Question

A solid sphere of radius $R$ is charged uniformly. At what distance from its surface is the electrostatic potential is half the potential at its centre?

• A. $R$
• B. $R / 2$
• C. $R / 3$
• D. $2 R$

Answer 1

Answer: (C)

The potential at the centre of the sphere
$V_{C}=\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q}{2 R}$
We require a point where
$V=\frac{V_{C}}{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q}{4 R}$
This point cannot be inside the sphere where
$V \geq \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}$
Let the point be outside the sphere at a distance $r$ from the centre. Then
$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q}{4 R} $
$\therefore r=\frac{4}{3} R$
Distance from the surface $= r - R$
$=\frac{4}{3} R-R=\frac{R}{3}$