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 $\alpha4\pi R^{2}$ (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 $\left(\alpha4\pi R^{2}\right)$ is $dq=\alpha Q$
Given that potential at surface before removing charge dq is $V_{0}=\frac{Q}{4\,\pi\,\varepsilon_{0}\,R}$
$\therefore V_{center}=V_{0}-V_{\left(dq\right)}$
$=\frac{Q}{4\pi\varepsilon_{0}R}-\frac{\alpha Q}{4\pi\varepsilon_{0}R}=V_{0}\left(1-\alpha\right)$
Also $V_{B}=V_{0}- \frac{\alpha Q}{4\pi\varepsilon_{0}\left(\frac{R}{2}\right)}=V_{0}\left(1-2\alpha\right)$
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 $\Delta E_{c}=\frac{\alpha V_{0}}{R}$