Question

Two similarly and equally charged identical metal spheres a and b repel each other with a force of 2×10^-5n.A third identical uncharged sphere c is touched with a and then placed at the mid point between a and



b.What is net electric force on



c.

Answer 1

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