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)

Since electric potential is scalar quantity therefore :
$\text{V }=\text{ V}_{\text{1}}+\text{ V}_{\text{2}}+\text{ V}_{\text{3}}$
$=\frac{\text{1}}{\text{4\pi } \text{ε}_{\text{0}}}\text{.}\frac{\text{Q}}{\text{R}}+\frac{\text{1}}{\text{4\pi ε}_{\text{0}}}\left[\frac{- \text{2Q}}{\text{R}}\right]+\frac{\text{1}}{\text{4\pi ε}_{\text{0}}}\left[\frac{\text{3Q}}{\text{R}}\right]$
$=\frac{1}{4 \pi \epsilon _{0}}.\left[\frac{2 Q}{R}\right]$