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Unformatted text preview: a / 2 ln . rb / ra / EXECUTE:! Using the given values gives 8 # 6.00 V % 2.00 & 2 . 9.00 " 10 N + m /C / ln ' ( ) 1.20 * 9 2 2 # 6.53 " 10!10 C/m 24-10 Chapter 24 (b) SET UP:! Q # 8 L EXECUTE:! Q # (6.53 " 10!10 C/m)(0.350 m) = 2.29 " 10!10 C (c) SET UP:! The definition of capacitance is C # Q / V . 2.29 " 1010 C # 3.81" 10!11 F 6.00 V (d) SET UP:! The energy stored in a capacitor is U # 1 CV 2 . 2 EXECUTE:! C # 24.35. EXECUTE:! U # 1 (3.81" 10!11 F)(6.00 V)2 # 6.85 " 10!10 J 2 EVALUATE:! The stored energy could be converted to heat or other forms of energy. IDENTIFY:! U # 1 QV . Solve for Q. C # Q / V . 2 SET UP:! Example 24.4 shows that for a cylindrical capacitor, EXECUTE:! (a) U # 1 QV gives Q # 2 C 2, !0 # . L ln(rb /ra ) 2U 2(3.20 " 10!9 J) # # 1.60 " 10!9 C. V 4.00 V C 2, !0 r # . b # exp(2, !0 L/C ) # exp(2, !0 LV/Q ) # exp(2, !0 (15.0 m)(4.00 V) (1.60 " 10!9 C)) # 8.05. L ln( rb ra ) ra The radius of the outer conductor is 8.05 times the radius of the inner conductor. EVALUATE:! When the ratio rb / ra increases, C / L decreases and less charge is stored for a given potential difference. IDENTIFY:! Apply Eq.(24.11). % rr & Q Q # ' a b (Vab . SET UP:! Example 24.3 shows that E # between the conducting shells and that 2 4, !0 r 4, !0 ) rb ! ra * (b) 24.36. % r r &V % [0.125 m][0.148 m] & 120 V 96.5 V + m EXECUTE:! E # ' a b ( ab # ' (2# 2 r2 ) 0.148 m ! 0.125 m * r ) rb ! ra * r (a) For r # 0.126 m , E # 6.08 " 103 V/m . u # 1 !0 E 2 # 1.64 " 10!4 J/m3 . 2 24.37. (b) For r # 0.147 m , E # 4.47 " 103 V/m . u # 1 !0 E 2 # 8.85 " 10!5 J/m3 . 2 EVALUATE:! (c) No, the results of parts (a) and (b) show that the energy density is not uniform in the region between the plates. E decreases as r increases, so u decreases also. IDENTIFY:! Use the rules for series and for parallel capacitors to express the voltage for each capacitor in terms of the applied voltage. Express U, Q, and E in term...
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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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