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Unformatted text preview: .7 nF . EXECUTE:! (a) so Ceq # 1 2 # #0 Ceq C1 C2 C1 0 C2 150 nF 0 120 nF Q # CV # (66.7 nF)(36 V) # 2.4 " 10!6 C # 2.4 $ C (b) Q # 2.4 $ C for each capacitor. (c) U # 1 CeqV 2 # 1 (66.7 " 10!9 F)(36 V)2 # 43.2 $ J 2 2 (d) We know C and Q for each capacitor so rewrite U in terms of these quantities. U # 1 CV 2 # 1 C (Q / C ) 2 # Q 2 / 2C 2 2 (2.4 " 10!6 C)2 (2.4 " 10!6 C)2 # 19.2 $ J ; 120 nF: U # # 24.0 $ J !9 2(150 " 10 F) 2(120 " 10!9 F) Note that 19.2 $ J 0 24.0 $ J # 43.2 $ J , the total stored energy calculated in part (c). 150 nF: U # Q 2.4 " 10!6 C Q 2.4 " 10!6 C # # 16 V ; 120 nF: V # # # 20 V !9 C 150 " 10 F C 120 " 10!9 F Note that these two voltages sum to 36 V, the voltage applied across the network. EVALUATE:! Since Q is the same the capacitor with smaller C stores more energy ( U # Q 2 / 2C ) and has a larger voltage ( V # Q / C ). (e) 150 nF: V # 24.33. Q . U # 1 CV 2 . 2 V SET UP:! For capacitors in parallel, the voltages are the same and the charges add. EXECUTE:! (a) Ceq # C1 0 C2 # 35 nF 0 75 nF # 110 nF . Qtot # CeqV # (110 " 10!9 F)(220 V) # 24.2 $ C IDENTIFY:! The two capacitors are in parallel. Ceq # C1 0 C2 . C # (b) V # 220 V for each capacitor. 35 nF: Q35 # C35V # (35 " 10!9 F)(220 V) # 7.7 $ C ; 75 nF: Q75 # C75V # (75 " 10!9 F)(220 V) # 16.5 $ C . Note that Q35 0 Q75 # Qtot . (c) U tot # 1 CeqV 2 # 1 (110 " 10!9 F)(220 V)2 # 2.66 mJ 2 2 (d) 35 nF: U 35 # 1 C35V 2 # 1 (35 " 10!9 F)(220 V)2 # 0.85 mJ ; 2 2 24.34. 75 nF: U 75 # 1 C75V 2 # 1 (75 " 10!9 F)(220 V)2 # 1.81 mJ . Since V is the same the capacitor with larger C stores more 2 2 energy. (e) 220 V for each capacitor. EVALUATE:! The capacitor with the larger C has the larger Q. IDENTIFY:! Capacitance depends on the geometry of the object. (a) SET UP:! The potential difference between the core and tube is V # density gives 8 # 8 ln . rb / ra / . Solving for the linear charge 2, !0 2, !0V 4, !0V . # ln . rb / r...
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