Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^–7 C? The radii of A and B are negligible compared to the distance of separation.

Given:

Distance between centres of spheres, r = 50 cm = .5m

Charge on each sphere, q = 6.5 × 10^-7 C

Mutual force of electrostatic repulsion

…(1)

Where, F= mutual force of attraction

q₁ = charge on sphere 1

q₂ = charge on sphere 2

r = distance between centres

Where, ε₀ is the permittivity of the free space.

Now, putting the values of q₁, q₂ and r in equation (1).

⇒ F = 1.52 × 10^-2 N

Hence the mutual force between two spheres is 1.52 × 10^-2N. Since the sign is positive so the force is repulsive in nature.