Assignment 7_ Potential.pdf

8/17/2019 Assignment 7_ Potential.pdf

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Assignment 7: Potential

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

Problem 28.2

The electric field strength is 2.20×104 inside a parallel-plate capacitor with a 1.50 spacing. An electron is

released from rest at the negative plate.

Part A

What is the electron's speed when it reaches the positive plate?

Express your answer with the appropriate units.

ANSWER:

Correct

Electric Potential Energy of Three Point Charges

Part A

Three equal point charges, each with charge 1.90 , are placed at the vertices of an equilateral triangle whosesides are of length 0.600 . What is the electric potential energy of the system? (Take as zero the potential

energy of the three charges when they are infinitely far apart.)

Use = 8.85×10−12 for the permittivity of free space.

Hint 1. How to approach the problem

Use the equation for the electric potential energy between two point charges to calculate the energy for each

interaction between two of the three point charges. The sum of these energies will be the total electric

potential energy. Be careful to avoid double counting.

Hint 2. Find the electric potential energy of one pair

Assume that one charge is interacting with a second charge, ignoring any effects from the third c harge.

What is the electric potential energy for this single interaction?

Express your answer in joules to three significant figures.

Hint 1. Electric potential energy of a pair of charges

Recall that the electric potential energy between two charges and separated by a distance

is given by the formula

N / C m m

3.40×106 m s

μ C

m U

ϵ

0

C

2

N ⋅ m

2

U

1 2

U

1

2

=

1

1

2


2/15

.

ANSWER:

Hint 3. How many interactions are there?

How many pair interactions are there for the three charges?

Hint 1. Double counting

It is important to keep in mind that a pair of charges can interact only once, so if the first charge is

interacting with the second charge for one pair, the interaction of the second charge with the first

charge cannot also be used, since the pair has already been counted.

ANSWER:

ANSWER:

Correct

The potential energy is usually written

.

This means that all pairs of charges (1-2, 1-3, and 2-3) will interact, but no charge can interact with itself (

), nor can any pair be counted twice as a result of the condition for all possible pairs. For example,

, will be counted, while , will not.

Electric Potential Ranking Task

In the figurethere are two point charges, and . There are also six posit ions, labeled A through F, at various

distances from the two point charges. You will be asked about the electric potential at the different points (A through F

U =

1

4 π ϵ

0

1

2

= 5.41×10−2 U 1 2 J

3

= 0.162U J

U = ∑


3/15

Part A

Rank the locations A to F on the basis of the electric potential at each point. Rank positive electric potentials ashigher than negative electric potentials.

Rank the locations from highest to lowest potential. To rank items as equivalent, overlap them.

Hint 1. Definition of electric potential

The electric potential surrounding a point charge is defined by

,

where is the source charge creating the electric potential and is the distance between the source charge

and the point of interest. If more than one source is present, determine the electric potential from eachsource and sum the results.

Hint 2. Conceptualizing electric potential

Because positive charges create positive electric potentials in their vicinity and negative charges create

negative potentials in their vicinity, electric potential is sometimes visualized as a sort of "elevation."

Positive charges represent mountain peaks and negative charges deep valleys. In this picture, when you

are close to a positive charge, you are "high up" and have a higher positive potential. Conversely, near a

negative charge, you are deep in a "valley" and have a negative potential. The utility of this picture

becomes clearer when we begin to think of charges moving through a region of space containing an electric

potential. Just as particles naturally roll downhill, converting gravitational potential energy into kinetic

energy, positively charged particles naturally "roll downhill" as well, toward regions of lower electricpotential, converting electrical potential energy into kinetic energy.

ANSWER:

V =


4/15

Correct

Problem 28.21

Part A

What is the electric potential at points A, B, and C in the figure? Assume that = 1.8 , = 1.2 , and

= 1.9 .Enter your answers numerically separated by

commas.

ANSWER:

n C

1

c m

2

c m

, , = 1350,1350,853V A

V

B

V

C

V

B > A > C = D > F > E


5/15

Correct

Part B

What is the potential difference ?

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

ANSWER:

Correct

Part C

What is the potential difference ?


ANSWER:

Correct

Enhanced EOC: Problem 28.26

The electric potential at the dot in the figure is 3020 .

You may want to review ( pages 828 - 830) .

