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 $xy$-plane with its center at $(b,0)$. All lengths are measured in meters.

The value of $R$ is _______ meter.

Answer 1

Answer: 1.73

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