A non-uniformly shaped conductor is charged. | vedclass

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(B) For a charged conductor in electrostatic equilibrium,the entire conductor is an equipotential surface,meaning the electric potential is the same a

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Electric field will be maximum

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(B) For a charged conductor in electrostatic equilibrium,the entire conductor is an equipotential surface,meaning the electric potential is the same at every point. However,the surface charge density $\sigma$ is inversely proportional to the radius of curvature $R$ $(\sigma \propto 1/R)$. At the sharpest point,the radius of curvature $R$ is minimum,which makes the surface charge density $\sigma$ maximum. Since the electric field near the surface is given by $E = \sigma / \epsilon_0$,the electric field is also maximum at the sharpest point.

$A$ non-uniformly shaped conductor is charged. Then,at its sharpest point:

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