For help with math skills, you may want to review:

Substituting Numbers into Mathematical Expressions

Rearrangement of Equations Involving Multiplication and

Division

Δ V

A B

= 0Δ V A B

V

Δ V

B C

= 497Δ V B C

V

V

https://session.masteringphysics.com/myct/eBookLaunch?pageID=828&bookCode=knight3


6/15

Part A

What is charge ?



What is the electrical potential due to a single point charge at a given distance from the charge? How do

you determine the electrical potential due to several point charges?

You should consider converting to coulombs and centimeters to meters for consistency of units in your

calculations.

What is the electrical potential at the dot due to the and charges? What is the electrical

potential at the dot due to the charge ?

What is the electrical potential at the dot due to all the charges in terms of ? What is the charge ?

ANSWER:

Correct

Problem 29.4

The figure is a graph of . The potential at the origin is -200

.

Part A

What is the potential at ?


n C

+ 5 n C − 5 n C

= 9.4 n C

E

V

= 3 . 0 m


7/15

ANSWER:

Correct

Potential of a Charged Disk

A disk of radius has a total charge uniformly distributed over its surface. The disk has negligible thickness and

lies in the xy plane. Throughout this problem, you may use

the variable in place of .

Part A

What is the electric potential on the z axis as a function of , for ?

Express your answer in terms of , , and . You may use instead of .


Potential is additive. Divide the disk into a series of rings with infinitesimal thickness, find an expression for

the potential of each ring, and then integrate over the disk to obtain the total potential.

Hint 2. Find the potential due to a ring

What is the potential at position due to a ring of the disk with radius and infinitesimal

thickness ?

Express your answer in terms of , , , , and .

Hint 1. The potential due to a ring

The potential at distance above the center due to an infinitesimally thin ring with radius is

= -700V ( 3 . 0 m ) V

a Q

1

4 π ϵ

0

V ( ) > 0

Q a

1

4 π ϵ

0

V ( ) ( 0 , 0 , )

Q a

V ( )


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,

where is the total charge on the ring and is the permittiv ity of free space.

Hint 2. The charge on a ring

What is the charge on a ring of the disk of radius and thickness ?

Express your answer in terms of , , , and .

Hint 1. Compute the charge density

Find , the charge per unit area of the disk.

Express your answer in terms of and .

ANSWER:

Hint 2. Compute the area of the ring

What is the intinitesimal area of a ring of the disk of radius and thickness ?

Express your answer in terms of and (to first order in only).

ANSWER:

ANSWER:

ANSWER:

Hint 3. A useful antiderivative

ANSWER:

V ( ) =

4 π ϵ

0

+

2

2

√

ϵ

0

Q a

σ

Q a

=σ Q

π a

2

A

= A 2

π

= 2

Q

a

2

= V

(

)

Q

1

2 π ϵ

0

a

2

+

2

2

√

∫ =

+

2

2

√

+

2

2

− − − − − − −

√


9/15

Correct

Part B

What is the magnitude of the electric field on the axis, as a function of , for ?

Express your answer in terms of some or all of the variables , , and . You may use instead of .

Hint 1. Direction of the electric field

In what direction does the electric field on the positive z axis point?

ANSWER:

Hint 2. Electric field from potential

You can obtain an expression for an electric field from a potential by the following relationship:

,

where is the electric field vector and is the electric potential.

ANSWER:

Correct

Since the magnitude of the electric field (and potential) must be symmetric about the plane, the general

expression for the magnitude of the electric field on the z axis for all is

.

Note the use of instead of .

=V ( )

−

2 Q

a

2

+

2

a

2

− − − − − − −

√

E > 0

Q a

1

4 π ϵ

0

̂

̂

̂

= − − − E

⃗

∂ V

∂

̂

∂ V

∂

̂

∂ V

∂

̂

E

⃗

V

=E 1 − 2

Q

a

2

+

2

a

2

√

= 0

E ( ) =

( 1 −

=

( 1 −

Q

2 π ϵ

0

a

2

| |

+

2

a

2

√

2 Q

a

2

| |

+

2

a

2

√

| |


10/15

Introduction to Capacitance

Learning Goal:

To understand the meaning of capacitance and ways of calculating capacitance

When a posit ive charge is placed on a conductor that is insulated from ground, an electric field emanates from the

conductor to ground, and the conductor will have a nonzero potential relative to ground. If more charge is placed on

the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance of

this conductor: .

Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are

essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they are

often called electrodes), and the relevant voltage is the voltage between these two electrodes.

This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a

parallel-plate capacitor, which consists of two plates each of area separated by a small distance with air or

vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in mind

that the voltage difference is the line integral of the electric field between the plates.

