Oil Drop Experiment

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Introduction

The oil drop experiment is an experiment that allows a precise meaurement of the elementary charge $ e $. The experiment was developed and performed in 1910 by the american physicist Robert Andrews Millikan. He measured the following value for the elementary charge:

$$ e = \SI{1.592e-19}{C} $$

Nowadays there are more precise methods to determine the elementary charge. The precise value is:

$$ \SI{1.602176e-19}{C} $$

Calculation using the method of hovering oil drops

The capacitor voltage $ U $ is increased during the oil drop experiment until an oil drop is not moving anymore (hovering). Then the electric field is switched off and the falling velocity $ v $ is measured.

Overview of the forces

Gravitational force (of a spheric oil drop in the homogenous gravitational field of the earth):
$ F_\mathrm{G} = m \cdot g = V \cdot \rho_\mathrm{Oil} \cdot g = \dfrac{4}{3} \pi \, r^3 \cdot \rho_\mathrm{Oil} \cdot g $
Buoyant force (of a sphere in air):
$ F_\mathrm{A} = \dfrac{4}{3} \pi \, r^3 \cdot \rho_\mathrm{Air} \cdot g $
Electric force in a homogenous electric field:
$ F_\mathrm{El} = q \cdot E = \dfrac{q \cdot U}{d} $
Stokes' drag (when the drop is falling):
$ F_\mathrm{R} = 6 \, \pi \, \eta \, r \, v $

With:

$ \rho_\mathrm{Oil} $ = Density of oil
$ \rho_\mathrm{Air} $ = Density of air

$ g $ = Gravitational acceleration

$ U $ = Capacitor voltage
$ d $ = Plate distance of the plate capacitor

$ \eta $ = Viscosity of air

$ v $ = Falling velocity of the oil drop

Effective density $ \rho^ \cdot $

The expression $ F_\mathrm{G} - F_\mathrm{A} $ can be calculated by using the "effective" density $ \rho^ \cdot = \rho_\mathrm{Oil} - \rho_\mathrm{Air} $:

\begin{aligned} F_\mathrm{G} - F_\mathrm{A} &= \dfrac{4}{3} \pi \, r^3 \cdot \rho_\mathrm{Oil} \cdot g \enspace - \enspace \dfrac{4}{3} \pi \, r^3 \cdot \rho_\mathrm{Air} \cdot g \\ & \\ F_\mathrm{G} - F_\mathrm{A} &= \dfrac{4}{3} \pi \, r^3 \cdot \left( \rho_\mathrm{Oil} - \rho_\mathrm{Air} \right) \cdot g \\ & \\ F_\mathrm{G} - F_\mathrm{A} &= \dfrac{4}{3} \pi \, r^3 \cdot \rho^ \cdot \cdot g \\ \end{aligned}

The experiment

The following values hold for the simulation "oil drop experiment":

$ \rho_\mathrm{Oil} = 973 \frac{Kg}{m^3} $, $ \rho_\mathrm{Air} = 1,29 \frac{Kg}{m^3} $, $ d = 5 mm $, $ \eta = 1,828 \cdot 10^{-5} \frac{Ns}{m^2} $,

The distance of two grey lines on the scale in the field of the capacitor is $ 1 mm $ and between two thin lines $ 0,25 mm $.

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$ U = $ -1 $ V$

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Electric charge $ q $ of the hovering drop

To determine the electric charge, the capacitor voltage $ U $ is increased until a drop is hovering. Then the electric force $ F_\mathrm{El} $ and the buoyant force $ F_\mathrm{A} $ are equal to the gravitational force $ F_\mathrm{G} $.

\begin{aligned} F_\mathrm{El} + F_\mathrm{A} &= F_\mathrm{G} \\ & \\ F_\mathrm{El} &= F_\mathrm{G} - F_\mathrm{A} \\ & \\ \dfrac{q \cdot U}{d} &= \dfrac{4}{3} \pi \, r^3 \cdot \rho^ \cdot \cdot g \\ & \\ q &= \dfrac{4}{3} \pi \, r^3 \cdot \rho^ \cdot \cdot g \cdot \dfrac{d}{U} \\ & \\ \end{aligned}

Radius $ r $ of the falling drop

When the capacitor is switched of the drops are falling down and accelerated by the gravitational force $ F_\mathrm{G} $. With increasing speed also the drag force $ F_\mathrm{R} $ increases. When this drag force and the buoyant force are equal to the gravitational force, the oil drop is not accelerated anymore, but falls with constant velocity $ v $. This force equality enables us to calculate the radius $ r $ of the drop.

\begin{aligned} F_\mathrm{G} &= F_\mathrm{R} + F_\mathrm{A} \\ & \\ F_\mathrm{G} - F_\mathrm{A} &= F_\mathrm{R} \\ & \\ \dfrac{4}{3} \cancel \pi \, r^{\cancel 3 2} \cdot \rho^ \cdot \cdot g &= 6 \, \cancel \pi \, \eta \, \cancel r \, v \\ & \\ \dfrac{4}{3} \, r^{2} \cdot \rho^ \cdot \cdot g &= 6 \, \eta \, v \\ & \\ r^{2} &= \dfrac{9 \, \eta \, v}{2 \, \rho^ \cdot g } \\ & \\ r &= \sqrt{ \dfrac{9 \, \eta \, v}{2 \, \rho^ \cdot g } } \\ & \\ \end{aligned}

Determining the elementary charge

Charge of oil drops

Every oil drop consists of many atoms and can carry not only one but multiple elementary charges. Therefore every calculated charge $ q $ of an oil drop is an integral multiple of the elementary charge $ e $. This can be seen in the diagram of oil charges.