A solid sphere of radius R is charged uniformly. At a point P, a distance r from the surface, the electrostatic potential is observed to be half of the potential at the centre. Then

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$r = \frac{4 R}{3}$

$r = \frac{R}{3}$

$r = 2 R$

$r = R$

answer is D.

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Detailed Solution

Since, $V = \frac{k Q}{2 R} (3 R^{2} - r^{2})$

So, at r = 0, potential at the centre is $V_{c} = \frac{3 k Q}{2 R}$

As per the question we require a point where $V = \frac{V_{c}}{2} = \frac{3 k Q}{4 R}$

This point cannot lie inside the sphere where $V \geq \frac{k Q}{R}$

Let the point lie outside the sphere, at a distance r+R from the centre. Then

$V = \frac{k Q}{(R + r)} = \frac{3 k Q}{4 R}$

$\Rightarrow r = \frac{R}{3}$

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