Mastering Physics

Assignment 2Due: 11:59pm on Sunday, September 7, 2014

To understand how points are awarded, read the Grading Policy for this assignment.

Electric Field due to Two Point Charges

Two point charges are placed on the x axis.The first charge, = 8.00 , is placed a distance 16.0 from the origin

along the positive x axis; the second charge, = 6.00 , isplaced a distance 9.00 from the origin along the negative xaxis.

Part A

Find the electric field at the origin, point O.

Give the x and y components of the electric field as an ordered pair. Express your answer in newtonsper coulomb to three significant figures. Keep in mind that an x component that points to the right ispositive and a y component that points upward is positive.

Hint 1. How to approach the problem

Find the contributions to the electric field at the origin separately for and , then add them together(using vector addition) to find the total electric field at the origin.

Hint 2. Determine the directions of the electric fields

Which of the following describes the directions for the electric fields and created by charges and , respectively?

ANSWER:

q1 nC mq2 nC

m

q1 q2

E⃗ O1 E⃗ O2q1 q2

Hint 3. Calculate the electric field due to

What is the magnitude of the electric field at the origin due to charge only?

Express your answer in newtons per coulomb to three significant figures.

ANSWER:

Hint 4. Calculate the electric field due to

What is the magnitude of the electric field at the origin due to charge only?

Express your answer in newtons per coulomb to three significant figures.

ANSWER:

ANSWER:

Correct

Part B

Now, assume that charge is negative; . What is the net electric field at the origin, point O?

Give the x and y components of the electric field as an ordered pair. Express your answer in newtonsper coulomb to three significant figures. Keep in mind that an x component that points to the right ispositive and a y component that points upward is positive.

Both and point to the right.

Both and point to the left.

points to the right and points to the left.

points to the left and points to the right.

E⃗ O1 E⃗ O2

E⃗ O1 E⃗ O2

E⃗ O1 E⃗ O2

E⃗ O1 E⃗ O2

q1

q1

= 0.281 EO1 N/C

q2

q2

= 0.666 EO2 N/C

= 0.385,0 ,EOx EOy N/C

q2 = −6 nCq2

Hint 1. How to approach the problem

Find the contributions to the electric field at the origin separately for and , then add them together(using vector addition) to find the total electric field at the origin.

Hint 2. Determine the directions of the electric fields

Which of the following describes the directions for the electric fields and created by charges and , respectively?

ANSWER:

ANSWER:

Correct

Problem 20.26

q1 q2

E⃗ O1 E⃗ O2q1 q2

Both and point to the right.

Both and point to the left.

points to the right whereas points to the left.

points to the left whereas points to the right.

E⃗ O1 E⃗ O2

E⃗ O1 E⃗ O2

E⃗ O1 E⃗ O2

E⃗ O1 E⃗ O2

= -0.947,0 ,EOx EOy N/C

Part A

What is the strength of an electric field that will balance the weight of a 1.0 plastic sphere that has beencharged to 3.0 ?

Express your answer using two significant figures.

ANSWER:

Correct

Part B

What is the direction of an electric field that will balance the weight of a 1.0 plastic sphere that has beencharged to 3.0 ?

ANSWER:

Correct

Problem 20.27

Part A

What is the strength of an electric field that will balance the weight of a proton?

ANSWER:

Correct

Part B

What is the direction of an electric field that will balance the weight of a proton?

ANSWER:

g− nC

= 3.3×106 E N/C

g− nC

upward

downward

= 1.02×10−7 E N/C

Correct

Part C

What is the strength of an electric field that will balance the weight of an electron?

ANSWER:

Correct

Part D

What is the direction of an electric field that will balance the weight of an electron?

ANSWER:

Correct

Problem 20.29

A parallel-plate capacitor is constructed of two square plates, size , separated by distance . The plates aregiven charge .

Part A

What is the ratio of the final electric field strength to the initial electric field strength if isdoubled?

ANSWER:

upward

downward

= 5.58×10−11 E N/C

upward

downward

L × L d±Q

/Ef Ei Ef Ei Q

Correct

Part B

What is the ratio of the final electric field strength to the initial electric field strength if isdoubled?

ANSWER:

Correct

Part C

What is the ratio of the final electric field strength to the initial electric field strength if is doubled?

ANSWER:

Correct

Problem 20.30

A parallel-plate capacitor is formed from two 1.0 1.0 electrodes spaced 2.3 apart. The electric fieldstrength inside the capacitor is .

Part A

What is the charge (in ) on positive electrode?


ANSWER:

= 2EfEi

/Ef Ei Ef Ei L

= 0.250EfEi

/Ef Ei Ef Ei d

= 1EfEi

cm × cm mm1.0 × N/C106

nC

= 0.89 Q nC

Correct

Part B

What is the charge (in ) on negative electrode?


ANSWER:

Correct

The Electric Field inside a Conductor

Learning Goal:

To understand how the charges within a conductor respond to an externally applied electric field.

