A solid sphere of radius $R$ is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre?
Diffcult
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Potential at the center of sphere is $V_{c}=\frac{3}{2} \frac{k Q}{R}$
Potential at some distance $\times$ from center $=\frac{K Q}{x}$
$\frac{k Q}{x}=\frac{1}{2}\left(\frac{3}{2} \frac{k Q}{R}\right)$ or $x=4 R / 3$
Distance from surface is $x-R=R / 3$
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