(f) For r ≥bwe have 2total/4Eqrε0=πor 320.3baErρ3−=Thus, for r= 3.00b = 6.00a, the electric field is 33932122200(2.00 )7(1.84 10C/m )(0.100 m)71.35 N/C.3(6.00 )3363(8.85 10C /N m )36aaaEaρρεε−−−×⎛⎞====⎜⎟×⋅⎝⎠50. The field is zero for 0 ≤r ≤aas a result of Eq. 23-16. Thus, (a) E= 0 at r= 0, (b) E= 0 at r= a/2.00, and (c) E= 0 at r= a. For a r bthe enclosed charge qenc(for a r b) is related to the volume by qraenc=−FHGIKJρππ4343. Therefore, the electric field is Eqrrrar−FHGIKJ=−144434020202περπερεencfor a ≤r ≤b. (d) For r=1.50a, we have 93222(1.50 )2.375(1.84 10
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.