Question

A cavity is made inside a solid conducting sphere and a charge $q$ is placed inside the cavity at the centre. A charge $q_1$ is placed outside the sphere, as shown in the figure. Point $A$ is inside the sphere and point $B$ is inside the cavity. Then

• A. Electric field at point $A$ is zero.
• B. Electric field at point $B$ is $\frac{1}{4\pi {ϵ}_o}\frac{q}{y^2}$.
• C. Potential at point $A$ is $\frac{1}{4\pi {ϵ}_o}[\frac{q_1}{a}+\frac{q}{R}]$.
• D. Potential at point $B$ is $\frac{1}{4\pi {ϵ}_o}[\frac{q_1}{a}+\frac{q}{R}+\frac{q}{y}-\frac{q}{r}]$

Answer 1

Answer: (A), (B), (C), (D)

Potential at point $B$ is $\frac{1}{4\pi {ϵ}_o}[\frac{q_1}{a}+\frac{q}{R}+\frac{q}{y}-\frac{q}{r}]$

The charge distribution is shown in figure.
Due to $q_1$, the charge induction is non-uniform, but the net induced charge due to $q_1$ is zero.

At point $A$, the electric field is zero.

At point $B$, the electric field is due to charge $q$ only.
$E_B=\frac{1}{4\pi {ϵ}_o}\frac{q}{y^2}$

At point $A$, potential is equal to the potential at centre.

$V_A=V_{\text{centre}}=\frac{1}{4\pi {ϵ}_o}\frac{q}{R}+\frac{1}{4\pi {ϵ}_o}\frac{q_1}{a}=\frac{1}{4\pi {ϵ}_o}[\frac{q_1}{a}+\frac{q}{R}]$

At point $B$, potential is due to charges, $q$ at centre of cavity, $-q$ induced on inner surface of cavity, in addition to the potential at the centre.
$V_B=\frac{1}{4\pi {ϵ}_o}[\frac{q_1}{a}+\frac{q}{R}+\frac{q}{y}-\frac{q}{r}]$

Therefore, all the options are correct.