corrections to problems

# corrections to problems - Errata Instructors Solutions...

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Unformatted text preview: Errata Instructors Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Griths Date: September 1, 2004 Page 4, Prob. 1.15 (b): last expression should read y + 2 z + 3 x . Page 4, Prob.1.16: at the beginning, insert the following figure Page 8, Prob. 1.26: last line should read From Prob. 1.18: v a = 6 xz x + 2 z y + 3 z 2 z ( v a ) = x ( 6 xz ) + y (2 z ) + z (3 z 2 ) = 6 z + 6 z = 0 . X 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 number in the box should be -12 (insert minus sign). Page 9, Prob. 1.31, line 2: change 2 x 3 to 2 z 3 ; first line of part (c): insert comma between dx and dz . Page 12, Probl 1.39, line 5: remove comma after cos . Page 13, Prob. 1.42(c), last line: insert z after ). Page 14, Prob. 1.46(b): change r to a . Page 14, Prob. 1.48, second line of J : change the upper limit on the r integral from to R . Fix the last line to read: = 4 ( e r ) R + 4 e R = 4 ( e R + e ) + 4 e R = 4 . X Page 15, Prob. 1.49(a), line 3: in the box, change x 2 to x 3 . 1 Page 15, Prob. 1.49(b), last integration constant should be l ( x, z ), not l ( x, y ). Page 17, Prob. 1.53, first expression in (4): insert , so d a = r sin dr d . Page 17, Prob. 1.55: Solution should read as follows: Problem 1.55 (1) x = z = 0; dx = dz = 0; y : 0 1 . v d l = ( yz 2 ) dy = 0; R v d l = 0 . (2) x = 0; z = 2 2 y ; dz = 2 dy ; y : 1 . v d l = ( yz 2 ) dy + (3 y + z ) dz = y (2 2 y ) 2 dy (3 y + 2 2 y )2 dy ; Z v d l = 2 Z 1 (2 y 3 4 y 2 + y 2) dy = 2 y 4 2 4 y 3 3 + y 2 2 2 y 1 = 14 3 . (3) x = y = 0; dx = dy = 0; z : 2 . v d l = (3 y + z ) dz = z dz. Z v d l = Z 2 z dz = z 2 2 2 = 2 . Total: H v d l = 0 + 14 3 2 = 8 3 . Meanwhile, Stokes thereom says H v d l = R ( v ) d a . Here d a = dy dz x , so all we need is ( v ) x = y (3 y + z ) z ( yz 2 ) = 3 2 yz. Therefore R ( v ) d a = R R (3 2 yz ) dy dz = R 1 R 2 2 y (3 2 yz ) dz o dy = R 1 3(2 2 y ) 2 y 1 2 (2 2 y ) 2 dy = R 1 ( 4 y 3 + 8 y 2 10 y + 6) dy = y 4 + 8 3 y 3 5 y 2 + 6 y 1 = 1 + 8 3 5 + 6 =...
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corrections to problems - Errata Instructors Solutions...

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