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Millikan's Oil Drop Experiment

 
 

General Idea

Millikan sprayed a lot of oil drops from a perfume bottle, and then watched them fall. A few of them would fall through a pinhole, then Millikan focused on one at a time. Once the drop hit terminal velocity (it's small because it's a very light oil drop), he was able to do some math and determine the apparent weight. Next, he turned on an electric field. Because the particles had a very slightly negative charge, the field would push the particles up and counteract the force of gravity. Once those two forces were equal, he was be able to get the charge of the drop. After conducting over 1,000 trials of this experiment, Millikan found the greatest common factor of all of the drop charges. Because every single charge is a multiple of the smallest charge, the charge of an electron, the greatest common factor is the charge of an electron.

Recall that before, J.J. Thomson calculated the charge to mass ratio of an electron, so with Millikan's experiment, scientists were able to calculate that mass of an electron.

 

 

Derivation

MathJax TeX Test Page Before we do this, we must keep in mind the big picture. Our goal is to find an equation for the charge of the oil drop. When the drop is falling, the viscosity of the air pushes up on the drop. The drag is determined by Stoke's Law: $$F_d = 6\pi\eta{}rv_T$$ $$\eta \text{ is the viscosity of the air, an experimentally determined constant.}$$ $$r \text{ is the radius}$$ $$v \text{ is the velocity of the particle}$$

By Archimedes' Principle (what he used in his famous Eureka story), the apparent weight = weight of the object - weight of the displaced fluid. Note that $m = \rho\cdot V$ $$w_{apparent} = \rho Vg - \rho_{air}V_{drop}g$$ Note that the volumes are equal, so this equals $$= (\rho - \rho_{air})Vg$$ Because the drop is spherical, the volume of a sphere is $\dfrac{4}{3}\pi r^3$. When the drop hits terminal velocity, $F_g = F_B + F_D \rightarrow F_g - F_B = F_D$. Remember the the apparent also equals $F_g - F_B$, so $W_{apparent} = F_D$. $$\dfrac{4}{3}\pi r^3(\rho - \rho_{air})g = 6\pi\eta rv_T$$ $$\dfrac{\frac{4}{18}r^2(\rho - \rho_{air})g}{\eta} = V_T$$ We want to isolate the radius, because that will vary from drop to drop, and that is difficult to calculate. $$r^2 = \dfrac{\frac{18}{4}\eta v_T}{g(\rho - \rho_{air})} \rightarrow r = \sqrt{\dfrac{18\eta v_T}{4g(\rho - \rho_{air})}}$$ Once the electric field is turned on, the drop becomes suspended in mid-air, and the force of gravity = the upward force of electricity + (the much smaller) buoyant force. $$F_g - F_B = \frac{Voltage}{d}q \rightarrow q = \frac{d}{V}\frac{4}{3}\pi r^3 (\rho - \rho_{air})g$$ $$q = \frac{d}{V}\frac{4}{3}\pi \sqrt{\dfrac{18^3\eta v_T^3}{4^3g^3(\rho - \rho_{air})^3}} (\rho - \rho_{air})g$$ $$q = 18\pi \frac{d}{V}\sqrt{\dfrac{\eta^3v_t^3}{2(\rho - \rho_{air})}}$$ We can go through this equation. We know $d$ because we can measure it. We know $V$ because we built the machine. We know $\eta$ because that has been experimentally determined. We know $\rho$ and $\rho_{air}$ as well because of that. We must measure $v_t$, and we get charge. Afterward, we get the charge of thousands of droplets and we find the greatest common factor of all of the differences.
David Witten

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