文档介绍：Chapter 2
Electrostatics
Problem
(a) I Zero.!
(b) IF =
_411"1=0r
1 q~, I where r is the distance from center to each numeral. F points toward the missing q.
Explanation: by superposition, this is equivalent to (a), with an extra -q at 6 o'clock-since the force of all
twelve is zero, force is that of -q only.
(c) I Zero. I
(d) 1-4
1r€01 q~,r I pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as
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 E
(a) "Horizontal" components cancel. Net vertical field is: Ez = 4;<02~ cos O.
1 2qz A
Here 1-2= Z2 + (~)2 ; cosO=i, so I E = 41r€0(Z2 + (~)2)3/2 z.
x
When z» d you're so far away it just looks like a single charge 2q; the field
should reduce to E
= _4 ?r<o1 z~ z. And it aoes (just setd -+ 0 in the formula).
(b) This time the "vertical" components cancel, leaving E
E = _4 1 2~ sinOx, or
?r<0 'V
E 1 qd A
= 41r€0 (z2 + (~)2)3/2 x. x
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, we get E = 0, as is appropriate; to the extent that this configurationlooks like a singlepoint charge
from far away, charge is zero, so E -+ 0.)
22
23
Problem
Ez 1 rL), dx () 2 2 2
411"fO JO ~ cos ; (1- = Z + x ; COS() = ~)
---1-)..z rL 1 dx
411"fO JO (z2+x2)3/2
L
---1-)..Z 1 x - ---1-~~
411"fO [Z2~ ]I 0 - 411"fOz~'
xdx
Ex 1- rL )'dx sin() - 1-)..
/Ldq = )"dx 411"fOJo ~ - 411"fO I (X2+z2)3/2
-- L
x x - ---1-).. - L- - - ---1-).. 1 - L-
411"fO [ ~ ]I 0 - 411"fO [Z ~. ]
1)" Z L
E - -- - 1+ x+A zA
- 4m:0 Z [ ( ..,jZ2 + L2 ) (..,jz2 + 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
1 ~kz.
term-t 0, andthe