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Unformatted text preview: 89:!$862! * * * $ D *D** D .! ! .! ! -25G<! =52$ # i+D* 0 +* * ! *+D+* k 0 D * # $ )+D ! +* , * 0 D * (! * *D 0 ** * $ $ D . )*(E " DE!H J,) ! F(E " DE!H J, (c)! =52$ # )*(E " DE!F 6:,)?EE G <,* 0 # !D(? N ! * E(EE?E G (d) O89/2!-32!2925:%!8<!$2<<!-309!X25.B!-32!<%<-2G!8<!r>.19A(s! .! ! (e) 732!G0]8G1G!<2#050-8.9!8<!4329!-32!=2$./8-%!8<!X25.V! !D(? N # D * !:8=2<! $ . )*(E " DE!H J,)!F(E " DE!H J, $# # E(EIP G (! !D(? N (f) L.4!1<89:! +D # IEE G < 09A! +* # DKEE G < B!42!;89A! =52$ # 0?(H N (!732!#05-8/$2<!A.!2</0#2B!09A!-32!;890$!52$0-8=2! =2$./8-%!8<! +D ! +* # * =52$ $ # *)?(H N, # ?KE G < (! *(E " DE!F 6: EVALUATE: C.5!09!8<.$0-2A!<%<-2G!-32!=2$./8-%!.;!-32!/29-25!.;!G0<<!8<!/.9<-09-!09A!-32!<%<-2G!G1<-!52-089!-32! 6892-8/!2925:%!0<<./80-2A!48-3!-32!G.-8.9!.;!-32!/29-25!.;!G0<<(! ! ! 24 CAPACITANCE AND DIELECTRICS 24.1. 24.2. 24.3. Q Vab SET UP: 1 $ F # 10 !6 F EXECUTE: Q # CVab # (7.28 " 10 !6 F)(25.0 V) # 1.82 " 10 !4 C # 182 $ C EVALUATE: One plate has charge 0 Q and the other has charge !Q . !A Q and V # Ed . IDENTIFY and SET UP: C # 0 , C # d V A 0.00122 m 2 (a) C # !0 # !0 # 3.29 pF d 0.00328 m !8 Q 4.35 " 10 C (b) V # # # 13.2 kV C 3.29 " 10 !12 F V 13.2 " 103 V (c) E # # # 4.02 " 106 V/m d 0.00328 m EVALUATE:! The electric field is uniform between the plates, at points that aren't close to the edges. IDENTIFY and SET UP:! It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1. Q Q 0.148 " 10 !6 C EXECUTE:! (a) C # so Vab # # # 604 V Vab C 245 " 10 !12 F IDENTIFY:! C # !12 !3 !0 A Cd . 245 " 10 F / . 0.328 " 10 m / so A # # # 9.08 " 10!3 m 2 # 90.8 cm 2 d !0 8.854 " 10 !12 C 2 / N + m 2 V 604 V (c) Vab # Ed so E # ab # # 1.84 " 106 V/m d 0.328 " 10!3 m (b) C # (d) E # = !0 so = # E!0 # .1.84 " 106 V/m / . 8.854 " 10!12...
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

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