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Millikan's oil-drop experiment is used to measure electronic charge. Its value is $e=1.602\times10^{-19}$ Coulomb.

Millikan's oil-drop appratus consists of two parallel metallic plates separated by a distance $d$. The plates are connected to a high voltage source. The electric field between the plates is controlled by varying the applied voltage.

An atomizer (sprayer) is used to create small oil drops. Some of these drops are charged. A drop having $n$ excess electrons have a charge $q=ne$. The drops are released through a hole in the upper plate. They start falling towards the lower plate. The drops falls with a terminal velocity due to viscous force.

The motion of the drop is observed through a microscope (or telescope). The time to travel a given distance is measured with the help of graduated graticule placed in the position of the crosswires of the eyepiece. The speed of the charged drop is measured (i) under gravity alone (ii) under gravity and known electric field $E$.

In the absence of electric field, a drop falls under the influence of its weight, buoyancy and viscous drag. The weight of a spherical drop of radius $r$ and oil-density $\rho$ is \begin{align} W=mg=\frac{4}{3}\pi r^3 \rho g \end{align} The buoyant force acting on the drop dro to air of density $\sigma$ is \begin{align} F_B=\frac{4}{3}\pi r^3 \sigma g \end{align} The viscous drag on the spherical drop of radius $r$ travelling with speed $v$ in the air of viscosity $\eta$ is \begin{align} F_D=6\pi\eta r v \end{align} The drop attains terminal velocity $v_1$ when net force on it is zero i.e., $W=F_B+F_D$. Simplify to get the radius of the drop as \begin{align} r=\left[\frac{9\eta v_1}{2(\rho-\sigma)g}\right]^{1/2} \end{align}

When drop comes close to the lower plate, the electric field is switched on. The electric field for applied voltage $V$ is $E=V/d$. The drop starts moving upward due to the electric force, \begin{align} F_E=qE=\frac{neV}{d} \end{align}

The drop attains terminal velocity $v_2$ when net force on its is zero i.e., $F_E+F_B=F_D+W$ (note that drag force is downward). Simplify to get \begin{align} q=ne=\frac{9d\sqrt{2\pi \eta^{3/2}}}{g(\rho-\sigma)}\frac{(v_1+v_2)\sqrt{v_1}}{V}. \end{align}

Above formula gaives the charge $q=ne$. The charge in integer multiples of $e$.

**Problem (IIT JEE 2010):** A tiny spherical oil drop carrying net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81\pi}{7} \times{10}^{5}\;\mathrm{V/m}$. When the field is switched off, the drop is observed to fall with terminal velocity $2\times{10}^{-3}\;\mathrm{m/s}$. Given $g=9.8\;\mathrm{m/s^2}$, viscosity of the air = $1.8\times{10}^{-5}\;\mathrm{N\cdot s/m^2}$ and the density of oil = $900\;\mathrm{kg/m^3}$, the magnitude of $q$ is,

- $1.6\times{10}^{-19}\;\mathrm{C}$
- $3.2\times{10}^{-19}\;\mathrm{C}$
- $4.8\times{10}^{-19}\;\mathrm{C}$
- $8.0\times{10}^{-19}\;\mathrm{C}$

**Solution: ** The forces acting on the oil drop are its weight, buoyant force, and electrostatic force. The buoyant force on the oil drop is very small as compared to other two forces. Thus, the weight of the spherical oil drop is balanced by electrostatic force,
\begin{align}
qE=\tfrac{4}{3}\pi r^3\rho\, g.
\end{align}
The drop attains terminal velocity in absence of electric field, when viscous force is equal to its weight i.e.,
\begin{align}
6\pi\eta rv=\tfrac{4}{3}\pi r^3\rho\, g.
\end{align}
Eliminate $r$ from above equations to get the charge,
\begin{align}
q=\frac{4\pi \rho g}{3E} \left[\frac{9\eta v}{2\rho g}\right]^{3/2}.
\end{align}
Substitute given parameters in above equation to get $q=8.0\times{10}^{-19}\;\mathrm{C}$. Note that charge on oil drop is integral multiple of charge on an electron. The readers are encouraged to find numerical value of buoyant force and compare it with values of other two forces.

**Problem (JEE Mains 2021):**
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0\times{10}^{-5}$ metre and density $1.2\times{10}^{3}$ kg/m^{3}? (Take viscosity of liquid $=1.8\times{10}^{-5}$ Ns/m^{2}, neglect buoyancy due to air.)

- $5.8\times10^{-10}$ N
- $1.8\times10^{-10}$ N
- $3.8\times10^{-11}$ N
- $3.9\times10^{-10}$ N