Figure shows three circular arcs- each of radius R and total charge as indicated- The net electric potential the center of curvature is -displaystyle frac-Q-2pi varepsilon-0R-none of thesedisplaystyle frac-3Q-32pi varepsilon-0R-displaystyle frac-5Q-12pi varepsilon-0R-

Question

Toppr Admin · Accepted Answer

Charge can be considered as located at a distance R from the-xA0-center- Total charge is -Q-x2212-2Q-3Q-2QHence V-14-x3C0-x3B5-02QR-Q2-x3C0-x3B5-0R

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

$\frac{Q}{2 π ε_{0} R}$
none of these
$\frac{3 Q}{32 π ε_{0} R}$
$\frac{5 Q}{12 π ε_{0} R}$

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

Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the center of curvature is :Q2πε0Rnone of these3Q32πε0R5Q12πε0R

Charge can be considered as located at a distance R from the center. Total charge is (Q−2Q+3Q)=2QHence V=14πε02QR=Q2πε0R

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

$\frac{Q}{2 π ε_{0} R}$
none of these
$\frac{3 Q}{32 π ε_{0} R}$
$\frac{5 Q}{12 π ε_{0} R}$

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