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jackson电动力学答案.pdf

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    A Companion to Classical Electrodynamics 3rd Edition by J.D. Jackson Rudolph J. Magyar February 5, 2002 cRudolph J. Magyar. No portion of this may be reproduced for profit without the expressed prior written consent of Rudolph J. Magyar. 1
    A lot of things can be said about Classical Electrodynamics, the third edition, by David J. Jackson. It’s seemingly exhaustive, well researched, and certainly popular. Then, there is a general consensus among teachers that this book is the definitive graduate text on the subject. In my opinion, this is quite unfortunate. The text often assumes familiarity with the material, skips vital steps, and provides too few examples. It is simply not a good introductory text. On the other hand, Jackson was very ambitious. Aside from some notable omissions (such as conformal mapping methods), Jackson exposes the reader to most of classical electro-magnetic theory. Even Thomas Aquinas would be impressed! As a reference, Jackson’s book is great! It is obvious that Jackson knows his stuff, and in no place is this more apparent than in the problems which he asks at the end of each chapter. Sometimes the problems are quite simple or routine, other times difficult, and quite often there will be undaunting amounts of algebra required. Solving these problems is a time consuming endevour for even the quickest reckoners among us. I present this Companion to Jackson as a motivation to other students. These problems can be done! And it doesn’t take Feynmann to do them. Hopefully, with the help of this guide, lots of paper, and your own wits; you’ll be able to wrestle with the concepts that challenged the greatest minds of the last millenium. solving these problems. Before I begin, I will recommend several things which I found useful in • Buy Griffiths’ text, an Introduction to Electrodynamics. It’s well writ- ten and introduces the basic concepts well. This text is at a more basic level than Jackson, and to be best prepared, you’ll have to find other texts at Jackson’s level. But remember Rome wasn’t build in a day, and you have to start somewhere. • Obtain other texts on the level (or near to it) of Jackson. I rec- ommend Vanderlinde’s Electromagnetism book or Eyges’ Electromag- netism book. Both provide helpful insights into what Jackson is talking about. But even more usefully, different authors like to borrow each others’ problems and examples. A problem in Jackson’s text might be an example in one of these other texts. Or the problem might be rephrased in the other text; the rephrased versions often provide insight into what Jackson’s asking! After all half the skill in writing a hard i
    physics problem is wording the problem vaguely enough so that no one can figure out what your talking about. • First try to solve the problem without even reading the text. More often than not, you can solve the problem with just algebra or only a superficial knowledge of the topic. It’s unfortunate, but a great deal of physics problems tend to be just turning the crank. Do remember to go back and actually read the text though. Solving physics problems is meaningless if you don’t try to understand the basic science about what is going on. • If you are allowed, compare your results and methods with other stu- dents. This is helpful. People are quick to tear apart weak arguments and thereby help you strengthen your own understanding of the physics. Also, if you are like me, you are a king of stupid algebraic mistakes. If ten people have one result, and you have another, there’s a good like- lihood that you made an algebraic mistake. Find it. If it’s not there, try to find what the other people could have done wrong. Maybe, you are both correct! • Check journal citations. When Jackson cites a journal, find the refer- ence, and read it. Sometimes, the problem is solved in the reference, but always, the references provide vital insight into the science behind the equations. A note about units, notation, and diction is in order. I prefer SI units and will use these units whenever possible. However, in some cases, the use of Jacksonian units is inevitable, and I will switch without warning, but of course, I plan to maintain consistency within any particular problem. I will set c = 1 and ¯h = 1 when it makes life easier; hopefully, I will inform the reader when this happens. I have tried, but failed, to be regular with my symbols. In each case, the meaning of various letters should be obvious, or else if I remember, I will define new symbols. I try to avoid the clumsy d3x symbols for volume elements and the d2x symbols for area elements; instead, I use dV and dA. Also, I will use ˆx,ˆy, and ˆz instead of ˆi,ˆj, and ˆk. The only times I will use ijk’s will be for indices. Please, feel free to contact me, rmagyar@eden.rutgers.edu, about any typos or egregious errors. I’m sure there are quite a few. ii
