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JEE Main 2025 (7 Apr, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2025 (7 Apr, Shift 1)

Q26JEE Main 2025 · 7 APR · SHIFT 1Medium

Electrostatic Potential Energy

Two charges q1q_1 and q2q_2 are separated by a distance of 30 cm. A third charge q3q_3 initially at ' C ' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of q3q_3 from C to D is given by q3K4πϵ0\frac{q_3 K}{4\pi\epsilon_0}, the value of K is:

Figure

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Answer: (d)

  1. q3q_3 moves on a circle centred at q1q_1, so the q1q3q_1q_3 energy does not change.
  2. Distance from q2q_2: at C, 302+402=50\sqrt{30^2+40^2} = 50 cm; at D, 40−30=1040 - 30 = 10 cm.
  3. ΔU=q2q34πϵ0(10.1−10.5)=q34πϵ0(8q2)\Delta U = \frac{q_2q_3}{4\pi\epsilon_0}\left(\frac{1}{0.1} - \frac{1}{0.5}\right) = \frac{q_3}{4\pi\epsilon_0}(8q_2)
  4. So K=8q2K = 8q_2, answer (d).
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