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기초electromagnetics 3판 정답 David J. Griffiths Introduction To Electrodynamics Solutions

INSTRUCTOR`S SOLUTIONS MANUAL

INTRODUCTION to ELECTRODYNAMICS
Third Edition

David J. Griffiths

Errata Instructor’s Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Gri?ths Date: September 1, 2004 ? Page 4, Prob. 1.15 (b): last expression should read y + 2z + 3x. ? Page 4, Prob.1.16: at the beginning, insert the following ?gure

? Page 8, Prob. 1.26: last line should read From Prob. 1.18: ∇ × va = ?6xz x + 2z y + 3z 2 ? ⇒ ? ? z ∇ · (∇ × va ) =
∂ ∂x (?6xz)

+

∂ ∂y (2z)

+

∂ 2 ∂z (3z )

= ?6z + 6z = 0.

? Page 8, Prob. 1.27, in the determinant for ∇×(∇f ), 3rd row, 2nd column: change y 3 to y 2 . ? Page 8, Prob. 1.29, line 2: the num…(생략(省略))




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