Question

A concentric spherical cavity is cut out from a solid conducting sphere and a charge $+Q$ is placed at the centre of the cavity. The magnitude of net electric field $E$ and net potential $V$ at point $A$ is
[$k=\frac{1}{4\pi {ϵ}_0}$]

• A. $E=\frac{4kQ}{R^2},V=\frac{3kQ}{2R}$
• B. $E=\frac{kQ}{R^2},V=0$
• C. $E=\frac{4kQ}{R^2},V=\frac{kQ}{R}$
• D. $E=\frac{kQ}{4R^2},V=\frac{kQ}{2R}$

Answer 1

The correct option is A $E=\frac{4kQ}{R^2},V=\frac{3kQ}{2R}$

Based on the concept of Gauss's Law, since electric field inside a conductor is zero, the charge distribution is as shown in figure.



Now, using principle of superposition
${\overrightarrow{E}}_A={\overrightarrow{E}}_{s_1}+{\overrightarrow{E}}_{s_2}+{\overrightarrow{E}}_Q$....$\left(1\right)$
where
${\overrightarrow{E}}_{s_1}\to$ electric field due to outer shell
${\overrightarrow{E}}_{s_2}\to$ electric field due to inner shell
${\overrightarrow{E}}_Q\to$ electric field due to point charge $Q$

We know, electric field inside a shell is zero, thus, ${\overrightarrow{E}}_{s_1}={\overrightarrow{E}}_{s_2}=0$

Further, electric field due to a point charge, $E_Q=\frac{kQ}{r^2}=\frac{kQ}{(R/2{)}^2}=\frac{4kQ}{R^2}$
So, from $\left(1\right)$,
Magnitude of net electric field, $E_A=0+0+E_Q=\frac{4kQ}{R^2}$

Also, net potential, $V_A=V_{s_1}+V_{s_2}+V_Q$

Potential inside a shell is constant and given by $V_s=\frac{kQ}{r}$
Also, $V_Q=\frac{kQ}{R/2}$
$\Rightarrow V_A=\frac{k(-Q)}{R}+\frac{k\left(Q\right)}{2R}+\frac{k\left(Q\right)}{R/2}$
$\Rightarrow V_A=\frac{3kQ}{2R}$