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Question

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 center?

A

$R$

B

$R / 2$

C

$R / 3$

D

$2 R$

Medium

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Solution

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Correct option is C)

Potential at the center of sphere is $V_{c} = \frac{3}{2} \frac{k Q}{R}$

Potential at some distance x from center $= \frac{K Q}{x}$

$\frac{k Q}{x} = \frac{1}{2} (\frac{3}{2} \frac{k Q}{R})$ or $x = 4 R / 3$

Distance from surface is $x - R = R / 3$

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