ch23_p50 - 50. The field is zero for 0 r a as a result of...

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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 33 . 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 21 2 00 (1.50 ) 2.375 (1.84 10 )(0.100) 2.375 7.32 N/C. 3 (1.50 ) 3 2.25 3(8.85 10 ) 2.25 aa a E a ρρ εε −× = = × (e) For r = b =2.00 a , the electric field is 9 2 (2.00 ) 7 (1.84 10 )(0.100) 7
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