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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鈥�1913. In his experiment, oil was sprayed in very fine drops (about 10-4 mm in diameter) into the space between two parallel horizontal plates separated by a distance d. A potential difference VAB was maintained between the plates, causing a downward electric field between them. Some of the oil drops acquired a negative charge because of frictional effects or because of ionization of the surrounding air by x rays or radioactivity. The drops were observed through a microscope. (a) Show that an oil drop of radius r at rest between the plates remained at rest if the magnitude of its charge was q=4饾洃饾洅gd/3VAB where r is oil鈥檚 density. (Ignore the buoyant force of the air.) By adjusting VAB to keep a given drop at rest, Millikan determined the charge on that drop, provided its radius r was known. (b) Millikan鈥檚 oil 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 vt of the drop as it fell. (We discussed terminal speed in Section 5.3.) The viscous force F on a sphere of radius r moving at speed v through a fluid with viscosity h is given by Stokes鈥檚 law: F = 6phrv. When a drop fell at vt, the viscous force just balanced the drop鈥檚 weight w = mg. Show that the magnitude of the charge on the drop was

$\mathbf{q}\mathbf{=}\mathbf{18}\mathbf{蟺}\frac{\mathbf{d}}{{\mathbf{V}}_{\mathbf{AB}}}\sqrt{\frac{{\mathbf{畏}}^{\mathbf{3}}{\mathbf{v}}_{\mathbf{t}}^{\mathbf{3}}}{\mathbf{2}\mathbf{蚁驳}}}$

(c) You repeat the Millikan oil drop experiment. Four of your measured values of VAB and vt are listed in the table:

In your apparatus, the separation d between the horizontal plates is 1.00 mm. The density of the oil you use is 824 kg/m3 . For the viscosity h of air, use the value 1.81 * 10-5 N . s/m2 . Assume that g = 9.80 m/s2 . Calculate the charge q of each drop. (d) If electric charge is quantized (that is, exists in multiples of the magnitude of the charge of an electron), then the charge on each drop is -ne, where n is the number of excess electrons on each drop. (All four drops in your table have negative charge.) Drop 2 has the smallest magnitude of charge observed in the experiment, for all 300 drops on which measurements were made, so assume that its charge is due to an excess charge of one electron. Determine the number of excess electrons n for each of the other three drops. (e) Use q = -ne to calculate e from the data for each of the four drops, and average these four values to get your best experimental value of e.

Step by step solution

01

Step 1:

According to Newton鈥檚 second law

$\begin{array}{ll}\mathbf{F}\mathbf{=}\mathbf{ma}& \\ =\mathrm{mg}& \\ =\mathrm{蚁痴驳}& \\ =\mathrm{蚁}\frac{4}{3}{\mathrm{蟺谤}}^3\mathrm{g}& \\ =\frac{4}{3}{\mathrm{蟺谤}}^3\mathrm{g}& \mathrm{蚁}\\ & \end{array}$

Electric field that affects the oil drop is given by

$$\begin{gathered} E=\frac{F_e}{q} \\ =\frac{V_{AB}q}{d} \\ =\frac{F}{q} \\ =\frac{{\displaystyle \frac{4}{3}}{\mathrm{蟺谤}}^3\mathrm{驳蚁}}{\mathrm{q}} \end{gathered}$$

Solve for the charge of the oil drop

$\begin{array}{ll}q=\frac{4}{3}{\mathrm{蟺谤}}^{3}g蚁& \frac{d}{V_{AB}}\\ =\frac{4\mathrm{蟺}}{3}\frac{蚁r^{3}gd}{V_{AB}}& \end{array}$

02

Step 2:

According to Stokes鈥檚 law:

