PHYS320 Week 5 Homework Answers

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Assignment 5 Due: 12:00pm on Tuesday, April 28, 2009

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Potential Energy of a Battery

Description: The electric potential and potential energy associated with a battery.

Learning Goal: To understand electrical potential, electrical potential energy, and the relationship between them.

Electric potential and electric potential energy are related but different concepts. Be careful not to confuse the terms. Electrical potential energy is the potential energy that a charge has due to its position relative to other charges. The electric potential at a specific

position is a measure of the amount of potential energy per unit charge a particle of net charge would have at that position. In other words, if a charge has an electric potential energy , the electric potential at the location of is

.

Recall that the gravitational potential energy ( ) of an object of mass depends on where you define . The difference in gravitational potential energy between two points is the physically relevant quantity. Similarly, for electric potential energy, the

important quantity is the change in electric potential energy: . This is why we often just measure the potential difference . When we say that the potential of a car battery is 12 , we mean that the potential difference between the positive and negative

terminals of the battery is 12 .

Consider dropping a ball from rest. This ball moves from a state of high gravitational potential energy to one of low gravitational potential energy as it falls to the ground. Similarly, charges move from a state of high electric potential energy to one of low electric potential energy.

Since potential difference is the energy per unit charge, it is measured in units of energy divided by charge. Specifically, potential difference is generally measured in volts (whose symbol is ). One volt is equal to one joule per coulomb: .

Part A

Mustang Sally just finished restoring her 1965 Ford Mustang car. To save money, she did not get a new battery. When she tries to start the car, she discovers that the battery is dead and she needs a jump start. While unhooking the jumper cables, the positive and negative cables almost touch and a spark jumps between the ends of the cables. This spark is caused by the movement of electrons through the air between the battery terminals. In what direction are the electrons traveling?

The positive terminal is at a higher potential than the negative terminal. Unless provided with energy, positive charges will flow from a high to a low potential, and negatively charged electrons will flow from a low to a high potential. The table below summarizes this movement.

Hint A.1 Another way to think about the movement of charge

You can think of the movement of charges in terms of Coulomb's force. A positive (high) potential is created by positive charges and a low (negative) potential is created by negative charges. To understand which way electrons will flow across a potential difference, think about the forces on an electron. An electron will be repelled by a negative charge and attracted to a positive charge.

The negative terminal of a battery can be viewed as having a negative charge.

ANSWER: The electrons are traveling from the negative to the positive terminal.

Direction of motion

high to low potential

low to high potential

Part B

There is a 12 potential difference between the positive and negative ends of the jumper cables, which are a short distance apart. An electron at the negative end ready to jump to the positive end has a certain amount of potential energy. On what quantities does this

electrical potential energy depend?

Hint B.1 The expression for electric potential energy

The electric potential energy difference is given by the potential difference times the charge: .

ANSWER: the distance between the ends of the cables

the potential difference between the ends of the cables

the charge on the electron

the distance and the potential difference

the distance and the charge

the potential difference and the charge

the potential difference, charge, and distance

Part C

Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a negative ion with a charge at this terminal. What are the electric potential and electric potential energy of the negative ion relative to the

electron?

ANSWER: The electric potential and the electric potential energy are both twice as much.

The electric potential is twice as much and the electric potential energy is the same.

The electric potential is the same and the electric potential energy is twice as much.

The electric potential and the electric potential energy are both the same.

The electric potential is the same and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron.

The electric potential is twice as much and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron.

Part D

What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable? In other words, assume that the electric potential of the positive terminal is 0 and that of the negative terminal is . Recall that

.

Enter your answer numerically in joules.

ANSWER: =

Part E

At the negative terminal of the battery the electron has electric potential energy. What happens to this energy as the electron jumps from the negative to the positive terminal?

Just as gravitational potential energy is converted to kinetic energy when something falls, electrical potential energy is converted to kinetic energy when a charge goes from a high potential energy state to a low potential energy state.

ANSWER: It disappears.

It is converted to kinetic energy.

It heats the battery.

It increases the potential of the battery.

Part F

If you wanted to move an electron from the positive to the negative terminal of the battery, how much work would you need to do on the electron?

Enter your answer numerically in joules.

Because moving a negative charge from the positive to the negative terminal of the battery would increase its electric potential energy, it would take positive work to move the charge. This is simliar to lifting a ball upward. You do positive work on the ball to increase its gravitational potential energy.

Hint F.1 Formula for work

The work done on a charge is equal to the product .

ANSWER: =

Electric Potential and Potential Energy

Description: Short quantitative problem on the motion of a charged particle along a line of varying potential and a line of constant potential. Based on Young/Geller Quantitative Analysis 18.2.

A particle with charge 4.80×10−19 is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction.

Part A

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 1.44×10−18 . In what direction and through what potential difference does the particle move?

