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Test 1 Due: 11:59pm on Sunday, February 13, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy [Switch to Standard Assignment View]
Problem 21.86: Operation of an Inkjet Printer In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The pattern on the paper is controlled by an electrostatic valve that determines at each nozzle position whether ink is squirted onto the paper or not. The ink drops have a radius of 17.0 and leave the nozzle and travel toward the paper at a velocity of 22.0 . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length 2.50 where there is a uniform vertical electric field with a magnitude of 8.50×104 . Part A If a drop is to be deflected a distance of 0.290
by the time it reaches the end of the deflection plate,
what magnitude of charge must be given to the drop? (Assume that the density of the ink drop is ).
ANSWER:
−13 = 1.09×10 All attempts used; correct answer displayed
Exercise 21.17 Three point charges are arranged along the x-axis. Charge -5.00
is at
= 0.200
. Charge
= -8.00
= +3.00
is at the origin, and charge
=
.
Part A Where is
ANSWER:
located if the net force on
is 7.00
in the
direction?
= -0.144 Correct
Exercise 21.23 Four identical charges
are placed at the corners of a square of side
.
Part A
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Find the magnitude total force exerted on one charge by the other three charges. Express your answer in terms of the variables Q, L and appropriate constants. ANSWER: = Correct
Exercise 21.24 Two charges, one of 2.50 other at
= 0.600
and the other of -3.50
, are placed on the x-axis, one at the origin and the
, as shown in the figure .
Part A Find the position on the ANSWER:
-axis where the net force on a small charge
would be zero.
-3.27 Correct
Exercise 21.50 A point charge is at the point
= -4.00 = 0.60
is at the point ,
= 0.60
,
= 0.80
, and a second point charge
= +6.00
= 0.
Part A Calculate the magnitude of the net electric field at the origin due to these two point charges. ANSWER:
= 132 Correct
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Part B Calculate the direction of the net electric field at the origin due to these two point charges. ANSWER:
= 167 Correct
counterclockwise from + -axis
Exercise 21.35 An electron is projected with an initial speed 1.40×106
into the uniform field between the parallel plates in
the figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates.
Part A If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field? ANSWER:
6 = 1.57×10 Correct
Exercise 21.46 Two particles having charges
= 0.700
and
= 4.50
are separated by a distance of 1.90
.
Part A At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero? ANSWER:
= 0.537 Correct
from the charge
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Test Your Understanding 22.2: Calculating Electric Flux Part A Each of the surfaces listed below is a flat plate with area vector electric field
that lies in a region of uniform
. Rank the surfaces in order of the electric flux through them, from most positive to
most negative. ANSWER:
View Correct
In each case the electric field is uniform over the area of the plate, so the flux through each plate is given by In the five cases we have
Exercise 22.17 On a humid day, an electric field of 2.00×104
is enough to produce sparks about an inch long. Suppose
that in your physics class, a van de Graaff generator (see the Figure 22.27 in the textbook) with a sphere radius of 11.0 is producing sparks 4 inches long. Part A Use Gauss's law to calculate the amount of charge stored on the surface of the sphere before you bravely discharge it with your hand.
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ANSWER:
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−8 = 9.95×10 Correct
Part B Assume all the charge is concentrated at the center of the sphere, and use Coulomb's law to calculate the electric field at the surface of the sphere. ANSWER:
4 = 7.40×10 Correct
Exercise 22.25 The electric field at a distance of 0.146 1710
from the surface of a solid insulating sphere with radius 0.352
is
.
Part A Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ANSWER:
−7 = 2.58×10 Correct
Part B Calculate the electric field inside the sphere at a distance of 0.219 ANSWER:
from the center.
= 2130 Correct
Problem 22.41 A small sphere with a mass of 3.00×10−3
and carrying a charge of 5.20×10−8
hangs from a thread near
a very large, charged insulating sheet, as shown in the figure . The charge density on the sheet is . −2.60×10−9
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Part A Find the angle of the thread. ANSWER:
= 14.6 Correct
Problem 22.35 The electric field
at one face of a parallelepiped is uniform over the entire face and is directed out of the
face. At the opposite face, the electric field
is also uniform over the entire face and is directed into that
face (the figure ). The two faces in question are inclined at 30.0 from the horizontal, while and are both horizontal; , and
has a magnitude of 2.20×104
has a magnitude of 7.40×104
.
