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QJEEAdv21Ph89a9: JEE Advanced 2021 Physics - Class 12 JEEAdv21Ph89a9 Solution: Two point charges Q and +Q/ are placed in the xy-plane at the…

Two point charges −- Q and +Q/ 3\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.

JEE Advanced 2021 Paper 1 Online Physics - Electrostatics Question 26 English

The value of b is __________ meter.

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 1 Online Physics - Electrostatics Question 26 English Explanation
V at B is zero if

kQ(2+R+x)=kQ3x+R{{kQ} \over {(2 + R + x)}} = {{{{kQ} \over {\sqrt 3 }}} \over {x + R}} ( ∵\because k=14πε0k = {1 \over {4\pi {\varepsilon _0}}} )

3(x+R)=2+R+x\sqrt 3 (x + R) = 2 + R + x

(3−1)x+(3−1)R=2(\sqrt 3 - 1)x + (\sqrt 3 - 1)R = 2 .....(i)

V at A is zero if

kQ2−x′=kQ3x′{{kQ} \over {2 - x'}} = {{{{kQ} \over {\sqrt 3 }}} \over {x'}}

3x′=2−x′\sqrt 3 x' = 2 - x'

x′=23+1x' = {2 \over {\sqrt 3 + 1}}

x′+x=Rx' + x = R

23+1+x=R{2 \over {\sqrt 3 + 1}} + x = R

2+(3+1)x=(3+1)R2 + (\sqrt 3 + 1)x = (\sqrt 3 + 1)R

x=(3+1)R−23+1x = {{(\sqrt 3 + 1)R - 2} \over {\sqrt 3 + 1}}

(3+1)R−(3+1)x=2(\sqrt 3 + 1)R - (\sqrt 3 + 1)x = 2 .....(ii)

Using Eqs. (i) and (ii), we get

R=3m=1.73mR = \sqrt 3 m = 1.73m

x=(3+1)3−23+1=3+13+1=1x = {{(\sqrt 3 + 1)\sqrt 3 - 2} \over {\sqrt 3 + 1}} = {{\sqrt 3 + 1} \over {\sqrt 3 + 1}} = 1 m

Hence, the centre of circle is having x-coordinate

= b = 2 + x = 3.00 m.

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