Question

A thin spherical conducting shell of radius $R$ has a charge $q$. Another charge $Q$ is placed at the center of the shell. The electrostatic potential at a point $P$ at a distance $R/2$ from the center of the shell is

• A. $\frac{2Q}{4\pi {\epsilon }_{0}R}+\frac{q}{4\pi {\epsilon }_{0}R}$
• B. $\frac{2Q}{4\pi {\epsilon }_{0}R}$
• C. $\frac{2Q}{4\pi {\epsilon }_{0}R}-\frac{q}{4\pi {\epsilon }_{0}R}$
• D. $\frac{2Q}{4\pi {\epsilon }_{0}R}+\frac{2q}{4\pi {\epsilon }_{0}R}$

Answer 1

The correct option is A $\frac{2Q}{4\pi {\epsilon }_{0}R}+\frac{q}{4\pi {\epsilon }_{0}R}$


Potential at $P$ due to charge $Q$ is
$V_1=\frac{Q}{4\pi {\epsilon }_{0}R/2}=\frac{2Q}{4\pi {\epsilon }_{0}R}$

Due to spherical shell, potential at all point is same inside the shell and it is equal to potential at the centre of the shell.
Potential at $P$ due to sphere is $V_2=\frac{q}{4\pi {\epsilon }_{0}R}$

So, net potential at $P$,
$V=\frac{2Q}{4\pi {\epsilon }_{0}R}+\frac{q}{4\pi {\epsilon }_{0}R}$