Question

The electric field intensity at a point near and outside the surface of a charged conductor of any shape is $E_1$. The electric field intensity due to uniformly charged infinite thin plane sheet is $E_2$. The relation between $E_1$ and $E_2$ is :

• A. $2E_1=E_2$
• B. $E_1=E_2$
• C. $E_1=2E_2$
• D. $E_1=4E_2$

Answer 1

Answer: (C)

$E_1=2E_2$

The above figure shows an irregular shaped conductor having a charge density $\sigma$. We need to find the electric field intensity at a point p which lies very close to the surface.

Let a Gaussian cylinder of cross-section area $dA$.

The net charge enclosed by the Gaussian cylinder, $q_{enc}=\sigma dA$

$\therefore$ Net flux through the cylinder, $\varphi =\frac{\sigma dA}{{ϵ}_o}$ .........(1)

The electric field inside the conductor is Zero, i.e $E_M=0$

Flux through the cylinder, $\varphi =\int \overrightarrow{E}.\overrightarrow{dS}$

As flux through the cross-section and the curved surface of the cylinder is zero.

$⟹\varphi =E_{1}dA$ ...........(1)

From (1) and (2), $E_{1}dA=\frac{\sigma dA}{{ϵ}_o}$

$⟹$ Electric field at p, $E_1=\frac{\sigma }{{ϵ}_o}$

Electric field intensity due to uniformly charged infinite thin plane sheet, $E_2=\frac{\sigma }{2{ϵ}_o}$

$⟹$ $E_1=2E_2$