Two identical conducting spheres A and B carry an equal charges. They are separated by a distance much…

Two identical conducting spheres A and B carry an equal charges. 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:
  1. 3F4
  2. F2
  3. 3F8
  4.  F

Solution



F=kq2r2 when A and C are touched charge on both will be q2. Then when B and C are touched, qB= q2+q2=3q4
F=kqAqBr2= k ×q2 ×3q4r2=38kq2r2=38F.

Asked in: JEE Main 2018 (16 Apr Online)

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