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 $\alpha 4\pi R^2$ (a<<1) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct ?

• A. 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 $\frac{1-\alpha }{1-2\alpha }$
• B. The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha V_0}{2R}$
• C. 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 $\frac{\alpha V_0}{2R}$
• D. The potential at the center of the shell is reduced by $2\alpha V_0$

Answer 1

Answer: (A)

For uniformly distributed charged shell surface charge density $\left(\sigma \right)=\frac{Q}{4\pi R^2}$
$\therefore$ Charge of small area $\left(\alpha 4\pi R^2\right)$ is $dq=\alpha Q$
Given that potential at surface before removing charge dq is $V_0=\frac{Q}{4\pi {\epsilon }_{0}R}$
$\therefore V_{center}=V_0-V_{\left(dq\right)}$
$=\frac{Q}{4\pi {\epsilon }_{0}R}-\frac{\alpha Q}{4\pi {\epsilon }_{0}R}=V_0\left(1-\alpha \right)$
Also $V_B=V_0-\frac{\alpha Q}{4\pi {\epsilon }_0\left(\frac{R}{2}\right)}=V_0\left(1-2\alpha \right)$
Applying principle of superposition for $\vec{E}$
$E_A=\frac{kQ}{{\left(2R\right)}^2}-\frac{k\alpha Q}{{\left(R\right)}^2}=E_{shell}-\frac{\alpha V_0}{R}\Rightarrow \Delta E_A=-\frac{\alpha V_0}{R}$
Similarly $\Delta E_c=\frac{\alpha V_0}{R}$