Question

A solid sphere of radius $R$ has charge ${}^{\prime }{q}^{\prime }$ uniformly distributed over its volume. The distance from its surface at which the electrostatic potential is equal to half of the potential at the centre is :

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

Answer 1

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

Electric potential inside the sphere $V=$ $kq\frac{1}{2R}(3-\frac{r^2}{R^2})$

$V=$ $kq\frac{3}{2R}$ at the center

Potential out side the sphere is $V_1=Kq/r$

now $V_1=V$

gives $r=4R/3$

so distance from surface 4R/3-R=R/3