Consider a fixed uniformly charged insulating sphere with radius R… | NEET 2026 (Re-neet)

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Electrostatic Potential Energy

Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$ . A point charge $-q$ ( $q \ll Q$ ) with mass $m$ is released from rest at a distance of $3R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is: ( $\varepsilon_0$ is the permittivity of vacuum, neglect gravitational forces).

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Answer: (d)

Outside the sphere, $V = \frac{kQ}{r}$ . The charge $-q$ is attracted and loses PE.
$\frac{1}{2}mv^2 = kQq\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{2kQq}{3R}$
$v^2 = \frac{4Qq}{3\cdot4\pi\varepsilon_0 mR} = \frac{Qq}{3\pi\varepsilon_0 mR}$
So the answer is (d) $\sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$ .

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