oil drop experiment form 2 physics

• Structure of Atom
• Oil Drop Experiment

Milliken's Oil Drop Experiment

The Millikens Oil Drop Experiment was an experiment performed by Robert A. Millikan and Harvey Fletcher  in 1909 to measure the charge of an electron. This experiment proved to be very crucial in the physics community.

Millikens Oil Drop Experiment Definition

In the experiment, Milliken allowed charged tiny oil droplets to pass through a hole into an electric field. By varying the strength of the electric field the charge over an oil droplet was calculated, which always came as an integral value of ‘e.’

Apparatus of the Milliken’s Oil Drop Experiment

The apparatus for the experiment was constructed by Milliken and Fletcher. It incorporated two metal plates held at a distance by an insulated rod. There were four holes in the plate, out of which three were there to allow light to pass through them and one was there to allow viewing through the microscope.

Ordinary oil wasn’t used for the experiment as it would evaporate by the heat of the light and so could cause an error in the Millikens Oil Drop Experiment. So, the oil that is generally used in a vacuum apparatus which is of low vapour pressure was used.

Milliken’s Oil Drop Experiment Procedure

• Oil is passed through the atomizer from where it came in the form of tiny droplets. They pass the droplets through the holes present in the upper plate of the apparatus.
• The downward motions of droplets are observed through a microscope and the mass of oil droplets, then measure their terminal velocity.
• The air inside the chamber is ionized by passing a beam of X-rays through it. The electrical charge on these oil droplets is acquired by collisions with gaseous ions produced by ionization of air.
• The electric field is set up between the two plates and so the motion of charged oil droplets can be affected by the electric field.
• Gravity attracts the oil in a downward direction and the electric field pushes the charge upward. The strength of the electric field is regulated so that the oil droplet reaches an equilibrium position with gravity.
• The charge over the droplet is calculated at equilibrium, which is dependent on the strength of the electric field and mass of droplet.

Milliken’s Oil Drop Experiment Calculation

F up = F down

F up = Q . E

F down = m.g

Q  is  an  electron’s  charge,  E  is  the  electric  field,  m  is  the  droplet’s  mass,  and  g  is  gravity.

One can see how an electron charge is measured by Millikan. Millikan found that all drops had charges that were 1.6x 10 -19 C multiples.

Milliken’s Oil Drop Experiment Conclusion

The charge over any oil droplet is always an integral value of e (1.6 x 10 -19 ). Hence, the conclusion of  Millikens Oil Drop Experiment is that the charge is said to be quantized, i.e. the charge on any particle will always be an integral multiple of e.

Frequently Asked Questions – FAQs

What did millikan’s oil drop experiment measure.

Millikan oil-drop test, the first simple and persuasive electrical charge calculation of a single electron. It was first conducted by the American physicist Robert A. in 1909. He discovered that all the drops had charges that were simple multiples of a single integer, the electron’s fundamental charge.

What is the importance of Millikan’s oil drop experiment?

The experiment with Millikan is important since it defined the charge on an electron. Millikan used a very basic, very simple system in which the behaviour of gravitational, electrical, and (air) drag forces were controlled.

What did Millikan conclude after performing his oil drop experiment?

An integral multiple of the charge on an electron is the charge on every oil decrease. About an electric force. In a relatively small amount, the charge and mass of the atom must be condensed.

Why charges are quantized?

Charges are quantized since every object’s charge (ion, atom, etc.) Charge quantization, therefore, implies that no random values can be taken from the charge, but only values that are integral multiples of the fundamental charge (proton / electron charge).

Can charge be created or destroyed?

The Charge Conservation Law does not suggest that it is difficult to generate or remove electrical charges. It also means that any time a negative electrical charge is produced, it is important to produce an equal amount of positive electrical charge at the same time so that a system’s overall charge does not shift.

