# Millikan Oil Drop Experiment Question

Kennedy111

## Homework Statement

A 1.50 x 10^-14 kg oil drop accelerates downward at a rate of 1.80 m/s^2 when placed between two horizontal plates that are 9.40 ch apart. The potential difference between the two plates is 980 V. Determine the magnitude of the charge on the oil drop.

m = 1.50 x 10^-14 kg
a = 1.80 m/s^2 (down)
d = 9.40 cm = 9.40 x 10^-2 m
V = 980 V

Fnet = Fg + Fe
Fnet = ma
Fe = Eq
Fg = mg
E = V/d

## The Attempt at a Solution

Fnet = Fg + Fe
Fe = Fnet - Fg
= ma - mg
= ((1.50 x 10^-14 kg)*1.80 m/s^2)) - ((1.50 x 10^-14 kg)(9.81 m/s^2))
= -1.2015 x 10^-13 N

E = V/d
= (980 V) / (9.40 x 10^-2 m)
= 10425.53191 V/m

Fe = Eq
q = Fe / E
= (-1.2015 x 10^-13 N) / (10425.53191 V/m)
= -1.1524592 x 10^-17 C

The charge on the oil drop is approx. -1.15 x 10^-17 C ?

Homework Helper
Gold Member
Looks ok to me. Does the charge correspond to an integer number of electrons on the drop?

{Accurate analysis of the Millikan Oil Drop Experiment would take into account the viscosity of the air and the buoyant force acting on the drop, and involves measuring terminal velocities rather than accelerations. See http://en.wikipedia.org/wiki/Oil_drop_experiment . But what you did looks correct based on the information given.}

Kennedy111
If I divide that charge by 1.60 x 10^-19 C I get 72.0207 electrons. Does it have to be a whole number?

Homework Helper
Gold Member
Does it have to be a whole number?

Yes! At least if the data is good! That's one reason the experiment is considered a classic. Millikan found the net charge of the drops to be "quantized" as integer multiples of a fundamental amount of charge e. 72.02 is close to an integer. The discrepancy can be chalked up as "experimental uncertainty in the data".

[EDIT: Millikan reported net charges on the drops ranging from 1*e to 136*e. See http://www.aip.org/history/gap/PDF/millikan.pdf ]

Last edited:
Kennedy111
Okay, I understand! Thank you for your help!