Explanation

(b) [image:1_image268.emf]
According to coulomb’s law, the force of repulsion between them is
$F = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r^{2}}$
When the charged spheres $A$ and $B$ are brought in contact, each sphere will attain equal charge $q''''''''$
$q^{''''''''} = \ \frac{q_{1} + q_{2}}{2}$
Now, the force of repulsion between them at the same distance $r$ is
[image:1_image269.emf]
$F^{''''''''} = \frac{q'''''''' \times q''''''''}{4\pi\varepsilon_{0}r^{2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{\left( \frac{q_{1} + q_{2}}{2} \right)\left( \frac{q_{1} + q_{2}}{2} \right)}{r^{2}} = \frac{\left( \frac{q_{1} + q_{2}}{2} \right)^{2}}{4\pi\varepsilon_{0}r^{2}}$
As $\left( \frac{q_{1} + q_{2}}{2} \right)^{2} > q_{1}q_{2}\ \therefore F^{''''''''} > F$