The millikan experiment, carried out by Robert Millikan (1868-1953) together with his student Harvey Fletcher (1884-1981), began in 1906 and aimed to study the properties of electric 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. In addition, the mass of the electron was found.
Previously the physicist J.J. Thompson had found experimentally the charge-mass relationship of this elementary particle, which he called "corpuscle", but not the values of each magnitude separately..
From this charge - mass relationship and the charge of the electron, the value of its mass was determined: 9.11 x 10-31 Kg.
To achieve their purpose, Millikan and Fletcher used an atomizer with which a fine mist of oil droplets was sprayed. Some of the droplets were electrically charged due to friction in the nozzle.
The charged drops were slowly settling on parallel plane plate electrodes, where a few passed through a small hole in the upper plate, as shown in the diagram in figure 1.
Inside the parallel plates it is possible to create a uniform electric field perpendicular to the plates, whose magnitude and polarity were controlled by modifying the voltage..
The behavior of the drops was observed by illuminating the interior of the plates with a bright light..
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If the drop has a charge, the field created between the plates exerts a force on it that counteracts gravity..
And if it also manages to remain suspended, it means that the field exerts an upward vertical force, which exactly balances gravity. This condition will depend on the value of what, the burden of the drop.
Indeed, Millikan observed that after turning on 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 - a drop could be made to remain suspended within the plates. Although in practice it is not easy to achieve, if it happens, only the force exerted by the field and gravity will act on the drop..
If the mass of the drop is m and its load is what, knowing that the force is proportional to the applied field of magnitude AND, Newton's second law states that both forces must be balanced:
mg = q.E
q = mg / E
The value of g, the acceleration of gravity is known, as well as the magnitude AND of the field, which depends on the voltage V established between the plates and the gap between them L, What:
E = V / L
The question was to find the mass of the tiny drop of oil. Once this is accomplished, determine the load what it is perfectly possible. Naturally that m Y what are respectively the mass and charge of the oil drop, not the electron.
But ... the drop is charged because it loses or gains electrons, so its value is related to the charge of said particle.
Millikan and Fletcher's problem was to determine the mass of a drop, not an easy task due to its small size..
Knowing the density of the oil, if you have the volume of the drop, the mass can be solved. But the volume was also very small, so conventional methods weren't helpful..
However, the researchers knew that such small objects do not fall freely, since the resistance of the air or of the environment, intervenes by slowing down their movement. Although the particle, when released with the field off, experiences an accelerated vertical movement and downwards, it ends up falling with constant speed.
This speed is called "terminal velocity" or "limiting velocity", which, in the case of a sphere, depends on its radius and the viscosity of the air..
In the absence of a field, Millikan and Fletcher measured the time it took for the drops to fall. Assuming that the drops were spherical and with the value of the viscosity of the air, they managed to determine the radius indirectly from the terminal velocity.
This speed is found by applying Stokes' law and here is its equation:
-vt is the terminal velocity
-R is the radius of the drop (spherical)
-η is the viscosity of air
-ρ is the density of the drop
Millikan's experiment was crucial, because it revealed several key aspects in Physics:
I) The elemental 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 electric charge comes in multiples of the fundamental charge.
III) Knowing the charge of the electron and the charge-mass relationship of J.J. Thomson, it was possible to determine the mass of the electron.
III) At the level of particles as small as elementary particles, the gravitational effects are negligible compared to electrostatics.
Millikan received the Nobel Prize in Physics in 1923 for these discoveries. His experiment is also relevant because he determined these fundamental properties of electric charge, starting from a simple instrumentation and applying laws well known to all..
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 make them more "presentable".
Millikan measured many, many drops in his experiment and not all of them were oil. He also tried 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 time, he obtained a variety of charged drops by striking X-rays through the plates, to ionize the air between them. In this way charged particles are released that the drops can accept.
Furthermore, he did not focus solely on the suspended droplets. Millikan observed that when the drops ascended, the rate of rise also varied according to the load supplied..
And if the drop descended, this extra charge added thanks to the intervention of the X-rays, did not change the speed, because any mass of electrons added to the drop is tiny, compared to the mass of the drop itself..
Regardless of how much charge he added, Millikan found that all the drops acquired charges integer multiples of a certain value, which is and, the fundamental unit, which as we have said is the charge of the electron.
Millikan initially scored 1,592 x 10-19 C for this value, slightly less than currently accepted, which is 1,602 x 10-19 C. The ratio may have been the value you gave to the viscosity of the air in the equation to determine the terminal velocity of the drop..
We see the following example. An oil droplet has a density ρ = 927 kg / m3 and is released in the middle of the electrodes with the electric field turned off. The droplet quickly reaches the terminal velocity, by which the radius is determined, the value of which turns out to be R = 4.37 x10-7 m.
The uniform field is turned on, directed vertically upward, and has magnitude 9.66 kN / C. In this way, the drop is suspended at rest..
It asks:
a) Calculate the droplet charge
b) Find how many times the elemental charge is contained in the charge of the drop.
c) Determine if possible, the sign of the load.
Previously, the following expression was derived for a drop at rest:
q = mg / E
Knowing the density and radius of the drop, the mass of the drop is determined:
ρ = m / V
V = (4/3) πR3
Therefore:
m = ρ.V = ρ. (4/3) πR3= 927 kg / m3. (4/3) π. (4.37 x10-7 m)3= 3.24 x 10-16 kg
Therefore, the charge of the drop is:
q = mg / E = 3.24 x 10-16 kg x 9.8 m / stwo/ 9660 N = 3.3 x 10-19 C
Knowing that the fundamental charge is e = 1.6 x 10 -19 C, the load 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 = 2.05
The result is that the charge on the drop is approximately twice (n≈2) the elemental charge. It is 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..
It is possible to determine the sign of the charge, thanks to the fact that the statement gives information about the direction of the field, which is directed vertically upwards, as well as the force.
The electric field lines always start from positive charges and end in negative charges, therefore the lower plate is charged with a + sign and the upper plate with a - sign (see figure 3).
Since the drop is directed towards the plate above driven by the field, and since charges of opposite sign attract, the drop must have a positive charge.
In reality, keeping the drop suspended is not easy to achieve. So Millikan used the vertical displacements (ups and downs) that the drop experienced by turning the field off and on, plus the changes in X-ray charge and travel times, to estimate how much extra charge the drop had acquired..
This acquired charge is proportional to the charge of the electron, as we have already seen, and can be calculated with the rise and fall times, the mass of the drop and the values of g Y AND.
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