A charged oil drop is suspended in a uniform electric field of \( 3\times10^{4}\,\mathrm{V\,m^{-1}} \) so that it neither falls nor rises. The charge on the drop will be (take the mass of the drop \( =9.6\times10^{-15}\,\mathrm{kg} \) and \( g=10\,\mathrm{m\,s^{-2}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.2\times10^{-18}\,\mathrm{C} \)
The drop neither falls nor rises when the electric force balances its weight, so \( qE=mg \) and the charge is \( q=mg/E \).
\( m=9.6\times10^{-15}\,\mathrm{kg} \)
\( g=10\,\mathrm{m\,s^{-2}} \)
\( E=3\times10^{4}\,\mathrm{V\,m^{-1}} \)
\( \begin{aligned} q &= \frac{mg}{E} \\ &= \frac{9.6\times10^{-15}\,\mathrm{kg}\times10\,\mathrm{m\,s^{-2}}}{3\times10^{4}\,\mathrm{V\,m^{-1}}} \\ &= 3.2\times10^{-18}\,\mathrm{C} \end{aligned} \)