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MHT CET 2026 (13 May, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2026 (13 May, Shift 2)

Q6MHT CET 2026 · 13 MAY · SHIFT 2Medium

Electrostatic Potential Energy

Two charges of equal magnitude 'qq' are placed in air at a distance '2a2a' apart and third charge −2q-2q is placed at mid-point. The potential energy of the system is (ϵ0\epsilon_0 = permittivity of free space)

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Answer: (d)

  1. Pairs: (q,q)(q, q) at 2a2a; two pairs (q,−2q)(q, -2q) at aa.
  2. U=k[q22a−2q2a−2q2a]=−7kq22aU = k\left[\dfrac{q^2}{2a} - \dfrac{2q^2}{a} - \dfrac{2q^2}{a}\right] = -\dfrac{7kq^2}{2a}
  3. U=−7q28πϵ0aU = -\dfrac{7q^2}{8\pi\epsilon_0 a}. So the answer is (d).
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