Question

A hollow charged metal sphere has a radius $r$. If the potential difference between its surface and a point at a distance $3r$ from the centre is $V$, then electrical intensity at distance $3r$ from the centre is

• A. $\frac{V}{2r}$
• B. $\frac{V}{3r}$
• C. $\frac{V}{4r}$
• D. $\frac{V}{6r}$

Answer 1

Answer: (D)

The potential at a distance $r$, due to charge $q$ is $V=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{r}$
Potential at a distance $(3r)$ is
$V=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{3r}$
Difference in potential $=\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{r}-\frac{1}{3r}\right]$
$\Rightarrow V=\frac{2q}{4\pi {\epsilon }_0\times 3r}$
Intensity of electric field $E=\frac{q}{4\pi {\epsilon }_0}\frac{1}{(3r{)}^2}$
$\therefore \frac{E}{V}=\frac{q}{4\pi {\epsilon }_{0}q{r}^2}\times \frac{4\pi {\epsilon }_{0}3r}{2q}$
$\Rightarrow \frac{E}{V}=\frac{1}{6r}$
$\Rightarrow E=\frac{V}{6r}$