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Description This simulation is a simplified version of an experiment done by Robert Milliken in the early 1900s. Hoping to learn more about charge, Milliken sprayed slightly ionized oil droplets into an electric field and made observations of the droplets. When the voltage is zero and the run button is pressed, the drop will fall due to the force of gravity. It will reach a terminal velocity (v t ) as it falls. Pause the simulation while you record the terminal velocity. This terminal velocity can be used to determine the mass of the drop. Use the equation: mass = kv t 2 to determine the mass of the particle. The value of k in this simulation is 4.086 x 10 -17 kg s 2 /m 2 . Once the terminal velocity is recorded and the mass calculated, with the simulation still paused increase the voltage between the plates until the two force vectors are approximately equal length. This will produce an upward field and an upward force on the positive droplets. If the upward force of the electric field is equal to the downward force of gravity, and the drag force is zero, the particle will not accelerate. To be sure that the lack of acceleration is not related to drag forces, the velocity must also be zero as well as the acceleration in order to be sure that the two forces are balanced. Increase and decrease the voltage (use the left/right arrow keys) until both the acceleration and velocity are at zero. The velocity may not stay at exactly zero, but find the voltage that has the velocity changing most slowly as it passes v = 0. Use the methods discussed above to ultimately determine the charge on ten (or more) different oil-drops. Use V = Ed to calculate the field strength (d = 5 cm = 0.05 m). Use Eq = mg when the velocity is zero to determine the charge q on the droplet. Record all your data in a table or spreadsheet. After you get each q, create a new particle and start again. When you have the table filled in, look at the various values for q. Is there any pattern to them, or are they seemingly random? Can you draw any conclusions from the Q measurements?

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Who Did the Oil Drop Experiment?

The Oil Drop Experiment was performed by the American physicist Robert A Millikan in 1909 to measure the electric charge carried by an electron . Their original experiment, or any modifications thereof to reach the same goal, are termed as oil drop experiments, in general.

What is the Oil Drop Experiment?

In the original version, Millikan and one of his graduate students, Harvey Fletcher, took a pair of parallel horizontal metallic plates. A uniform electric field was created in the intermediate space by applying a potential difference between them. The plates were held apart by a ring of insulating material. The ring had four holes, three for allowing light to illuminate the setup, and the fourth one enabled a microscope for viewing. A closed chamber with transparent walls was fitted above the plates.

At the beginning of the experiment, a fine mist of oil droplets was sprayed into the chamber. In modern setups, an atomizer replaces the oil droplets. The oil was so chosen such that it had a low vapor pressure and capable of charging. Some of the oil drops became electrically charged by friction as they forced their way out of the nozzle. Alternatively, charging could also be induced by incorporating a source of ionizing radiation , such as an X-Ray tube, in the apparatus. The droplets entered the space between the plates and raised or fell, according to the requirement, by varying plate voltage.

In terms of the present-day arrangement, when the electric field is turned off, the oil drops fall between the plates under the action of gravity only. The friction with the oil molecules in the chamber makes them reach their terminal velocity fast. The terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration . Once the field is turned on, the charged drops start to rise. This motion happens since the electric force directed upwards is stronger than the gravitational force acting downwards. One charged drop is selected and kept at the center of the field of view of the microscope after allowing all other drops to fall by alternately switching off the voltage source. The experiment is conducted with this drop.

Theory and Calculations

First, the oil drop is allowed to fall in the absence of an electric field, and its terminal velocity, say v 1 , is found out. Using Stokes’ law, the drag force acting on the drop is calculated using the following formula.

Here r is the radius of the drop and ɳ, the viscosity of air.

The weight of the drop, w’, which is the product of its mass and acceleration due to gravity g, is given by the equation,

where ρ is the density of the oil.

However, what we need here is the apparent weight w of the drop in the air given by the difference of the actual weight and the upthrust of the air. We can express w  by the following formula.

Here ρ air denotes the density of air.

When the drop attains terminal velocity, then it has no acceleration. Hence, the total force acting on it must be zero. That means,

The above equation can be used to find out the value of r. Once r is calculated, the value of w can easily be found out from equation (i) marked above.

Now after turning on the electric field between the plates, the electric force F E acting on the drop is,

Where E is the electric field and q the charge on the oil drop. For parallel plates, the formula for E is,

Here V is the potential difference and d the distance between the plates. That implies,

Now if we adjust V to make the oil drop remain steady at a point, then

Thus, the value of q can be calculated.  By repeatedly applying this method to multiple oil droplets, the electric charge values on individual drops were always found to be integer multiples of the smallest value. This lowest charge could be nothing but the charge on the elementary particle, electron. By this method, the electronic charge was calculated to be approximate, 1.5924×10 −19  C, making an error of 1% of the currently accepted value, 1.602176487×10 −19 C. All subsequent research pointed to the same value of charge on the fundamental particle.

