Question

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

• A. 260 volt $/ cm$
• B. 260 newton/coulomb
• C. 130 volt $/ cm$
• D. 130 newton/coulomb

Answer 1

Answer: (B)

For balance, $m g=e E \Rightarrow E=\frac{m g}{e}$
Also, $m=\frac{4}{3} \pi r^{3} d=\frac{4}{3} \times \frac{22}{7} \times\left(10^{-7}\right)^{3} \times 1000\, kg$
$\Rightarrow E=\frac{(4 / 3) \times(22 / 7) \times\left(10^{-7}\right)^{3} \times 1000 \times 10}{1.6 \times 10^{-19}}=260\, N / C$
$\left(g=10\right.$ newton $/ kg , e=1.6 \times 10^{-19}$ coulomb $)$