A small charged water drop with charge \( 1.5 \times 10^{-6}\,\mathrm{C} \) and mass 10 g is suspended in air in electric field E. The value of E is given by (Take acceleration due to gravity = \( 10\,\mathrm{m\,s^{-2}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 6.67 \times 10^{4}\,\mathrm{N/C} \)
The drop stays suspended when the upward electric force \( qE \) balances its weight \( mg \).
\( \displaystyle qE = mg \;\Rightarrow\; E = \frac{mg}{q} \)
\( m = 10\,\mathrm{g} = 0.010\,\mathrm{kg} \)
\( q = 1.5\times10^{-6}\,\mathrm{C} \)
\( \displaystyle E = \frac{0.010 \times 10}{1.5\times10^{-6}} = \frac{0.1}{1.5\times10^{-6}} = 6.67\times10^{4}\,\mathrm{N\,C^{-1}} \)