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(A) The surface of a charged conductor is an equipotential surface because there is no potential difference between any two points on it. Therefore,th

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$< \sigma, V, < E$

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(A) The surface of a charged conductor is an equipotential surface because there is no potential difference between any two points on it. Therefore,the potential at point $Q$ is the same as at point $P$,i.e.,$V_Q = V_P = V$.The surface charge density $\sigma$ is inversely proportional to the radius of curvature $r$ $(\sigma \propto 1/r)$. Since the radius of curvature at $Q$ is larger than at $P$,the charge density at $Q$ is less than at $P$ $(\sigma_Q The electric field strength $E$ at the surface of a conductor is given by $E = \sigma / \epsilon_0$. Since $\sigma_Q Thus,at point $Q$,the values are: charge density $< \sigma$,potential $= V$,and electric field $< E$.

The figure shows a charged conductor resting on an insulating stand. If at point $P$ the charge density is $\sigma$,the potential is $V$,and the electric field strength is $E$,what are the values of these quantities at point $Q$?

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