Millikan's Oil Drop Experiment

In 1909, Robert A. Millikan and Harvey Fletcher performed an experiment to measure the eelementary electric charge ( the charge of the electron ).

The experiment consists of two metal circular plates A and B about 20 cm in diameter amd 1.5 cm apart with a small hole H in the centre of the upper plate A.

The upper plate is connected to a high potential tension battery whereas the lower plate is earthed.

With the help of Atomizer clock oil which is non volatile in nature is sprayed through hole H.

While the oil is moving through the pipe from Atomizer to hole H, it will gain few electronic charge due to the friction.

Inorder to provide enough light to view the motion of the oil drops W1 is provided and W2 is for passing x-rays to ionize the oil drop incase the oil drops arenot ionized by the friction.

Microscope with a crosswire and a micrometer scale is provided to observe and measure the motion of the oil drops.

There are two theories within the experiment.

Theory 1 : Motion of oil drop under gravity alone.

At first the electric field is not applied. In this case the oil drop falls under gravity and its velocity goes on increasing. At some stage when the viscous force on the oil drop becomes equal to its resultant weight then the oil drop moves with terminal velocity ( constant velocity ) v1.

Let r = radius of oil drop

     m = mass of the oil drop

      ρ = density of the oil

      σ = density of air

then

volume of the oil drop = 4/3πr³

weight of the oil drop ( W ) = 4/3πr³ρg

upthrust to the air,  U = weight of the air displaced by the drop

=4/3πr³σg

The viscous force on the oil drop in upward direction, F = 6πηra

When the oil drop is moving with terminal velocity a, then

F + U = W

⇒F = W - U

⇒6πηra = 4/3πr³ρg - 4/3πr³σg

∴  r = {9ηa/2(ρ-σ)g}½

Theory 2 : Motion of the oil drop under electric field

Now a strong electric field is applied between the plates in such a direction that force of negatively charged oil due to electric field acts in the vertically upward direction.

Here, the oil drop moves vertically upward and attains terminal velocity b.

Let E be the strength of electric field. As the drop carriers charge Q, then, electrostatic force on oil drop is

upward direction ( Fe ) = QE

viscous force in downward ( F ) = 6πηrb

W - U = 4/3πr(ρ-σ)g

When the oil drop attains terminal velocity b

Fe + U = F + W

⇒Fe = (W - U) + F

⇒QE = 4/3πr³(ρ - σ) + 6πηrb

∴ QE = 6πηra + 6πηrb

⇒ Q = 6πηr( a+b)/E

If the oil drop falls down with a small terminal velocity b even thoughthe electric field is applied, then the viscous force acts upward and charge on the oil drop will be

Hence through this experiment it is found that the charge of the drops is always approximately equal to integral multiple of 1.6 × 10^-19 C.

The decrease in the terminal velocity of the drop when positive potential is given to the upper plate shows that the drop has negative charge.

Thus, the charge of electron is -1.6×10^-19 C.