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Q.
The intensity of the electric field required to keep a water drop of radius $10^{-5} cm$ just suspended in air when charged with one electron is approximately :-
Solution:
For balance $m g=e E$
$\Rightarrow E=\frac{m g}{e}$
Also $m=\frac{4}{3} \pi r^{3} d=\frac{4}{3} \times \frac{22}{7} \times\left(10^{-7}\right)^{3} \times 1000\, kg$
$\Rightarrow E=\frac{\frac{4}{3} \times \frac{22}{7} \times\left(10^{-7}\right)^{3} \times 1000 \times 10}{1.6 \times 10^{-19}}$
$=260\, N / C$
$\left( g =10\right.$ newton $/ kg , e =1.6 \times 10^{-19}$ coulomb)