By David J. Griffiths

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**Read or Download Instructor’s Solution Manuals to Introduction to Electrodynamics PDF**

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This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 Here f (0) tan−1 (1) − π4 = π4 − CHAPTER 2. = ELECTROSTATICS π 4 = 0; f ′ (x) = 1√1 1 1+(1+x) 2 1+x = 1√ 411+x , 2(2+x) so f ′ (0) = 14 , so 1 x + ( )x2 + ( )x3 + · · · 4 ⇢ ✓ ◆ ✓ ◆ ✓ ◆ 1 @ @ 1 σa2k 1 @ k sin ✓ cos 2 sin ✓ cos ✓ sin 2 3k 2σ q a2 1 1 a2 1 ⇢ = ✏(since r·E = ✏ r + sin ✓ + = Thus = x ≪ 1), E ≈ = .

All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. III - -y 32 R III E·dl = 2y0 k V (x0 , y0 , z0 ) = (x0 ,y R0 ,z0 ) R z0 CHAPTER 2. ELECTROSTATICS z dz = ky0 z02 . 0 E·dl = k(x0 y02 + y0 z02 ), or V (x, y, z) = 0 k(xy 2 + yz 2 ). @ @ @ rV =k[ @x (xy 2 +yz 2 ) x ˆ+ @y (xy 2 +yz 2 ) y ˆ+ @z (xy 2 +yz 2 ) ˆ z]=k[y 2 x ˆ+(2xy+z 2 ) y ˆ+2yz ˆ z]=E.

Where β= = βV −1/2 = − ρ = = − ⇒ dt dt 2 dx ϵ0 ϵ0 Av ϵ0 A 2qV ϵ0 A 2q . ) 2 V 1/2 I m I m (d) ddxV2 = ✏10 ⇢ = ✏10 Av = ✏0 A 2qV ) = V , where = ✏ A 2q . ) ′ dV 1 ′ 0 dV dV (e) Multiply by V V ′ = dx=:βV −1/2 ⇒ V ′ dV ′ = β V −1/2 dV ⇒ V 2 = 2βV 1/2 + constant. dx dx 2 Z Z 0 0 dV dV 1 V 1/2 is ) V 0 dV 0 = V 1/2 dVat)cathode V 2 =is2zero), V 1/2 so + constant. But V (0) = VV ′0(0) ==0 (cathode at potential zero, and field the constant is zero, and dx dx 2 ′ dV 2 V 0 (0) = 1/4 But V (0) (cathode zero,dV and V = = 4βV 1/20⇒ β Vpotential ⇒ V −1/4 = field 2 βat dx;cathode is zero), so the constant is zero, and = 2is at dx p 4 3/4 p 0 dV −1/4 1/2 x+ V )= = V 2V = 4 V dV )2 β = 2dx ⇒V 1/4 V 21/4βdV =constant.