Two equally charged, indentical metal spheres `A` and `B` repel each other with a force `F"`. The spheres are kept fixed with a distance `r` between them. A third identical, but uncharged sphere `C` is brought in contact with `A` and removed. The magnitude of the net electric force on `A` is

Explanation

Let, initially the charge on each sphere be q and separation between their centres be r. Then according to given problem :
`=1/ (4pi epsi_(0)) (qxxq)/r^(2)=2xx10^(-5) N`
When sphere C touches B, the charge of B i.e. q will distribute equally on B and C as sphere are identical how charges on spheres,
`q_(B)=q_(C)=(q//2)`
So sphere c will experience a force
`F_(CA)=1/(4pi epsi_(0)) (q(q//2))/((r//2)^(2))=2F` along `vec(AB)` due to charge on A.
and `F_(CB)=1/(4pi epsi_(0)) ((q//2)(q//2))/((r//2)^(2))= F`, along `vec(BA)` due to charge on B :
So the net forve Fc on C due to charges on A and B
`F_(C)=F_(CA)=F_(CB)=2F-F=2xx10^(-5) N` along `vec(AB)`.