# Millikan’s Oil Drop Experiment

Robert Andrew (R.A.) Millikan determined the value of the basic unit of charge $(e)$ in 1909 by the use of oil drop. This method is called Millikan’s oil drop experiment.

### Principle

The main principle of this experiment is the Stoke’s law in viscosity. In this experiment, the motion of a nonvolatile and viscous small spherical drop is observed under following two cases;
1. Motion under the effect of gravity alone.
2. Motion under the combined effect of gravity and electric field.
​Nonvolatile and viscous materials having low vapour pressure like clock oil, glycerine, etc. is used because Stoke’s law holds good for them at constant temperature.

​The Millikan’s oil drop experiment consists of a doubled wall chamber as shown in figure.

​The apparatus consists of an atomizer which sprays fine oil droplets of density $ρ$. It consists two metal plates $A$ and $B$ separated by a small distance by means of an insulator. A high p.d. of order $10,000 V$ is established between the plates by connecting the plates to a high tension battery. The upper plate $A$ has a central hole $H$ from which the droplets pass.

A source of light is allowed to throw light from the window $W_2$ to illuminate he falling drops. X-rays are allowed to enter through the window $W_1$ to ionize the air inside so that ions can stick on the droplets. A travelling microscope $(T)$ is used to study the motion of an oil drop.

### Case I: The Motion Under The Effect Of Gravity Alone

​Initially, the electric field between the plates $A$ and $B$ is not establish. Under this condition, the oil drop falls under gravity experiencing upthrust. Let the radius of the observed oil drop of density $ρ$ be $r$. Then, $\text{Volume of the oil drop}=\frac{4}{3}πr^3$ and, $\text{Weight of the oil drop (W)}=\frac{4}{3}πr^3ρg$

If $σ$ is the density of air, then, according to Archimedes’ principle, Upthrust of air on the drop $=$ Weight of air displaced $\text{Upthrust on the drop (U)}=\frac{4}{3}πr^3σg$

Due to air resistance, after some time, the falling drop will attain terminal velocity $v_1$. If $η$ is the coefficient of viscosity of air then from Stoke’s law, the viscous force $F_1$ on the drop in upward direction is, $F_1=6πηrv_1$

Under Equilibrium condition of the drop, $W=U+F_1$ $\frac{4}{3}πr^3ρg=\frac{4}{3}πr^3σg+6πηrv_1$ $6πηrv_1=\frac{4}{3}πr^3(ρ-σ)g\text{ __(1)}$ $r=\sqrt{\frac{9ηv_1}{2(ρ-σ)g}}\text{ __(2)}$

### Case II: The Motion Under The Combined Effect Of The Gravity And The Electric Field

​An electric field $E$ is applied between the plates $A$ and $B$. If $V$ is the p.d. between the plates separated by distance $d$, then, $E=\frac{V}{d}$ Under electric field, the drop moves upward with terminal velocity $v_2$.

If $q$ is the total charge on the drop, then the upward electric force is, $F_e=qE$ The viscous force $F_2$ acting downward on the drop is, $F_2=6πηrv_2$

Under equilibrium condition of the liquid drop, $W+F_2=F_e+U$ $\frac{4}{3}πr^3ρg+6πσrv_2=qE+\frac{4}{3}πr^3σg$ $qE-\frac{4}{3}πr^3(ρ-σ)g=6πηrv_2$ $qE-6πηrv_1=6πηrv_2$ $qE=6πηr(v_1+v_2)$ $q\frac{V}{d}=6πηr(v_1+v_2)$ $q=\frac{6πηr(v_1+v_2)d}{V}$ $q=\frac{6πηd(v_1+v_2)}{V}\sqrt{\frac{9ηv_1}{2(ρ-σ)g}}\text{ __(3)}$

By using this formula, the charge $q$ on the drop can be calculated. Millikan repeated the experiment for many different droplets and found that all the charges were integral multiple of a unique value of $1.6×10^{-19}$ $C$ which is the lowest charge in Nature. Hence, Millikan concluded that the charge of an electron is $-1.6×10^{-19}$ $C$.

• In Millikan’s original experiment, he allowed drops to move down with velocity $v_2$ on applying electric field. In this case, $v_1$ and $v_2$ has opposite direction and equation $(3)$ becomes, $q=\frac{6πηd(v_1-v_2)}{V}\sqrt{\frac{9ηv_1}{2(ρ-σ)g}}$
• Later on, he adjusted electric field so that the drop remained stationary i.e. $v_2=0$, then equation $(3)$ becomes, $q=\frac{6πηd(v_1)}{V}\sqrt{\frac{9ηv_1}{2(ρ-σ)g}}$