Question

Two concentric spherical shells have charges $+q$ and $-q$ as shown in the figure. Choose the correct options.

• A. At $\text{A}$ electric field is zero, but electric potential is non- zero
• B. At $\text{B}$ electric field and electric potential both are non-zero.
• C. At $\text{C}$ electric field is zero but electric potential is non- zero.
• D. At $\text{C}$ electric field and electric potential both are zero.

Answer 1

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

At $\text{C}$ electric field and electric potential both are zero.

Let the radii of inner and outer shells be $r_1$ and $r_2$ respectively.


$\text{Case I}:$ At point $\text{A}$

Taking the Gaussian surface around $\text{A}$ and using Gauss law the electric field at this point,

$\int \overrightarrow{E}\cdot \overrightarrow{ds}=\frac{q_{en}}{{ϵ}_0}$

Charge enclosed by the Gaussian surface, $q_{en}=0$

$\Rightarrow E=0$

So, Electric field at $\text{A}$ is zero.

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

$V_A=\frac{kq}{r_1}....\left(1\right)$

The electric potential at $\text{A}$ due charge $-q$ at outer sphere is

$V_A^{\prime }=\frac{-kq}{r_2}....\left(2\right)$

The net electric potential at $\text{A}$ is

$V_A=V_A+V_A^{\prime }$

Substituting the values from $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow V_A=\frac{kq}{r_1}-\frac{kq}{r_2}$

$V_A$ is non-zero at $\text{A}$.

$\text{Case II:}$ At point $\text{B}$.

Let the distance between centre of shell and point $\text{B}$ is $r$.

Taking the Gaussian surface enclosing $\text{B}$ and using Gauss law

$\int \overrightarrow{E}\cdot \overrightarrow{ds}=\frac{q_{en}}{{\epsilon }_0}$

$\Rightarrow E\times 4\pi r^2=\frac{q}{{\epsilon }_0}$

$\Rightarrow E=\frac{q}{4\pi {\epsilon }_0{r}^2}$

So, the electric field at $\text{B}$ is non-zero.

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

$V_B=\frac{kq}{r}....\left(3\right)$

The electric potential at $B$ due charge $-q$ at outer sphere is

$V_B^{\prime }=\frac{-kq}{r_2}....\left(4\right)$

The net electric potential at $\text{B}$ is

$V_{net}=V_B+V_B^{\prime }$

From equation $\left(3\right)$ and $\left(4\right)$
$\Rightarrow V_{net}=\frac{kq}{r}-\frac{kq}{r_2}$

$V_{net}$ is non-zero.

$\text{Case III:}$ At point $\text{C}$.

Let the distance between the centre of the shell and point $\text{C}$ is $r^{\prime }$.

Taking the Gaussian surface around $\text{C}$ and using Gauss law,

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

$\Rightarrow \overrightarrow{E}\times 4\pi r^{\prime 2}=\frac{q-q}{{\epsilon }_0}$

$\Rightarrow E=0$

So, electric field is zero at $\text{C}$.

The electric potential at $\text{C}$ due to charge $+q$ on the inner shell is

$V_C=\frac{kq}{r^{\prime }}....\left(5\right)$

The electric potential at $\text{C}$ due charge $-q$ at outer sphere is

$V_C^{\prime }=\frac{-kq}{r^{\prime }}....\left(6\right)$

The net electric potential at $\text{C}$ is

$V_{net}=V_C+V_C^{\prime }$

Using equation $\left(5\right)$ and $\left(6\right)$, we have

$\Rightarrow V_{net}=\frac{kq}{r^{\prime }}-\frac{kq}{r^{\prime }}$

$\Rightarrow V_{net}=0$

$V_{net}$ is zero at $\text{C}$.

Hence, options (a) , (b) and (d) are the correct answers.