The excess of electron that must be placed on each of two small spheres spaced 3 cm apart if the force of repulsion between the spheres is to be 10^-19 N 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

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin