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

\frac{Q}{4 π ϵ_{0} R}

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\frac{Q}{2 π ϵ_{0} R}

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\frac{2 Q}{π ϵ_{0} R}

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\frac{Q}{π ϵ_{0} R}

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Solution

The correct option is B $\frac{Q}{2 π ϵ_{0} R}$
The potential at centre
$V = V_{1} + V_{2} + V_{3} = \frac{K Q}{R} + \frac{K (- 2 Q)}{R} + \frac{K (3 Q)}{R} = \frac{2 k Q}{R} = \frac{2 Q}{4 Π ε_{0} R} = \frac{Q}{2 Π ε_{0} R}$
Hence (A) option is correct

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

The correct option is B Q2πϵ0RThe potential at centre V=V1+V2+V3=KQR+K(−2Q)R+K(3Q)R=2kQR=2Q4Πε0R=Q2Πε0RHence (A) option is correct

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

The correct option is B $\frac{Q}{2 π ϵ_{0} R}$
The potential at centre
$V = V_{1} + V_{2} + V_{3} = \frac{K Q}{R} + \frac{K (- 2 Q)}{R} + \frac{K (3 Q)}{R} = \frac{2 k Q}{R} = \frac{2 Q}{4 Π ε_{0} R} = \frac{Q}{2 Π ε_{0} R}$
Hence (A) option is correct