A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. 1.7(a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.7(b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. 1.7(c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.

Explanation

Let the origanal repulsive force between A and B be
`F = (k q_(1) q_(2))/(r^(2))`
As A and C have same size, their charges are also shared equally.
As charges on A and B are halved and distance between them is also halved from 10 cm to 5 cm, therefore,
`F = k((q_(1)//2) (q_(2)//2))/((r//2)^(2)) = (k q_(1) q_(2))/(r^(2)) = F`