsilon_{0}} \frac{32 \mathrm{e}^{2}}{3 \mathrm{a}^{2}}14πε032e23a2
(4) zero
Q.20 Four point + ve charges of same magnitude (Q) are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to Z Z ZZZ axis. If a - ve point charge is placed at a distance z z z\mathrm{z}z away from the above frame ( z << L ) ( z << L ) (z<<L)(\mathrm{z}<<\mathrm{L})(z<<L) then
[AIIMS - 2005]
(1) - ve charge oscillates along the Z-axis.
(2) It moves away from the frame
(3) It moves slowly towards the frame and stays in the plane of the frame
(4) It passes through the frame only once

89. ELECTROSTATICS

Q.21 Two infinitely long parallel conducting plates having surface charge densities + σ + σ +sigma+\sigma+σ and − σ − σ -sigma-\sigma−σ respectively, are separated by a small distance. The medium between the plates is vacuum. If ε 0 ε 0 epsi_(0)\varepsilon_{0}ε0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is:
[AIIMS - 2005]
(1) 0 volts/meter
(2) σ 2 ε 0 σ 2 ε 0 (sigma)/(2epsi_(0))\frac{\sigma}{2 \varepsilon_{0}}σ2ε0 volts/meter
(3) σ ε 0 σ ε 0 (sigma)/(epsi_(0))\frac{\sigma}{\varepsilon_{0}}σε0 volts/meter
(4) 2 σ ε 0 2 σ ε 0 (2sigma)/(epsi_(0))\frac{2 \sigma}{\varepsilon_{0}}2σε0 volts/meter
Q.22 Two concentric conducting thin spherical shells A A AAA, and B B BBB having radii r A r A r_(A)r_{A}rA and r B ( r B > r A ) r B r B > r A r_(B)(r_(B) > r_(A))r_{B}\left(r_{B}>r_{A}\right)rB(rB>rA) are charged to Q A Q A Q_(A)Q_{A}QA and − Q B [ | Q B | > | Q A | ] − Q B Q B > Q A -Q_(B)[|Q_(B)| > |Q_(A)|]-Q_{B}\left[\left|Q_{B}\right|>\left|Q_{A}\right|\right]−QB[|QB|>|QA|]. The electrical field along a line, (passing through the centre) is :
[AIIMS - 2005]
Q.23 Two infinitely long parallel conducting plates having surface charge densities + σ + σ +sigma+\sigma+σ and − σ − σ -sigma-\sigma−σ respectively, are separated by a small distance. The medium between the plates is vacuum. It ε 0 ε 0 epsi_(0)\varepsilon_{0}ε0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is :
[AIIMS 2005]
(1) zero
(2) σ / 2 ε 0 V / m σ / 2 ε 0 V / m sigma//2epsi_(0)V//m\sigma / 2 \varepsilon_{0} \mathrm{~V} / \mathrm{m}σ/2ε0 V/m
(3) σ / ε 0 V / m σ / ε 0 V / m sigma//epsi_(0)V//m\sigma / \varepsilon_{0} \mathrm{~V} / \mathrm{m}σ/ε0 V/m
(4) 2 σ / ε 0 V / m 2 σ / ε 0 V / m 2sigma//epsi_(0)V//m2 \sigma / \varepsilon_{0} \mathrm{~V} / \mathrm{m}2σ/ε0 V/m
Q.24 The spatial distribution of the electric field due to charges (A, B) is shown in figure. Which one of the following statements is correct ?
[AIIMS 2006]

(1) A A A\mathrm{A}A is + ve and B − B − B-\mathrm{B}-B− ve and | A | > | B | | A | > | B | |A| > |B||\mathrm{A}|>|\mathrm{B}||A|>|B|
(2) A A A\mathrm{A}A is - ve and + ve B ; | A | = | B | B ; | A | = | B | B;|A|=|B|\mathrm{B} ;|\mathrm{A}|=|\mathrm{B}|B;|A|=|B|
(3) both are + ve but A > B A > B A > B\mathrm{A}>\mathrm{B}A>B
(4) both are - ve but A > B A > B A > B\mathrm{A}>\mathrm{B}A>B
Q.25 In the figure, a proton moves a distance d d d\mathrm{d}d in a uniform electric E → E → vec(E)\overrightarrow{\mathrm{E}}E→ as shown in the figure. Does the electric field do a positive or negative work on the proton ? Does the electric potential energy of the proton increase or decrease?
[AIIMS 2007]

