Q198 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
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 $]$
- A$Qq/4\pi\epsilon_0 D$
- Correct answerB$Q/4\pi\epsilon_0 ED$
- C$qQ/2\pi\epsilon_0 D^2$
- D$Q/4\pi\epsilon_0 E$
Solution
Work done by the uniform electric field on the particle is given by $W = qED$.
By the work-energy theorem, the kinetic energy gained by the particle is $K = qED$.
When the particle approaches the fixed static charge $Q$, its kinetic energy is converted into electrostatic potential energy.
At the distance of closest approach $r$, the velocity of the particle becomes zero and the entire kinetic energy is stored as potential energy.
The electrostatic potential energy at distance $r$ is $U = \dfrac{1}{4\pi\epsilon_0} \dfrac{qQ}{r}$.
Equating the kinetic energy and potential energy:
$qED = \dfrac{1}{4\pi\epsilon_0} \dfrac{qQ}{r}$
$r = \dfrac{Q}{4\pi\epsilon_0 ED}$
Answer: $Q/4\pi\epsilon_0 ED$
By the work-energy theorem, the kinetic energy gained by the particle is $K = qED$.
When the particle approaches the fixed static charge $Q$, its kinetic energy is converted into electrostatic potential energy.
At the distance of closest approach $r$, the velocity of the particle becomes zero and the entire kinetic energy is stored as potential energy.
The electrostatic potential energy at distance $r$ is $U = \dfrac{1}{4\pi\epsilon_0} \dfrac{qQ}{r}$.
Equating the kinetic energy and potential energy:
$qED = \dfrac{1}{4\pi\epsilon_0} \dfrac{qQ}{r}$
$r = \dfrac{Q}{4\pi\epsilon_0 ED}$
Answer: $Q/4\pi\epsilon_0 ED$
Hence, the correct answer is option: 'B'.