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Electrostatics question

2021 · Shift 1 · Q48
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Electrostatics question

2021 · Shift 1 · Q48

JEE AdvancedPhysicsElectrostaticsNumerical+2 / −1
Two point charges −-− Q and +Q/3\sqrt 33​ 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 32 English The value of b is ‾\underline{\hspace{2cm}}​ meter.
Numerical answer
View written solutionFree

Correct answer: 3.00

  1. Write the condition for zero potential

The charges are:

  • −Q-Q−Q at (0,0)(0,0)(0,0)
  • +Q3+\dfrac{Q}{\sqrt{3}}+3​Q​ at (2,0)(2,0)(2,0)

At any point P(x,y)P(x,y)P(x,y) in the plane, the potential is

V=k(−Qr1+Q/3r2),V = k\left(\frac{-Q}{r_1} + \frac{Q/\sqrt{3}}{r_2}\right),V=k(r1​−Q​+r2​Q/3​​),

where

  • r1=x2+y2r_1 = \sqrt{x^2+y^2}r1​=x2+y2​ is distance from (0,0)(0,0)(0,0),
  • r2=(x−2)2+y2r_2 = \sqrt{(x-2)^2+y^2}r2​=(x−2)2+y2​ is distance from (2,0)(2,0)(2,0).

For the equipotential with V=0V=0V=0,

−Qr1+Q/3r2=0.\frac{-Q}{r_1} + \frac{Q/\sqrt{3}}{r_2}=0.r1​−Q​+r2​Q/3​​=0.

Cancel QQQ:

−1r1+13 r2=0.-\frac{1}{r_1} + \frac{1}{\sqrt{3}\,r_2}=0.−r1​1​+3​r2​1​=0.

So,

13 r2=1r1⇒r1=3 r2.\frac{1}{\sqrt{3}\,r_2} = \frac{1}{r_1} \quad\Rightarrow\quad r_1 = \sqrt{3}\,r_2.3​r2​1​=r1​1​⇒r1​=3​r2​.
  1. Substitute distances

Thus,

x2+y2=3 (x−2)2+y2.\sqrt{x^2+y^2} = \sqrt{3}\,\sqrt{(x-2)^2+y^2}.x2+y2​=3​(x−2)2+y2​.

Squaring both sides,

x2+y2=3((x−2)2+y2).x^2+y^2 = 3\big((x-2)^2+y^2\big).x2+y2=3((x−2)2+y2).

Expand:

x2+y2=3(x2−4x+4+y2).x^2+y^2 = 3(x^2-4x+4+y^2).x2+y2=3(x2−4x+4+y2). x2+y2=3x2−12x+12+3y2.x^2+y^2 = 3x^2 - 12x + 12 + 3y^2.x2+y2=3x2−12x+12+3y2.

Bring all terms to one side:

0=2x2−12x+12+2y2.0 = 2x^2 - 12x + 12 + 2y^2.0=2x2−12x+12+2y2.

Divide by 2:

0=x2−6x+6+y2.0 = x^2 - 6x + 6 + y^2.0=x2−6x+6+y2. x2−6x+y2+6=0.x^2 - 6x + y^2 + 6 = 0.x2−6x+y2+6=0.

Complete the square in xxx:

(x−3)2−9+y2+6=0.(x-3)^2 - 9 + y^2 + 6 = 0.(x−3)2−9+y2+6=0. (x−3)2+y2=3.(x-3)^2 + y^2 = 3.(x−3)2+y2=3.
  1. Identify the circle

Comparing with the standard form

(x−b)2+y2=R2,(x-b)^2 + y^2 = R^2,(x−b)2+y2=R2,

we get

b=3,R=3.b=3, \qquad R=\sqrt{3}.b=3,R=3​.
  1. Final answer

The value of bbb is

3\boxed{3}3​

meters.

  1. Comparison with stored answer

Stored correct answer: 3.003.003.00

Our derived answer is 333, which matches 3.003.003.00.

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