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Q.

The electric field intensity at a point P due to a point charge q kept at a point Q is 40 NC-1 and the electric potential at a point P due to the same charge is 20 JC-1. The order of magnitude of charge q is

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a

10-6 C

b

10-7 C

c

10-12 C

d

10-9 C

answer is D.

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Detailed Solution

Electric field of a point charge,

E=14πε0qr2=40 NC-1

Electric potential of a point charge,

V=14πε0qr=20 JC-1

The distance PQ is r=VE=2040=0.5 m

∴ Magnitude of charge=4πε0Vr

=19×109×20×0.5=1.11×10-9 C≈10-9 C

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