Question

Figure shows three circular arcs, each of radius $R$ and total charge as indicated. The net electron potential at the centre of curvature is

• A. $\frac{Q}{2\pi {ϵ}_{0}R}$
• B. $\frac{Q}{4\pi {ϵ}_{0}R}$
• C. $\frac{2Q}{\pi {ϵ}_{0}R}$
• D. $\frac{Q}{\pi {ϵ}_{0}R}$

Answer 1

The correct option is A $\frac{Q}{2\pi {ϵ}_{0}R}$

Given three circular arcs each of radius $R$ and total charge as shown in the figure. We have to find the net electron potential at the center of curvature.

So, potential at center is $V_1+V_2+V_3$ is sum of potential due to three charges (i.e, $Q$, $-2Q$, $3Q$).

So, $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}+\frac{1}{4\pi {\epsilon }_0}\left(\frac{-2Q}{R}\right)+\frac{1}{4\pi {\epsilon }_0}\left(\frac{3Q}{R}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{2Q}{R}=\frac{Q}{2\pi {\epsilon }_{0}R}$