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JEE Advanced Physics Electrostatics 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Physics Question (2021) — Solution

Question

Two point charges -Q and +Q / 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 x y-plane with its center at (b, 0). All lengths are measured in meters. The value of  R  is ________ meter.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Let  P h , k  be be a general point with potential to be zero. V 1 + V 2 = 0    (as potential given  = 0 ) K − Q h 2 + k 2 + K Q 3 h − 2 2 + k 2 = 0 1 3 h − 2 2 + k 2 = 1 h 2 + k 2 3 h − 2 2 + k 2 = h 2 + k 2 3 h 2 − 4 h + 4 + k 2 = h 2 + k 2 2 h 2 + 2 k 2 − 12 h + 12 = 0 h 2 + k 2 − 6 h + 6 = 0 h 2 − 6 h + 6 + k 2 = 0 h 2 − 6 h + 9 + k 2 = 3 h − 3 2 + k 2 = 3 Standard equation of circle is given by x − x 1 2 + y − y 1 2 = R 2 ∴     x 1 = 3 ,   y 1 = 0  and  R = 3 Hence,  R = 3   m and value of  b = 3   m  

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