Errata Instructor s Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Griffiths Date: September, 004 Page 4, Prob..5 (b): last expression should read y +z +3x. Page 4, Prob..6: at the beginning, insert the following figure Page 8, Prob..6: last line should read From Prob..8: v a = 6xz ˆx +zŷ +3z ẑ ( v a )= x ( 6xz)+ y (z)+ z (3z )= 6z +6z =0. Page 8, Prob..7, in the determinant for ( f), 3rd row, nd column: change y 3 to y. Page 8, Prob..9, line : the number in the box should be - (insert minus sign). Page 9, Prob..3, line : change x 3 to z 3 ; first line of part (c): insert comma between dx and dz. Page, Probl.39, line 5: remove comma after cos θ. Page 3, Prob..4(c), last line: insert ẑ after ). Page 4, Prob..46(b): change r to a. Page 4, Prob..48, second line of J: change the upper limit on the r integral from to R. Fix the last line to read: = 4π ( e r) R 0 +4πe R =4π ( e R + e 0) +4πe R =4π. Page 5, Prob..49(a), line 3: in the box, change x to x 3.
Page 5, Prob..49(b), last integration constant should be l(x, z), not l(x, y). Page 7, Prob..53, first expression in (4): insert θ, soda = r sin θdrdφˆθˆθˆθ. Page 7, Prob..55: Solution should read as follows: Problem.55 () x = z =0; dx = dz =0; y :0. v dl =(yz ) dy =0; v dl =0. () x =0; z = y; dz = dy; y : 0. v dl =(yz ) dy +(3y + z) dz = y( y) dy (3y + y) dy; v dl = 0 [ ] y (y 3 4y 4 + y ) dy = 4y3 3 + y 0 y = 4 3. (3) x = y =0; dx = dy =0; z : 0. v dl =(3y + z) dz = zdz. v dl = 0 zdz= z 0 =. Total: v dl =0+ 4 3 = 8 3. Meanwhile, Stokes thereom says v dl = ( v) da. Here da = dy dz ˆx, so all we need is ( v) x = y (3y + z) z (yz )=3 yz. Therefore { } y ( v) da = (3 yz) dy dz = (3 yz) dz dy 0 0 = [ 0 3( y) y ( y)] dy = 0 ( 4y3 +8y 0y +6) dy = [ y 4 + 8 3 y3 5y +6y ] = + 8 0 3 5+6= 8 3. Page 8, Prob..56: change (3) and (4) to read as follows: (3) φ = π ; r sin θ = y =, so r = sin θ, dr = sin θ cos θdθ, θ : π θ 0 tan ( ). v dl = ( r cos θ ) ( (dr) (r cos θ sin θ)(rdθ)= cos θ cos θ ) cos θ sin θ sin θ sin dθ θ sin dθ θ ( cos 3 θ = sin 3 θ + cos θ ) dθ = cos θ ( cos θ +sin ) θ sin θ sin θ sin dθ = cos θ θ sin 3 θ dθ. Therefore v dl = θ 0 π/ cos θ sin 3 θ dθ = sin θ θ 0 π/ = (/5) () = 5 =.
(4) θ = θ 0,φ= π ; r : 5 0. v dl = ( r cos θ ) (dr) = 4 5 rdr. 0 v dl = 4 rdr= 4 5 5 5 r 0 = 4 5 5 5 =. Total: v dl =0+ 3π + = 3π. Page, Probl.6(e), line : change = z ẑ to +z ẑ. Page 5, Prob..: last line should read Since Q tot = 4 3 πr3 ρ, E = Q 4πɛ 0 R r (as in Prob..8). 3 Page 6, Prob..5: last expression in first line of (ii) should be dφ, not d phi. Page 8, Prob.., at the end, insert the following figure V(r).6.4. 0.8 0.6 0.4 0. 0.5.5.5 3 r In the figure, r is in units of R, and V (r) is in units of Page 30, Prob..8: remove right angle sign in the figure. q 4πɛ 0R. Page 4, Prob. 3.5: subscript on V in last integral should be 3, not. Page 45, Prob. 3.0: after the first box, add: q F = 4πɛ 0 { (a) ˆx } (b) ŷ + ( [cos θ ˆx +sin θ ŷ], a + b ) where cos θ = a/ a + b, sin θ = b/ a + b. {[ F = q a 6πɛ 0 (a + b ) ] 3/ a [ ˆx + b (a + b ) 3/ b ] } ŷ. 3
W = 4 4πɛ 0 [ q (a) + q (b) + q ] ( = q a + b ) 3πɛ 0 [ a + b a b Page 45, Prob. 3.0: in the second box, change and to an. Page 46, Probl 3.3, at the end, insert the following: [Comment: Technically, the series solution for σ is defective, since term-by-term differentiation has produced a (naively) non-convergent sum. More sophisticated definitions of convergence permit one to work with series of this form, but it is better to sum the series first and then differentiate (the second method).] Page 5, Prob. 3.8, midpage: the reference to Eq. 3.7 should be 3.7. Page 53, Prob. 3.(b), line 5: A should be R. σ 4ɛ 0R ; next line, insert r after Page 55, Prob. 3.3, third displayed equation: remove the first Φ. Page 58, Prob. 3.8(a), second line, first integral: R 3 should read R. Page 59, Prob. 3.3(c): change first V to W. Page 64, Prob. 3.4(a), lines and 3: remove ɛ 0 in the first factor in the expressions for E ave ; in the second expression change ρ to q. Page 69, Prob. 3.47, at the end add the following: Alternatively, start with the separable solution V (x, y) =(C sin kx + D cos kx) ( Ae ky + Be ky). Note that the configuration is symmetric in x, soc = 0, and V (x, 0) = 0 B = A, so (combining the constants) V (x, y) =A cos kx sinh ky. But V (b, y) =0,socoskb = 0, which means that kb = ±π/, ±3π/,, or k =(n )π/b α n, with n =,, 3,... (negative k does not yield a different solution the sign can be absorbed into A). The general linear combination is V (x, y) = A n cos α n x sinh α n y, n= and it remains to fit the final boundary condition: V (x, a) =V 0 = A n cos α n x sinh α n a. n= ]. 4
