JEE Main 2020 — Electrostatics Question with Solution
From: JEE Main 2020 (Online) 8th January Evening Slot
Question
A particle of mass m and charge q is released
from rest in a uniform electric field. If there is
no other force on the particle, the dependence
of its speed v on the distance x travelled by it
is correctly given by (graphs are schematic and
not drawn to scale)
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
As F = qE
ma = qE
a =
Also we know, v2 = u2 + 2as
v2 = 0 + 2s
v =
ma = qE
a =
Also we know, v2 = u2 + 2as
v2 = 0 + 2s
v =
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 2020, 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.