JEE Advanced 2021 — Paper 1 — Question 36

Lets take two points $\left( a,0 \right)$ and $\left( C,0 \right)$ on equipotential circle. Net potential at $\left( C,0 \right)=0$

\begin{array}{*{35}{r}}{} & \frac{K\left( -q \right)}{C}+\frac{Kq}{\frac{\sqrt{3}}{\left( C-2 \right)}}=0 \\{} & \frac{1}{C}=\frac{1}{\sqrt{3}\left( C-2 \right)} \\{} & ~\Rightarrow \sqrt{3C-2\sqrt{3}=C} \\\end{array}

$\Rightarrow \left( \sqrt{3}-1 \right)\text{C}=2\sqrt{3}$ $\Rightarrow \text{C}=\frac{2\sqrt{3}}{\sqrt{3}-1}$ Potential net at $\left( a,0 \right)=0$ $\frac{K\left( -q \right)}{a}+\frac{K\frac{q}{\sqrt{3}}}{\left( 2-a \right)}=0$ $\Rightarrow \frac{1}{\text{a}}=\frac{1}{\sqrt{3}\left( 2-\text{a} \right)}$ $\Rightarrow 2\sqrt{3}-\sqrt{3}a=a$ $\Rightarrow \text{a}=\frac{2\sqrt{3}}{1+\sqrt{3}}$ 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 $=\sqrt{{}}3$