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(B) The electric field on the surface of a charged spherical conductor is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}}$.Since the cha

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$4E, 2V$

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(B) The electric field on the surface of a charged spherical conductor is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}}$.Since the charge $q$ is constant,$E \propto \frac{1}{R^{2}}$.When the radius is halved $(R' = R/2)$,the new electric field $E'$ becomes $E' = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(R/2)^{2}} = 4 	imes \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}} ight) = 4E$.The electric potential of the conductor is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}$.Since the charge $q$ is constant,$V \propto \frac{1}{R}$.When the radius is halved $(R' = R/2)$,the new potential $V'$ becomes $V' = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(R/2)} = 2 	imes \left( \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R} ight) = 2V$.Thus,the new values are $4E$ and $2V$.

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

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