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 $\frac{1}{4\pi {ϵ}_0}\frac{q}{y^2}$
• C. Potential at point $A$ is $\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{a}+\frac{q}{R}]$
• D. Potential at point $B$ is $\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{a}+\frac{q}{R}+\frac{q}{y}-\frac{q}{r}]$

Answer 1

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

The correct options are
A Electric field at point $A$ is zero.
B Electric field at point $B$ is $\frac{1}{4\pi {ϵ}_0}\frac{q}{y^2}$
C Potential at point $A$ is $\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{a}+\frac{q}{R}]$
D Potential at point $B$ is $\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{a}+\frac{q}{R}+\frac{q}{y}-\frac{q}{r}]$

Charge distribution is as shown.


Due to $q_1$, the charge induction is non-uniform but net induced charge due to $q_1$ is zero.
At point $A$ field is zero (field inside a conducting sphere) and potential is equal to potential at the centre.
$V_A=V_C=\frac{1}{4\pi {ϵ}_0}\frac{q}{R}+\frac{1}{4\pi {ϵ}_0}\frac{q_1}{a}$

At point $B$, field is due to $q$ only.
$E_B=\frac{1}{4\pi {ϵ}_0}\frac{q}{y^2}$

The potential at $B$ is due to the charge $q$ at the centre of cavity, $-q$ induced in the surface of cavity and potential at the the centre of sphere.
$V_B=\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{a}+\frac{q}{R}+\frac{q}{y}-\frac{q}{r}]$