Figure shows a charged conductor resting on an insulating stand. If at the 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 $Q$
Charge density potential Electric intensity
Medium
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(d) The surface of the conductor is an equipotential surface since there is free flow of electrons within the conductor. Thus potential at $Q$ is the same as that at $ P$. That is ${V_P} = {V_Q} = V.$ The electric field $E$ at a point on the equipotential surface of the conductor is inversely proportional to the square of the radius of curvature $r$ at that point. That is $E \propto {r^{ - 2}}$
Since point $Q$ has a larger radius of curvature than that at point $P$, the electric field at $Q$ is less than that at $P$. That is ${E_Q} < {E_P} = E$
Since point $Q$ has a larger radius of curvature than that at point $P$, the electric field at $Q$ is less than that at $P$. That is ${E_Q} < {E_P} = E$
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