Two similarly and equally charged identical metal spheres A and B repel each other with a force of `2xx10^(-5)N`. A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Find the net electric force on C.

Explanation

Let initially the charge on each sphere be q and separation between their centres be r, then according to given problem.
`F = (1)/(4pi epsilon_(0)) (q xx q)/(r^(2)) = 2 xx 10^(-5)N`

When sphere C touches B, the charge of B, q will distribute equally on B and C as sphere are identical conductors, i.e., now charges on spheres ,
`q_(B) = q_(C ) = (q//2)`
So sphere C will experience a force

`F_(CA) = (1)/(4pi epsilon_(0))(q(q//2))/((r//2)^(2)) = 2F` along `vec(AB)` due to charge on A and, `F_(CB) = (1)/(4pi epsilon_(0))((q//2)(q//2))/((r//2)^(2)) = F` along `vec(BA)` due to charge on B So the net force `F_(C )` on C due to charges on A and B,
`F_(C ) = F_(CA) - F_(CB) = 2F - F = 2 xx 10^(-5) N` along `vec(AB)`.