Thus, if no forces other than the electric force act on a positively charged particle, the particle always moves toward a point at lower potential.

Hint A.1 How to approach the problem

Because no forces other than the electric force act on the particle, the positively charged particle must move in the direction parallel to the electric field, and the field must do positive work on the particle. Recall that when the electric field does positive work on a charged particle, the potential energy of the particle decreases. Thus, the particle must move in the direction in which its potential energy decreases (which is consistent with the fact that the particle's kinetic energy increases as it moves from a to b). Moreover, from the definition of potential and the energy conservation equation, you can directly calculate the potential difference .

Hint A.2 Electric potential

The electric potential at any point in an electric field is the electric potential energy per unit charge associated with a test charge at that point:

.

Hint A.3 Find the change in potential energy of the particle

What is the change in potential energy of the particle, , as it moves from a to b?

Express your answer in joules.

Hint A.3.1 Energy conservation

Recall that the total mechanical energy (kinetic plus potential) is conserved. That is,

,

where the subscripts refer to points a and b, and and are the corresponding kinetic and potential energies.

Hint A.3.2 Find the change in kinetic energy of the particle

What is the change in kinetic energy of the particle, , as it moves from a to b? Recall that particle is initially at rest, and its kinetic energy at b is 1.44×10−18 .

Express your answer in joules.

ANSWER: =

ANSWER: =

ANSWER: The particle moves to the left through a potential difference of 3.00 .

The particle moves to the left through a potential difference of 3.00 .

The particle moves to the right through a potential difference of 3.00 .

The particle moves to the right through a potential difference of 3.00 .

The particle moves to the left through a potential difference of 30.0 .

The particle moves to the right through a potential difference of 30.0 .

Part B

If the particle moves from point b to point c in the y direction, what is the change in its potential energy, ?

Hint B.1 How to approach the problem

Recall that the electric potential increases in the +x direction but does not change in the y or z direction.

ANSWER:

Every time a charged particle moves along a line of constant potential, its potential energy remains constant and the electric field does no work on the particle.

1.44×10−18

1.44×10−18

0

Equipotential Surfaces in a Capacitor

Description: Find the work done by a charge moving along an equipotential surface, and use that idea to find the distance between the charged plates in a parallel-plate capacitor such that there is a given potential with a given electric field between the plates.

Part A

Is the electric potential energy of a particle with charge the same at all points on an equipotential surface?

For a particle with charge on an equipotential surface at potential , the electric potential energy has a constant value .

Hint A.1 Formula for electric potential energy

For a particle with charge at potential , the electric potential energy is .

ANSWER: Yes

No

Part B

What is the work required to move a charge around on an equipotential surface at potential with constant speed?

Since the speed of the charge is constant as it moves along the equipotential surface, and the electric potential energy is constant on that surface, there is no change in the total energy of the charge. This also means that no work is done by the charge as it moves along the equipotential surface.

Hint B.1 A formula for work

The total work done on an object is equal to the change in its energy (potential + kinetic).

ANSWER: Work =

Part C

What is the work done by the electric field on a charge as it moves along an equipotential surface at potential ?

Just as in Part B, since there is no change in the electric potential energy, no work is done by the electric field as the charge moves along the equipotential surface.

Hint C.1 Work done by an electric field on a charged particle

The force due to an electric field is a conservative force. As such, the work done by such a force is equal to the change in the potential energy of the particle it is acting on.

ANSWER: Work done by the electric field =

Part D

The work done by the uniform electric field in displacing a particle with charge along the path is given by

,

where is the angle between and . Since in general, is not equal to zero, for points on an equipotential surface, what must be for to equal 0?

Express your answer in radians.

ANSWER:

Now assume that a parallel-plate capacitor is attached across the terminals of a battery as shown in the figure. The electric field in the region between the plates points in the negative z direction, from higher to lower voltage.

You have shown that equipotential surfaces are always perpendicular to the electric field at their surface.

=

Part E

Find the electric potential at a point inside the capacitor if the origin of the coordinate system is at potential .

Express your answer in terms of some or all of the variables , , , and .

Therefore, the equation of an equipotential surface at a potenial is given by

.

This is the equation of a plane that is parallel to the plates of the capacitor and perpendicular to the electric field. In particular, the lower plate, which is at zero potential, corresponds to the surface .

Hint E.1 The relation between electric potential and the electric field

ield in a region of space is

,

where the line integral may be taken along any path .

Hint E.2 Expressing an infinitesimal length element

In general, a small length vector along the path of choice can be written as

.

Substitute this expression into the integral for .

Hint E.3 Analysis of the equation

Recall that if and are perpendicular (where is one of the Cartesian coordinate axes), then , since . According to the setup of this part, only one of the directions ( , , ) will not be perpendicular to the electric field as defined.

