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.
You need to login to perform this action.
You will be redirected in
3 sec