Question

A small droplet of oil with mass of 2.22×10-15 kg is held suspended in a region of uniform electric field directed upward with a magnitude of 5225 N/C. Is the excess charge on the droplet positive or negative? How many excess elementary charges reside on the droplet? (The elementary charge is e = 1.60×10-19 C.)

Answer 1

First you have to determine if the electric force is upwards or downwards. It has to be upwards because gravity pulls the charge downwards.


Drawing a free body diagram and direction of electric field allows us to deduce that the charge must be positive.

The next part of the question is an application of newton's second law:
\[\Sigma F = 0 = F_{g}+F_{E}\]
Letting downwards be negative this yields:
\[0=-mg+qE\]
Rearranging,
\[qE=mg\]
\[q=\frac{mg}{E}\]
Substituting the values gives us:
\[q=4.168*10^{-18}C\]
How many elementary charges e are there?
\[q=ne\]
where n is the number of elementary charge and e is the elementary charge 1.6*10^-19.
\[n=\frac{q}{e}=26\]
There are 26 excess positive charges.