Answer
(a) $ V_{A} = 1.8*10^{3} $v
$ V_{B} = 1.8*10^{3} $v
$ V_{C} = 9*10^{2}$v
(b)$Δ_{AB} =0$
$Δ_{BC} = V_{B} - V_{C}= 9*10^{2}$v
Work Step by Step
(a) Potential at a point is given by
$ V = \frac{k*q}{r}$
$ V_{A} = \frac{9*10^{9}*2*10^{-9}}{10^{-2}} $v
$ V_{A} = 1.8*10^{3} $v
$ V_{B} = \frac{9*10^{9}*2*10^{-9}}{10^{-2}} $v
$ V_{B} = 1.8*10^{3} $v
$ V_{C} = \frac{9*10^{9}*2*10^{-9}}{2*10^{-2}} $v
$ V_{C} = 9*10^{2}$v
(b)$Δ_{AB} = V_{A} - V_{B}=1.8*10^{3}v -1.8*10^{3}v =0$
It is obvious as A and B lie on equipotential surface.
$Δ_{BC} = V_{B} - V_{C}=1.8*10^{3}v - 9*10^{2}v = 9*10^{2}$v
