Solved Problems in Classical Electromagnetism

By Jerrold Franklin

Solved Problems in Classical Electromagnetism is a valuable tool to help students learn to do physics while using concepts they learn in the courses. Students who are taking or have already taken an advanced EM course will find the book to be a useful adjunct to their textbook, giving added practice in applying what they are learning. For students who are taking an undergraduate EM course and want to get more depth, this book can help them achieve that aim and also help them prepare for graduate work. Beginning students, or those not even taking a course at the moment, can benefit from these problems and learn just from working on them with the help of the solutions. In each chapter, the problems start out relatively easy and then get progressively more advanced, helping students to go just as far as they can at their present level.
The book includes a number of review sections to assist students without previous advanced training in working out the problems. The first review section is a comprehensive development of vector calculus that will prepare students to solve the problems and provide a strong foundation for their future development as physicists. The problems are drawn from the topics of electrostatics, magnetostatics, Maxwell's equations, electromagnetic radiation, and relativisitc electromagnetism.

• Language: English
• Publisher: Dover Publications
• Release date: Sep 12, 2018
• ISBN: 9780486834764

Book preview

Preface

This book of solved problems is meant to serve and help several types of students. If you are taking, or have already taken, an advanced EM course, this book is meant to be a useful adjunct to your textbook, giving you added practice in working out problems. If you have taken an undergraduate EM course, and want to prepare yourself for graduate school, I hope this book can help you achieve that aim by preparing you for the GRE, or whatever test, and getting you ready for graduate work. I also think beginners, or students not even taking a course at the moment, can get into these problems and learn just from working on them with the help of my solutions.

I think this book can be useful for students on several levels. At least, that is my intention. If you just want to develop your skills, even on your own, I believe with serious work on your part, that this book can help you in that. The book includes a number of review sections to prepare students without previous advanced training to work out the problems. I also make reference to sections of my Dover textbook, Classical Electromagnetism. Second Edition (0-486-81371-1), (referred to as [CEM]) if you want more detailed development of any topic. Of course, you can also use any book of your choice as an additional reference.

A bit of advice on how to use this book to your benefit:

There are two types of problems included. Some are derivations that often will be done from a point of view you may not have seen before. You should do these problems by thinking about them, then start to do them on your own, but look over my approach before you get too bogged down.

Most of the problems are real problems with calculations to work out. Work long and hard on these problems before going to my solutions. You learn to do them on your own by doing them on your own. For these types of problems, my solutions are meant to be backups, either to look at to check your own answer or to run to if you really reach a dead end. Running to read my solutions too early will just make this a textbook instead of a workbook. There are two aspects to being a physicist: learning and doing. The learning is done in the standard course with a textbook to follow. Doing is done by doing. That is what this book is about.

One last comment, when I say Work, I don’t mean you can’t enjoy it. My greatest satisfaction has always been working out hard problems, even if they are done as work.

The book uses Gaussian units because the problems are simpler to work out without extraneous constants. The only constant in Gaussian units is c, which was originally introduced as the ratio between emu charge and esu charge, but some time later was found to be the speed of light. In Gaussian units, electric currents or the velocity of moving charges are divided by the constant c. Calculations are simpler in the Gaussian system because you don’t have to worry about where to put

It is important to be conversant with both Gaussian units and SI units, and how to go back and forth between them. There is an appendix showing the various conversions between Gaussian quantities and the SI counterparts, but you won’t need it too often to do the problems.

The procedure to convert SI equations into Gaussian equations is: and

To go from Gaussian to SI:

and a in an electricity case is usually replaced by I find it easiest to convert any SI units into Gaussian to do the calculation, and then if desired, convert back to SI units for the answer.

I hope you enjoy and learn from my book.

Jerrold Franklin

Chapter 1

Electrostatics

1.1 Static fields and forces

1.1.1 Applications of Coulomb’s law

1.Four point charges, each of charge q and mass m, are located at the four corners of a square of side L.

(a)Find the magnitude of the force on one of the charges.

(b)Use the force in part (a) to find the velocity of one of the charges a long time after the four charges are released from rest in the original configuration.

