the excess(equal in no) of electrons that must be placed on each of two spheres spaced 3 cm apart, with force of repulsion between sphere to be 10*-19 is

Dear pallavi

let the charge on spheres is q

so F= 1/4∏ε qq/.03²

  given F= 10^-19

       1/4∏ε qq/.03² =10^-19

 9 *10⁹ * q² /.03² =10^-19 

               q=10^-16

1 electron has  1.6 * 10^-19 C  

 ne =q

 n=q/e =10^-16/1.6 * 10^-19 C

           =625

Regards

Arun