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A spherical charged conductor has surface density of charge $= σ,$ and electric field intensity on its surface is E. If radius of surface is doubled, keeping $σ$ unchanged, what will be electric field intensity on the new sphere?

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E/2

2 E

E/4

answer is D.

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Detailed Solution

$E = σ / \in_{0},$ the value of E does not depend upon radius of the sphere.

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