# Millikan’s experiment: procedure, explanation, importance

The **Millikan experiment** , conducted by Robert Millikan (1868-1953) with his student Harvey Fletcher (1884-1981), began in 1906 and aimed to study the properties of electrical charge, analyzing the movement of thousands of drops. of oil in the middle of a uniform electric field.

The conclusion was that the electric charge did not have an arbitrary value, but came in multiples of 1.6 x ^{10-19} C, which is the fundamental charge of the electron. Also, the electron mass was found.

Previously, physicist JJ Thompson had experimentally found the charge-to-mass relationship of this elementary particle, which he called a “corpuscle,” but not the values of each magnitude separately.

From this charge-mass and charge ratio of the electron, the value of its mass was determined: 9.11 x ^{10-31} kg.

To achieve their goal, Millikan and Fletcher used an atomizer with which a fine mist of oil droplets was sprayed. Some drops were electrically charged thanks to friction in the sprayer.

The charged drops were slowly deposited onto flat parallel plate electrodes, where some passed through a small hole in the top plate, as shown in the diagram in Figure 1.

Within the parallel plates, it is possible to create a uniform electric field perpendicular to the plates, whose magnitude and polarity were controlled by voltage modification.

The behavior of the drops was observed by illuminating the interior of the plates with a strong light.

__explanation of experience__

__explanation of experience__

If the fall has a charge, the field created between the plates exerts a force on it that counteracts gravity.

And if it also manages to be suspended, it means that the field exerts an upward vertical force, which exactly balances gravity. This condition will depend on the value of *q* , the drop load.

In fact, Millikan observed that after lighting the field, some drops were suspended, others began to rise or continued to descend.

By adjusting the value of the electric field – by means of a variable resistance, for example -, it is possible to get a drop to remain suspended inside the plates. Although in practice it is not easy to achieve, if this happens, only the force exerted by the field and gravity act on the drop.

If the mass of the fall is *m* and its charge is *q* , knowing that the force is proportional to the applied field of magnitude *E* , Newton’s second law states that both forces must be balanced:

*mg = qE*

*q = mg / E*

It is known the value of *g* , the acceleration of gravity, as well as the magnitude *E* of the field, which depends on the voltage *V* established between the plates and the separation between these *L* , such as:

*E = V / L*

The question was to find the mass of the small drop of oil. When this is achieved, determining the charge *q* is perfectly possible. Of course, *m* and *q* are, respectively, the mass and charge of the oil drop, not the electron.

But … the fall is charged because it loses or gains electrons; therefore, its value is related to the charge of that particle.

**The mass of the oil drop**

Millikan and Fletcher’s problem was to determine the mass of a drop, a task that is not easy given its small size.

Knowing the density of the oil, if you have the drop volume, the mass can be cleaned. But the volume was also very small, so conventional methods weren’t helpful.

However, the researchers knew that these small objects do not fall freely, as the resistance of the air or the environment interferes with the speed of their movements. Although the particle released with the field turned off undergoes an accelerated and downward vertical movement, it ends up falling at a constant velocity.

This speed is called the “terminal speed” or “speed limit”, which, in the case of a sphere, depends on the radius and viscosity of the air.

In the absence of the field, Millikan and Fletcher measured the time needed for the drops to fall. Assuming that the drops were spherical and with the value of the viscosity of the air, they were able to determine the radius indirectly from the terminal velocity.

This velocity is found by applying Stokes’ law and here is its equation:

– *v _{t}* is the terminal speed

– *R* is the radius of the drop (spherical)

– *η* is the viscosity of air

– *ρ* is the drop density

__Importance__

__Importance__

Millikan’s experiment was crucial because it revealed several key aspects in physics:

I) The elementary charge is that of the electron, whose value is 1.6 x ^{10-19} C, one of the fundamental constants of science.

II) Any other electrical charge comes in multiples of the fundamental charge.

III) Knowing the electron charge and JJ Thomson’s charge/mass ratio, it was possible to determine the electron’s mass.

III) At the level of particles as small as elementary particles, gravitational effects are negligible compared to electrostatic ones.

