Question

Two insulated charged copper spheres A and B
have their centres separated by a distance of
50cm. what is the mutual force of electrostatic
repulsion if the charge on each is 6.5x10-'C​

Answer 1

Given: Charge on A and B is 6.5 × 10−7 C

To find: The mutual force of electrostatic  repulsion.

Solution:

• Now we have given that the charge on both the sphere is same, that is 6.5 × 10−7 C and the distance between the spheres is 50 cm.

• Now 50 cm = 0.5 m

• So, the force of repulsion between the two spheres is

            F = ( 1 / 4π∈o) x ( q(A).q(B) / r² )

            where ∈o is permittivity of free space

            1 / 4π∈o = 9 x 10^9 Nm²C^-2

• Now putting the values in the given formula, we get:

            F = 9 x 10^9 x ( 6.5 × 10−7 )² / ( 0.5 )²

• After calculation, we get:

            F = 1.52 x 10^-2 N

Answer:

            So the force between the two spheres is 1.52 × 10−2 N.