A tiny spherical oil drop carrying a net charge q is balanced in...

$qE = mg$ ...(i)

$6\piηrv = mg$

$\dfrac{4}{3}\pir^{3}ρg = mg$ ...(ii)

$∴ r = {\left(\dfrac{3mg}{4\piρg}\right)}^{1 / 3}$ ...(iii)

Substituting the value of r in Eq. (ii), we get

$6\piηv{\left(\dfrac{3mg}{4\piρg}\right)}^{1 / 3} = mg$

or~${\left(6\piηv\right)}^{3}\left(\dfrac{3mg}{4\piρg}\right) = {\left(mg\right)}^{3}$

Again substituting $mg = qE,$we get

\left(qE\right)2 = \left(\dfrac{3}{4\piρg}\right)\left(6\piηv\right)3$<br /><br />Or$qE = {\left(\dfrac{3}{4\piρg}\right)}^{1 / 2}{\left(6\piηv\right)}^{3 / 2}$<br /><br />$∴ q = \dfrac{1}{E}\left(\dfrac{3}{4\piρg}\right)\dfrac{1}{2}\left(6\piηv\right)3 / 2$<br /><br />Substituting the values, we get<br /><br />$q = \dfrac{7}{81\pi \times {10}^{5}}\sqrt{\dfrac{3}{4\pi \times 900 \times 9.8} \times 216{\pi}^{3}}$<br /><br />$\times \sqrt{\left(1.8 \times {10}^{- 5} \times 2 \times {10}^{- 3}\right)3} = 8.0 \times {10}^{- 19}C