Question

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

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

Answer 1

Answer: (A)

The distance from each charge to center is $R$.
As the potential is a scalar quantity so net potential at center is
$ \begin{array}{l} V = V _{ Q }+ V _{-2 Q }+ V _{3 Q }=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{ Q }{ R }+\frac{-2 Q }{ R }+\frac{3 Q }{ R }\right] \\ =\frac{2 Q }{4 \pi \epsilon_{0} R }=\frac{ Q }{2 \pi \epsilon_{0} R } \end{array} $