The value of R is ______ meter. [2021]

Question

Two point charges $- Q$ and $+ \frac{Q}{\sqrt{3}}$ are placed in the $x y$ -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.

Q. The value of R is ______ meter. [2021]

Smart Achievers · Accepted Answer

(1.73)

Let us consider a point $P$ on the circle.

$V_{P} = 0 = \frac{k (- Q)}{r_{1}} + \frac{k Q / \sqrt{3}}{r_{2}}$ $\Rightarrow \frac{k Q}{r_{1}} = \frac{k Q / \sqrt{3}}{r_{2}}$

$\Rightarrow \frac{1}{\sqrt{x^{2} + y^{2}}} - \frac{1}{\sqrt{3} \sqrt{{(x - 2)}^{2} + y^{2}}}$

$\Rightarrow 3 {(x - 2)}^{2} + 3 y^{2} = x^{2} + y^{2}$

$\Rightarrow 3 (x^{2} + 4 - 4 x) - x^{2} + 2 y^{2} = 0$

$\Rightarrow 2 x^{2} + 12 - 12 x + 2 y^{2} = 0$

$\Rightarrow x^{2} + 6 - 6 x + y^{2} = 0$

$\Rightarrow {(x - 3)}^{2} + y^{2} = {(\sqrt{3})}^{2}$

or ${(x - b)}^{2} + y^{2} = {(\sqrt{3})}^{2} = R^{2}$

$∴$ $R = \sqrt{3} \approx 1.73$ and $b = 3$

Solution

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Solution

Ans. (1.73) Let us consider a point P on the circle. VP=0=k(-Q)r1+kQ/3r2⇒kQr1=kQ/3r2 ⇒1x2+y2-13(x-2)2+y2 ⇒3(x-2)2+3y2=x2+y2 ⇒3(x2+4-4x)-x2+2y2=0 ⇒2x2+12-12x+2y2=0 ⇒x2+6-6x+y2=0 ⇒(x-3)2+y2=(3)2 or (x-b)2+y2=(3)2=R2 ∴ R=3≈1.73 and b=3

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