Question

$\textbf{The Millikan Oil-Drop Experiment.}$ The charge of an electron was first measured by the American physicist Robert Millikan during 1909-1913. In his experiment, oil 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 $V_{AB}$ 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 $$q = \frac{4\pi}{3} \frac{\rho{r^3gd}}{V_{AB}}$$ where $\rho$ is oil's density. (Ignore the buoyant force of the air.) By adjusting $V_{AB}$ to keep a given drop at rest, Millikan determined the charge on that drop, provided its radius $r$ was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determined $r$ by cutting off the electric field and measuring the $terminal \space speed \space v_t$ 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 $\eta$ is given by Stokes's law: $F = 6\pi \eta rv$. When a drop fell at $v_t$, the viscous force just balanced the drop’s weight $w =$ mg. Show that the magnitude of the charge on the drop was 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 $\eta$ of air, use the value 1.81 $\times$ 10$^{-5}$ N $\cdot$ s/m$^2$. Assume that $g =$ 9.80 m/s$^2$. 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 $e$.

Name: Problem 87, Chapter 23: University Physics with Modern Physics
Uploaded: 2020-11-07T19:47:16-08:00
Duration: 15 min 52 s
Channel: Vidhi Bhatt

   $\textbf{The Millikan Oil-Drop Experiment.}$ The charge of an electron was first measured by the American physicist Robert Millikan during 1909-1913. In his experiment, oil 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 $V_{AB}$ 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 $$q = \frac{4\pi}{3} \frac{\rho{r^3gd}}{V_{AB}}$$ where $\rho$ is oil's density. (Ignore the buoyant force of the air.) By
adjusting $V_{AB}$ to keep a given drop at rest, Millikan determined the charge on that drop, provided its radius $r$ was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determined $r$ by cutting off the electric field and measuring the $terminal \space speed \space v_t$ 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 $\eta$ is given by Stokes's law: $F = 6\pi \eta rv$. When a drop fell at $v_t$, the viscous force just balanced the drop’s weight $w =$ mg. Show that the magnitude of the charge on the drop was




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 $\eta$ of air, use the value 1.81 $\times$ 10$^{-5}$ N $\cdot$ s/m$^2$. Assume that $g =$ 9.80 m/s$^2$. 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 $e$.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 23, Problem 87

Instant Answer

Solved by Expert Vidhi Bhatt

11/07/2020

Step 1

The forces acting on the drop are its weight and the electric force due to the electric field between the plates. When the drop is at rest, these forces balance each other out. Therefore, we have: \[F_{gravity} = F_{electric}\] \[mg = qE\] where $m$ is the mass Show more…

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$\textbf{The Millikan Oil-Drop Experiment.}$ The charge of an electron was first measured by the American physicist Robert Millikan during 1909-1913. In his experiment, oil 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 $V_{AB}$ 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 $$q = \frac{4\pi}{3} \frac{\rho{r^3gd}}{V_{AB}}$$ where $\rho$ is oil's density. (Ignore the buoyant force of the air.) By adjusting $V_{AB}$ to keep a given drop at rest, Millikan determined the charge on that drop, provided its radius $r$ was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determined $r$ by cutting off the electric field and measuring the $terminal \space speed \space v_t$ 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 $\eta$ is given by Stokes's law: $F = 6\pi \eta rv$. When a drop fell at $v_t$, the viscous force just balanced the drop’s weight $w =$ mg. Show that the magnitude of the charge on the drop was 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 $\eta$ of air, use the value 1.81 $\times$ 10$^{-5}$ N $\cdot$ s/m$^2$. Assume that $g =$ 9.80 m/s$^2$. 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 $e$.

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00:01 Solve for the charge of the oil drop in the millikan oil drop experiment.

00:07 According to newton's second law, we know that f is equal to m .a.

00:22 Is equal to, we can write mg also.

00:27 Is equal to we can write row vg because m is equal to row b.

00:35 But instead of volume, we can write 4 by 0 .2.

00:40 3 pi r cube g so finally we are having the formula for the f is equivalent to 4 by 3 pi r cube g row in order to evaluate the electric field that affects the oil drop we use the following relation so we can write e is equal to f e upon q is equal to vab upon d into q is equal to vab upon d into q is equal to 4 by 3 pi r cube g row upon q now rearrange and solve for the charge of the oil draw so we can right, q is equal to, from the above equation, 4 pi by 3, row r -cube gd upon bab.

02:45 Now, to solve for the charge of the oil drop in the millican oil drop experiment, as per the problem mentioned, stock's law assigns that f is equal to, 6 pi eta rv.

03:17 Substituting from the previous calculation we can get f is equal to 6 pi eta rv equivalent to 4 by 3 pi r cube 0.

03:55 Now rearrange and solve for the radius of the draw.

04:02 So we can write from this formula r square is equal to 9 by 2 eta v upon 0.

04:24 As we have equated these two equations and so we will have r is equal to square root of 9x2, eta v upon g, row.

05:03 From part 1, the charge of the oil drop is given by q is equal to 4 pi by 3, row r cubed, g, d upon vab.

05:14 Now, substituting from the previous calculation, then we get q is equal to 4 by 3 pi into row instead of.

05:37 R -cube we can write square root of 9x2 eta v upon g row cube into g d upon vab by b by simplifying this we will get 18 pi d upon vab square root of eta cube v cube v cube upon 2g row.

06:34 As the problem mentioned, the separation between the horizontal plates is 1 mm.

06:42 The density of the oil drop is 824 kilogram per meter cube.

06:48 The viscosity of the air is 1 .81 into 10 to power minus 5 newton into second per meter square and g is 9 .8 meter per second square.

07:00 So from part b, the charge of the oil drop is given by this equation.

07:09 Now, solve for the first drop.

07:13 So for the first drop, we can say that charge q1 is equivalent to 18 pi.

07:29 D is 1 into 10 to power minus 3 meter upon 9 .5.

07:43 16 volt and in the square root of 1 .81 into 10 to power minus 5 cube into 2 .54 into 10 to power minus 5 cube upon 2 into 9 .8 meter per second square into 8 .2 .252 .2 .2 .2 .2 .2 .2 .2 meter per second square into 8.

08:36 Kilogram per meter cube.

08:41 So the answer is 4 .789 into 10 to power minus 19 column, which is rq1.

09:06 The same way we can calculate q2 also using the same formula...

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