(1) Negative, increase
(2) Positive, decrease
(3) Negative, decrease
(4) Positive, increase ELECTROSTATICS
Q.26 A parallel plate capacitor is made by stacking n n n\mathrm{n}n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C C C\mathrm{C}C, then the resultant capacitance is :
[AIIMS-2007]
(1) ( n − 1 ) C ( n − 1 ) C (n-1)C(\mathrm{n}-1) \mathrm{C}(n−1)C
( 2 ) ( n + 1 ) C ( 2 ) ( n + 1 ) C (2)(n+1)C(2)(n+1) C(2)(n+1)C
(3) C C C\mathrm{C}C
(4) n C n C nC\mathrm{nC}nC
Q.27 When the key K K K\mathrm{K}K is pressed at t = 0 t = 0 t=0\mathrm{t}=0t=0, which of the following statements about the current I I I\mathrm{I}I in the resistor A B A B AB\mathrm{AB}AB of the given circuit is true ?
[AIIMS-2008]

(1) I = 2 m A I = 2 m A I=2mA\mathrm{I}=2 \mathrm{~mA}I=2 mA at all t t t\mathrm{t}t
(2) I oscillates between 1 m A 1 m A 1mA1 \mathrm{~mA}1 mA and 2 m A 2 m A 2mA2 \mathrm{~mA}2 mA
(3) I = 1 m a I = 1 m a I=1ma\mathrm{I}=1 \mathrm{ma}I=1ma at all t t t\mathrm{t}t
(4) at t = 0 , I = 2 m A t = 0 , I = 2 m A t=0,I=2mA\mathrm{t}=0, \mathrm{I}=2 \mathrm{~mA}t=0,I=2 mA and with time it goes to 1 m A 1 m A 1mA1 \mathrm{~mA}1 mA
Q.28 The voltage of clouds is 4 × 10 6 4 × 10 6 4xx10^(6)4 \times 10^{6}4×106 volt with respect to ground. In a lightening strike lasting 100 m 100 m 100m100 \mathrm{~m}100 m sec, a charge of 4 coulombs is delivered to the ground. The power of lightening strike is :
(1) 160 M W 160 M W 160MW160 \mathrm{MW}160MW
(2) 80 M W 80 M W 80MW80 \mathrm{MW}80MW
(3) 20 M W 20 M W 20MW20 \mathrm{MW}20MW
(4) 500 K W 500 K W 500KW500 \mathrm{KW}500KW
[AIIMS 2008]
Q.29 A charge Q Q Q\mathrm{Q}Q is divided into two parts of q q q\mathrm{q}q and Q − q Q − q Q-q\mathrm{Q}-\mathrm{q}Q−q. If the coulomb repulsion between them when they are separated is to be maximum, the ratio of Q q Q q (Q)/(q)\frac{\mathrm{Q}}{\mathrm{q}}Qq should be :
[AIIMS 2011]
(1) 2
(2) 1 / 2 1 / 2 1//21 / 21/2
(3) 4
(4) 1 / 4 1 / 4 1//41 / 41/4
Q.30 Assertion (A) If the bob of a simple pendulum kept in a horizontal electric field, its period of oscillation will remain same.
Reason (R) If bob is charged and kept in horizontal electric field, then the time period will be decreased.
(1) Both A A A\mathrm{A}A and R R R\mathrm{R}R are correct and R R R\mathrm{R}R is the correct explanation of A A A\mathrm{A}A
[AIIMS 2012]
(2) Both A A AAA and R R RRR correct but R R RRR is not the correct explanation of A A AAA
(3) A A AAA is correct but R R RRR is incorrect
(4) A A A\mathrm{A}A is incorrect but R R R\mathrm{R}R is correct
Q.31 Two charged spheres separated by a distance ' d d ddd ' exert some force on each other. If they are immersed in a liquid of dielectric constant 2 , then what is the force exerted, if all other conditions are same?
[AIIMS-2016]
(1) F / 2 F / 2 F//2F / 2F/2
(2) F F F\mathrm{F}F
(3) 2 F 2 F 2F2 \mathrm{~F}2 F
(4) 4 F 4 F 4F4 \mathrm{~F}4 F
[AIIMS-2016]
Q.32 Assertion : Lines of force are perpendicular to conductor surface.
Reason : Generally electric field is perpendicular to equipotential surface.
(1) If both assertion and reason are true and reason is the correct explanation of assertion.
(2) If both assertion and reason are true but reason is not the correct explanation of assertion.
(3) If assertion is true but reason is false.
(4) If both assertion and reason are false.

