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

Ans (a).
Given
Charge on sphere A, q_A = 6.5 x 10⁻ ⁷ C  and

Charge on sphere B, q_B = 6.5 x 10⁻ ⁷ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres is given by expression:

F=1 /4πε₀   x   q_Aq_B/r²

Where ε₀=Permittivity of free space and  1 /4πε_{0  =}  9 x 10⁹ Nm²C⁻ ²

Therefore

F  =  9 x 10⁹ x (6.5 x 10⁻ ⁷)² / (0.5)2

= 1.52 x 10⁻ ² N
Therefore, the force between the two spheres is 1.52 x 10 ² N.

Ans (b).

If charge of each sphere is doubled and distance between then is halved then

Charge on sphere A, q_A = 1.3 x 10^-6 C

Charge on sphere B, q_B = 1.3 x 10^-6 C

The distance between the spheres is halved i.e. r=(0.5)/2= 0.25 m

Now the new force of repulsion between the two spheres,

F  =  9 x 10⁹ x (1.3 x 10⁻⁶)² / (0.25)2

= 16 x 1.52 x 10⁻² = 0.243 N

Therefore, the force between the two spheres is 0.243 N.