Charge on an Electron by Millikan's Oil Drop Experiment

Problems from IIT JEE

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. $1.6\times{10}^{-19}\;\mathrm{C}$
  2. $3.2\times{10}^{-19}\;\mathrm{C}$
  3. $4.8\times{10}^{-19}\;\mathrm{C}$
  4. $8.0\times{10}^{-19}\;\mathrm{C}$

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.