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.
($1$) The value of $R$ is. . . . meter.
($2$) The value of $b$ is. . . . . .meter.
IIT 2021, Advanced
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$V_p=0=\frac{k(-Q)}{r_1}+\frac{ kQ / \sqrt{3}}{r_2}$
$\frac{ kQ }{r_1}=\frac{ kQ / \sqrt{3}}{r_2}$
$\frac{1}{\sqrt{ x ^2+ y ^2}}=\frac{1}{\sqrt{3} \sqrt{( x -2)^2+ y ^2}}$
$3(x-2)^2+3 y^2=x^2+y^2$
$3\left(x^2+4-4 x\right)-x^2+2 y^2=0$
$2 x^2+12-12 x+2 y^2=0$
$x^2+6-6 x+y^2=0$
$(x-3)^2+y^2=(\sqrt{3})^2$
$R=\sqrt{3}=1.73,$
$b=3$
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