Question

A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_o$. A hole with a small area α4πR² (α << 1) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

• A. The potential at the center of the shell is reduced by $2\alpha V_o$
• B. The magnitude of electric field at a point, located on a line passing through the hole and shell’s center, on a distance $2R$ from the center of the spherical shell will be reduced by $\frac{\alpha V_0}{2R}$
• C. The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha V_o}{2R}$
• D. The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2}R$ from center towards the hole will be $\frac{1-\alpha }{1-2\alpha }$

Answer 1

The correct option is D The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2}R$ from center towards the hole will be $\frac{1-\alpha }{1-2\alpha }$

Total charge on the shell = $Q$
For uniformly distributed charged shell surface charge density $\left(\sigma \right)=\frac{Q}{4\pi R^2}$
$\therefore$ Charge of small area $\left(dq\right)$ $=\sigma dA=\sigma \left(\alpha 4\pi r^2\right)=\left[\frac{Q}{4\pi r^2}\right]\times \alpha 4\pi r^2$
$dq=\alpha Q$

Given that potential at surface before removing charge dq is $V_0=\frac{Q}{4\pi {\epsilon }_{0}R}$
$\therefore V_{\text{center}}=V_0-V_{\left(dq\right)}=\frac{Q}{4\pi {\epsilon }_{0}R}-\frac{\alpha Q}{4\pi {\epsilon }_{0}R}=V_0(1-\alpha )$

Also $V_B=V_0-\frac{\alpha Q}{4\pi {\epsilon }_0\left(\frac{R}{2}\right)}=V_0(1-2\alpha )$


Applying principle of superposition for $\overrightarrow{E}$

$E_A=\frac{kQ}{(2R{)}^2}-\frac{k\alpha Q}{(R{)}^2}=E_{\text{shell}}-\frac{\alpha V_0}{R}\Rightarrow \Delta E_A=-\frac{\alpha V_0}{R}$

Electric field at centre of the shell = $E_A-\frac{kdq}{R^2}=0-\frac{k\left(\alpha Q\right)}{R^2}=-\frac{V_o\alpha }{R}$
Therefore electric field at centre gets reduced by $\Delta E_c=\frac{\alpha V_0}{R}$