JEE Main 2019 — Electrostatics Question with Solution
From: JEE Main 2019 (Online) 9th April Evening Slot
Question
Four point charges –q, +q, +q and –q are placed
on y-axis at y = –2d, y = –d, y = +d and
y = +2d, respectively. The magnitude of the
electric field E at a point on the x-axis at
x = D, with D >> d, will behave as :-
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
Electric field at p = 2E1cos1 –2E2cos2
=
Applying binomial approximation d << D
=
Applying binomial approximation d << D
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This is a previous-year question from JEE Main 2019, 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.