Question

There is a equipotential circle in xy-plane with potential $V=0$.
Find out radius of circle (in meter)

Answer 1

$V$ at $B$:
$\frac{KQ}{(2+R+x)}=\frac{\frac{KQ}{\sqrt{3}}}{x+R}$
$\sqrt{3}(x+R)=2+R+x$
$(\sqrt{3}-1)x+(\sqrt{3}-1)R=\mathrm{2....}\left(i\right)$
$V$ at $A$ :
$\frac{KQ}{2-x^{\prime }}=\frac{\frac{KQ}{\sqrt{3}}}{x^{\prime }}$
$\sqrt{3}{x}^{\prime }=2-x^{\prime }$
$x^{\prime }=\frac{2}{\sqrt{3}+1}$
$x^{\prime }+x=R$
$\frac{2}{\sqrt{3}+1}+x=R$
$2+(\sqrt{3}+1)x=(\sqrt{3}+1)R$
$x=\frac{(\sqrt{3}+1)R-2}{\sqrt{3}+1}$
$(\sqrt{3}+1)R-(\sqrt{3}+1)x=\mathrm{2....}\left(ii\right)$
using equation (i) and (ii)
$R=\sqrt{3}=1.732\text{m}$