JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Electrostatics

Two point charges

- Q

and

+ Q / 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

R

is ____ meter.

Answer: 1.73

Solution

For zero potential on the circle,

\frac{- Q}{r _{1}} + \frac{Q / 3}{r _{2}} = 0

, which gives

r_{1} = 3 r_{2}

, i.e.

r_{1}^{2} = 3 r_{2}^{2}

. With

r_{1}^{2} = x^{2} + y^{2}

and

r_{2}^{2} = (x - 2)^{2} + y^{2}

, this becomes

x^{2} + y^{2} = 3 [(x - 2)^{2} + y^{2}]

, which simplifies to

x^{2} + y^{2} - 6 x + 6 = 0

, or

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

. The radius is therefore

3 \approx 1.73

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer

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