JEE Main 2020 — Electrostatics Question with Solution
From: JEE Main 2020 (Online) 5th September Evening Slot
Question
Ten charges are placed on the circumference
of a circle of radius R with constant angular
separation between successive charges.
Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each.
The potential V and the electric field E at the
centre of the circle are respectively.
(Take V = 0 at infinity)
(Take V = 0 at infinity)
Choose an option
Show full solutionCorrect option: A
Correct answer
AV = 0; E = 0
Step-by-step explanation
Net charge = 5q - 5q = 0
Potential of centre = V =
VC = = 0
Let E be electric field produced by each charge at the centre, then resultant electric field will be
EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.( From symmetric property of vector)
Potential of centre = V =
VC = = 0
Let E be electric field produced by each charge at the centre, then resultant electric field will be
EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.( From symmetric property of vector)
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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.