By David J. Griffiths
ISBN-10: 0321856953
ISBN-13: 9780321856951
Project very unlikely
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Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright as they No portion this material may be law: Say the charge on the inner cylinderlaws is Q, for acurrently length exist. L. The field isofgiven by Gauss’s R reproduced, in1any form or1 by any means, without Q 1 permission in writing from the publisher. E·da = E · 2⇡s · L = ✏0 Qenc = ✏0 Q ) E = 2⇡✏0 L s ˆ s. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
284718 a. ) y a a _,_ ( ÷2 ÷2 ) b+ b_ q_ q+ r x (b) For the square: q Ex = 4⇡✏0 where cos ✓± = Thus ✓ cos ✓+ 2 2 b+ p (a/ 2) ± r ; b± ◆ cos ✓ 2 2 b b2± = ✓ a p 2 ◆2 =0 + ✓ ) a p ±r 2 cos ✓+ cos ✓ = , b2+ b2 ◆2 = a2 ± p 2 ar + r2 . p p (a/ 2) + r (a/ 2) r p p = . (a2 + 2 ar + r2 )3/2 (a2 2 ar + r2 )3/2 To simplify, let w ⌘ p 2 r/a; then 1+w = (2 + 2w + w2 )3/2 (2 1 w , 2w + w2 )3/2 or (1 + w)2 (2 2w + w2 )3 = (1 w)2 (2 + 2w + w2 )3 . Multiplying out the left side: 8 8w 4w2 + 16w3 10w4 2w5 + 7w6 4w7 + w8 = (same thing with w !
But V V (0) = 0, 3 4/3 2/3 1/3 3 constant is also zero. 3 9 81I 2 m But V (0) = 0, so this V 3/4 = β x, so V (x) = β x4/3 , or V (x) = β x4/3 = x4/3 . 2 A2 q 2 2 4 32ϵ 0 ✓ ◆4/3 ✓ ◆2/3 ✓ ◆1/3 3p 3p 9 81I 2 m 4/3 4/3 V 3/4 = x, so V (x) = x , or V (x) = x = x4/3 . x 4/3 2 2 4 32✏20 A2 q (see graph). Interms of V0 (instead of I): V (x) = V0 d ⇣ x ⌘4/3 V✻ Without space-charge, V would increase linearly: V (x) = V0 xd . Interms of V0 (instead of I): V (x) = V0 (see graph). d V0 4ϵ0 V d2 V 4 1 −2/3 Without space-charge, V would1 increase linearly: V (x) =0 V0 xd .
Instructor’s Solution Manuals to Introduction to Electrodynamics by David J. Griffiths
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