JEE Main 2022PhysicsElectrostaticsMediumMCQ

JEE Main 2022 — Electrostatics Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1×105 N C-1. If the charge on the particle is 40 μC and the initial velocity is 200 m s-1, how much distance it will travel before coming to the rest momentarily

Choose an option

Show full solutionCorrect option: A
Correct answer
A0.5 m

Step-by-step explanation

Force experienced by a particle with charge q in an electric field F=qE.

Now, the acceleration produced is given as a=Fm=qEm=40×10-6×1051×10-4

=4×104 m  s-2  (As the particle is projected against the electric field, hence it is decelerated)

Using, v2=u2-2as

We have, 02=u2-2as

s=v22a=20022×4×104=0.5 m

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Electrostatics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.