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2. Electric Potential and Capacitance
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Consider a sphere of radius $R$ having charge $q$ uniformly distributed inside it. At what minimum distance from its surface the electric potential is half of the electric potential at its centre?
A
$R$
B
$\frac{R}{2}$
C
$\frac{4 R}{3}$
D
$\frac{R}{3}$
Solution

(d)
$V_c=\frac{3 k Q}{2 R}$
$V_s=\frac{k Q}{R+x}=\frac{3 k Q}{4 R}$
$4 R=3 R+3 x$
$3 x=R$
$x=\frac{R}{3}$
Standard 12
Physics