Millikan was able to measure both the amount of electric force and magnitude of electric field on the tiny charge of an isolated oil droplet and from the data determine the magnitude of the charge itself. Millikan’s oil drop experiment proved that the electric charge is quantized in nature. The electric charge appears in quanta of magnitude 1.6 X 10 -19 C in oil droplets.

Robert Millikan’s Oil Drop Experiment Animation

Millikan’s oil drop experiment and the atomic theory.

Until the time of the Oil Drop Experiment, the world had little or no knowledge of what is present inside an atom . Earlier experiments by the English Physicist J.J. Thomson had shown that atoms contain some negatively charged particles of masses significantly smaller than that of the hydrogen atom. Nevertheless, the exact value of the charge carried by these subatomic particles remained in the dark. The very existence of these particles was not accepted by many due to a lack of concrete evidence. Thus, the atomic model was shrouded in mystery. In this scenario, with Millikan’s groundbreaking effort to quantify the charge on an electron, the atomic theory came of age in the early years of the twentieth century.

Controversy about the Oil Drop Experiment and Discovery

Robert Millikan was the sole recipient of the Nobel Prize in Physics in 1923 for both his work in this classic experiment and his research in the photoelectric effect . Fletcher’s work on the oil drop project, however, was not recognized. Many years later, the writings of Fletcher revealed that Millikan wished to take the sole credit for the discovery in exchange for granting him a Ph.D. and helping him secure a job after his graduation.

The beauty of the oil drop experiment lies in its simple and elegant demonstration of the quantization of charge along with measuring the elementary charge on an electron that finds widespread applications to this day. With the progress of time, considerable modifications have been made to the original setup resulting in obvious perfection in the results. Still, no substantial deviation from the results of the classical experiment could yet be found.

• Robert Millikan and Harvey Fletcher conducted the oil drop experiment to determine the charge of an electron. The experiment was the first direct and riveting measurement of the electric charge of a single electron.
• They suspended tiny charged droplets of oil between two metal electrodes by balancing downward gravitational force with upward drag and electric forces.
• They later used their findings to determine the mass of the electron.
• Kentchemistry.com
• Physics.utah.edu
• Nobelprize.org
• Ffden-2.phys.uaf.edu
• Chem.libretexts.org

Article was last reviewed on Thursday, February 2, 2023

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Measurement II | High School Physics Form 2

Physics revision questions.

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Measuring length using vernier callipers.

• The main scale is read at zero mark of the vernier scale. The values given in cm.
• The vernier is read at the position where a mark on the vernier scale is exactly lined up with a mark on the main scale. The values are given as a two decimal of a cm.

Practice Example 1

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Micrometer screw gauge

Calculating the Size of a Molecule

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Measurement II Questions and Answers - Physics Form 2 Topical Revision

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• What is the diameter of the ball bearing?
• Find the density of the ball bearing giving your answer correct to three significant

• Water in a dam falls through a height 24.5m. If we assume that there are no energy losses,  calculate the new temperature of the water as it strikes the lower end, given that its initial  temperature at the top of the dam is 18.9°C
• Lycopodium powder is lightly sprinkled on a clean water surface in a large tray. A red hot  needle is plunged at the centre of the water surface. State and explain the observation
• A micrometer screw gauge has a negative zero error of 0.06mm. Show on a micrometer screw  gauge, including the essential parts only a reading of 5.99mm
• Determine the volume of one drop of oil
• Calculate the thickness of a molecule
• State any two assumptions made in this experiment
• In the experiment,  lycopodium powder is used on the water surface. What is the role  of the lycopodium powder?
• A molecule of a liquid occupies a space about 1.5 x 10 -9 m high and about 0.6 x 10 -9 m in  thickness and breadth. Calculate the number of molecules in a litre of the liquid

• State the reading taken when the cylinder is grasped by the jaws
• In the space below, sketch the scale that gives the reading in (a) above if it has a pitch

• What is the diameter of the marble?
• Determine the density of the marble give your answer to three significant figure (assume that the  marble is spherical)

• The volume of the olive oil drop in m 3 (Take π = 22 / 7 )
• Using the value of (a) above, estimate the thickness of the film.
• Explain why lycopodium powder is sprinkled on the surface of water before the  oil is dropped on it.
• State two assumptions made when finding the thickness of the film formed.

