Answer
a. $6.4 \times 10^{-19} C$.
b. 4 electrons.
Work Step by Step
a. The gravitational force, mg, equals the electric force, qE.
We use the data given in the problem.
$$mg = (1.1 \times 10^{–14}kg)(9.8 \frac{m}{s^2}) = q(1.68 \times 10^{5} \frac{N}{C})$$
$q = 6.4 \times 10^{-19} C$.
b. Divide the amount of charge on the droplet by the magnitude of the charge of one electron.
$$n = q/e = \frac{6.4 \times 10^{-19} C}{1.6 \times 10^{-19} C} = 4$$
We find that this oil drop carries 4 extra electrons.
