Question

In an experiment, the tiny droplets of the negatively electrified oil are dropping through a vacuum. Magnitude of the electric field is 5.92×104 NC−1. Direction of the electric field is assumed to be downward. (a)Anexperimenterobservesaoneparticulardropletandthatdropletremainssuspendedagainst gravity. Assume the mass of the droplet to be 2.93×10−15 kg. What is the charge carried by the droplet? (b) Now another droplet of same mass falls, in the vacuum, from rest to 15.3cm in 0.275s. What is the charge on the droplet in this case?

Answer 1

a) 2nd Newtons law says

$$qE + mg = 0 \Rightarrow q = -\frac{mg}{E}=-4.86 \cdot 10^{-19}\text{ C}$$

As electric field directed downwards, this droplet has negative charge - therefore electric force directed upwards.

b) Acceleration of droplet depends on passed distance and time:

$$s=\frac{1}{2}at^2 \Rightarrow a = \frac{2s}{t^2} = 4.05 \text{ms}^{-2}$$

2nd Newtons law says

$$mg+qE=ma \Rightarrow q= \frac{m(a-g)}{E}=-2.85 \cdot 10^{-19}\text{ C}$$