90. ELECTROSTATICS

Q.33 If a charge q q q\mathrm{q}q is placed at the centre of the line joining two equal charges Q Q Q\mathrm{Q}Q such that the system is in equilibrium then the value of q q qqq is
[AIIMS 2017]
(1) Q / 2 Q / 2 Q//2\mathrm{Q} / 2Q/2
( 2 ) − Q / 2 ( 2 ) − Q / 2 (2)-Q//2(2)-Q / 2(2)−Q/2
(3) Q / 4 Q / 4 Q//4\mathrm{Q} / 4Q/4
(4) − Q / 4 − Q / 4 -Q//4-\mathrm{Q} / 4−Q/4
Q.34 Which of the following figure shows the correct equipotential surfaces of a system of two positive charges?
[AIIMS 2017]
( 4 ) ( 4 ) (4)(4)(4)
Q.35 Gauss's law states that
[AIIMS 2017]
(1) the total electric flux through a closed surface is 1 ε 0 1 ε 0 (1)/(epsi_(0))\frac{1}{\varepsilon_{0}}1ε0 times the total charge placed near the closed surface.
(2) the total electric flux through a closed surface is 1 ε 0 1 ε 0 (1)/(epsi_(0))\frac{1}{\varepsilon_{0}}1ε0 times the total charge enclosed by the closed surface.
(3) the total electric flux through an open surface is 1 ε 0 1 ε 0 (1)/(epsi_(0))\frac{1}{\varepsilon_{0}}1ε0 times the total charge placed near the open surface.
(4) the line integral of electric field around the boundary of an open surface is 1 ε 0 1 ε 0 (1)/(epsi_(0))\frac{1}{\varepsilon_{0}}1ε0 times the total charge placed near the open surface.
Q.36 Figure shows the electric lines of force emerging from a charged body. If the electric field at A and B B BBB are E A E A E_(A)E_{A}EA and E B E B E_(B)E_{B}EB respectively and if the displacement between A A AAA and B B BBB is r r rrr, then [AIIMS 2017]

(1) E A > E B E A > E B E_(A) > E_(B)\mathrm{E}_{\mathrm{A}}>\mathrm{E}_{\mathrm{B}}EA>EB
(2) E A < E B E A < E B E_(A) < E_(B)\mathrm{E}_{\mathrm{A}}<\mathrm{E}_{\mathrm{B}}EA<EB
(3) E A = E B / r E A = E B / r E_(A)=E_(B)//r\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{B}} / \mathrm{r}EA=EB/r
(4) E A = E B / r 2 E A = E B / r 2 E_(A)=E_(B)//r^(2)\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{B}} / \mathrm{r}^{2}EA=EB/r2

91. ELECTROSTATICS

92. EXERCISE-5

93. (CBSE Previous Year's Questions)

