Explanation

(c) Net electric field due to both charges $q/3$, will get cancelled. Electric field due to $\left( \frac{- 2q}{3} \right)$ will be directed in $-$ve axis
[image:1_image255.emf]
$E = \frac{k\left( \frac{2q}{3} \right)}{R^{2}} \Rightarrow E = \frac{q}{6\pi\varepsilon_{0}R^{2}}$
P.E. of system $= \frac{K\left( \frac{q}{3} \right)^{2}}{2R} + \frac{K\frac{q}{3}\left( - \frac{2q}{3} \right)}{2R\sin 60^{o}} + \frac{K\frac{q}{3}\left( - \frac{2q}{3} \right)}{2R\cos 60^{o}}$
P.E. of system $\neq 0$
Force between $B$ and $C$
$F = \frac{K\left( \frac{2q}{3} \right)\left( \frac{q}{3} \right)}{\left( 2R\sin 60^{o} \right)^{2}} = \frac{4 \times 2Kq^{2}}{9 \times 4 \times 3R^{2}} = \frac{2q^{2}}{9 \times 3 \times 4\pi\varepsilon_{0}R^{2}}$
$(attractive) = \frac{1}{54}\frac{q^{2}}{{\pi \in}_{0}R^{2}}$
Potential at O, $V = \frac{K\left( \frac{q}{3} + \frac{q}{3} - \frac{2q}{3} \right)}{R} = 0$