# Question #94aa8

##### 1 Answer
Feb 15, 2017

Let the charge on each identical conducting sphere A and B be $q$ and they are separated by a distance $r$ which is much larger than their diameter.

Assuming their charges are situated at their centers as point charge the force of repulsion as per Coulomb's law becomes

$F = {k}_{c} {q}^{2} / {r}^{2} , \text{ where " k_c" "" Coulomb's const}$

Another uncharged identical conducting sphere C when comes in contact of A ,the total charge $q$ will be equally distributed between them as their capacitances are same because they are identical .

So Charge on both A and C will then be $\frac{q}{2}$
Now C with charge $\frac{q}{2}$ is touched with $B$ carrying charge $q$. As B and C are also identical they will have same capacitance and the total $q + \frac{q}{2} = \frac{3 q}{2}$ charge will be equally divided between them so as to make their potetial same.

Now after removing C from B , the sphere B will carry charge $\frac{3 q}{4}$

So finally the force between A and B will be

$F ' = {k}_{c} \frac{\frac{q}{2} \times \frac{3 q}{4}}{r} ^ 2 = \frac{3}{8} \left({k}_{c} {q}^{2} / {r}^{2}\right) = \frac{3}{8} F$