Q.1 Define the term electric dipole moment. Is it a scalar or a vector quantity? [1; CBSE-2006]
Q.2 A point charge ' q q qqq ' is placed at O O O\mathrm{O}O as shown in the figure.
[ 2 ; [ 2 ; [2;[2 ;[2; CBSE-2006]
Is V P − V Q V P − V Q V_(P)-V_(Q)V_{P}-V_{Q}VP−VQ positive or negative when (i) q > 0 q > 0 q > 0q>0q>0, (ii) q < 0 q < 0 q < 0q<0q<0 ? Justify your answer.
Q.3 Using Gauss's theorem, show mathematically that for any point outside the shell, the field due to a uniformly charged thin spherical shell is the same as if the entire charge of the shell is concentrated at the centre. Why do you expect the electric field inside the shell to be zero according to this theorem?
[3; CBSE-2006]
Q.4 Two point charges 4 μ C 4 μ C 4muC4 \mu \mathrm{C}4μC and − 2 μ C − 2 μ C -2muC-2 \mu \mathrm{C}−2μC are separated by a distance of 1 m 1 m 1m1 \mathrm{~m}1 m in air. Calculate at what point on the line joining the two charges is the electric potential zero.
[1; CBSE-2007]
Q.5 State Gauss's theorem in electrostatics. Apply this theorem to derive an expression for electric field intensity at a point near an infinitely long straight charged wire.
[2; CBSE-2007]
Q.6 A 500 μ C 500 μ C 500 muC500 \mu \mathrm{C}500μC Charge is at the centre of a square of side 10 c m 10 c m 10cm10 \mathrm{~cm}10 cm. Find the work done in moving a charge of 10 μ C 10 μ C 10 muC10 \mu \mathrm{C}10μC between two diagonally opposite points on the square.
[ 1 ; [ 1 ; [1;[1 ;[1; CBSE-2008]
Q.7 Derive the expression for the electric potential at any point along the axial line of an electric dipole.
[1; CBSE-2008]
Q.8 (i) Can two equi - potential surfaces intersect each other? Give reasons.
(ii) Two charges − q − q -q-\mathrm{q}−q and + q + q +q+\mathrm{q}+q are located at points A ( 0 , 0 , − a ) A ( 0 , 0 , − a ) A(0,0,-a)\mathrm{A}(0,0,-\mathrm{a})A(0,0,−a) and B ( 0 , 0 , + a ) B ( 0 , 0 , + a ) B(0,0,+a)\mathrm{B}(0,0,+\mathrm{a})B(0,0,+a) respectively. How much work is done in moving a test charge from point P ( 7 , 0 , 0 ) P ( 7 , 0 , 0 ) P(7,0,0)\mathrm{P}(7,0,0)P(7,0,0) to Q ( − 3 , 0 , 0 ) Q ( − 3 , 0 , 0 ) Q(-3,0,0)\mathrm{Q}(-3,0,0)Q(−3,0,0) ?
[2; CBSE-2009]
Q.9 State Gauss's law in electrostatics. Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet.
[3; CBSE-2009]
Q.10 Name the physical quantity whose S.I. unit is J C − 1 J C − 1 JC^(-1)\mathrm{JC}^{-1}JC−1. Is it a scalar or a vector quantity?
[1; CBSE-2010]

94. ELECTROSTATICS

Q.11 Define electric dipole moment. Write its S.I. unit.
[1; CBSE-2011]
Q.12 A hollow metal sphere of radius 5 c m 5 c m 5cm5 \mathrm{~cm}5 cm is charged such that the potential on its surface is 10 V 10 V 10V10 \mathrm{~V}10 V. What is the potential at the centre of the sphere?
[1; CBSE-2011]
Q.13 A thin straight infinitely long conducting wire having charge density λ λ lambda\lambdaλ is enclosed by a cylindrical surface of radius r r rrr and length l l lll, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.
[2; CBSE-2011]
Q.14 Plot a graph showing the variation of coulomb force ( F ) ( F ) (F)(\mathrm{F})(F) versus ( 1 r 2 ) 1 r 2 ((1)/(r^(2)))\left(\frac{1}{\mathrm{r}^{2}}\right)(1r2), where r r r\mathrm{r}r is the distance between the two charges of each pair of charges : ( 1 μ C , 2 μ C ) ( 1 μ C , 2 μ C ) (1muC,2muC)(1 \mu \mathrm{C}, 2 \mu \mathrm{C})(1μC,2μC) and ( 2 μ C , − 3 μ C ) ( 2 μ C , − 3 μ C ) (2muC,-3muC)(2 \mu \mathrm{C},-3 \mu \mathrm{C})(2μC,−3μC), interpret the graphs obtained.
[2; CBSE-2011]
Q.15 Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker?
[1; CBSE-2012]
Q.16 A charge ' q q qqq ' is moved without acceleration from A A AAA to C C CCC along the path from A A AAA to B B BBB and then from B B B\mathrm{B}B to C C C\mathrm{C}C in electric field E E E\mathrm{E}E as shown in the figure. (i) Calculate the potential difference between A A A\mathrm{A}A and C. (ii) At which point (of the two) is the electric potential more and why?
Q.17 An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180 ∘ 180 ∘ 180^(@)180^{\circ}180∘.
[ 2 ; [ 2 ; [2;[2 ;[2; CBSE-2012]
Q.18 Two charges of magnitudes − 2 Q − 2 Q -2Q-2 \mathrm{Q}−2Q and + Q + Q +Q+\mathrm{Q}+Q are located at points (a, 0 ) ) ))) and ( 4 a , 0 ) ( 4 a , 0 ) (4a,0)(4 \mathrm{a}, 0)(4a,0) respectively.

origin?
[CBSE-2013]
Q.19 (a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.

