JEE Advanced 2021 · Paper 1 · Physics Q9

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PhysicsNumerical+2Question 9 of 57

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Question Stem for Question Nos. 9 and 10

Two point charges $-Q$ and $+Q/\sqrt3$ 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.

xy axes with charge −Q at the origin and charge +Q/√3 at (2, 0) on the x-axis. A dashed circle (the V = 0 equipotential) surrounds the +Q/√3 charge, centred on the x-axis slightly to the right of it.

The value of $R$ is ____ meter.

Your answer

Answer and worked solution

Official answer: 1.7

Zero-potential condition. $V=0$ where
$\frac{Q}{r_1}=\frac{Q/\sqrt3}{r_2},\quad\text{i.e. } r_1=\sqrt3\,r_2,$
with $r_1$ the distance from the origin and $r_2$ the distance from $(2,0)$ .

The circle.
$x^2+y^2=3\left[(x-2)^2+y^2\right]\implies x^2+y^2-6x+6=0\implies(x-3)^2+y^2=3.$

So $R=\sqrt3\approx1.73$ m.

Topics:Equipotential surfaces Electric potential

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JEE Advanced 2021 · Paper 1, Physics question 9