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 3 R from the centre of the cha

Question

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 $3 R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is :
( $ϵ_{0}$ is the permittivity of vacuum, neglect gravitational forces).NEET 2026 (Re-NEET)

TestHub · Accepted Answer

Initial potential energy of the system when the charge $- q$ is at a distance of $3 R$ from the centre is:

$U_{i} = - \frac{1}{4 π ϵ _{0}} \frac{Qq}{3 R}$

Final potential energy of the system when the charge reaches the surface of the sphere (distance $R$ from the centre) is:

$U_{f} = - \frac{1}{4 π ϵ _{0}} \frac{Qq}{R}$

According to the principle of conservation of mechanical energy:

$K_{i} + U_{i} = K_{f} + U_{f}$

$0 - \frac{Qq}{12 π ϵ _{0} R} = \frac{1}{2} m v^{2} - \frac{Qq}{4 π ϵ _{0} R}$

$\frac{1}{2} m v^{2} = \frac{Qq}{4 π ϵ _{0} R} - \frac{Qq}{12 π ϵ _{0} R}$

$\frac{1}{2} m v^{2} = \frac{Qq}{4 π ϵ _{0} R} (1 - \frac{1}{3})$

$\frac{1}{2} m v^{2} = \frac{2 Qq}{12 π ϵ _{0} R}$

$v^{2} = \frac{Qq}{3 π ϵ _{0} m R}$

$v = \frac{Qq}{3 π ϵ _{0} m R}$

Answer: $\frac{Qq}{3 π ϵ _{0} m R}$

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