Question

A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_0$. A hole with a small area $\alpha 4\pi R^2$ $\left(\alpha <<<1\right)$ is made in the shell without affecting the rest of the shell. Which one of the following is correct.

• A. The magnitude of $\overrightarrow{E}$ at a point located on a line passing through the hole and shell’s centre at a distance $2R$ from the centre of the spherical shell will be reduced by $\frac{\alpha V_0}{2R}$
• B. The potential at the centre of the shell is reduced by $2\alpha V_0$
• C. The magnitude of $\overrightarrow{E}$ at the centre of the shell reduced by $\frac{\alpha V_0}{R}$
• D. The ratio of potential at the centre of the shell to that of the point at $\frac{1}{2}R$ from the centre toward the hole will be $\frac{\left(1-\alpha \right)}{\left(1-2\alpha \right)}$

Answer 1

The correct option is D

The ratio of potential at the centre of the shell to that of the point at $\frac{1}{2}R$ from the centre toward the hole will be $\frac{\left(1-\alpha \right)}{\left(1-2\alpha \right)}$

The explanation for the correct option(s):

Option D: The ratio of potential at the center of the shell to that of the point at $\frac{1}{2}R$ from the center towards the hole will be $\frac{\left(1-\alpha \right)}{\left(1-2\alpha \right)}$

We know that the potential at a surface is given as,
$V_0=\frac{KQ}{R}{E}_C=\frac{K\alpha Q}{R^2}=\frac{\alpha V_0}{R}$

Where, $K$ is a constant equal to $9.0\times {10}^{9}N\cdot m^2/C^2$

$\alpha$ is the radius of the hole

$Q$ is the charge on the sphere

$R$ is the radius of the sphere
And also the potential at the centre $C$ will be,
$V_C={\displaystyle \frac{KQ}{R}}-{\displaystyle \frac{K\alpha Q}{R}}=V_0\left(1-\alpha \right)$ …….$\left(i\right)$
Potential at the point B will be,
$V_B={\displaystyle \frac{KQ}{R}}-{\displaystyle \frac{K\alpha Q}{{\displaystyle \frac{R}{2}}}}=V_0\left(1-2\alpha \right)$ ……..$\left(ii\right)$
Taking the ratio between the equation $\left(i\right)$ and $\left(ii\right)$ then,
${\displaystyle \frac{V_c}{V_B}}={\displaystyle \frac{\left(1-\alpha \right)}{\left(1-2\alpha \right)}}$

Hence, The ratio of potential at the center of the shell is $\frac{\left(1-\alpha \right)}{\left(1-2\alpha \right)}$

The explanation for the incorrect option(s):

Option A: The magnitude of $\overrightarrow{E}$ at a point located on a line passing through the hole and shell’s centre at a distance $2R$ from the centre of the spherical shell will be reduced by $\frac{\alpha V_0}{2R}$

The electric field experienced at $C$ is given as,
$$E_C={\displaystyle \frac{K\alpha Q}{R^2}}={\displaystyle \frac{\alpha V_0}{R}}$$
The electric field at $C$ is increasing by$${\displaystyle \frac{\alpha V_0}{R}}$$.

Option B: Potential at the centre of the shell is reduced by $2\alpha V_0$

The potential at the centre $C$ will be,
$V_C={\displaystyle \frac{KQ}{R}}-{\displaystyle \frac{K\alpha Q}{R}}=V_0\left(1-\alpha \right)$

Option C: The magnitude of $\overrightarrow{E}$ at the centre of the shell is reduced by $\frac{\alpha V_0}{R}$

The electric field acting at the point $A$ is,
$$E_A={\displaystyle \frac{KQ}{{\left(2R\right)}^2}}-{\displaystyle \frac{K\alpha Q}{R^2}}={\displaystyle \frac{KQ}{4R^2}}-{\displaystyle \frac{\alpha V_0}{R}}$$
The electric field will be reduced by $${\displaystyle \frac{\alpha V_0}{R}}$$

Therefore, the ratio of potential at the centre of the shell to that of the point at $\frac{1}{2}R$ from the centre towards the hole will be $\frac{\left(1-\alpha \right)}{\left(1-2\alpha \right)}$

Hence, Option 'D' is Correct.