Answer:
Considering the data, drop C contained 5 extra electrons.
Explanation:
Robert Millikan and his colleague Harvey Fletcher puplished in 1911 a work in which they discribed an experiment that lead to the measurement of the elementary charge (e). These scientists determined that each electron has a charge equal to 1.60217662 × 10⁻¹⁹ coulombs.
Considering that each oil drop is neutral, the charge observed is due to extra electrons. Therefore, in order to determine the number of extra electrons (n) we need do divide the drop charge C by the elementary charge (e):
[tex]n=\frac{C}{e} \\[/tex]
n = 8.00x10⁻¹⁹/1.60217662 × 10⁻¹⁹
n = 5 extra electrons.