    Now, the fun begins... iii
    Problem 1.1 Use Gauss’ theorem to prove the following: a. Any excess charge placed on a conductor must lie entirely on its surface. In Jackson’s own words, “A conductor by definition contains charges capable of moving freely under the action of applied electric fields”. That implies that in the presence of electric fields, the charges in the conductor will be accelerated. In a steady configuration, we should expect the charges not to accelerate. For the charges to be non-accelerating, the electric field must vanish everywhere inside the conductor, E = 0. When E = 0 everywhere inside the conductor 1, the divergence of E must vanish. By Gauss’s law, we see that this also implies that the charge density inside the conductor vanishes: 0 = ∇ · E = ρ 0 . b. A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it. The charge density within the conductor is zero, but the charges must be located somewhere! The only other place in on the surfaces. We use Gauss’s law in its integral form to find the field outside the conductor. qi E · d A = 1 0 Where the sum is over all enclosed charges. Evidently, the field outside the conductor depends on the surface charges and also those charges concealed deep within the cavities of the conductor. c. The electric field at the surface of a conductor is normal to the surface and has a magnitude σ 0 , where σ is the charge density per unit area on the surface. We assume that the surface charge is static. Then, E at the surface of a con- ductor must be normal to the surface; otherwise, the tangential components of the E-field would cause charges to flow on the surface, and that would contradict the static condition we already assumed. Consider a small area. ∇ · EdV = E · d A = ρ 0 dV 1excluding of course charge contained within any cavities 1
    But ρ = 0 everywhere except on the surface so ρ should more appropriately be written σδ(f (x)). Where the function f (x) subtends the surface in ques- tion. The last integral can then be written rearranged. 0 ˆn · d A. Our equation can be σ E · d A = ˆn · d A → σ 0 E − σ 0 ˆn · d A = 0 And we conclude E = σ 0 ˆn 2
    In spherical coordinates, a charge Q distributed over spherical shell of Problem 1.3 Using Dirac delta functions in the appropriate coordinates, express the following charge distributions as three-dimensional charge den- sities ρ(x). a. radius, R. The charge density is zero except on a thin shell when r equals R. The charge density will be of the form, ρ ∝ δ(r − R). The delta function insures that the charge density vanishes everywhere except when r = R, the radius of the sphere. Integrating ρ over that shell, we should get Q for the total charge. Aδ(r − R)dV = Q A is some constant yet to be determined. Evaluate the integral and solve for A. Aδ(r − R)dV = Aδ(r − R)r2d(cos θ)dφdr = 4πR2A = Q So A = Q 4πR2 , and ρ(r) = Q 4πR2 δ(r − R) b. In cylindrical coordinates, a charge λ per unit length uniformly distributed over a cylindrical surface of radius b. Aδ(r − b)dA = λ Since we are concerned with only the charge density per unit length in the axial direction, the integral is only over the plane perpendicular to the axis of the cylinder. Evaluate the integral and solve for A. Bδ(r − b)dA = Bδ(r − b)rdθdr = 2πbB = λ So B = λ 2πb, and ρ(r) = δ(r − b) λ 2πb c. In cylindrical coordinates, a charge, Q, spread uniformly over a flat circular disc of negligible thickness and radius, R. AΘ(r − R)δ(z)dV = Q 3
    The Θ function of x vanishes when x is negative; when x is positive, Θ is unity. AΘ(R − r)δ(z)dV = AΘ(R − r)δ(z)rdθdzdr = πR2A = Q So A = Q πR2 , and ρ(r) = d. The same as in part c, but using spherical coordinates. Q πR2 Θ(R − r)δ(z) θ − π 2 dV = Q AΘ(R − r)δ Evaluate the integral and solve for A. AΘ(R − r)δ θ − π 2 dV = AΘ(R − r)δ θ − π 2 r2d(cos θ)dφdr = 2πR2A = Q So A = Q 2πR2 , and ρ(r) = Q 2πR2 Θ(R − r)δ θ − π 2 4
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