Millikan received the Nobel Prize in Physics in 1923 for these discoveries. His experiment is also relevant because he determined these fundamental properties of electrical charge, from a simple instrumentation and application of well-known laws.

However, Millikan was criticized for having discarded many observations in his experiment, for no apparent reason, to reduce the statistical error of the results and that they were more “presentable”.

**Drops with a variety of charges**

Millikan measured many drops in his experiment and not all of them were oil. He also tested it with mercury and glycerin. As stated, the experiment started in 1906 and lasted for a few years. Three years later, in 1909, the first results were published.

During this period, he obtained a variety of charged droplets, making X-rays pass through the plates, to ionize the air between them. In this way, charged particles are released that the drops can accept.

Also, he didn’t just focus on the suspended drops. Millikan observed that as the falls increased, the rate of increase also varied according to the load provided.

And if the drop were to fall, this additional cost would increase thanks to the intervention of X-rays, the velocity would not change, because any mass of electrons added to the drop is small compared to the mass of the drop itself.

Regardless of how much charge he added, Millikan found that all droplets took on multiple integer charges of a certain value, which is *e* , the fundamental unit, which, as we said, is the charge on the electron.

Millikan initially obtained 1.592 x 10 ^{-19} C for this value, slightly less than the currently accepted, which is 1.602 x 10 ^{-19} C. The reason may have been the value that he attributed to the viscosity of the air in the equation Determine terminal speed of the fall.

**Example**

**Levitating a drop of oil**

We see the following example. A drop of oil has a density ρ = 927 kg / m ^{3} and is released in the middle of the electrodes with the electric field turned off. The drop quickly reaches the terminal velocity, through which the radius is determined, whose value turns out to be R = 4.37 x10 ^{-7} m.

The uniform field lights up, is directed vertically upwards and has a magnitude of 9.66 kN / C. In this way, it is possible for the drop to be suspended at rest.

He asks:

a) Calculate the drop charge

b) Find how many times the elementary charge is contained in the delivery rate.

c) Determine, if possible, the sign of the charge.

Figure 3. A drop of oil in the middle of a constant electric field. Source: Fundamentals of Physics. Rex-Wolfson

**Solution for**

Previously, the following expression was deduced for a fall to rest:

*q = mg / E*

Knowing the density and radius of the fall, the mass of it is determined:

*ρ = m / V*

*V = (4/3) πR ^{3}*

Therefore:

*m = ρ.V = ρ. (4/3) πR ^{3} = *

*927 kg / m*

^{3}. (4/3) π. (*4.37 x10*

^{-7}m)^{3}= 3.24 x 10^{-16}kgTherefore, the charge for the fall is:

*q = mg / E = 3.24 x 10 ^{-16} kg x 9.8 m / s ^{2} / N = 3.3 x 9660 ^{10-19} C*

**Solution b**

Knowing that the fundamental charge is e = 1.6 ^{10-19} C, the charge obtained in the previous section is divided by this value:

*n = q / e = **3.3 x 10 ^{-19} C / *

*1.6 x 10*

^{-19}C = 2.05The result is that the charge on the droplet is approximately twice (n≈2) the elementary charge. It’s not exactly double, but this slight discrepancy is due to the inevitable presence of experimental error, as well as rounding in each of the previous calculations.

**Solution c**

It is possible to determine the sign of the charge, thanks to the fact that the statement provides information about the direction of the field, which is directed vertically upwards, as well as the force.

Electric field lines always start with positive charges and end with negative charges; therefore, the bottom plate is loaded with a + sign and the top plate with a – sign (see Figure 3).

As the drop is directed towards the field-triggered top plate and as charges of the opposite sign attract, the drop must have a positive charge.

In fact, keeping gout suspended is not easy to achieve. Thus, Millikan used the vertical displacements (high and low) that the drop experienced when turning off and going into the field, as well as changes in X-ray charge and travel times, to estimate how much extra charge the drop had acquired.

That acquired charge is proportional to the charge on the electron, as we have seen, and can be calculated with rise and fall times, the mass of the fall and the values of *g* and *E* .