Figure shows three circular arcs- each of radius R and total change as indicated- The net electric potential the center of curvature is -dfrac-Q-4 pi epsilon -0 R-dfrac-Q-2 pi epsilon -0 R-dfrac-2Q-pi epsilon -0 R-dfrac-Q- pi epsilon -0 R-

Question

Toppr Admin · Accepted Answer

The potential at centre-xA0-V-V1-V2-V3-KQR-K-x2212-2Q-R-K-3Q-R-2kQR-2Q4-x3A0-x3B5-0R-Q2-x3A0-x3B5-0RHence -A- option is correct-xA0-

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

$\frac{Q}{4 π ϵ_{0} R}$
$\frac{Q}{2 π ϵ_{0} R}$
$\frac{2 Q}{π ϵ_{0} R}$
$\frac{Q}{π ϵ_{0} R}$

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

Figure shows three circular arcs, each of radius R and total change as indicated. The net electric potential at the center of curvature is :-Q4πϵ0RQ2πϵ0R2Qπϵ0RQπϵ0R

The 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 :-

$\frac{Q}{4 π ϵ_{0} R}$
$\frac{Q}{2 π ϵ_{0} R}$
$\frac{2 Q}{π ϵ_{0} R}$
$\frac{Q}{π ϵ_{0} R}$

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