A charged metallic sphere $A$ is suspended by a riylon 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.

Let the original charge on sphere $A$ be $q$ and that on $B$ be $q^{\prime} .$ At a distance $r$ between their centres, the magnitude of the electrostatic force on each is given by

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q q^{\prime}}{r^{2}}$

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 $q\prime /2$. Now, if the separation between $A$ and $B$ is halved, the magnitude of the electrostatic force on each is

${F^\prime } = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{[q/2][{q^\prime }/2]}}{{{{(r/2)}^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {q{q^\prime }} \right)}}{{{r^2}}} = F$

Thus the electrostatic force on $A$, due to $B$, remains unaltered.