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Question

A spherical conductor of radius R carries charge Q on its surface. The value of electric field just outside the surface of the conductor is E. If the radius of the counductor is doubled while keeping the charge same, what will be the value of new electric field just outside the surface of the conductor?

A
E2
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B
2E
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C
E4
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D
E
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Solution

The correct option is C E4
We know that electric field just outside the surface of a conductor is given by :
E=σϵo

Surface charge density(σ) for spherical surface is given by :
σ=Q4πR2

Therefore,
E=Q4πϵoR2

When the radius of the sphere is doubled, the electric field just outside the surface of the sphere is given by :
E=Q4πϵo(2R)2

E=Q16πϵoR2

E=E4

Hence, option (c) is the correct answer.

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