Question

Figure shows three circular arcs, each of radius $R$ and total change as indicated. The net electric potential at the center of curvature is :-

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

Answer 1

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

The potential at centre

$V=V_1+V_2+V_3=\frac{KQ}{R}+\frac{K(-2Q)}{R}+\frac{K\left(3Q\right)}{R}=\frac{2kQ}{R}=\frac{2Q}{4\Pi {\epsilon }_{0}R}=\frac{Q}{2\Pi {\epsilon }_{0}R}$

Hence (A) option is correct