Question

Two point charges $- Q$ and $+ Q / \sqrt{3}$ are placed in the $xy$-plane at the origin $(0,0)$ and a point $(2,0)$, respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V=0$ in the $x y$-plane with its center at $(b, 0)$. All lengths are measured in meters.

The value of $R$ is _______ meter.

Answer 1

Lets take two points $( a , 0)$ and $( C , 0)$ on equipotential circle.
Net potential at $( C , 0)=0$
$\frac{K(-q)}{C}+\frac{K q}{\frac{\sqrt{3}}{(C-2)}}=0$
$\frac{1}{C}=\frac{1}{\sqrt{3}(C-2)} $
$\Rightarrow \sqrt{3} C-2 \sqrt{3}=C$
$\Rightarrow (\sqrt{3}-1) C =2 \sqrt{3} $
$\Rightarrow C =\frac{2 \sqrt{3}}{\sqrt{3}-1}$
Potential net at $(a, 0)=0$
$\frac{K(-q)}{a}+\frac{K \frac{q}{\sqrt{3}}}{(2-a)}=0 $
$\Rightarrow \frac{1}{a}=\frac{1}{\sqrt{3}(2-a)} $
$\Rightarrow 2 \sqrt{3}-\sqrt{3} a=a $
$\Rightarrow a=\frac{2 \sqrt{3}}{1+\sqrt{3}}$
So, Radius $=\frac{C-a}{2}=\frac{\frac{2 \sqrt{3}}{\sqrt{3}-1}-\frac{2 \sqrt{3}}{\sqrt{3}+1}}{2}$
$=\sqrt{3}\left(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\right)=\sqrt{3}\left(\frac{\sqrt{3}+1-\sqrt{3}+1}{3-1}\right) $
Radius $=\sqrt{3}$