Question

A point charge $q$ is placed at a distance r from the centre $O$ of an uncharged spherical shell of inner radius R and outer radius $2R$. The distance r $<$ $R$. The electric potential at the centre of the shell will be.

• A. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{r}-\frac{1}{2R}\right)$
• B. $\frac{q}{4\pi {\epsilon }_{0}r}$
• C. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{r}+\frac{1}{2R}\right)$
• D. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{r}-\frac{1}{R}\right)$

Answer 1

Answer: (A)

$\frac{q}{4\pi {\epsilon }_0}\left(\frac{1}{r}-\frac{1}{2R}\right)$

The potential due to the inner radius of the shell

$V=\frac{1}{4\pi {ϵ}_0}\frac{q}{r}$ (for inner radius)

The potential due to the charge on outer radius

$V^1=\frac{1}{4\pi {ϵ}_0}\frac{q}{2R}$ (for outer radius)

Since,

$V$ $\alpha$ $\frac{1}{R}$

$2R>R$

$V>V^1$

Hence, the difference of electric potential $V-V^1=\frac{q}{4\pi {ϵ}_0}(\frac{1}{r}-\frac{1}{2R})$