Part A Assuming that no other electric field lines cross the surfaces of the parallelepiped, determine the net charge contained within. ANSWER:
−10 = −6.91×10 Correct
Part B Is the electric field produced only by the charges within the parallelepiped, or is the field also due to charges outside the parallelepiped? How can you tell? Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: My Answer:
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Exercise 23.10 Four electrons are located at the corners of a square 10.0
on a side, with an alpha particle at its
midpoint. Part A How much work is needed to move the alpha particle to the midpoint of one of the sides of the square? ANSWER:
−21 = −6.08×10 Correct
Problem 23.71Self-Energy of a Sphere of Charge A solid sphere of radius
contains a total charge
distributed uniformly throughout its volume.
Part A Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge in a sphere of radius , how much energy would it take to add a spherical shell of thickness having charge ? Then integrate to get the total energy.) Express your answer in terms of the given quantities and appropriate constants. ANSWER: = Correct
Problem 23.72 Part A Positive electric charge
is distributed uniformly throughout the volume of an insulating sphere with radius
.From the expression for electric potential
for
as a function of
and
for
, find the expression for the
both inside and outside the uniformly charged sphere. Assume that
= 0 at infinity. Express your answer in terms of the given quantities and appropriate constants. ANSWER: = Correct
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Part B Express your answer in terms of the given quantities and appropriate constants. ANSWER: = Correct
Part C Graph
as function of
from
= 0 to
=
.
ANSWER:
View All attempts used; correct answer displayed
Part D Graph
as function of
from
= 0 to
=
.
ANSWER:
View Correct
Test Your Understanding 23.5: Electric Potential Gradient The electric potential in a certain region of space is given by In this expression
and
are both positive constants.
Part A
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Which statement about the electric field at a point in the first quadrant ( > 0, > 0) of the
-plane is
correct? ANSWER:
not enough information given to decide Correct
The electric field components are
A > 0 and B > 0. If x > 0 and y > 0, then
and
.
Problem 23.90 A hollow, thin-walled insulating cylinder of radius paper) has charge
and length
(like the cardboard tube in a roll of toilet
uniformly distributed over its surface.
Part A Calculate the electric potential at any point
along the axis of the tube. Take the origin to be at the center
of the tube, and take the potential to be zero at infinity. Express your answer in terms of the given quantities and appropriate constants. ANSWER: = All attempts used; correct answer displayed
Part B
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Show that if
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, the result of part A reduces to the potential on the axis of a ring of charge of radius
. Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: My Answer:
Part C Use the result of part A to find the electric field at any point
along the axis of the tube.
Express your answer in terms of the given quantities and appropriate constants. ANSWER: = All attempts used; correct answer displayed
Exercise 23.36 A very long insulating cylinder of charge of radius 3.00
carries a uniform linear density of 13.0
.
Part A If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 195 ?
ANSWER:
= 3.90 Correct
Problem 23.89 An alpha particle with kinetic energy 14.5
makes a collision with lead nucleus, but it is not "aimed" at the
center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude , where is the magnitude of the initial momentum of the alpha particle and =1.30×10−12
. (Assume that the lead nucleus remains stationary and that it may be treated as a point
charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.) Part A What is the distance of closest approach? ANSWER:
10 of 12
−12 = 1.31×10
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All attempts used; correct answer displayed
Part B Repeat for =1.30×10−13
ANSWER:
.
−13 = 1.38×10 All attempts used; correct answer displayed
Part C Repeat for =1.40×10−14
ANSWER:
.
−14 = 2.43×10 All attempts used; correct answer displayed
Problem 23.91 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 in diameter) into the space between two parallel horizontal plates separated by a distance
. A potential difference
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. Part A Show that an oil drop of radius
at rest between the plates will remain at rest if the magnitude of its
charge is where
is the density of the oil. (Ignore the buoyant force of the air.) By adjusting
to
keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: My Answer:
Part B Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determined cutting off the electric field and measuring the terminal speed on a sphere of radius
moving with speed
. When the drop is falling at
by
of the drop as it fell. The viscous force
through a fluid with viscosity
is given by Stokes's law:
, the viscous force just balances the weight
of the drop.
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Show that the magnitude of the charge on the drop is
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 . That is, they found drops with charges of , and so on, but none with values such as 0.76 or 2.49 . A drop with charge one extra electron; if its charge is
has acquired
, it has acquired two extra electrons, and so on.
Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: My Answer:
Part C A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.30
at constant speed in 51.1
= 0. The same drop can be held at rest between two plates separated by 1.50 How many excess electrons has the drop acquired? The viscosity of air is density of the oil is 824
ANSWER:
= 6
if
= 6.87
if .
, and the
.
All attempts used; correct answer displayed
Part D What is the radius of the drop? ANSWER:
−7 = 5.07×10 Correct
Score Summary: Your score on this assignment is 80.7%. You received 80.73 out of a possible total of 100 points.
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