Consider two identical metallic spheres of radius R each having… | JEE Main 2026 (21 Jan, Shift 2)

Q19JEE Main 2026 · 21 JAN · SHIFT 2Difficult

Electrostatic Potential Energy

Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$ . Their centers have an initial separation of $4R$ . 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 $kQ^2 > Gm^2$ where $G$ is the Gravitational constant)

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Answer: (d)

Total KE $= 2\cdot\frac{1}{2}mu^2 = mu^2$ ; they just touch when the centres are $2R$ apart and both stop.
Electric PE rises by $kQ^2\left(\frac{1}{2R} - \frac{1}{4R}\right) = \frac{kQ^2}{4R}$ ; gravitational PE falls by $\frac{Gm^2}{4R}$ .
$mu^2 = \frac{kQ^2 - Gm^2}{4R}$
$u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$ , answer (d).

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