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A spherical charged conductor has surface charge density $\sigma $ . 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
$2E,\, 2V$
$4E,\, 2V$
$4E,\, 4V$
$2E,\,4V$
Solution
$\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{R}^{2}}$
As $\mathrm{q}$ is constant, so $E \propto \frac{1}{R^{2}}$
Radius is halved. Therefore, electric field will becomes $4$ times or $4 \mathrm{\,E}.$
Further, $\quad \mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{R}}.$
As $\mathrm{q}$ is constant, so $\mathrm{V} \propto \frac{1}{\mathrm{R}}.$
Radius is halved, so potential will becomes two time or $2 \mathrm{\,V}.$
