First slide

Pontial due to

Question

A solid sphere of radius R is charged uniformly through out the volume. At what distance from its surface is the electrostatic potential half of the potential at the centre.

Moderate

A

R
B

R/2
C

R/3
D

2R

Solution

Potential at centre $V_{c} = K \cdot \frac{3 Q}{2 R}$
$\begin{array}{l} V_{1} = K \cdot \frac{Q}{r} \\ K \cdot \frac{3 Q}{4 R} = \frac{KQ}{r} \\ r = \frac{4 R}{3} \end{array}$
Distance from surface $\frac{4 R}{3} - R = \frac{R}{3}$

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