Two point charges -Q and +Q / √{3} are placed in xy-plane at origin (0,0) and a point (2,0) , resp

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2. Electric Potential and Capacitance

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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.

A

$1.70,5$

B

$1.75,4$

C

$1.73,3$

D

$1.76,6$

(IIT-2021)

Solution

$V_p=0=\frac{k(-Q)}{r_1}+\frac{ kQ / \sqrt{3}}{r_2}$

$\frac{ kQ }{r_1}=\frac{ kQ / \sqrt{3}}{r_2}$

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

$3(x-2)^2+3 y^2=x^2+y^2$

$3\left(x^2+4-4 x\right)-x^2+2 y^2=0$

$2 x^2+12-12 x+2 y^2=0$

$x^2+6-6 x+y^2=0$

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

$R=\sqrt{3}=1.73,$

$b=3$

Standard 12

Physics

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