Introduction to Electrodynamics - ch02 - Chapter 2...

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Chapter 2 Electrostatics Problem 2.1 (a) I Zero.! (b) IF = _ 4 1 q~, I where r is the distance from center to each numeral. F points toward the missing q. 11"1=0 r Explanation: by superposition, this is equivalent to (a), with an extra -q at 6 o'clock-since the force of all twelve is zero, the net force is that of -q only. (c) I Zero. I (d) 1- 4 1 q~, I pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as 1r€0r a cancellation in pairs of opposite charges (1 o'clock against 7 o'clock; 2 against 8, etc.), with one unpaired q doing the job, then you'll need a different explanation for (d). Problem 2.2 (a) "Horizontal" components cancel. Net vertical field is: Ez = 4;<02~ cos O. q I -q From far away, (z » d), the field goes like E ~ 4;<0 ~ z, which, as we shall see, is the field of a .dipole. (If we set d -+ 0, weget E = 0, as is appropriate; to the extent that this configurationlookslike a singlepoint charge from far away, the net charge is zero, so E -+ 0.) 1 2qz A Here 1-2 = Z2 + (~)2 ; cosO = i, so I E = 41r€0(Z2 + (~)2)3/2 z. When d you're so far away it just looks like a single charge 2q; the field should reduce to E = _ 4 1 ~ z. And it aoes (just setd -+ 0 in the formula). ?r<o z (b) This time the "vertical" components cancel, leaving E = _ 4 1 2~ sinOx, or ?r<0 'V E 1 qd A = 41r€0 (z2 + (~)2)3/2 x. E x E x 22
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Problem 2.3 Ez 1 rL), dx () ( 2 2 2 411"fO JO ~ cos ; 1- = Z + x ; COS() = ~) /Ldq = )"dx Ex ---1-). .z rL 1 dx 411"fO JO (z2+x2)3/2 [ ] I L ---1-). .Z 1 x - ---1-~~ 411"fO Z2~ 0 - 411"fO z~' 1- rL )'dx sin() - 1-). . I xdx 411"fOJo ~ - 411"fO (X2+z2)3/2 [ ] I L [ ] - ---1- ).. - L- - - ---1- ).. 1 - L- 411"fO ~ 0 - 411"fO Z ~. x -- x E 1)" [ ( 1 Z ) A ( L ) A ] - -- - + x+ z - 4m:0 Z ..,j Z2 + L2 ..,j z2 + L2 . , For z » L you expect it to look like a point char ge q = )"L: E -+ _ 4 1 ¥-z. It checks, for with z » L the x 1I"fO Z term-t 0,andthe z term -+ _ 4 1 ~kz. 1I"fOz z Problem 2.4 From Ex. 2.1, with L -+ ~ and z -+ vz2 + (~)2 (distance from center of edge to F), field of one edge is: El = ~ )..a . 41f1:0 vz2 + a2 vz2 + a2 + a2 4 4 4 There are 4 sides, and we want vertical components only, so multiply by 4 cos () = 4 !i+?- : z2+ a4 E = ~ 4). .az z. 47r£0 (Z2 + a42) V Z2 + ~2 Problem 2.5 "Horizontal" components cancel, leaving: E = 4;fO {I ~ cos ()}z. Here, 1-2= r2 + z2, cos() = ~ (both constants), while I dl = 27rr. So E = ~ )"(27rr)z A 47r£0 (r2 + Z2)3/2 z. Problem 2.6 Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the field of each ring. Total charge of a ring is 0" . 27rr . dr = ).. . 27rr, so ).. = O"dr is the "line charge" of each ring. 1 (O"dr)27rrz 1 l R r Ering= _ 4 3/2; Edisk = _ 4 27rO"z 3/2 dr. 7r£0
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This note was uploaded on 02/17/2010 for the course PHYSICS mae142 taught by Professor Kim during the Spring '10 term at American College of Computer & Information Sciences.

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Introduction to Electrodynamics - ch02 - Chapter 2...

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