Electrostatics

JEE Main level, single correct, medium

198 of 311

Question 198 of 311

In free space, a particle AA of charge 1μC1\mu C is held fixed at point PP . Another particle BB of the same charge and mass 4μg4\mu g is kept at a distance of 1mm1mm from P.P. If BB is released, then its velocity at a distance of 9mm9mm from PP is:
[Take 14πϵ0=9×109Nm2C−2\dfrac{1}{4\pi {ϵ}_{0}}=9\times {10}^{9}N{m}^{2}{C}^{-2} ]

Show answer and solution

Answer: (C) 2.0×103ms−12.0\times {10}^{3}m{s}^{-1}

From conservation in mechanical energy:
−ΔP.E.=ΔK.E.-\Delta P.E.=\Delta K.E.
14πϵ0(1×106)2[110−3−19×10−3]=12×4×10−6×V2\dfrac{1}{4\pi {\epsilon }_{0}}{\left(1\times {10}^{6}\right)}^{2}\left[\dfrac{1}{{10}^{-3}}-\dfrac{1}{9\times {10}^{-3}}\right]=\dfrac{1}{2}\times 4\times {10}^{-6}\times {V}^{2}
9×109×10−12[89×10−3]=2×10−6×V29\times {10}^{9}\times {10}^{-12}\left[\dfrac{8}{9\times {10}^{-3}}\right]=2\times {10}^{-6}\times {V}^{2}
V2=4×106{V}^{2}=4\times {10}^{6}
V=2×103ms−1V=2\times {10}^{3}m{s}^{-1}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electric field lines simulationDrag the sliders and watch the numbers in this question change.Revise Electrostatics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Electrostatics (Electric Potential and Potential Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.