95. OR

Using Gauss' law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R R RRR at a point (i) outside and (ii) inside the shell.
Plot a graph showing variation of electric field as a function of r > r > r >r>r> Rand r < R r < R r < Rr<Rr<R. (r being the distance from the centre of the shell)
[CBSE-2013]

96. ELECTROSTATICS

Q.20 Why do the electrostatic field lines not form closed loops?
[CBSE-2014]
Q.21 Draw a labelled diagram of Van de Graaff generator. State its working principle to show how by introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to the outer sphere state the use of this machine and also point out its limitations.
OR
(a) Deduce the expression for the torque acting on a dipole of dipole moment p → p → vec(p)\vec{p}p→ in the presence of a uniform electric field E → E → vec(E)\overrightarrow{\mathrm{E}}E→.
(b) Consider two hollow concentric spheres, S 1 S 1 S_(1)S_{1}S1 and S 2 S 2 S_(2)S_{2}S2, enclosing charges 2 Q 2 Q 2Q2 Q2Q and 4 Q 4 Q 4Q4 Q4Q respectively as shown in the figure. (i) Find out the ratio of the electric flux through them. (ii) How will the electric flux through the sphere S 1 S 1 S_(1)S_{1}S1 change if a medium of dielectric constant ' ε r ′ ε r ′ epsi_(r)^(')\varepsilon_{r}^{\prime}εr′ is introduced in the space inside S 1 S 1 S_(1)S_{1}S1 in place of air ? Deduce the necessary expression.
[CBSE-2014]
Q.22 (a) State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge ' q q qqq ', in vacum within a closed surface, is independent of its size or shape and is given by q / ε 0 q / ε 0 q//epsi_(0)\mathrm{q} / \varepsilon_{0}q/ε0.
(b) Two parallel uniformly charged infinite plane sheets, ' 1 ' and '2', have charge densities + σ + σ +sigma+\sigma+σ and − 2 σ − 2 σ -2sigma-2 \sigma−2σ respectively. Give the magnitude and direction of the net electric field at a point.
(i) in between the two sheets and
(ii) outside near the sheet ' 1 '.
[5; CBSE-2015]
OR
(a) Define electrostatic potential at a point. Write its S.I. unit.
Three point charges q 1 , q 2 q 1 , q 2 q_(1),q_(2)\mathrm{q}_{1}, \mathrm{q}_{2}q1,q2 and q 3 q 3 q_(3)\mathrm{q}_{3}q3 are kept respectively at points A , B A , B A,B\mathrm{A}, \mathrm{B}A,B and C C C\mathrm{C}C as shown in the figure. Derive the expression for the electrostatic potential energy of the system.
(b) Depict the equipotential surfaces due to
(i) an electric dipole,
(ii) two identical positive charges separated by a distance.
[5; CBSE-2015]
Q.23 What is the amount of work done in moving a point charge Q Q Q\mathrm{Q}Q around a circular arc of radius ' r r r\mathrm{r}r ' at the centre of which another point charge ' q q qqq ' is located?
[1; CBSE-2016] ANSWER REY

97. EXERCISE-1

Q.1 3
Q.6 4
Q.11 1
Q.16 3
Q.21 1
Q.26 3
Q.31 3
Q.36 3
Q.41 4
Q.46 1
Q.51 2
Q.56 2
Q.61 1
Q.66 2
Q.71 3
Q.76 1
Q.81 2
Q.86 1
Q.91 2
Q.96 3
Q.101 2
Q.106 4
Q.111 2
Q.116 2
Q.121 1
Q.1 3
Q.6 3
Q.11 3
Q.16 3
Q.21 1
Q.26 4
Q.31 2
Q.36 2
Q.41 2
Q.46 2
Q.51 1
Q.56 3
Q.61 3
Q.66 4
Q.1 2
Q.6 1
Q.11 2
Q.16 1 Q.2 4
Q.7 2
Q.12 3
Q.17 2
Q.22 1
Q.27 4
Q.32 1
Q.37 2
Q.42 2
Q.47 3
Q.52 3
Q.57 2
Q.62 4
Q.67 1
Q.72 2
Q.77 4
Q.82 2
Q.87 3
Q.92 3
Q.97 3
Q.102 3
Q.107 2
Q.112 3
Q.117 2