Solution:

(a)The force on the upper right-hand charge due to the other three charges at the corners of a square with sides

of length L is

The magnitude of this force on any of the four charges is

(b)If the four charges are released from rest, we can write for the acceleration of any of the charges

where L' is the side length of the square at any time during the motion. The center of the square remains fixed, and the distance, r, of a charge from the center is related to L' by

and the squared velocity after a long time is given by

The velocity is the square root of this:

2.Four point charges, each of charge q and mass m, are located at the four corners of a square of side L.

(a)Find the potential energy of this configuration.

(b)Use conservation of energy to find the velocity of one of the charges a long time after the four charges are released from rest in the original configuration.

Solution:

(a)In the configuration of four point charges q at the four corners of a square of side L, each of the four charges is a distance L from two other charges, and a distance from a third charge. Consequently, the potential energy is given by

(b)The kinetic energy of the four charges (each of mass m) at a long time after their release (so that is given by

3.Four point charges, each of charge q, are fixed at the four corners of a square of side L. Find the electric field a distance z above the plane of the square on the perpendicular axis of the square.

Solution:

The four point charges q are located at the corners of a square with sides of length L. The distance from each charge to a point z above the square, on the perpendicular axis of the square, is The horizontal fields cancel, and the

magnitude of the vertical field is given by

4.Show that the electric field a distance r from a long straight wire with a uniform linear charge density λ is given by

Solution:

By symmetry, the electric field of a long straight wire is perpendicular to the wire. The field a distance r from the wire is given by

We made the substitution z = r tan θ in doing the integral.

5.Two long parallel wires, each with uniform linear charge density λ, are a distance a apart. The origin of coordinates is the midpoint between the two wires.

(a)Find the electric field in terms of the vectors r and a.

(b)Write down the x and y components of E, taking a in the x direction.

Solution:

(a)For the configuration of two wires a distance a apart, the electric field is

(b)The electric field in Cartesian coordinates is

6.A straight wire of length L has a uniformly distributed charge Q. Find the electric field on the axis of the wire a distance z from the center of the wire for the cases

(a)z > L/2.

(b)–L/2 < z < L/2. (Hint: Use symmetry for part (b).)

Solution:

(a)The field on the axis of the uniformly charged wire, a distance z from the center of the wire, is given for z > L/2 by

(b)For – L/2 < z < L/2, the point z is a distance (L/2 – z) from the end of the wire. The wire can be thought of as two parts.

The part of the wire from z' = z – (L/2 – z) = 2z – L/2 to z' = L/2 is symmetric about the point z. This means that the field due to that portion of the wire will cancel. The remaining part of the wire has a length L' = L – 2(L/2 – z) = 2z and a charge Q'< = 2zQ/L. The midpoint of this part of the wire is at z0 = (2z – L/2 – L/2)/2 = z – L/2. Thus the electric field from this part of the wire is

7.(a)Find the electric field a distance z along the axis of a uniformly charged ring of charge Q and radius R.

(b)Integrate the field for a ring to find the electric field a distance z on the axis of a uniformly charged disk of charge Q and radius R.

(c)Find the electric field of the charged disk for

(1)z = 0+ (just above the disk).

(2)z R. (Expand the square root.)

Solution:

(a)Every point on the uniformly charged ring is the same distance from a point a distance z along the axis of the ring, and a line from any point on the ring makes the same angle θ with the z-axis. Thus the electric field at z is

(b)The disk has a surface charge density It can be considered as a collection of rings, each of radius r' with a charge

The electric field a distance z along the axis of the disk is given as (using part (a))

(c)(i) For z = 0+,

(ii) For z R, we write Ε as

Using the binomial theorem, we get

This limit, equal to the field of a point charge Q, can be used as a check on the original result.

8.A straight wire of length L has a uniformly distributed charge Q. Find Ex and Ey a distance d from the wire in the configuration shown in the figure on the next page. Express your answers in terms of the angles θ1 and (θ2.

Solution:

The parallel component of the electric field a distance d from a uniformly charged straight wire of length L is given by the integral

where x1 and x2 are the two endpoints of the wire. Let

Then

where θ1 and θ2 are the angles shown.

For the perpendicular component of E, the same substitution for z, leads to

9.Four