$\mathbf{F}\mathbf{=}\mathbf{6}\mathbf{蟺畏谤惫}$

Equating with$\mathrm{F}=\frac{4}{3}{\mathrm{蟺谤}}^3\mathrm{驳蚁}$

$$\begin{gathered} 6\mathrm{蟺畏谤惫}=\frac{4}{3}{\mathrm{蟺谤}}^3\mathrm{驳蚁} \\ {\mathrm{r}}^3=\frac{3脳6}{4}\frac{\mathrm{畏惫}}{\mathrm{驳蚁}} \\ =\frac{9}{2}\frac{\mathrm{畏惫}}{\mathrm{驳蚁}}\mathrm{r} \\ \mathrm{r}=\sqrt{\frac{9}{2}\frac{\mathrm{畏惫}}{\mathrm{驳蚁}}} \end{gathered}$$

03

Step 3:

Substitute r to find q

$$\begin{gathered} \mathrm{q}=\frac{4\mathrm{蟺}}{3}\frac{{\mathrm{蚁谤}}^3\mathrm{gd}}{{\mathrm{V}}_{\mathrm{AB}}} \\ =\frac{4\mathrm{蟺}}{3}\frac{\mathrm{蚁}{\left(\sqrt{{\displaystyle \frac{9}{2}\frac{\mathrm{畏惫}}{\mathrm{驳蚁}}}}\right)}^3\mathrm{d}}{{\mathrm{V}}_{\mathrm{AB}}} \\ =\frac{4\mathrm{蟺}}{3}\frac{\mathrm{蚁}{\left(\sqrt{{\displaystyle \frac{9}{2}\frac{{\mathrm{畏惫g}}^3}{\mathrm{驳蚁}}}}\right)}^3\mathrm{d}}{{\mathrm{V}}_{\mathrm{AB}}} \\ =18\mathrm{蟺}\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{AB}}}\left(\sqrt{\frac{{\mathrm{畏}}^3{\mathrm{v}}^3}{2\mathrm{驳蚁}}}\right) \end{gathered}$$

That the magnitude of the charge on the drop was $18\mathrm{蟺}\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{AB}}}\left(\sqrt{\frac{{\mathrm{畏}}^3{\mathrm{v}}^3}{2\mathrm{驳蚁}}}\right)$.

04

Step 4:

The separation

Solve for the first drop

$$\begin{gathered} q_1=18\mathrm{蟺}\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{AB}}}\sqrt{\frac{{\mathrm{畏}}^3{\mathrm{v}}^3}{2\mathrm{驳蚁}}} \\ =18\mathrm{蟺}\frac{1脳{10}^{-3}}{9.16\mathrm{V}}\sqrt{\frac{\left(1.81脳{10}^{-5}\mathrm{N}.\mathrm{s}/{\mathrm{m}}^2\right){\left(2.54脳{10}^{-5}\mathrm{m}/\mathrm{s}\right)}^3}{2脳9.8\mathrm{m}/{\mathrm{s}}^3脳824\mathrm{kg}/{\mathrm{m}}^3}} \\ =4.7839脳{10}^{-19}\mathrm{C} \end{gathered}$$

Solve for the second drop

$$\begin{gathered} q_2=18\mathrm{蟺}\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{AB}}}\sqrt{\frac{{\mathrm{畏}}^3{\mathrm{v}}^3}{2\mathrm{驳蚁}}} \\ =18\mathrm{蟺}\frac{1脳{10}^{-3}}{9.16\mathrm{V}}\sqrt{\frac{\left(1.81脳{10}^{-5}N.s/m^2\right){\left(0.767脳{10}^{-5}m/s\right)}^3}{2脳9.8\mathrm{m}/{\mathrm{s}}^3脳824\mathrm{kg}/{\mathrm{m}}^3}} \\ =1.593脳{10}^{-19}C \end{gathered}$$

Solve for the third drop

$$\begin{gathered} {\mathrm{q}}_3=18\mathrm{蟺}\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{AB}}}\sqrt{\frac{{\mathrm{畏}}^3{\mathrm{v}}^3}{2\mathrm{驳蚁}}} \\ =18\mathrm{蟺}\frac{1脳{10}^{-3}}{9.16\mathrm{V}}\sqrt{\frac{\left(1.81脳{10}^{-5}\mathrm{N}.\mathrm{s}/{\mathrm{m}}^2\right){\left(4.39脳{10}^{-5}\mathrm{m}/\mathrm{s}\right)}^3}{2脳9.8\mathrm{m}/{\mathrm{s}}^3脳824\mathrm{kg}/{\mathrm{m}}^3}} \\ =8.089脳{10}^{-19}\mathrm{C} \end{gathered}$$

