Question

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 field intensity
a	$ > \sigma$	$> V$	$> E$
b	$ > \sigma$	$V$	$> E$
c	$ < \sigma$	$V$	$E$
d	$ < \sigma$	$V$	$< E$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (D)

The surface of the conductor is an equipotential surface (i.e., potential is same everywhere on the 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 $E_{Q} < E_{P}=E$.