What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Force if charges on each sphere is doubled and the distance is halved.

Given:

Q₁ = 2 × q

Q₂ = 2 × q

R = 0.5 × r

Using Coulomb’s law, we get,

…(1)

Where, ε₀ is the permittivity of the free space.

⇒

⇒

⇒ F = 0.243 N

The mutual force is 0.243N and it is repulsive in nature.

The new force is found to be 16 times the force in case I.