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Test 1Due: 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

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Problem 21.86: Operation of an Inkjet PrinterIn an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. Thepattern on the paper is controlled by an electrostatic valve that determines at each nozzle position whether inkis squirted onto the paper or not. The ink drops have a radius of 17.0 and leave the nozzle and traveltoward 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 platesof 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: = 1.09×10−13

All attempts used; correct answer displayed

Exercise 21.17Three point charges are arranged along the x-axis. Charge = +3.00 is at the origin, and charge =

-5.00 is at = 0.200 . Charge = -8.00 .

Part A

Where is located if the net force on is 7.00 in the direction?

ANSWER: = -0.144

Correct

Exercise 21.23Four identical charges are placed at the corners of a square of side .

Part A

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrintView?assignme...

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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 21.24Two charges, one of 2.50 and the other of -3.50 , are placed on the x-axis, one at the origin and the

other at = 0.600 , as shown in the figure .

Part A

Find the position on the -axis where the net force on a small charge would be zero.

ANSWER: -3.27

Correct

Exercise 21.50A point charge = -4.00 is at the point = 0.60 , = 0.80 , and a second point charge = +6.00

is at the point = 0.60 , = 0.

Part A

Calculate the magnitude of the net electric field at the origin due to these two point charges.

ANSWER: = 132Correct


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Part B

Calculate the direction of the net electric field at the origin due to these two point charges.


counterclockwise from + -axis

Exercise 21.35An 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 isuniform and directed vertically downward, and that thefield outside the plates is zero. The electron enters thefield 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 itemerges from the field?

ANSWER: = 1.57×106

Correct

Exercise 21.46Two 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 chargesequal 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 that lies in a region of uniform

electric field . 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 isgiven by

In the five cases we have

Exercise 22.17On 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 sphereradius 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 bravelydischarge it with your hand.


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ANSWER: = 9.95×10−8

Correct

Part B

Assume all the charge is concentrated at the center of the sphere, and use Coulomb's law to calculate theelectric field at the surface of the sphere.

ANSWER: = 7.40×104

Correct

Exercise 22.25The electric field at a distance of 0.146 from the surface of a solid insulating sphere with radius 0.352 is1710 .

Part A

Assuming the sphere's charge is uniformly distributed, what is the charge density inside it?

ANSWER: = 2.58×10−7

Correct

Part B

Calculate the electric field inside the sphere at a distance of 0.219 from the center.


Problem 22.41A 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: = 14.6Correct

Problem 22.35The 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 areinclined at 30.0 from the horizontal, while and

are both horizontal; has a magnitude of 2.20×104

, and 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 chargecontained within.

ANSWER: = −6.91×10−10

Correct

Part B

Is the electric field produced only by the charges within the parallelepiped, or is the field also due tocharges 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 23.10Four electrons are located at the corners of a square 10.0 on a side, with an alpha particle at itsmidpoint.

Part A

How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

ANSWER: = −6.08×10−21

Correct

Problem 23.71Self-Energy of a Sphere of ChargeA 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. Thisenergy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge ina 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 for and for , find the expression for the

electric potential as a function of both inside and outside the uniformly charged sphere. Assume that

= 0 at infinity.


ANSWER: =

Correct


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Part B


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 GradientThe 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 iscorrect?

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 23.90A hollow, thin-walled insulating cylinder of radius and length (like the cardboard tube in a roll of toilet

paper) has charge 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 centerof the tube, and take the potential to be zero at infinity.


ANSWER:

=


Part B


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Show that if , the result of part A reduces to the potential on the axis of a ring of charge of radius

.


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.


ANSWER:

=


Exercise 23.36A 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 placedso that the voltmeter reads 195 ?

ANSWER: = 3.90Correct

Problem 23.89An 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 leadnucleus) 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 pointcharge. 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: = 1.31×10−12


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Part B

Repeat for =1.30×10−13 .

ANSWER: = 1.38×10−13


Part C

Repeat for =1.40×10−14 .

ANSWER: = 2.43×10−14


Problem 23.91 The Millikan Oil-Drop ExperimentThe charge of an electron was first measured by the American physicist Robert Millikan during 1909-1913. Inhis 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 negativecharge 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 itscharge 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.


ANSWER: My Answer:

Part B

Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determined bycutting off the electric field and measuring the terminal speed of the drop as it fell. The viscous force

on a sphere of radius moving with speed through a fluid with viscosity is given by Stokes's law:. When the drop is falling at , 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 acharge 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 has acquired

one extra electron; if its charge is , it has acquired two extra electrons, and so on.


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 if = 0. The same drop can be held at rest between two plates separated by 1.50 if = 6.87 .

How many excess electrons has the drop acquired? The viscosity of air is , and the

density of the oil is 824 .

ANSWER: = 6All attempts used; correct answer displayed

Part D

What is the radius of the drop?

ANSWER: = 5.07×10−7

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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Documents

correct answer

net electric field

uniform field

total electric field

correct exercise

assignm answer

point charges

magnitude of charge