JEE Main 2017 — Electrostatics Question with Solution
From: JEE Main 2017 (Online) 8th April Morning Slot
Question
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60o with the direction of the field ?
Choose an option
Show full solutionCorrect option: C
Correct answer
C589.4 V
Step-by-step explanation
Potential gradient,
V = E. d
589.8 589.0 = (E d)max
(E d)max = 0.8
V = E d cos
= 0.8 cos60o
= 0.4
Maximum potential on the sphere = 589.4 V
V = E. d
589.8 589.0 = (E d)max
(E d)max = 0.8
V = E d cos
= 0.8 cos60o
= 0.4
Maximum potential on the sphere = 589.4 V
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This is a previous-year question from JEE Main 2017, 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.