ANSWER: =

Part F

What is the distance between two surfaces separated by a potential difference ?

Express your answer in terms of and .

Hint F.1 How to approach the problem

Use the equation you found in Part E to find equations that represent the potentials and of the planes located at and . Use these expressions to find an equation for .

ANSWER:

=

The Fate of an Electron in a Uniform Electric Field

Description: Review concept of electric potential and electric potential energy for an electron between two capacitor plates. Compute the speed of the electron when it reaches the upper plate after being released from rest in the middle of the capacitor. Compute how this speed changes when the mass or charge of the "electron" is changed.

In this problem we will study the behavior of an electron in a uniform electric field. Consider a uniform electric field (magnitude ) as shown in the figure within a parallel plate capacitor in vacuum.

First, let us review the relationship between an electric field and its associated electric potential . For now, ignore the electron located between the plates.

Part A

Calculate the electric potential inside the capacitor as a function of height . Take the potential at the bottom plate to be zero.

Hint A.1 Relationship of field and potential

The general relation between the electric potential in an electric field is . Because electric fields are gradient (or conservative) fields, we can write this relation in an integral form: .

Hint A.2 Limits of integration

Now an electron of mass and charge (where is a positive quantity) is placed within the electric field (see the figure) at height .

Express in terms of and .

Integrate from the bottom plate to arbitrary height , where . You know the value of the potential at the bottom plate: It is zero. Be careful with signs! You are integrating against the direction of , which affects the sign.

ANSWER: =

Part B

Calculate the electon's potential energy , neglecting gravitational potential energy.

Express your answer in terms of , , and .

Hint B.1 Definition of electric potential energy

The definition of the electric potential is that it is equal to the potential energy per unit charge. Therefore, for a charge .

ANSWER: =

Part C

The electron, having been held at height , is now released from rest. Calculate its speed (i.e., ) when it reaches the top plate.

Hint C.1 How to approach the problem

Although this problem can be done in several different ways, the easiest (that is, with the least amount of calculation) is to use which of the following principles?

The only two forms of energy that we are considering in this problem (since we have excluded gravity) are the electric potential energy and the kinetic energy.

ANSWER: conservation of energy with gravitational potential energy and kinetic energy

conservation of energy with electrostatic potential energy and kinetic energy

conservation of energy with gravitational and electrostatic potential energy Newton's second law ( ) followed by kinematics

Hint C.2 Find the initial energy

Find an expression for the initial energy (kinetic plus potential) of the electron at height .

Express in terms of , , , and .

ANSWER: =

Hint C.3 Find the final energy

Assume that the speed of the electron when it reaches the upper plate is . Find an expression for the final energy (kinetic plus potential) of the electron when it reaches the upper plate (at position ).

Express in terms of , , , , , and .

Hint C.3.1 Kinetic energy of a point particle

The kinetic energy of a point particle of mass moving with velocity is .

ANSWER: =

Now we consider the effect of changing either the charge or the mass of the charged particle that is released from rest at height .

Express in terms of , , , , , and other given quantities and constants.

ANSWER:

=

Part D

Imagine a particle that has three times the mass of the electron. All other quantities given above remain the same. What is the ratio of the kinetic energy that this heavy particle would have when it reaches the upper plate to the kinetic energy that the electron would

have? That is, what is ?

Hint D.1 Energy conservation

In the motion of both particles, total energy is conserved. Does the potential energy of the particles depend on their mass?

ANSWER: 3

1

none of the above: the heavy particle will never reach the upper plate

Part E

Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron ( ). It has a positive charge that is three times the magnitude ( ) of the charge on an electron. What is the ratio of the speed that the cyberon would have when it reaches

the upper plate after being released from rest at position to the speed that the electron would have? That is, what is ?

Because it has positive instead of negative charge, the cyberon will accelerate downward, toward the lower plate.

ANSWER: 3

1

none of the above: the cyberon will never reach the upper plate

Introduction to Capacitance

Description: Introduces the concept of capacitance, and the basic formula for air-filled parallel-plate capacitance

Learning Goal: To understand the meaning of capacitance and ways of calculating capacitance

When a positive 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?

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 .

ANSWER: ability to conduct electric current

ability to distort an external electrostatic field

ability to store charge

ability to store electrostatic energy

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

What is the easiest way to obtain ?

ANSWER: 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.

Hint B.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 B.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.

Hint B.3.1 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 B.4 Recall Gauss's law

Gauss's law states that .

ANSWER: =

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 C.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:

=

Part D

Now find the capacitance of the parallel-plate capacitor.

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

You have derived the general formula for the capacitance of a parallel-plate capacitor with plate area and plate separation . It is worth remembering.

ANSWER: =

Part E

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

Hint E.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: 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.

Part F

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

ANSWER: 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.

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