Question

$\text{A}$ and $\text{B}$ are two concentric spherical shells. If $\text{A}$ is given a charge $+q$ while $\text{B}$ is earthed as shown in figure, then $[k=\frac{1}{4\pi {\epsilon }_0}]$

• A. charge on the outer surface of shell $\text{B}$ is zero.
• B. the charge on $\text{B}$ is equal and opposite to that of $\text{A}$.
• C. the field inside $\text{A}$ and outside $B$ is zero .
• D. the field outside $\text{A}$ has a finite non-zero value.

Answer 1

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

the field inside $\text{A}$ and outside $B$ is zero .

Given,
Charge on the shell $\text{A}$ is $+q$.

Let,
Radius of shell $\text{B}$ is $r_B$ and
Charge on the shell $\text{B}$ is $q_B$,

$\text{Case I:}$ Analysis of shell $\text{B}$
When it is earthed, all the charges from outer surface flow to the earth and the charge on outer surface becomes zero.

According to given problem, shell $\text{B}$ earthed so its electric potential at this, $\left(V_B\right)$ will be zero. i.e.,

$V_B=0...\left(1\right)$

The electric potential at shell $\text{B}$ due to charge $+q$ on $\text{A}$ is

$V_{BA}=\frac{kq}{r_B}...\left(2\right)$

The electric potential at shell $\text{B}$ due to charge on $\text{B}$ is given by

$V_{BB}=\frac{k{q}_B}{r_B}...\left(3\right)$

The net electric potential of $\text{B}$ is calculated as

$V_B=V_{BA}+V_{BB}$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we have

$\Rightarrow \frac{kq}{r_B}+\frac{k{q}_B}{r_B}=0$

$\Rightarrow q_B=-q$

So, $-q$ charge will be induced on inner surface of shell $\text{B}$.


$\text{Case II:}$ Analysis of shell $\text{A}$.

Taking Gaussian surface inside the shell $\text{A}$, we can write

$\Rightarrow \int \overrightarrow{E}.ds=\frac{q_{enclosed}}{{\epsilon }_0}$

Since, the charge enclosed by this Gaussian surface is zero.i.e, $q_{enclosed}=0$

$\Rightarrow \overrightarrow{E}=0$

Thus, electric field inside the shell $\text{A}$ is zero.

Let us consider a point $\text{P}$ outside the shell at distance $r$ from the centre of shell as shown in figure.


Considering Gaussian spherical surface with radius more than $r$, $\overrightarrow{E}$ outside $\text{B}$, at point $\text{P}$:

${\overrightarrow{E}}_{net}=\overrightarrow{E_A}+\overrightarrow{E_B}=\frac{kq}{r^2}-\frac{kq}{r^2}=0$

So, options (a), (b) and (c) are the correct answers.