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JEE Main 2024 (6 Apr, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2024 (6 Apr, Shift 2)

Q31JEE Main 2024 · 6 APR · SHIFT 2Medium

Electric Charge & Coulomb's Law

Two identical conducting spheres P and S with charge Q on each, repel each other with a force $16\text{ N}$ . A third identical uncharged conducting sphere $R$ is successively brought in contact with the two spheres. The new force of repulsion between $P$ and $S$ is:

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Answer: (b)

R touches P: $P = \dfrac{Q}{2}$ , $R = \dfrac{Q}{2}$ . R touches S: $S = \dfrac{Q/2 + Q}{2} = \dfrac{3Q}{4}$ .
$F' = \dfrac{k(Q/2)(3Q/4)}{r^2} = \dfrac{3}{8}F$
$F' = \dfrac{3}{8} \times 16 = 6\text{ N}$ . So the answer is (b) 6 N.

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