j0104 - 1 Case 1 - Charge on Surface So7 we have a...

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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 - Non-Uniform 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.

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j0104 - 1 Case 1 - Charge on Surface So7 we have a...

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