Question

In 1909, Millikan performed a famous experiment in which he measured the charge of the electron (and for which he won the Nobel prize). In his experiment, oil drops carrying excess electrons are made to float between two charged large parallel, horizontal plates, as shown, by adjusting a uniform vertical electric field between the plates so that the drops do not fall down. Suppose each drop has a mass of 3.3 × 10−15kg, and one of the drops floats (the net force on it is zero) when the electric field magnitude is 39,200 N/C. (a) What is the net charge on this drop, and how many excess electrons does that correspond to? Note that the number of electrons is an integer, but due to experimental uncertainty, the number you initially calculate might not be an integer, so you will need to round! (b) How many electrons total are there in the drop (including all the electrons in the neutral atoms)? Assume that most of the weight of the oil is due to carbon atoms; Each carbon atom has 6 electrons. You will need to look up the atomic weight of carbon.

Answer 1

a)

q = net charge on the drop

m = mass of the drop = 3.3 x 10^-15 kg

E = magnitude of electric field = 39200 N/C

for the drop to float , the electric force due to electric field in upward direction must balance the weight of the drop in downward direction

electric force = weight

qE = mg

q (39200) = (3.3 x 10^-15) (9.8)

q = 8.25 x 10^-19 C

n = number of excess electron = ?

e = magnitude of charge on each electron = 1.6 x 10^-19 C

using the equation

q = ne

8.25 x 10^-19 = n (1.6 x 10^-19)

n = 5

b)

m = mass of drop = 3.3 x 10^-15 kg

M = atomic mass of carbon = 12 x 10^-3 g

N = number of atoms of carbon

N = (6.023 x 10²³)m/M

total number of electrons are given as

n_total = 6 N + n

n_total = 6 (6.023 x 10²³)m/M + n

n_total = 6 (6.023 x 10²³)(3.3 x 10^-15/(12 x 10^-3)) + 5 = 9.94 x 10¹¹