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

$(g = 10newton/kg,e = 1.6 \times 10^{- 19}coulomb)$

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

Answer 1

Answer: (B)

$260newton/coulomb$

Answer 2

For balance $mg = eE$ ⇒ $E = \frac{mg}{e}$

Also$m = \frac{4}{3}\pi r^{3}d = \frac{4}{3} \times \frac{22}{7} \times (10^{- 7})^{3} \times 1000kg$

⇒ $E = \frac{4/3 \times 22/7 \times (10^{- 7})^{3} \times 1000 \times 10}{1.6 \times 10^{- 19}}$= 260 N/C