page 78, Problem 2.19:... of Sect Refer to Prob if you get stuck.

Transcription

1 Some corrections in blue to Pearson New International Edition Introduction to Electrodynamics David J. Griffiths Fourth Edition Chapter 2 page 78, Problem 2.19:... of Sect Refer to Prob if you get stuck. page 78, line 3 from below:... problem. The first theorem in section page 85, footnote 7:... Helmholtz theorem (Appendix B starting on page 583) page 88, Problem 2.29:... Laplacian and using Eq page 96, line 14 from below:... the classical. We shall return to this problem in Chapter 11. page 97, line 4:... radiation theory (Chapter 11) page 101, line 14 from below:... cancellation. Nope: As we ll see in the uniqueness theorems in Chapter 3,... Chapter 3 page 113, line 3: form of the following equation form of Eq. 2.8 page 113, line 7: equation Eq page 113, line 13: Poisson s equation Poisson s equation (2.24) page 120, Example 3.1, line 7 from below:... constant V 0 (that s item (iv), in Sect ) page 120, Example 3.1, line 3 from below:... is zero the same result we found in Sect on rather different grounds page 125, last line: Consider the following equation According to Eq page 130, Problem 3.10a... [Hint: Refer to Prob. 2.52] page 130, Problem [Hint: Exploit the result of Prob. 2.52] page 147, last line:... r = R (Eq. 2.34) page 148, line 4:... at the surface: (Eq. 2.36) page 150, Problem 3.22: In Prob. 2.25, you found the potential on the axis of a uniformly charged disk: V (r, 0) = σ ( ) r2 + R 2ɛ 2 r. 0 1

2 (a) Use this, together with the fact that P l (1) = 1, to evaluate the first three terms in the expansion (Eq. 3.72) for the potential of the disk at points off the axis, assuming r > R. (b) Find the potential for r < R by the same method, using Eq [Note: You must break the interior region up into two hemispheres, above and below the disk. Do not assume the coefficients A l are the same in both hemispheres.] page 155, line 3:... usual way (Sect ) page 163, Problem 3.47b:... Gauss s law (see Prob. 2.12) page 163, Problem 3.48, line 3:... before integrating. If you don t understand why, reread the discussion in Sect Chapter 4 page 168, last line: (Prob. 2.12) At equilibrium... page 172, Problem 4.8:... Use Prob. 4.7 and Eq page 173, line 6 from below:... single dipole p (Eq. 3.99) page 174, line 1: where (unlike Prob. 1.13) the differentiation... page 174, line 2:... in the front cover (page 594) page 175, line 2+3: The potential of such a configuration is But we already computed the potential of such a configuration, in Ex. 3.9 page 178, line 5 from below: The field in the region In Prob. 2.18, you calculated the field in the region page 180, line 8 from below: The average field You proved in Prob. 3.47(d) that the average field page 180, line 3 from below:... and that, according to Eq , is page 183, footnote 6:... delta function (see Prob. 1.46) page 184, Problem 4.15a:... use Gauss s law (Eq. 2.13) page 185, line 8: boundary conditions can be represented in boundary conditions of Sect can be recast in page 185, line 12:... boundary conditions on E (Eqs and 2.32) page 192, last line:... of course, continuous (Eq. 2.34) 2

