Two conducting solid spheres A & B, of radii a & b respectively, are set at a di

Question

Two conducting solid spheres A & B, of radii a & b respectively, are set at a distance d from each other, with d > > a, b. Initially both spheres are charged with equal and opposite charge Q. so that QA = Q and QB = - Q. After they are connected by a thin metal wire and reach electrostatic equilibrium, what are the final charges on each sphere? You may use the fact that the electrostatic potential at the surface of an isolated conducting sphere of radius R is kQ/R, but explain why its use in this case is justified.

Explanation / Answer

electric potential is same for A and B spheresafter connecting thin wire between them VA = kQA/a          VB = k QB/b by above statement VA =VB        QA/a = QB/b ........................1 but we know total charge remains same    Total charge = Q-Q=0 (before connectingwire) after connecting wire    total charge =QB+ QA =0 ....................2 this means solving above equations we get QA = QB = 0

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