Q.11. Derive an expression for electric field intensity at a point close to and outside the surface of charged conductor of any shape.

Ans: Expression for electric intensity at a point close to and outside the surface of a charged conductor of any shape:

i. Suppose a charged conductor of any shape is placed in a medium of dielectric constant ' \(\mathrm{k}\) '. \(\mathrm{P}\) is a point near the charged conductor as shown in the figure. We have to find electric field intensity at \(P\).

ii. Let ' \(\sigma\) ' be the surface charge density of conductor.

iii. Draw a Gaussian surface which is cylindrical, half inside and half outside the conductor surface. It is oriented transverse to the surface of the conductor with its outer circular flat end passing through \(P\).

iv. Let 'ds' be the cross-sectional area of the cylinder.

v. Contribution to T.N.E.I from the surface lying within the conductor is zero, as electric field inside the conductor is always zero. vi. Contribution to T.N.E.I from the cylindrical surface lying outside the conductor is zero, since there area vector \(\overrightarrow{\mathrm{ds}}\) and \(\overrightarrow{\mathrm{E}}\) are transverse, i.e., \(\theta=90^{\circ}\) and \(\cos 90^{\circ}=0\).

vii. Thus, the only contribution to T.N.E.I is from the circular flat end passing through \(P\).

\(\therefore \quad\) T.N.E.I \(=\varepsilon E d s \cos \theta\)

 T.N.E.I \(=\varepsilon E d s\) \[ \begin{equation*} \left[\because \theta=0^{\circ} \therefore \cos \theta=1\right] \tag{i} \end{equation*} \] 

viii. According to Gauss' theorem,

T.N.E.I \(=\mathrm{q}\)

T.N.E.I \(=\sigma\) ds

From equations (i) and (ii), we have, \(\varepsilon\) Eds \(=\sigma \mathrm{ds}\)

\(\mathrm{E}=\frac{\sigma}{\varepsilon}\)

\(\therefore \quad \mathrm{E}=\frac{\sigma}{\mathrm{k} \varepsilon_{0}}\)

Equation (iii) represents expression for the electric intensity near a charged conductor. If surrounding medium is air or vacuum then, \(\mathrm{k}=1\).

In this case, \(\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}\)

Note:

1. If the surface is enclosed by a number of point charges \(q_{1}, q_{2}, q_{3}, q_{4} \cdots \cdots q_{n}\), then the resultant charge \(\mathrm{q}\) is given by

\(\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2}-\mathrm{q}_{3}-\mathrm{q}_{4} \ldots \ldots . \mathrm{q}_{\mathrm{n}}\)

Since total charge enclosed by the surface is equal to T.N.E.I,

T.N.E.I \(=q_{1}=q_{1}+q_{2}-q_{3}-q_{4} \ldots \ldots . . q_{n}\)