Example

Question:

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 \(10\,\text{cm}\), as shown in Fig. 1.4(a). The repulsion is noted (for example, by shining a beam of light and observing the deflection of its shadow on a screen). Spheres A and B are then touched by uncharged spheres C and D respectively, as shown in Fig. 1.4(b). C and D are then removed and B is brought close to A at a distance of \(5.0\,\text{cm}\) between their centres, as shown in Fig. 1.4(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.

Solution:

Let the original charge on sphere A be \(q\) and that on B be \(q'\). At a distance \(r\) between their centres, the magnitude of the electrostatic force on each is given by \[ F = \frac{1}{4\pi\varepsilon_0} \frac{qq'}{r^2} \] neglecting the sizes of spheres A and B in comparison to \(r\). When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge \(q/2\).
Similarly, after D touches B, the redistributed charge on each is \(q'/2\). Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is \[ F' = \frac{1}{4\pi\varepsilon_0} \frac{(q/2)(q'/2)}{(r/2)^2} = \frac{1}{4\pi\varepsilon_0} \frac{qq'}{r^2} = F \] Thus the electrostatic force on A, due to B, remains unaltered.