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MHT CET 2026 (13 May, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2026 (13 May, Shift 1)

Q14MHT CET 2026 · 13 MAY · SHIFT 1Medium

Electrostatic Potential Energy

Three point charges +q,−2q+q, -2q and −2q-2q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to '2a2a' will be (K=14πε0)\left(K = \frac{1}{4\pi\varepsilon_0}\right)

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Answer: (d)

  1. Initial PE: Ui=Ka[(q)(−2q)+(q)(−2q)+(−2q)(−2q)]=Ka(−2q2−2q2+4q2)=0U_i = \frac{K}{a}\left[(q)(-2q) + (q)(-2q) + (-2q)(-2q)\right] = \frac{K}{a}(-2q^2 - 2q^2 + 4q^2) = 0
  2. At side 2a2a the same bracket gives Uf=0U_f = 0.
  3. W=Uf−Ui=0W = U_f - U_i = 0
  4. So the answer is (d) zero.
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