Physics - Electrostatics Question with Solution | TestHub

Original charges on spheres$A$and$B$be$q_1 \mathrm{a}\mathrm{n}\mathrm{d} q_2$respectively.
Distance between the two spheres$=r$
Since, both the spheres are of same size, they will possess equal charges on being brought in contact.

$\therefore q_1^{\prime }=\frac{q_1+q_2}{2}$

Similarly$q_{2 }^{\prime }=\frac{q_1+q_2}{2}$
Therefore, new force of repulsion between spheres$A \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }B$is

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{\left[\frac{q_1+q_2}{2}\right]\left[\frac{q_1+q_2}{2}\right]}{r^2}$

$=\frac{{\left[\frac{q_1+q_2}{2}\right]}^2}{4\pi {\epsilon }_0{r}^2}$

As${\left[\frac{q_{1+q_2}}{2}\right]}^2>q_1{q}_2$

$\therefore F^{\prime }>F$