Question

two identical metal spheres carry of +q and -q respectivally. When the spheres are separated by a large distance r the force between them is F. Now the spueres are allowed to touch and then moved back to the same separation .find the new force of repulsion between them.​

Answer 1

F/4 bcoz charge will be q/2 after touching each other.

Answer 2

Answer:

The new force between them will be zero.

Explanation:

Given that,

Two identical metal spheres carry of +q and -q respectively. When the spheres are separated by a large distance r the force between them is F.

The force between them,

$F=\dfrac{kq_{1}q_{2}}{d^2}$

Here, $q_{1}$ = first charge

$q_{2}$ = second charge

$d$ = distance between two charges

The force between them

$F=\dfrac{kq^2}{r^2}$....(I)

If the spheres are allowed to touch and then moved back to the same separation then the charges on spheres will be same.

Therefore, The charges on the sphere will be

$Q=\dfrac{q_{1}+q_{2}}{2}$

$Q=\dfrac{q-q}{2}$

$Q=0$

After contact, the charge will be zero on the spheres.

Therefore, The new force will be zero.

Hence, The new force between them will be zero.