Two identical conducting spheres A and B, carry equal charge. They… | JEE Main 2018 (16 Apr, Shift 1)

Q66JEE Main 2018 · 16 APR · SHIFT 1Medium

Electric Charge & Coulomb's Law

Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to:

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Answer: (c)

C touches A: $A = \dfrac{q}{2}$ , $C = \dfrac{q}{2}$ .
C touches B: $B = \dfrac{q/2 + q}{2} = \dfrac{3q}{4}$ .
$F' = \dfrac{k(q/2)(3q/4)}{r^2} = \dfrac{3}{8}F$
So the answer is (c) $\dfrac{3F}{8}$ .

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