DATA 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 = 4 π 3 ρ r 3 gd V AB where ρ 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 speed υ 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 υ through a fluid with viscosity η is given by Stokes’s law: F = 6πηrυ. When a drop fell at υ t , the viscous force just balanced the drop’s weight w = mg. Show that the magnitude of the charge on the drop was q = 18 π d V AB η 3 υ t 3 2 ρg (c) You repeat the Millikan oil-drop experiment. Four of your measured values of V AB and υ t 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/m 3 . For the viscosity η of air, use the value 1.81 × 10 −4 N · 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.
DATA 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 = 4 π 3 ρ r 3 gd V AB where ρ 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 speed υ 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 υ through a fluid with viscosity η is given by Stokes’s law: F = 6πηrυ. When a drop fell at υ t , the viscous force just balanced the drop’s weight w = mg. Show that the magnitude of the charge on the drop was q = 18 π d V AB η 3 υ t 3 2 ρg (c) You repeat the Millikan oil-drop experiment. Four of your measured values of V AB and υ t 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/m 3 . For the viscosity η of air, use the value 1.81 × 10 −4 N · 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.
DATA 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 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
π
3
ρ
r
3
gd
V
AB
where ρ is oil’s 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’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 speed υt 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 υ through a fluid with viscosity η is given by Stokes’s law: F = 6πηrυ. When a drop fell at υt, the viscous force just balanced the drop’s weight w = mg. Show that the magnitude of the charge on the drop was
q
=
18
π
d
V
AB
η
3
υ
t
3
2
ρg
(c) You repeat the Millikan oil-drop experiment. Four of your measured values of VAB and υt 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 η of air, use the value 1.81 × 10−4 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.
Draw the velocity vectors starting at the black dots and the acceleration vectors including those equal to zero.
You toss a ball straight up by giving it an initial upward velocity of 18 m/s. What is the velocity of the ball 0.50 s after you released it? Define the positive y direction to be upward,
the direction that you toss the ball.
10:44 AM Fri Jan 31
O Better endurance
Limb end points travel less
D
Question 2
Take Quiz
1 pt:
Two springs are arranged in series, and the whole arrangement is pulled a vertical distance of 2
cm. If the force in Spring A is 10 N, what is the force in Spring B as a result of the
displacement?
05N
5 N
0.2 N
10 N
O2N
Question 3
1 pts
Chapter 23 Solutions
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