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The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during $1909-1913 .$ In his experiment, oil is sprayed in very fine drops (around $10^{-4} \mathrm{mm}$ in diameter) into the space between two parallel horizontal plates separated by a distance $d . \mathrm{A}$ potential difference $V_{A B}$ is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by $x$ rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius $r$ at rest between the plates will remain at rest if the magnitude of its charge is $$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where $\rho$ is the density of the oil. (Ignore the buoyant force of the air.) By adjusting $V_{A B}$ to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined $r$ by cutting off the electric field and measuring the terminal speed $v_{t}$ of the drop as it fell. (We discussed the concept of terminal speed in Section $5.3 . )$ The viscous force $F$ on a sphere of radius $r$ moving with speed $v$ through a fluid with viscosity $\eta$ is given by Stokes's law: $F=6 \pi \eta r v .$ When the drop is falling at $v_{1}$ , the viscous force just balances the weight $w=m g$ of the drop. Show that the magnitude of the charge on the drop is $$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$ Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge $e$ . That is, they found drops with charges of $\pm 2 e, \quad \pm 5 e,$ and so on, but none with values such as 0.76$e$ or 2.49$e .$ A drop with charge $-e$ has acquired one extra electron; if its charge is $-2 e,$ it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if $V_{A B}=0 .$ The same drop can be held at rest between two plates separated by 1.00 $\mathrm{mm}$ if $V_{A B}=9.16 \mathrm{V} .$ How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is $1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},$ and the density of the oil is 824 $\mathrm{kg} / \mathrm{m}^{3} .$

          The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during $1909-1913 .$ In his experiment, oil is sprayed in very fine drops (around $10^{-4} \mathrm{mm}$ in diameter) into the space between two parallel horizontal plates separated by a distance $d . \mathrm{A}$ potential difference $V_{A B}$ is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by $x$ rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius $r$ at rest between the plates will remain at rest if the magnitude of its charge is
$$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where $\rho$ is the density of the oil. (Ignore the buoyant force of the air.) By adjusting $V_{A B}$ to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined $r$ by cutting off the electric field and measuring the terminal speed $v_{t}$ of the drop as it fell. (We discussed the concept of terminal speed in Section $5.3 . )$ The viscous force $F$ on a sphere of radius $r$ moving with speed $v$ through a fluid with viscosity $\eta$ is given by Stokes's law: $F=6 \pi \eta r v .$ When the drop is falling at $v_{1}$ , the viscous force just balances the weight $w=m g$ of the drop. Show that the magnitude of the charge on the drop is
$$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$
Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge $e$ . That is, they found drops with charges of $\pm 2 e, \quad \pm 5 e,$ and so on, but none with values such as 0.76$e$ or 2.49$e .$ A drop with charge $-e$ has acquired one extra electron; if its charge is $-2 e,$ it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if $V_{A B}=0 .$ The same drop can be held at rest between two plates separated by 1.00 $\mathrm{mm}$ if $V_{A B}=9.16 \mathrm{V} .$ How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is $1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},$ and the density of the oil is 824 $\mathrm{kg} / \mathrm{m}^{3} .$

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