98. EXERCISE-2

Q.2 1
Q.7 2
Q.12 3
Q.17 2
Q.22 1
Q.27 1
Q.32 1
Q.37 4
Q.42 1
Q.47 4
Q.52 2
Q.57 2
Q.62 1
EXERCISE-3
Q.3 3
Q.2 1
Q.7 3
Q.12 4
Q.17 2 Q.3 1
Q.8 1 1 quad1\quad 11
Q.13 3
Q.18 4
Q.23 1
Q.28 2
Q.33 4
Q.38 1 1 quad1\quad 11
Q.43 3
Q.48 1
Q.53 2
Q.58 1
Q.63 3
Q.68 3
Q.73 2
Q.78 2
Q.83 1
Q.88 3
Q.93 1
Q.98 1
Q.103 1
Q.108 1
Q.113 3
Q.118 1
Q.3 2
Q.8 3
Q.13 2
Q.18 1
Q.23 3
Q.28 2
Q.33 3
Q.38 3
Q.43 2
Q.48 3
Q.53 1
Q.58 2
Q.63 3
Q.8 4
Q.13 2
Q.18 2 Q.4 1
Q.9 1
Q.14 1
Q.19 4
Q.24 4
Q.29 3
Q.34 4
Q.39 2
Q.44 4
Q.49 4
Q.54 2
Q.59 3
Q.64 3
Q.69 1
Q.74 1
Q.79 1
Q.84 2
Q.89 2
Q.94 1
Q.99 4
Q.104 1
Q.109 4
Q.114 2
Q.119 2
Q.4 2
Q.9 2
Q.14 3
Q.19 1
Q.24 2
Q.29 2
Q.34 4
Q.39 2
Q.44 2
Q.49 4
Q.54 2
Q.59 3
Q.64 3
Q.4 1 Q.5 4 Q.9 1 Q.10 1 Q.14 1 Q .15 1  Q.4       1       Q.5       4  Q.9       1       Q.10       1  Q.14       1      Q .15      1 {:[" Q.4 ",1," Q.5 ",4],[" Q.9 ",1," Q.10 ",1],[" Q.14 ",1,Q.15,1]:}\begin{array}{llll}\text { Q.4 } & 1 & \text { Q.5 } & 4 \\ \text { Q.9 } & 1 & \text { Q.10 } & 1 \\ \text { Q.14 } & 1 & Q .15 & 1\end{array} Q.4 1 Q.5 4 Q.9 1 Q.10 1 Q.14 1Q.151
Q.5 1
Q.10 3
Q.15 1
Q.20 2
Q.25 1
Q.30 2
Q.35 2
Q.40 2
Q.45 3
Q.50 4
Q.55 1
Q.60 3
Q.65 4
Q.70 1
Q.75 1
Q.80 2
Q.85 3
Q.90 4
Q.95 1
Q.100 3
Q.105 3
Q.110 3
Q.115 1
Q.120 3
Q.5 1 1 quad1\quad 11
Q.10 3
Q.15 4
Q.20 1
Q.25 2
Q.30 1
Q.35 2
Q.40 1
Q.45 4
Q.50 4
Q.55 2
Q.60 2
Q.65 4
Q.19 1

99. ELECTROSTATICS

100. EXERCISE-4 (A)

Q.1 2
Q.6 1
Q.11 1
Q.16 3
Q.21 4
Q.26 4
Q.31 4
Q.36 1
Q.41 2
Q.46 2
Q.2 1
Q.7 2
Q.12 2
Q.17 2
Q.22 3
Q.27 3
Q.32 3
Q.37 2
Q.42 3
Q.1 3
Q.6 4
Q.11 4
Q.16 2
Q.21 3
Q.26 1
Q.31 1
Q.36 1
Q.2 1 Q.7 3  Q.2       1  Q.7       3 {:[" Q.2 ",1],[" Q.7 ",3]:}\begin{array}{ll}\text { Q.2 } & 1 \\ \text { Q.7 } & 3\end{array} Q.2 1 Q.7 3
Q.12 3
Q.17 2
Q.22 1
Q.27 4
Q.32 1 Q.3 2
Q.8 4
Q.13 3
Q.18 4
Q.23 3
Q.28 1
Q.33 3
Q.38 1
Q.43 4
Q.4 3
Q.9 2
Q.14 2
Q.19 1
Q.24 1
Q.29 3
Q.34 3
Q.39 4
Q.44 3

101. EXERCISE-4 (B)

Q.4 1
Q.9 3
Q.14 1
Q.19 4
Q.24 1
Q.29 1
Q.34 3 Q.5 2
Q.10 3
Q.15 1
Q.20 4
Q.25 3
Q.30 3
Q.35 1
Q.40 3
Q.45 3
Q.5 3
Q.10 2
Q.15 3
Q.20 1
Q.25 1
Q.30 1
Q.35 2
Last modified: Monday, 6 March 2023, 11:53 PM