Question

A charged water droplet hangs from ceiling of a room which has a uniform vertically upwards electrical field of strength 4.00 $\times {10}^{5}N{C}^{-1}$ . This field levitates and prevents a droplet of mass 1.0 $\times {10}^{-4}$kg from falling. Can you find the net charge on the droplet?

• A. 4.00 × 10^–9 C
• B. 2.45 × 10^–9 C
• C. 5.40 × 10^–9 C
• D. 1.75 × 10^–9 C

Answer 1

The correct option is B

2.45 × 10^–9 C

The forces acting on the droplet are

(i) Electrical force q $\overrightarrow{E}$ upwards

(ii) Force of gravity $m\overrightarrow{g}$ downwards

So, if in order to prevent the drop from falling, the two external forces should balance each other out.

$q\overrightarrow{E}=m\overrightarrow{g}q(4.00\times {10}^{8}Kg-m{s}^{-2}{C}^{-1})=(1.00\times {10}^{-4}kg)\times 9.8m-s^{-2}q=2.45\times {10}^{-9}C$