Solve for the forth drop

$$\begin{gathered} q_4=18\mathrm{蟺}\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{AB}}}\sqrt{\frac{{\mathrm{畏}}^3{\mathrm{v}}^3}{2\mathrm{驳蚁}}} \\ =18\mathrm{蟺}\frac{1脳{10}^{-3}}{9.16\mathrm{V}}\sqrt{\frac{\left(1.81脳{10}^{-5}N.s/m^2\right){\left(1.25脳{10}^{-5}m/s\right)}^3}{2脳9.8\mathrm{m}/{\mathrm{s}}^3脳824\mathrm{kg}/{\mathrm{m}}^3}} \\ =3.23脳{10}^{-19}C \end{gathered}$$

Therefore, the charge of first drop is 4.789 x 10-19C

The charge of second drop is 1.593 x 10-19C

The charge of third drop is 8.089 x 10-19C

The charge of fourth drop is 3.233 x 10-19 C

05

Step 4:

Number of electrons is given by

$\mathrm{n}=\frac{{\mathrm{Q}}_{\mathrm{Total}}}{{\mathrm{q}}_{\mathrm{e}}}$

Where qeis the standard charge of electron=1.62 x 10-19C

Solve for the first drop

$\begin{array}{l}\mathrm{n}=\frac{{\mathrm{q}}_1}{{\mathrm{q}}_{\mathrm{e}}}\\ =\frac{4.789脳{10}^{-19}\mathrm{C}}{1.62脳{10}^{-19}\mathrm{C}}\\ =3\end{array}$

Solve for the second drop

localid="1664877132809" $\begin{array}{l}\mathrm{n}=\frac{{\mathrm{q}}_2}{{\mathrm{q}}_{\mathrm{e}}}\\ =\frac{15.93脳{10}^{-19}\mathrm{C}}{1.62脳{10}^{-19}\mathrm{C}}\\ =1\end{array}$

Solve for the third drop

$\begin{array}{l}\mathrm{n}=\frac{{\mathrm{q}}_3}{{\mathrm{q}}_{\mathrm{e}}}\\ =\frac{8.089脳{10}^{-19}\mathrm{C}}{1.62脳{10}^{-19}\mathrm{C}}\\ =5\end{array}$

Solve for the forth drop

$\begin{array}{l}\mathrm{n}=\frac{{\mathrm{q}}_4}{{\mathrm{q}}_{\mathrm{e}}}\\ =\frac{3.233脳{10}^{-19}\mathrm{C}}{1.62脳{10}^{-19}\mathrm{C}}\\ =2\end{array}$

The number of excess electrons in first drop is 3

The number of excess electrons in second drop is 1

The number of excess electrons in third drop is 5

The number of excess electrons in fourth drop is 2

06

Step 5:

According to the problem, the charge of the oil-drop:

$q=-ne$

Evaluate the charge of the first drop

role="math" localid="1664877646904" $$\begin{gathered} e_1=\frac{q_1}{n_1} \\ =\frac{1.593脳{10}^{-19}C}{1} \\ =1.593脳{10}^{-19}C \end{gathered}$$

Evaluate the charge of the second drop

role="math" localid="1664877693309" $$\begin{gathered} e_2=\frac{q_2}{n_2} \\ =\frac{1.593脳{10}^{-19}C}{5} \\ =1.593脳{10}^{-19}C \end{gathered}$$

e for first drop is 1.597 x 10^-19 C

e for second drop is 1.593 x 10^-19 C

e for third drop is 1.618 x 10^-19 C

e for fourth drop is 1.617 x 10^-19 C

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