Q.
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 Potential Electric
Density intensity
A) $> \sigma\ \ \ \ \ \ \ \ \ \ \ \ \ \ > V\ \ \ \ \ \ \ \ \ \ \ \ \ > E$
B) $> \sigma\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V\ \ \ \ \ \ \ \ \ \ \ \ \ \ > E$
C) $< \sigma\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E$
D) $< \sigma\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V\ \ \ \ \ \ \ \ \ \ \ \ < E$
Explanation
(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 $E_{Q} < E_{P} = E$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.