Part A

What property of objects is best measured by their capacitance?

ANSWER:

V

C

C = / V

A

ability to conduct electric current

ability to distort an external electrostatic field

ability to store charge


11/15

Correct

Capacitance is a measure of the ability of a system of two conductors to store electric charge and energy.

Capacitance is defined as . This ratio remains constant as long as the system retains its geometry

and the amount of dielectric does not change. Capacitors are special devices designed to combine a large

capacitance with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has

some capacitance. Even an isolated electrode has a small capacitance. That is, if a charge is placed on it,

its potential with respect to ground will change, and the ratio is its capacitance .

Part B

Assume that charge is placed on the top plate, and is placed on the bottom plate. What is the magnitude

of the electric field between the plates?

Express in terms of and other quantities given in the introduction, in addition to and any other

constants needed.

Hint 1. How do you find the magnitude of the electric field?

What is the easiest way to obtain ?

ANSWER:

Hint 2. What is the electric flux integral due to the electric field?

Apply Gauss's law to a small box whose top side is just above the lower plate and whose bottom is just

below it, where . Start by finding the electric flux integral .

Express this integral in terms of the area of the top side of the box and the magnitude of the

electric field between the plates.

ANSWER:

Hint 3. Find the enclosed charge

Find the amount of charge enclosed in a small box whose top side is just above the lower plate and

whose bottom is just below it, where .

Express the enclosed charge in terms of the cross-sectional area of the box and other quantities

given in the introduction.

C =

Q

V

Q

V

Q

V

C

− +

E

E ϵ

0

E

Use Gauss's law and the fact that outside the capacitor.

Use Gauss's law and the symmetry of the lower plate.

Use Coulomb's law integrating over all charge on the bottom plate.

Use Coulomb's law integrating over all charge on both plates.

E = 0

E = 0 Φ

E

a E

=Φ E a E

e n c l

E = 0

a


12/15

Hint 1. Find the surface charge

What is , the charge per unit area on the lower plate?

Express in terms of any necessary constants and quantities given in the introduction.

ANSWER:

ANSWER:

Hint 4. Recall Gauss's law

Gauss's law states that .

ANSWER:

Correct

Part C

What is the voltage between the plates of the capacitor?

Express in terms of the quantities given in the introduction and any required physical constants.

Hint 1. The electric field is the derivative of the potential

The voltage difference is the integral of the electric field from one plate to the other; in symbols, .

ANSWER:

Correct

σ

σ

=σ

A

= e n c l

a

A

= Φ

E

e n c l

ϵ

0

=E

A ϵ

0

V

V

V = E ⋅

=V

A ϵ

0


13/15

Part D

Now find the capacitance of the parallel-plate capacitor.

Express in terms of quantities given in the introduction and constants like .

ANSWER:

Correct

You have derived the general formula for the capacitance of a parallel-plate capacitor with plate area and

plate separation . It is worth remembering.

Part E

Consider an air-filled charged capacitor. How can its capacitance be increased?

Hint 1. What does capacitance depend on?

Capacitance depends on the inherent properties of the system of conductors, such as its geometry and the

presence of dielectric, not on the charge placed on the conductors.

ANSWER:

Correct

Part FConsider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?

ANSWER:

C

C ϵ

0

=C A ϵ

0

A

Increase the charge on the capacitor.

Decrease the charge on the capacitor.Increase the spacing between the plates of the capacitor.

Decrease the spacing between the plates of the capacitor.

Increase the length of the wires leading to the capacitor plates.


14/15

Correct

Problem 29.17

Two 5.0- -diameter aluminum electrodes are spaced 0.50 apart. The electrodes are connected to a 200

battery.

Part A

What is the capacitance?


ANSWER:

Correct

Part B

What is the magnitude of the charge on each electrode?


ANSWER:

Correct

Problem 29.7

Part A

Double the charge and double the plate area.

Double the charge and double the plate separation.

Halve the charge and double the plate separation.

Halve the charge and double the plate area.

Halve the plate separation; double the plate area.

Double the plate separation; halve the plate area.

c m m m V

= 35C p F

= 7.0Q n C

V J


15/15

How much charge does a 9.0 battery transfer from the negative to the positive terminal while doing 39 of

work?


ANSWER:

Correct

Score Summary:

Your score on this assignment is 102%.

You received 102.19 out of a possible total of 100 points.

V J

= 4.3 C

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