Two point charges $- Q$ and $\frac{+ Q}{\sqrt{3}}$ 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 $b$ in meter is

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Solution

Let a point $P$ on circle

Since, $V_{P} = 0 = \frac{k (- Q)}{r_{1}} + \frac{k (\frac{Q}{\sqrt{3}})}{r_{2}}$
$\Rightarrow \frac{k Q}{r_{1}} = \frac{k (\frac{Q}{\sqrt{3}})}{r_{2}} . . . . (i)$
From the figure we have,
$r_{1} = \sqrt{x^{2} + y^{2}}, r_{2} = \sqrt{(x - 2)^{2} + y^{2}}$
Putting these values in eq $(i)$ we have
$\frac{1}{r_{1}} = \frac{1}{\sqrt{3} r_{2}} \Rightarrow \frac{1}{\sqrt{x^{2} + y^{2}}} = \frac{1}{\sqrt{3} \sqrt{(x - 2)^{2} + y^{2}}}$
$\Rightarrow 3 (x - 2)^{2} + 3 y^{2} = x^{2} + y^{2}$
On solving above equation we get
$(x - 3)^{2} + y^{2} = (\sqrt{3})^{2}$
This is the equation of circle whose radius is $R = \sqrt{3} = 1.732$ and center is at $(3, 0)$
Thus, $b = 3$

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