Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

In Exercise 1.12 the distance between the spheres, A and B was r = 0.5 m and the charge on each sphere, q = 6.5 x 10⁻⁷ C

When sphere A is touched with an neutral (uncharged) sphere C, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, becomes q/2.

When sphere C with charge q/2 is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given by
1/2 x (q+q/2)=1/2  x 3q/2

=3q/4

Therefore the Force of repulsion between sphere A with charge q/2 and sphere B with charge 3q/4 will be

F= 1 /4πε₀   x   q_Aq_B/r²

= 1 /4πε₀ x (q/2)(3q/4)/r²

=1 /4πε₀ x (3q²/8)/r²

=9 x 10⁹ x 3 x (6.5 x 10⁻⁷)²/(8 x (0.5)²

=5.703 x 10⁻³N

Therefore ,the force of attraction between the two spheres is 5.703 x 10⁻³N.