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Unformatted text preview: 1 Case 1  Charge on Surface So7 we have a conductor with charge on the surface. or1<a :>E:0,7’<a or2<a So7 :>E: Q T>Cl 47reor2 7 2 Case 2  Uniform Charge Density Fiji For both inside and out, 7% : f and or—1<a Jackson Problem 6.1
B. J. Mattson 1 E  A = —p V (6)
60
2 4 3
E(47r7" ) : fra01€0p(§7rr ) (7)
But p = 3% So,
3
=> E = Mgagr, 7" < a
0 T2 > T
—p A 1 _’ 3
ng 77,:— p(:c)d:1c (8)
60
2 1
Eew ) = —Q (9)
60 =>E Q 7‘77‘>a — 47r€0r2 3 Case 3  NonUniform Charge Density NOW we have a sphere with charge density, p(7') = 77"". orl<a Start with the Charge density. / p(7°) dV 2 77"”47rr2d7' (10) V 0
= 47W Trn+2dr (11) 0
7'” + 3 = 4 12
7W n + 3 ( )
n > —3 (13)
(14) Now to ﬁnd 7. Set the above integral equal to Q for 7" = a. 2 4
H
g;
>1©
A
933
3+
We»
V So,
Tnn3
[V W) W = QW
And then,
—» 1
jg f1, = — p(3?)d3:v
60
1 Tn+3
2 _
E(47TT ) — gQan+3
=>E=$r“+l,forr<a,n>—3
o 7" > a Outside, the ﬁeld is the same as the outside ﬁeld of the previous two cases. :Ezj—Zr r>a
41mm" 7 ...
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This note was uploaded on 01/10/2012 for the course PHYSIS 506 taught by Professor Smith during the Spring '08 term at University of Arkansas Community College at Morrilton.
 Spring '08
 smith

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