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?

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

Tardigrade Admin · Accepted Answer

The potential at the centre of the sphere

V_{C} = \frac{1}{4 π ε _{0}} \frac{3 Q}{2 R}

We require a point where

V = \frac{V _{C}}{2} = \frac{1}{4 π ε _{0}} \frac{3 Q}{4 R}

This point cannot be inside the sphere where

V \geq \frac{1}{4 π ε _{0}} \frac{Q}{R}

Let the point be outside the sphere at a distance

r

from the centre. Then

V = \frac{1}{4 π ε _{0}} \frac{Q}{r} = \frac{1}{4 π ε _{0}} \frac{3 Q}{4 R}

∴ r = \frac{4}{3} R

Distance from the surface

= r - R

= \frac{4}{3} R - R = \frac{R}{3}

Q. 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?

$R$

$R /2$

$R /3$

$2 R$

Q. 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?

R

R/2

R/3

2R

Q. 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?

$R$

$R /2$

$R /3$

$2 R$