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 -

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 - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-7tygzvdooiru.png" /> (A) (Q/2 π Îµ0 R) (B) (Q/4 π Îµ0 R) (C) (2 Q/π Îµ0 R) (D) (Q/π Îµ0 R)

Tardigrade Admin · Accepted Answer

The distance from each charge to center is

R

.
As the potential is a scalar quantity so net potential at center is

V = V_{Q} + V_{- 2 Q} + V_{3 Q} = \frac{1}{4 π ϵ _{0}} [\frac{Q}{R} + \frac{- 2 Q}{R} + \frac{3 Q}{R}] = \frac{2 Q}{4 π ϵ _{0} R} = \frac{Q}{2 π ϵ _{0} R}

Q. 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 -

$\frac{Q}{2 π ε _{0} R}$

$\frac{Q}{4 π ε _{0} R}$

$\frac{2 Q}{π ε _{0} R}$

$\frac{Q}{π ε _{0} R}$

The distance from each charge to center is $R$ .
As the potential is a scalar quantity so net potential at center is
$V = V_{Q} + V_{- 2 Q} + V_{3 Q} = \frac{1}{4 π ϵ _{0}} [\frac{Q}{R} + \frac{- 2 Q}{R} + \frac{3 Q}{R}] = \frac{2 Q}{4 π ϵ _{0} R} = \frac{Q}{2 π ϵ _{0} R}$

Q. 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 -

2πε0​RQ​

4πε0​RQ​

πε0​R2Q​

πε0​RQ​