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Q.

A spherical charged conductor has σ  as the surface density of charge. The electric field on its surface is ‘E’. If the radius of the sphere is doubled, keeping surface density of charge unchanged, What will be the electric field on the surface of the new spheres?

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a

E/2

b

E

c

E/4

d

E/3

answer is C.

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Detailed Solution

E=K.4πR2.σR2=4πkσE'=k.4π(2R)2.σ4R2=E

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