question_answer

A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B carried by an insulating handle is brought close to A such that the distance between their centres is 10 cm as shown in Fig. (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 calibrated screen). Spheres A and B are touched by uncharged spheres C and D, respectively as shown in Fig. (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. (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.                 

Answer:

                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\[(q\text{ }/\text{2})\]. Similarly, after D touches B, the redistributed charge on each is\[(q'/2)\]. If now 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.