A liquid drop having charge 8 x 10−19 C is suspended in a vertical electric field of 2.45 x 103 V/m. If gravitational acceleration is 9.8 m/s2, the mass of the drop is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 x 10−16 kg
The drop is suspended, so the electric force balances its weight.
\( \displaystyle qE = mg \)
\( \displaystyle m = \frac{qE}{g} \)
\( \displaystyle m = \frac{8\times10^{-19} \times 2.45\times10^{3}}{9.8} \)
\( \displaystyle m = \frac{19.6\times10^{-16}}{9.8} \)
\( \displaystyle m = 2\times10^{-16}\,\mathrm{kg} \)