Two similarly and equally charged identical metal spheres A and B repel each other with a force of  0.000002 N. A third identical sphere C is touched with A and then placed at the mid point between A nd B. What is the net electric force on C?

Let the charge on A and B be = Q each.
Let the distance of separation between them be = 2R
According to Coloumb’s law of electrostats,
Force between them F1 = KQ2/(2R)2
F1 = KQ2/4R2 this force is given in the problem as 2*10-5N.
When an uncharged sphere C is brought in contact with A first, the charges get equally shared between them , hence each sphere acquires a charge Q/2.
Now force between A &C after it is placed at mid point ie at distance R from A is:
F2 = K(Q/2 )*( Q/2)/R2
F2 = KQ2/4R2
Force between C and B will be :
F3 = (K(Q/2) * Q)/(R)2
F3 = KQ2/2R2
Clearly, F3>F2 therefore F3-F2:-
Hence KQ2/R2( 1/2 - 1/4 )
= KQ2/R2( ¼ )
= KQ2/4R2
But this is the given force F1 from eqn 1 hence the net force on C = 2*10-5N

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