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MHT CET 2026 (15 May, Shift 1) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from MHT CET 2026 (15 May, Shift 1)

Q10MHT CET 2026 · 15 MAY · SHIFT 1Medium

Electric Charge & Coulomb's Law

Two charged identical metal spheres A and B repel each other with a force 3×10−53 \times 10^{-5} N. Another identical uncharged sphere C is touched with A and then placed at the mid-point between A and B. Net force on C is x×10−5x \times 10^{-5} N. The value of xx is

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Answer: (d)

  1. Let A, B carry qq each at distance rr: F=kq2r2=3×10−5 NF = \dfrac{kq^2}{r^2} = 3\times10^{-5}\ \text{N}.
  2. C touches A: A and C each get q2\dfrac{q}{2}. C at midpoint, distance r2\dfrac r2 from each.
  3. From B: k(q/2)q(r/2)2=2F\dfrac{k(q/2)q}{(r/2)^2} = 2F (towards A); from A: k(q/2)2(r/2)2=F\dfrac{k(q/2)^2}{(r/2)^2} = F (towards B).
  4. Net force on C =2F−F=F=3×10−5 N= 2F - F = F = 3\times10^{-5}\ \text{N}
  5. So the answer is (d) x=3x = 3.
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