To illustrate the behavior of charge inside conductors, consider a long conducting rod that is suspended byinsulating strings (see the figure). Assume that the rod is initially electrically neutral, and that it remains so for thisdiscussion. The rod is positioned along the x axis, and an external electric field that points in the positive x direction(to the right) can be applied to the rod and the surrounding region. The atoms in the rod are composed of positivenuclei (indicated by plus signs) and negative electrons (indicated by minus signs). Before application of the electricfield, these atoms were distributed evenly throughout the rod.

Part A

What is the force felt by the electrons and the nuclei in the rod when the external field described in the problemintroduction is applied? (Ignore internal fields in the rod for the moment.)

nC

= -0.89 Q nC

Hint 1. Formula for the force on a charge in an electric field

The force on a charge in an electric field is given by

.

ANSWER:

Correct

Part B

What is the motion of the negative electrons and positive atomic nuclei caused by the external field?

Hint 1. How to approch this part

Newton's 2nd Law tells you that an object at rest will move in the direction of the force applied on it.

Hint 2. Masses and charges of nuclei and electrons

A nucleus contains as many protons as the atom does electrons. So if the atom has N electrons, thenucleus will contain N protons. This means that the force on the nucleus will be N times as much as thaton the electron. This N is of the order of 10-100.However, the mass of a nucleus is roughly 2N times the mass of a proton, since it contains both protonsand neutrons. Each proton itself weighs about 1836 times as much as an electron! So a typical nucleusreally does weigh a lot more than an electron.

Given this information, how would the distance moved by a nucleus compare with that moved by anelectron?

ANSWER:

F ⃗ q E⃗

= qF ⃗ E⃗

Both electrons and nuclei experience a force to the right.

The nuclei experience a force to the right and the electrons experience a force to the left.

The electrons experience a force to the left but the nuclei experience no force.

The electrons experience no force but the nuclei experience a force to the right.

Both electrons and nuclei move to the right.

The nuclei move to the right and the electrons move to the left through equal distances.

The electrons move to the left and the nuclei are almost stationary.

The electrons are almost stationary and the nuclei move to the right.

Correct

The nuclei of the atoms of a conducting solid remain almost in their places in the crystal lattice, while theelectrons relatively move a lot. In an insulator, the electrons are constrained to stay with their atoms (ormolecules), and at most, the charge distribution is displaced slightly.

The motion of the electrons due to the external electric field constitutes an electric current. Since thenegatively charged electrons are moving to the left, the current, which is defined as the "flow" of positivecharge, moves to the right.

Part C

Imagine that the rightward current flows in the rod for a short time. As a result, what will the net charge on theright and left ends of the rod become?

Hint 1. How to approach this part

Remember that the rod as a whole must remain electrically neutral even if the charges are redistributed.This is because applying an electric field does not change the charge on the rod, only redistributes it.

ANSWER:

Correct

Given that the positively charged nuclei do not move, why does the right end of the rod become positivelycharged? The reason is that some electrons have moved to the left end, leaving an excess of stationarynuclei at the right end.

Part D

The charge imbalance that results from this movement of charge will generate an additional electric field in theregion within the rod. In what direction will this field point?

Hint 1. Direction of the electric field

The electric field point away from positive charges and towards negative ones.

ANSWER:

left end negative and right end positive

left end negative and right end negative

left end negative and right end nearly neutral

left end nearly neutral and right end positive

both ends nearly neutral

Correct

An electric field that exists in an isolated conductor will cause a current flow. This flow sets up an electricfield that opposes the original electric field, halting the motion of the charges on a nanosecond time scalefor meter-sized conductors. For this reason, an isolated conductor will have no static electric field inside it,and will have a reduced electric field near it. This conclusion does not apply to a conductor whose endsare connected to an external circuit. In a circuit, a rod (or wire) can conduct current indefinitely.

Problem 21.1

Moving a charge from point A, where the potential is 310 , to point B, where the potential is 200 , takes5.0×10−4 of work.

Part A

What is the value of the charge?


ANSWER:

Correct

Problem 21.9

An electron with an initial speed of 4.00×105 is brought to rest by an electric field.

Part A

Did the electron move into a region of higher potential or lower potential?

ANSWER:

It will point to the right and enhance the initial applied field.

It will point to the left and oppose the initial applied field.

V VJ

= −4.5×10−6 q C

m/s

region of higher potential

region of lower potential

Correct

Part B

What was the potential difference that stopped the electron?

ANSWER:

Correct

Part C

What was the initial kinetic energy of the electron, in electron volts?

ANSWER:

Correct

Problem 21.10

A proton with an initial speed of 800000 is brought to rest by an electric field.

Part A

Did the proton move into a region of higher potential or lower potential?

ANSWER:

Correct

Part B

What was the potential difference that stopped the proton?

