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.


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