ch23-p050

Ch23-p050 - 50 The field is zero for 0 r a as 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 b the

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(f) For r b we have 2 total /4 Eq r ε 0 or 3 2 0 . 3 ba E r ρ 3 = Thus, for r = 3.00 b = 6.00 a , the electric field is 33 9 3 21 2 2 2 00 (2.00 ) 7 (1.84 10 C/m )(0.100 m) 7 1.35 N/C. 3 (6.00 ) 3 36 3(8.85 10 C /N m ) 36 aa a E a ρρ εε −× ⎛⎞ == = = ⎜⎟ ×⋅ ⎝⎠ 50. The field is zero for 0 r a as 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 b the enclosed charge q enc (for a r b ) is related to the volume by q ra enc =− F H G I K J ρ ππ 4 3 4 3 . Therefore, the electric field is E q rr r a r F H G I K J = 1 44 4 3 4 0 2 0 2 0 2 πε ρ πε ρ ε enc for a r b . (d) For r =1.50 a , we have 9 3 2 2 2 (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.

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