First slide

Electrostatic potential

Question

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

Moderate

A

$\frac{Q}{2 {πε}_{0} R}$
B

$\frac{5 Q}{12 {πε}_{0} R}$
C

$\frac{3 Q}{32 {πε}_{0} R}$
D

none of these

Solution

Charge can be considered as located at a distance R from the center. Total charge is $(Q - 2 Q + 3 Q) = 2 Q$
Hence $V = \frac{1}{4 {πε}_{0}} \frac{2 Q}{R} = \frac{Q}{2 {πε}_{0} R}$

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