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Q.

A spherical charged conductor has surface charge density $σ$ . The electric field on its surface is E and electric potential of conductor is V. Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be

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a

2E, 2V

b

4E, 2V

c

4E, 4V

d

2E, 4V

answer is B.

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Detailed Solution

$E = \frac{1}{4 π ε_{0}} \frac{q}{R^{2}} = \frac{σ}{ε_{0}}$

$V = \frac{1}{4 π ε_{0}} \frac{q}{R} = \frac{σ R}{ε_{0}} (q is constant)$

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A spherical charged conductor has surface charge density σ. The electric field on its surface is E and electric potential of conductor is V. Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be