3 page 193, line 4: Solution This is reminiscent of Ex. 3.8, in which an uncharged conducting sphere was introduced into a uniform field. In that case, the field of the induced charge canceled E 0 within the sphere; in a dielectric, the cancellation (from bound charge) is incomplete. Our problem... page 193, line 3 from below:... Inside the sphere Eq says page 194, footnote 14:... Legendre polynomials (Eq. 3.68). page 195, line 8:... meanwhile, is σ b /2ɛ 0 (see footnote after Eq. 2.33). page 195, line 2 from below:... similar circumstances (Eq. 3.10). page 195, footnote 15: This method mimics Prob page 197, line 2:... work to charge up a capacitor (Eq. 2.55). page 198, footnote 17:... more simply by the method of Sect , starting with... is untrue, in general. Study the derivation of Eq. 2.42, and you will see that it applies only to the total charge. For linear dielectrics it happens to hold for the free charge alone, but this is scarcely obvious a priori and, in fact, is most easily confirmed by working backward from Eq page 202, line 6 :... into an electric field (Eq. 2.51). page 206, Problem 4.32, line 1: Earnshaw s theorem (Prob. 3.2) page 206, Problem 4.32b, last line: [Hint: Use Prob. 3.4(a).] Chapter 5 page 222, line 7 from below:... use the average field, just as we did in Sect page 232, line 4:... = 0 (Prob. 1.63), so. page 232, line 7 from below:... involves the divergence we were at pains to calculate in Chapter 1 (Eq ): page 236, line 3:... Stokes theorem (Sect ): page 245, line 3:... authorized by Theorem 1 (of Sect ): page 246, line 2:... Poisson s equation (2.24), page 246, line 5:... the solution is Eq. 2.29), page 246, footnote 19:... Cartesian components (see Sect ) page 247, line 9 from below:... theoretical importance as we shall see in Chapter 10. 3

4 page 251, Problem 5.31b, line 1: Complete the proof of Theorem 2, Sect That is, show that any... page 254, line 6:... can be ignored. As we found in Sect (Eq. 3.94), page 255, line 8:... more illuminating way if we invoke Eq , page 255, line 5 from below:... of the loop (Prob. 1.62); page 256, line 8:... the total charge vanishes (Sect ). page 264, Problem 5.61, line 1:... at r = 0. (If you get stuck, refer to Prob ) page 264, Problem 5.61, extra line: Compare the electrostatic analog, Eq Chapter 6 page 272, line 2:... F = (p E). (See footnote to Eq. 4.5.) page 273, Problem 6.2, line 1:... the Lorentz force law, in the form of Eq. 5.16, page 275, line 2:... this is not magnetostatics, however, and the details will have to await Chapter 7 (see Prob. 7.52). page 297, Problem 6.26: Compare Eqs. 2.15, 4.9, and Notice that if ρ, P, and M are uniform, the same integral is involved in all three: r r r r 3 dτ Therefore, if you happen to know the electric field of a uniformly charged object, you can immediately write down the scalar potential of a uniformly polarized object, and the vector potential of a uniformly magnetized object, of the same shape. Use this observation to obtain V inside and outside a uniformly polarized sphere (Ex. 4.2), and A inside and outside a uniformly magnetized sphere (Ex. 6.1). Chapter 8 page 364, Problem 8.1: Calculate the power (energy per unit time) transported down the cables of Ex and Prob. 7.62, assuming the two conductors are held at potential difference V, and carry current I (down one and back up the other). page 364, Problem 8.2: Consider the charging capacitor in Prob (a) Find the electric and magnetic fields in the gap, as functions of the distance s from the axis and the time t. (Assume the charge is zero at t = 0.) (b) Find the energy density u em and the Poynting vector S in the gap. Note especially the direction of S. Check that Eq is satisfied. 4

5 (c) Determine the total energy in the gap, as a function of time. Calculate the total power in the gap, by integrating the Poynting vector over the appropriate surface. Check that the power input is equal to the rate of increase of energy in the gap (Eq. 8.9 in this case W = 0, because there is no charge in the gap). [If you re worried about the fringing fields, do it for a volume of radius b < a well inside the gap.] Chapter 9 page 429, Problem 9.24: If you take the model in Ex. 4.1 at face value, what natural frequency do you get? Put in actual numbers. Where, int the electromagnetic spectrum does this lie, assuming the radius of the atom is 0.5 Å? Find the coefficients of refraction and dispersion, and compare them with the measured values for hydrogen at 0 degree C and atmospheric pressure: A = , B = m 2. 5