Introduction to electrodynamics solution manual 3rd

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Introduction to electrodynamics solution manual 3rd

INSTRUCTOR’S SOLUTIONS MANUALINTRODUCTION to ELECTRODYNAMICSThird EditionDavid J. Griffithswww.pdfgrip.comErrataInstructor’s Solutions Manual Introduc

Introduction to electrodynamics solution manual 3rd ction to Electrodynamics. 3rd ed Author: David Griffiths Date: September 1. 2004Page 4. Prob. 1.15 (b): last expression should read y + 2; + 3x.Page 4

, Prob. 1.16: al the beginning, insert the following figurePage 8, Prob. 1.26: last line should readHom Prob. 1.18: V X v| = -6-c;x + 2s ỳ + 3ỉ2Ể = Introduction to electrodynamics solution manual 3rd

(V X V.) = £(-teí) + ẳ(2z) + £(32*) = -6z + 6z = 0. ✓Page 8. Prob. 1.27, in the determinant for Vx(?/), 3rd row, 2nd column: change y3 to If2.Page 8,

Introduction to electrodynamics solution manual 3rd

Prob. 1.29, line 2: the number in the box should be -12 (insert minus sign).Page 9. Prob. 1.31, line 2: change 2z3 to 2;3: first line of part (c): ins

INSTRUCTOR’S SOLUTIONS MANUALINTRODUCTION to ELECTRODYNAMICSThird EditionDavid J. Griffithswww.pdfgrip.comErrataInstructor’s Solutions Manual Introduc

Introduction to electrodynamics solution manual 3rd .46(b): change r' to a.Page 11. Prob. 1.48. second line of./: change the upper limit on the r integral from 00 to /Í. Fix the last line to read:= Iff

(-e-r)|u + 4ffe~* = 4x (-e-ft + e-0) + 4ffe-W = 4ff. /Page 15, Prob. 1.49(a), line 3: in the 1>OX. change z2 to z3.1www.pdfgrip.comPage 15, Prob. 1.49 Introduction to electrodynamics solution manual 3rd

(b), last integration “constant” should be l(x. 2), not /(*71/).Page 17. Prob. 1.53, first expression in (4): insert 0. so da = rsin Odr dộẻ.Page 17,

Introduction to electrodynamics solution manual 3rd

Prob. 1.55: Solut ion should read as follows:Problem 1.55(1) X = z = 0; dr = d-i = 0; Ĩ/: 0 —• 1. V • (fl = (pi2) dy = 0; f V ■ (fl = 0.-2X = 0; z = 2

INSTRUCTOR’S SOLUTIONS MANUALINTRODUCTION to ELECTRODYNAMICSThird EditionDavid J. Griffithswww.pdfgrip.comErrataInstructor’s Solutions Manual Introduc

Introduction to electrodynamics solution manual 3rd ) dz = zdz.f . *2I° v<« = /^=y|2=-2.Total: f V r/1 = 0 + y - 2 = [Ĩ?!Meanwhile, Stokes’ thercoin says f V • (fl = f (Vxv) ■ da. Here da = dydzx, so al

l we need isI(Vxv)z = ^(3y + z) - ±(yz2} = 3 - 2yz. Therefore Introduction to electrodynamics solution manual 3rd

INSTRUCTOR’S SOLUTIONS MANUALINTRODUCTION to ELECTRODYNAMICSThird EditionDavid J. Griffithswww.pdfgrip.comErrataInstructor’s Solutions Manual Introduc