A pith ball A of mass 9$\times$10^-5 kg carries a charge of 5 $\mu$C. What must be the magnitude and sign of the charge on a pith ball B held 2 cm directly above the pith ball A, such that the pith ball A, remains stationary?

Solution:

Here, charge on pitch ball A, ${\mathrm{q}}_1=5\mathrm{\mu C}=5\times {10}^{-6}\mathrm{C}$
Mass of pitch ball A, ${\text{m}}_1=9\times {10}^{-5}\mathrm{kg}$
The weight m₁  of the pitch ball A acts vertically downwards.

Let q₂ be charge on the pitch ball B held 2 cm above the pitch ball A, so that the pitch ball A remains stationary. It can be possible only, if the charges on two pitch balls are of opposite signs,
i.e if charge on pitch ball A is positive, charge on B must be negative. Then, the force on pitch ball A due to B, i.e. F_AB will act vertically upwards (figure). For charge q₁ to remain stationary,

${\mathrm{F}}_{\mathrm{AB}}={\mathrm{m}}_1\mathrm{g}\text{ or }\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{AB}}^2}={\mathrm{m}}_1\mathrm{g}$

Here, AB = 2 cm = 0.02 m

$\begin{array}{l}\therefore 9\times {10}^9\times \frac{5\times {10}^{-6}\times {\mathrm{q}}_2}{(0.02{)}^2}=9\times {10}^{-5}\times 9.8\\ \text{ or }{\mathrm{q}}_2=7.84\times {10}^{-12}\mathrm{C}\end{array}$