Use Fourier s trick, multiplying by cos α n x and integrating: b b V 0 cos α n xdx= A n sinh α n a cos α n x cos α n xdx b So A n = V 0 b V 0 sin α n b α n = n= b A n sinh α n a (bδ n n)=ba n sinh α n a; n= sin α n b α n sinh α n a. But sin α nb = sin V (x, y) = V 0 b ( n ( ) n sinh α ny α n sinh α n a cos α nx. n= ) π = ( ) n, so Page 74, Prob. 4.4: exponent on r in boxed equation should be 5, not 3. Page 75, Prob. 4.7: replace the (defective) solution with the following: If the potential is zero at infinity, the energy of a point charge Q is (Eq..39) W = QV (r). For a physical dipole, with q at r and +q at r+d, [ ] r+d U = qv (r + d) qv (r) =q [V (r + d) V (r)] = q E dl. For an ideal dipole the integral reduces to E d, and U = qe d = p E, since p = qd. If you do not (or cannot) use infinity as the reference point, the result still holds, as long as you bring the two charges in from the same point, r 0 (or two points at the same potential). In that case W = Q [V (r) V (r 0 )], and U = q [V (r + d) V (r 0 )] q [V (r) V (r 0 )] = q [V (r + d) V (r)], as before. Page 75, Prob. 4.0(a): r should be 3 r. Page 79, Prob. 4.9: in the upper right box of the Table (σ f for air) there is a missing factor of ɛ 0. Page 9, Problem 5.0(b): in the first line µ 0 I /π should read µ 0 I a/πs; in the final boxed equation the first should be a s. Page 9, Prob. 5.5: the signs are all wrong. The end of line should read right (ẑ), the middle of the next line should read left ( ẑ). In the first box it should be (n n ), and in the second box the minus sign does not belong. r 5
Page 4, Prob. 6.4: last term in second expression for F should be +ẑ Bx z (plus, not minus). Page 9, Prob. 6.(a): replace with the following: The magnetic force on the dipole is given by Eq. 6.3; to move the dipole in from infinity we must exert an opposite force, so the work done is U = r F dl = r (m B) dl = m B(r)+m B( ) (I used the gradient theorem, Eq..55). As long as the magnetic field goes to zero at infinity, then, U = m B. If the magnetic field does not go to zero at infinity, one must stipulate that the dipole starts out oriented perpendicular to the field. Page 5, Prob. 7.(b): in the box, c should be C. Page 9, Prob. 7.8: change first two lines to read: Φ= B da; B = µ 0I πs ˆφˆφˆφ; Φ= µ s+a 0Ia ds π s s = µ ( ) 0Ia s + a π ln ; s E = I loop R = dq dt R = dφ dt = µ 0a ln( + a/s)di π dt. dq = µ 0a ln( + a/s) di πr Q = µ 0aI ln( + a/s). πr Page 3, Prob. 7.7: in the second integral, r should be s. Page 3, Prob. 7.3(c), last line: in the final two equations, insert an I immediately after µ 0. Page 40, Prob. 7.47: in the box, the top equation should have a minus sign in front, and in the bottom equation the plus sign should be minus. Page 4, Prob. 7.50, final answer: R should read R. Page 43, Prob. 7.55, penultimate displayed equation: tp should be. Page 47, Prob. 8., top line, penultimate expression: change a to a 4 ;in (c), in the first box, change 6 to 8. Page 49, Prob. 8.5(c): there should be a minus sign in front of σ in the box. Page 49, Prob. 8.7: almost all the r s here should be s s. In line change a <r<r to s<r ; in the same line change dr to ds; in the next line change dr to ds (twice), and change ˆr to ŝ; in the last line change r to s, dr to ds, and ˆr to ŝ (but leave r as is). 6
Page 53, Prob. 8., last line of equations: in the numerator of the expression for R change.0 to.0. Page 75, Prob. 9.34, penultimate line: α = n 3 /n (not n 3 /n 3 ). Page 77, Prob. 9.38: half-way down, remove minus sign in k x + k y + k z = (ω/c). Page 8, Prob. 0.8: first line: remove. Page 84, Prob. 0.4: in the first line, change (9.98) to (0.4). Page 03, Prob..4: at beginning of second paragraph, remove. Page, Prob..5, end of first sentence: change comma to period. Page 5, Prob..3. The figure contains two errors: the slopes are for v/c = / (not 3/), and the intervals are incorrect. The correct solution is as follows: Page 7, Prob..33: first expression in third line, change c to c. 7