(c) From Coulomb’s law, the force of attraction between two charged particles ($q$), kept at a distance $r$ apart is
$F = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{r^{2}}$
[image:1_image300.emf]
When two identical spheres are brought in contact, charge on them is equalised, hence total charge on $C$ is equally shared when brought in contact with sphere $B$ having a charge $q$.
Therefore, charge on $B\ and\ C$ is $\frac{q}{2}.$
From Coulomb’s law, the force on $C$ is
$F_{c} = \frac{q \times q/2}{4\pi\varepsilon_{0}(r/2)^{2}} - \frac{(q/2)(q/2)}{4\pi\varepsilon_{0}(r/2)^{2}}$
$\ \ \ \ \ = \frac{qq}{4\pi\varepsilon_{0}r^{2}}(2 - 1) = F$