
ΔV=Ed
AlongdiameterΔV=589.8−589.0V
=0.8V
0.8=Ed
Now

ΔV=E2dcosθ=0.4×cos60o
=0.2
∴589.8−V=0.2
V=589.6V
JEE Main 2017 — Physics Electromagnetism
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between the limits 589.0V to 589.8V. What is the potential at a point on the sphere whose radius vector makes an angle of 60∘ with the direction of the field?
Held on 8 Apr 2017 · Verified 6 Jul 2026.
589.4V
589.5V
589.2V
589.6V
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