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MHT CET 2026 (19 Apr, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2026 (19 Apr, Shift 2)

Q15MHT CET 2026 · 19 APR · SHIFT 2Medium

Electrostatic Potential Energy

A particle of mass 'm' and charge 'q', initially at rest, is accelerated by a uniform electric field 'E' through a distance 'D' and is then allowed to approach a fixed static charge 'Q' of the same sign. The distance of the closest approach of the charge q is

[ $\epsilon_0$ = permittivity of free space ]

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Answer: (b)

KE gained in the field: $K = qED$ .
At closest approach all KE becomes PE: $\frac{qQ}{4\pi\epsilon_0 r} = qED$
$r = \frac{Q}{4\pi\epsilon_0 ED}$
So the answer is (b).

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