Question

A solid sphere if radius $R$ is charged uniformly through out the volume. At what distance from its surface is the electric potential $\frac{1}{4}$ of the potential at the centre ?

• A. $\frac{8R}{3}$
• B. $\frac{R}{3}$
• C. $\frac{5R}{3}$
• D. $\frac{2R}{3}$

Answer 1

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

We know that potential outside the sphere is $V_1=\frac{KQ}{x}$

where $x$ is the distance from the centre $x>r$ and potential inside the sphere $V_1=\frac{KQ}{2r}[3-x^2/r^2]$

for centre put $x=0$ $V_0=\frac{3KQ}{2r}$

condition given is, $V_1=\frac{V_0}{2}$

$\frac{KQ}{x}=\left(\frac{3KQ}{2r}\right)/2$

$x=4r/3$

distance from the surface is $x-r$ $=\frac{4r}{3}-r=\frac{r}{3}$