Q 12-02-003NEETNEET 2026Top questionMedium
Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$. A point charge $-q$ ($q << Q$) 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 :
($\epsilon_0$ is the permittivity of vacuum, neglect gravitational forces).
Answer: (C) $\sqrt{\dfrac{Qq}{3\pi\epsilon_0 mR}}$
Concept. Loss of PE $=$ gain of KE.
Step 1. Energy. $\dfrac12mv^2 = \dfrac{Qq}{4\pi\epsilon_0}\left(\dfrac1R - \dfrac{1}{3R}\right) = \dfrac{Qq}{4\pi\epsilon_0}\cdot\dfrac{2}{3R}$.
Step 2. Speed. $v = \sqrt{\dfrac{Qq}{3\pi\epsilon_0mR}}$.
Answer. $\sqrt{\dfrac{Qq}{3\pi\epsilon_0mR}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics