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JEE Advanced 2021 Paper 1

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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 xy-plane with its center at

(b, 0)

. All lengths are measured in meters.

Section: Physics

Question : 9 of 57

Marks: +1, -0

The value of

R

Solution:

Let us consider a point

P

on the circle

V_{P}=0=\;\frac{k(-Q)}{r_{1}}+\;\frac{\frac{kQ}{\sqrt{3}}}{r_{2}}

\Rightarrow \;\frac{kQ}{r_{1}}=\;\frac{\frac{kQ}{\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}+3y^{2}=x^{2}+y^{2}

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

\Rightarrow 2x^{2}+12-12x+2y^{2}=0 \Rightarrow x^{2}+6-6x+y^{2}=0

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

\therefore \;\; R=\sqrt{3}=1.73

and

b=3

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