Question

two identical metal spheres carrying charges of + q and minus 2 q respectively when the spheres are separated back by a large distance are the force between them is f now the spheres are allowed to touch and then move back to the same separation find the new force of repulsion between them

Answer 1

The force between +q and -2q at a distance r is given as F Now
F= K ((+q)×(-2q))/r^2 = -2kq^2/r^2 When the chages +q and –2q comes in contact, than it neutralises the chage+q by –q charge and lefts only –q. These -q charge will be equally distributed both the sphare. If it is seperated the two sphare again than each sphare will aquirea mount of charge of –q/2 and –q/2 .Now the new force between them will be F_n= K ((-q^2/2)(-q^2/2))/r^2 = 1/4 ((kq^2)⁄r^2 ) = (-F)/8So
the force will be 1/8 part of previous force and it will be repulsive foce while F was attractive.

Answer 2

Answer:

Explanation:

F= K ((+q)×(-2q))/r^2 = -2kq^2/r^2 When the chages +q and –2q comes in contact, than it neutralises the chage+q by –q charge and lefts only –q. These -q charge will be equally distributed both the sphare. If it is seperated the two sphare again than each sphare will aquirea mount of charge of –q/2 and –q/2 .Now the new force between them will be F_n= K ((-q^2/2)(-q^2/2))/r^2 = 1/4 ((kq^2)⁄r^2 ) = (-F)/8So