I am currently doing the Millikan oil drop experiment. Where we use an oil atomizer and spray oil between two capacitors of opposite polarity. I have two questions to ask.

  1. We have been asked to take a set of measurements and recorded the data and then reverse the polarity. Is the reason for reversing the polarity because once I have worked out the actually value of the charge of a particle from my data, it will show that for the opposite charge that the value is exactly the same?

  2. We have been given the following formulas and I quote:

    The Stokes law must be modified for very small objects, i.e. $r<1$ μm, such as the oil droplets. For such small objects the friction force is not given by $F=γv$ but by $F=γv(1-a/r)$, where $a=0.07776μm$. This means that the correct radius $r_c$ is related to the measured radius $r$ by $$r_c = r (1 - a/r) ^{1/2}\tag{1}$$ and the correct charge $q_c$ is related to the measured charge q by $$q_c = q (1 - a/r)^{3/2}\tag{2}$$

It is the chrage correction that is bothering me, why has is the radius of the oil droplet without the correction to the radius in (2) and not $r_c$ instead?


1 Answer 1


If the radius of the drop is not known then you need two sets of data as there are two unknowns: the radius of the droplet and the charge on the droplet.
The two sets of readings can be for the droplet falling with the electric field aiding the fall and then reversing the electric field so that the droplet moves slower or starts moving upwards which is what Millikan did or observing the droplet with and without an electric field being present.

As well as worrying about the correction for the radius of the droplet you must also correct for the fact that the viscosity of air is temperature dependent.
There are numerous formulae and calculators available on the Internet.

Millikan knew about this correction and one of the more recent papers relating to it was written by Allen and Raabe.
An alternative version of the correction is the correction to the Stokes' law equation of $6 \,\pi \, r \, v\, \eta \, \left ( 1+ \frac{7.9 \times 10^{-3}}{r P}\right )^{-1} $ where $r$ is the radius of a sphere in metres and $P$ is the atmospheric pressure in pascals.
If you put in standard atmospheric pressure into this formula and have the radius of the droplet $2\,\rm \mu m$ the correction is approximately $4\%$ and the correction becomes larger as the radius of the droplet decreases.

The correction arises because the uncorrected Stokes's law formula is for a continuous fluid where particles interact with their neighbours immediately whereas in a real gas the gas molecules on average travel a mean free path before interacting with their neighbours.
This means that the resistance to the motion of a droplet in a real gas is going to be less that in a continuous fluid and the effect becomes significant when the droplet size becomes comparable with the mean free path of the gas molecules $\approx 0.07\,\rm \mu m$ at one atmosphere.

To see how one might get the correction term for the charge it is easiest to start with the no electric field condition which can be solved as a quadratic equation in $r$.

$$\frac 43 \,\pi\; r^3\;\rho'\;g = 6 \,\pi\, r \, v_{\rm f} \,\eta \left ( 1+ \dfrac Kr\right )^{-1} \Rightarrow r = r_0\, \left (1 +\frac{K^2}{4r_0^2}\right )^{\frac 12} - \frac K2$$

where $K = 7.82 \times 10^{-8}\, \rm m$ at standard atmospheric pressure and $r_0 = \left (\frac{9\, \eta\, v_f}{4\, \rho' \,g}\right )^{\frac 12}$ is the radius of the droplet evaluated without the correction to Stokes' law.

This is equivalent to solving your equation (1) for $r$.

Your equation (2) was obtained saying that the correction to the Stokes' law equation is equivalent to making a correction of $\left ( 1+ \frac Kr \right )^{-1} $ to the viscosity and because the viscosity appears as $\eta^{\frac 32}$ in the final equation for the charge on the droplet the correction term for the charge must be $\left ( 1+ \frac Kr\right )^{-\frac 3 2} $.

  • $\begingroup$ May I ask how the correction factor you have displayed has a positive sign and a negative exponent, where as I have a positive exponent and an negative sign $\endgroup$
    – user157589
    Oct 15, 2017 at 17:34
  • $\begingroup$ @JasonTaylor I suspect that after a binomial expansion they are similar expressions? $\endgroup$
    – Farcher
    Oct 15, 2017 at 19:47

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