Oil Drop Experiment

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 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.

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 \( 0,25   mm \).


ResetStart
Spray oil

Electric field:

On Off
  
 \( U = \) -1 \(   V\)

Timer:

00:00.00

Start Reset

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

© Ixitixel, Wikipedia
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.

Sources