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 figure (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 figure (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 figure (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

Coulombian force between A and B initially,
`F = k(q_(A)q_(B))/r^(2)`…….(1)
When A and C are brought in contact and then separated charge on each of them will be,
`q_(A)^(.) = q_( C)^(.) =(q_(A) + 0)/2 =q_(B)/2`……..(2)
Similarly when B and D are brought in contact and then separated charge on each of them will be,
`q_(B)^(.) = q_(D)^(.) = (q_(B) + 0)/2 = q_(B)/2`.......(3)
Now, when A and B are kept at `r. = r/2` if new Coulombian force between them is F. then,
`F. = k(q_(A)^(.).q_(B)^(.))/r^(2) = k(q_(A//2))(q_(B//2))/(r/2)^(2) = k(q_(A).q_(B))/r^(2)`........(4)
From equation (1) and (4),
`F.=F`