Identical isolated conducting spheres A and B have the same excess charges and are separated by a distance that is large compared with their diameters (Fig. 22-20a). The electrostatic force acting on sphere B due to sphere A is vector FAB. Suppose now that a third identical sphere C, having an insulating handle and initially neutral, is touched first to sphere AWhat is the magnitude of the electrostatic force F' that now acts on sphere B?

If the spheres A and B. are separated by a distance that is large compared with their diameters, this means that we can treat to both spheres as if they were point charges.

If they are point charges, the magnitude of the force between them (that it will be repulsive as the spheres has the same excess charge) can be expressed by Coulomb's Law, as follows:

[tex]FAB = \frac{k*qa*qb}{rab^{2} }[/tex]

Now, if we make the sphere A be touched by a third identical sphere C, just due to symmetry and the behavior of charges in conductors, once separated, the excess charge on sphere A, will have been reduced exactly to the half of the original value, as the other half will have been passed to the sphere C.

So, if we replace in the expression of FAB, qa by qa/2, the new force is as follows:

[tex]F'AB = \frac{k*(qa/2)*qb}{rab^{2} } = \frac{1}{2} *\frac{k*qa*qb}{rab^{2}} = FAB *\frac{1}{2} =\frac{FAB}{2}[/tex]

As it can be seen, the magnitude of the electrostatic force F' that now acts on sphere B, is just the half of the original force FAB.