• Draw a sketch of a micrometer screw gauge showing a reading of 8.53mm.

• Sketch a vernier calipers scale reading 3.41 cm.

• Volume of the metal cylinder.
• Mass of the liquid displaced by the cylinder.
• Density of the liquid

• Exact diameter reading – 0.11

• Determine the extension of the system 6N purchase 2cm extension 50N = 2 × 50 = 16.667 cm                6 Total extension = 16.667 × 2 + 16.667 / 2 33.33 +8.33 41.66 cm
• Take specific heat capacity of water = 4200Jkg -1 K -1 m hg = m CΔθ 24.5 × 10= 4200 J/kgK × (θ − 18.9) 24.5 = θ − 18.9 420 θ= 18.9583 0 c
• The high temperature of the needle lowers the surface tension of the water around it.
• High surface tension on the sided pits the powder away

• Volume of one drop = 26.00 – 25.2                                         50 = 0.8 = 0.016cm 3     50
• Thickness of oil molecule = vol. of drop                                        Area of drop =          0.016            3.142 x 3.5 x 3.5 = 4.15 x 10 -4 cm
• – The patch is even - Oil drop forms a monolayer
• – To show the circular patch formed by the oil drop
• Vol. of molecule = 1.5 x 10 -9 x 0.6 x 10 -9 x 0.6 x 10 -9 = 0.54 x 10 -27 = 5.4 x 10 -28  m 3 1 litre = 1000cm 3 = 1.0 x 10 -3 m 3 No. of molecules in 1litre = 1.0 x 10 -3                                         5.4 x 10 -28 = 0.18515 x 10 25 = 1.8519 x 10 24 molecules
• Actual reading = 3.21 + 0.06; = 3.27cm;
• Reading = actual reading = zero error = 2.32 mm + -0.01 = 2.31 mm (1mk)

• Reading, mass = 2.75Kg ✓ 1 Density = mass / volume = 2.75Kg /3 x 10 -4 m 3 ✓ 1 = 9.167 x 10 3  Kg/m 3 ✓1 = 9167 Kg/m 3
• Main scale reading = 3.1cm = 3.1cm Vernier scale reading = (4x 0.01) = 0.04cm Diameter f the marble = 3.13 x 10 -2 m
• Volume of the marble = 4 / 3 πr 3 = 4 x 3.14 x 1.565 x 1.565 x 1.565 x 10 -6 = 10.0476 x 10 -6 m 3 Mass of the marble = 2.0 x 10 -3 kg Density of the disc = mass / volume = 2.0 x 10 -3    16.0476 x 10 -6 = 0.1246 x 10 3 = 12.46 x 10 2 Kg/m 3
• ½ X 3 ✓ = 1.5cm ✓ 1
• Volume of drop = 4 / 3 r 3 V = 4 / 3 × 22 / 7 × ( 1.36 / 1000 ) 3 ✓ 1 = 1.054 × 10 8 m 3
• 4 / 3 r 3 = R 2 t ✓1 t= 4 / 3 ×(1.36 × 10 -3 ) 3 (4.0 × 10 -1 ) 2 ✓1 t = 4 / 3 × 1.36 3 × 10 -9               4.0 2 × 10 -2 = 0.2096 × 10 -7 = 2.096 × 10 -8  m ✓
• Lycopodium powder makes the film outline clearly visible  1 1 Mk
•  - The film/ patch is a perfect circle - The film is a monolayer - There is no space between the molecules An
•  Zero error + 0.04 Reading diameter = 0.93 – 0.04
• Main scale = 5.5mm Head scale coincidence = 23 / 100 mm Reading = 5.50              - 0.23                5.73mm Actual reading = 5.73mm – 0.01mm = 5.72mm

• Main scale =             6.50 mm Thimble scale =          0.34 mm Micrometer reading = 6.84 mm

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