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 F

_{1}= KQ^{2}/(2R)^{2}F

_{1}= KQ^{2}/4R^{2}this force is given in the problem as 2*10^{-5}N.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:

F

_{2}= K(Q/2 )*( Q/2)/R^{2}F

_{2}= KQ^{2}/4R^{2}Force between C and B will be :

F

_{3}= (K(Q/2) * Q)/(R)^{2}F

_{3}= KQ^{2}/2R^{2}Clearly, F

_{3}>F_{2}therefore F_{3}-F_{2}:-Hence KQ

^{2}/R^{2}( 1/2 - 1/4 )= KQ

^{2}/R^{2}( ¼ )= KQ

^{2}/4R^{2}But this is the given force F

_{1}from eqn 1 hence the net force on C = 2*10^{-5}N
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