﻿ Oil Drop Experiment - College Physics

# 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

Electric field:

On Off

$$U =$$ -1 $$V$$

Timer:

00:00.00

### 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

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

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