Question

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

• A. $\frac{4R}{3}$
• B. $\frac{R}{2}$
• C. $\frac{R}{3}$
• D. $2\text{R}$

Answer 1

The correct option is C $\frac{R}{3}$


For a uniformly charged solid sphere, the electric potential inside the surface, at a distance $r^{\prime }$ from centre is given by

$V_{inside}=\frac{kq}{2R}\{3-\frac{r^{\prime 2}}{R^2}\}$

Potential at the centre of the sphere is obtained by substituting $r^{\prime }=\text{0}$.
$\Rightarrow V_{centre}=\frac{3kq}{2R}\dots \left(i\right)$

Let the electric potential becomes half at the point $\text{P}$ with respect to the centre.


$\Rightarrow V_P=\frac{1}{2}{V}_{centre}$
$\Rightarrow V_P=\frac{3kQ}{4R}\dots \left(ii\right)$

The electric potential at point $\text{P}$ is
$V_P=\frac{kq}{r}$
Substituting $V_P$ in equation $\left(ii\right)$
$\Rightarrow \frac{kq}{r}=\frac{3kq}{4R}$
$\Rightarrow \frac{1}{r}=\frac{3}{4R}$
$\Rightarrow r=\frac{4R}{3}$

Then the distance of point $\text{P}$ from the surface will be
$r-R=\frac{4R}{3}-R=\frac{R}{3}$

Hence, option (c) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept: Potential inside a solid sphere}\\ \text{is given by}\\ V=\frac{kq}{2R}\{3-\frac{r^2}{R^2}\}\end{array}$