Question

Question #94aa8

Answer 1

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_cq^2/r^2," 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 `q/2`
Now C with charge `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+q/2=(3q)/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 `(3q)/4`

So finally the force between A and B will be

`F'=k_c(q/2xx(3q)/4)/r^2=3/8(k_cq^2/r^2)=3/8F`