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Electrostatics

JEE Main level, single correct, medium

158 of 311

Question 158 of 311

Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$ . Their centers have an initial separation of $4 R$ . Both the spheres are given an initial speed of $u$ towards each other. The minimum value of $u$ , so that they can just touch each other is :
(Take $k=\frac{1}{4 \pi \epsilon_{0}}$ and assume $k Q^{2}\gt G m^{2}$ where $G$ is the Gravitational constant)

Show answer and solution

Answer: (B) $\sqrt{\frac{k Q^{2}}{4 m R}\left(1-\frac{G m^{2}}{k Q^{2}}\right)}$

Using energy conservation, the system's kinetic energy must overcome the increase in electrostatic potential energy as the spheres approach.
Initial state: separation 4R, speed u each.
Final state: touching (separation 2R), speed zero.
Change in potential energy:

\Delta U = k\frac{Q^2}{2R} - k\frac{Q^2}{4R} = k\frac{Q^2}{4R}

.
Initial kinetic energy:

KE_i = mu^2

.
Energy equation:

mu^2 = k\frac{Q^2}{4R}\left(1 - \frac{Gm^2}{kQ^2}\right)

.
Therefore,

u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}

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