A pith ball of mass 9 × 10^–5 kg carries a charge of 5 $\mathrm{\mu }$C. What must be charge in another pith ball held directly 2 cm below the given pith ball such that the first pith ball be in equilibrium?  [g = 9.8 m/s²]

If we have to place another pith ball 2 cm above the first one, then its weight should be balanced and equal to the force of repulsion existing between the two pith balls.

$F=mg$

According to Coulomb’s law, the force between two charges is directly proportional to the product of the magnitude of their charges and inversely proportional to the square of the distance between them.

Charge on the first body, $\left({\mathrm{q}}_1\right)=5\mathrm{\mu C}=5\times {10}^{-6}\mathrm{C} \left(\because 1\mathrm{C}={10}^6 \mathrm{\mu C}\right)$

Charge on second body, $q_2$= q

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}=\mathrm{mg}$

Substituting the values, we get:

$\Rightarrow \frac{9\times {10}^9\times 5\times {10}^{-6}\times \mathrm{q}}{{\left(2\times {10}^{-2}\right)}^2}=9\times {10}^{-5}\times 9.8$

$\Rightarrow \mathrm{q}=\frac{9\times {10}^{-5}\times 9.8\times {\left(2\times {10}^{-2}\right)}^2}{9\times {10}^9\times 5\times {10}^{-6}}$

$\Rightarrow \mathrm{q}=\frac{9\times {10}^{-5}\times 9.8\times {\left(2\times {10}^{-2}\right)}^2}{9\times {10}^9\times 5\times {10}^{-6}}$

$\Rightarrow \mathrm{q}=\frac{9\times {10}^{-5}\times 9.8\times 4\times {10}^{-4}\times {10}^6\times {10}^{-9}}{9\times 5}$

$\Rightarrow \mathrm{q}=7.84\times {10}^{-12} \mathrm{C}$