Electrostatics

JEE Main level, single correct, medium

229 of 311

Question 229 of 311

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2−V1=20 VV_2-V_1=20 \mathrm{~V}. (i.e. plate 2 is at a higher potential). The plates are separated by d=0.1 md=0.1 \mathrm{~m} and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1 . What is its speed when it hits plate 2?2 ?
(e=1.6×10−19C,me=9.11×10−31 kg)\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.11 \times 10^{-31} \mathrm{~kg}\right)

Show answer and solution

Answer: (B) 2.65×106 m/s2.65 \times 10^6 \mathrm{~m} / \mathrm{s}

12mv2=eV\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}
v=2eVm=2.65×106 m/s\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=2.65 \times 10^6 \mathrm{~m} / \mathrm{s}

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.