JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Work, Power & Energy

A small piece carrying charge $q$ breaks off the bottom of the sphere, which sits a distance $R$ from the centre, and falls to a distance $R+y$ from the centre. Two forces do work on it as it falls: gravity, and the Coulomb repulsion from the remaining charge $Q$, which may be treated as concentrated at the centre. The gravitational work over the drop is $mgy$, while the electrostatic work is ${\int }_R^{R+y}\frac{Qq}{4\pi {ϵ}_0{x}^2}dx=\frac{Qq}{4\pi {ϵ}_0}\left(\frac{1}{R}-\frac{1}{R+y}\right)=\frac{Qqy}{4\pi {ϵ}_{0}R(R+y)}$. By the work-energy theorem these together equal the kinetic energy gained, $\frac{1}{2}m{v}^2$, so $\frac{1}{2}m{v}^2=mgy+\frac{Qqy}{4\pi {ϵ}_{0}R(R+y)}$. Solving for the speed gives $v^2=2y\left[\frac{qQ}{4\pi {ϵ}_{0}R(R+y)m}+g\right]$.