A spherical charged conductor has surface charge density . The intensity of electric field and potential on its surface are E and V. Now radius of sphere is halved keeping the charge density as constant. The new electric field on the surface and potential at the centre of the sphere are

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4E, V

E, V

E, V/2

2E, 4V

answer is B.

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Detailed Solution

$E = \frac{1}{4 π ε_{0}} \frac{q}{R^{2}} = \frac{σ}{ε_{0}} and V = \frac{1}{4 π ε_{0}} \frac{q}{R} = \frac{σ R}{ε_{0}}$
$\Rightarrow E$ remains same, but potential will be halved.

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