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

Initial charges were q and -2q. Suppose the distance between them is r.
The force between them , F = 2kq^2/r^2

Now it is touched and again kept at same distance, so the charge will be redistributed = q-2q/2= - q/2.
Now the force between them ,F’=kq^2/4r^2

Now divide both the forces
F/F’ = 2*4/1
F/F’ = 8/1
Thus F’ = F/8
So, new force of repulsion is equal to 1/8 times of initial force.


Hope it helps.

Answer 2

Answer:

Explanation: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.