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(D) The electric potential at the centre of a uniformly charged non-conducting sphere is given by $V_c = \frac{3 k q}{2 R}$.We want to find the distan

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$\frac{R}{3}$

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(D) The electric potential at the centre of a uniformly charged non-conducting sphere is given by $V_c = \frac{3 k q}{2 R}$.We want to find the distance $x$ from the surface such that the potential $V_s$ at that point is half of $V_c$. The distance from the centre of the sphere to this point is $r = R + x$.Since the point is outside the sphere $(r > R)$,the potential is $V_s = \frac{k q}{r} = \frac{k q}{R + x}$.According to the problem,$V_s = \frac{1}{2} V_c$.Substituting the expressions:$\frac{k q}{R + x} = \frac{1}{2} \left( \frac{3 k q}{2 R} \right)$$\frac{k q}{R + x} = \frac{3 k q}{4 R}$$\frac{1}{R + x} = \frac{3}{4 R}$$4 R = 3(R + x)$$4 R = 3 R + 3 x$$3 x = R$$x = \frac{R}{3}$

Consider a sphere of radius $R$ having charge $q$ uniformly distributed inside it. At what minimum distance from its surface is the electric potential half of the electric potential at its centre?

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