Question

Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B, which was originally 3.0 N, becomes:

Answer 1

What a delightful little problem !
Here's how I see it:

When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged.  So now, 'A' has half of its original charge, and 'C' has the other half.

Then, when 'C' is touched to 'B', charge flows to it until the two of them are equally charged.  How much is that ?  Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.

To review what we have now:  'A' has 1/2 of its original charge, and 'B' has 3/4 of it.

The force between any two charges is:

F = (a constant) x (one charge) x (the other one) / (the distance between them)².

For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.

The original force between them was  3 = (some constant) x (1 charge) x (1 charge).

The new force between them is  F = (the same constant) x (1/2) x (3/4) .

Divide the first equation by the second one, and you have a proportion:

3 / F  =  1 / ( 1/2 x 3/4 )

Cross-multiply this proportion:

3 (1/2 x 3/4)  = F

F  =  3/2 x 3/4  =  9/8  =  1.125 newton.


That's my story, and I'm sticking to it.