Explanation

(d) $qE\ = mg\ \ \ \ \ \ \ \ \ \ \ldots(i)$ $6\pi\eta\ rv = mg\ \ \ $
$\frac{4}{3}\pi r^{3}\ \rho g = mg\ \ldots(ii)$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r = \left( \frac{3mg}{4\pi\rho g} \right)^{1/3}\ \ \ \ldots(iii)$ Substituting the value of r in Eq. (ii), we get
$6\pi\eta v\left( \frac{3mg}{4\pi pg} \right)^{1/3} = mg$
or $\ (6\pi\eta v)^{3}\ \left( \frac{3mg}{4\pi\rho g} \right)\ \ \ \ = (mg)^{3}$
Again substituting $mg = qE,\ $we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (qE)^{2} = \left( \frac{3}{4\pi\rho g} \right)(6\pi\eta v)^{3}$
Or $qE = \left( \frac{3}{4\pi\rho g} \right)^{1/2}\ \ (6\pi\eta g)^{3/2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = \frac{1}{E}\left( \frac{3}{4\pi\rho g} \right)^{\frac{1}{2}}(6\pi\eta v)^{3/2}$
Substituting the values, we get
$q = \frac{7}{81\pi \times 10^{5}}\sqrt{\frac{3}{4\pi \times 900 \times 9.8} \times 216\pi^{3}}\ \ \ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \sqrt{\left( 1.8 \times 10^{- 5} \times 2 \times 10^{- 3} \right)^{3}\ } = 8.0 \times 10^{- 19}$ C