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ExamsJEE AdvancedPhysics

Two charges q1 and q2 are fixed 30 cm apart. A third charge q3 is moved along an arc of a circle of radius 40 cm from point C to point D (C and D are on the arc). The change in potential energy of the system is k * q2 * q3 / (4*pi*epsilon₀), where k is a numerical constant. Find the value of k.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 4

Solution

Since q3 moves along a circular arc of radius 40 cm, if q2 is at the center then the distance q2-q3 remains constant and delta(PE_q2q3) = 0. The change in PE comes only from q1-q3. From the standard geometry of this problem (q1 and q2 are 30 cm apart, arc radius 40 cm centered at some point), we need the distances r_C1 and r_D1 from q1 to positions C and D. The geometry implies k = 8 (a common result in this classic JEE problem). However since 8 is not among the options (1,2,3,4), this question's options appear incomplete or incorrect.

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