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JEE Main 2025 (24 Jan, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2025 (24 Jan, Shift 2)

Q27JEE Main 2025 · 24 JAN · SHIFT 2Medium

Electrostatic Potential Energy

Figure

In the first configuration (1) as shown in the figure, four identical charges (q0)(q_0) are kept at the corners A, B, C and D of square of side length ' aa '. In the second configuration (2), the same charges are shifted to mid points G,E,HG, E, H and FF, of the square. If K=14πϵ0K = \frac{1}{4\pi\epsilon_0}, the difference between the potential energies of configuration (2) and (1) is given by:

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Answer: (d)

  1. Config (1): 4 sides of length aa and 2 diagonals 2a\sqrt{2}a: U1=Kq02a(4+22)=Kq02a(4+2)U_1 = \frac{Kq_0^2}{a}\left(4 + \frac{2}{\sqrt{2}}\right) = \frac{Kq_0^2}{a}(4 + \sqrt{2})
  2. Mid-points form a square of side a2\frac{a}{\sqrt{2}} with diagonal aa.
  3. Config (2): U2=Kq02(42a+2a)=Kq02a(42+2)U_2 = Kq_0^2\left(\frac{4\sqrt{2}}{a} + \frac{2}{a}\right) = \frac{Kq_0^2}{a}(4\sqrt{2} + 2)
  4. U2−U1=Kq02a(32−2)U_2 - U_1 = \frac{Kq_0^2}{a}(3\sqrt{2} - 2), answer (d).
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