= -0.456 ∆V V

= 0.456 Ki eV

m/s

region of higher potential

region of lower potential

ANSWER:

Correct

Part C

What was the initial kinetic energy of the proton, in electron volts?

ANSWER:

Correct

Electric Fields and Equipotential Surfaces

The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 increments.

Part A

What is the work done by the electric force to move a 1 charge from A to B?

Express your answer in joules.

Hint 1. Find the potential difference between A and B

What is the potential difference between point A and point B?

= 3340 ∆V V

= 3340 Ki eV

V

WAB C

−VA VB

Express your answer in volts.

Hint 1. Equipotential surfaces

Recall that an equipotential surface is a surface on which the electric potential is the same atevery point.

ANSWER:

Hint 2. Potential difference and work

Recall that the potential difference (in volts) between a point and a point equals the work (in joules)done by the electric force to move a 1 charge from to .

ANSWER:

Correct

Part B

What is the work done by the electric force to move a 1 charge from A to D?

Express your answer in joules.

Hint 1. Find the potential difference between A and D

What is the potential difference between point A and point D?

Express your answer in volts.

ANSWER:

Hint 2. Potential difference and work

Recall that the electric potential energy difference between any two points is equal to the negative of thework done by the electric force as a charged object moves between those two points. If we combine thiswith the relationship between electric potential energy and electric potential we have:

= 0 −VA VB V

a bC a b

= 0 WAB J

WAD C

−VD VA

= -1 −VD VA V

= −∆ QWAD, by electric force VAD

ANSWER:

Correct

Part C

The magnitude of the electric field at point C is

Hint 1. Electric field and equipotential surfaces

Since the diagram shows equal potential differences between adjacent surfaces, equal amounts of workare done to move a particular charge from one surface to the next adjacent one. It follows then that if theequipotentials are closer together, the electric force does the same amount of work in a smallerdisplacement than if the equipotentials were farther apart. Therefore, the electric force, as well as thecorresponding electric field, has a larger magnitude.

ANSWER:

Correct

Problem 21.12

A 1.0 1.0 parallel-plate capacitor has a 3.0 spacing. The electric field strength inside the capacitor is1.2×105 .

Part A

What is the potential difference across the capacitor?


ANSWER:

= 1 WAD J

greater than the magnitude of the electric field at point B.

less than the magnitude of the electric field at point B.

equal to the magnitude of the electric field at point B.

unknown because the value of the electric potential at point C is unknown.

cm × cm mmV/m

= 360 ∆V V

Correct

Part B

How much charge is on each plate?

Enter your answers numerically separated by a comma. Express your answer using two significantfigures.

ANSWER:

Answer Requested

Problem 21.14

Part A

In the figure, which capacitor plate, left or right, is the positive plate?

ANSWER:

Correct

= 1.1×10−10,−1.1×10−10 ,q1 q2 C

left

right

Part B

What is the electric field strength inside the capacitor?


ANSWER:

Correct

Part C

What is the potential energy of a proton at the midpoint of the capacitor?


ANSWER:

Correct

Problem 21.16

Part A

What is the electric potential at points , , and in the figure?

Enter your answers numerically separated bycommas. Express your answer using two significantfigures.

ANSWER:

= 1.0×105 E V/m

= 2.4×10−17 U J

A B C

Correct

Part B

What is the potential energy of an electron at point in the figure?


ANSWER:

Correct

Part C



ANSWER:

Correct

Part D



ANSWER:

Correct

Part E

= 1800,1800,900 , ,VA VB VC V

A

= −2.9×10−16 UA J

B

= −2.9×10−16 UB J

C

= −1.4×10−16 UC J

∆VAB

What is the potential difference ?


ANSWER:

Correct

Part F

What is the potential difference ?


ANSWER:

Correct

Problem 21.17

A 3.0 -diameter ball bearing has 3.0×109 excess electrons.

Part A

What is the ball bearing's potential?


ANSWER:

Correct

Problem 21.19

Part A

∆VAB

= 0 −VB VA V

∆VBC

= -900 −VC VB V

mm

= -2900 V V

What is the potential difference between the terminals of an ordinary AA or AAA battery? (If you’re not sure, findone and look at the label.)

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Correct

Part B

An AA battery is connected to a parallel-plate capacitor having 6.00- -diameter plates spaced 4 apart.How much charge does the battery move from one plate to the other?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Answer Requested

Problem 21.20

Part A

In the figure, which point, or , has a higher electric potential?

ANSWER:

= 1.5 V V

cm mm

= −9.4×10−12 Q C

A B

Correct

Part B

What is the potential difference between and ?


ANSWER:

Correct

Problem 21.22

Part A

What is the potential difference between = 20 and = 40 in the uniform electric field = 1500?


ANSWER:

Correct

Score Summary:Your score on this assignment is 88.8%.You received 8.87 out of a possible total of 10 points.

point

point

A

B

=VA VB

A B

= 70 −VA VB V

xi cm xf cm Ex V/m

= 300 ∆V V

Mastering Physics

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