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Unformatted text preview: (1.50 " 10!3 m) # 4.5 " 103 V (d) Energy # 1 QV # 1 (0.0180 " 10!6 C)(200 V) # 1.80 " 10!6 J # 1.80 $ J 2 2 EVALUATE:! We could also calculate the stored energy as 24.27. Q 2 (0.0180 " 10!6 C) 2 # # 1.80 $ J . 2C 2(9.00 " 10!11 F) IDENTIFY:! The energy stored in a charged capacitor is 1 CV 2 . 2 SET UP:! 1 $ F # 10!6 F EXECUTE:! 1 2 CV 2 # 1 (450 " 10!6 F)(295 V)2 # 19.6 J 2 EVALUATE:! Thermal energy is generated in the wire at the rate I 2 R , where I is the current in the wire. When the capacitor discharges there is a flow of charge that corresponds to current in the wire. 24-8 Chapter 24 24.28. IDENTIFY:! After the two capacitors are connected they must have equal potential difference, and their combined charge must add up to the original charge. Q2 1 SET UP:! C # Q / V . The stored energy is U # # CV 2 2C 2 EXECUTE:! (a) Q # CV0 . Q Q2 QQ C Q 3 (b) V # 1 # 2 and also Q1 0 Q2 # Q # CV0 . C1 # C and C2 # so 1 # and Q2 # 1 . Q # Q1 . C (C 2) C1 C2 2 2 2 2 Q1 2 Q 2 # # V0 . Q1 # Q and V # 3 C 3C 3 1 % Q 2 Q 2 & 1 9 ( 2 Q) 2 2( 1 Q) 2 : 1 Q 2 1 03 # CV0 2 (c) U # ' 1 0 2 ( # > 3 ?# C < 3C 3 2 ) C1 C2 * 2 ; C 1 (d) The original U was U # 1 CV0 2 , so BU # ! CV0 2 . 2 6 (e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation. EVALUATE:! The original charge of the charged capacitor must distribute between the two capacitors to make the potential the same across each capacitor. The voltage V for each after they are connected is less than the original voltage V0 of the charged capacitor. IDENTIFY and SET UP:! Combine Eqs. (24.9) and (24.2) to write the stored energy in terms of the separation between the plates. Q2 !A xQ 2 EXECUTE:! (a) U # ; C # 0 so U # 2C x 2!0 A 24.29. (b) x E x 0 dx gives U # . x 0 dx / Q 2 2!0 A . x 0 dx / Q 2 ! xQ 2 # % Q 2 & dx dU # ' ( 2!0 A 2!0 A ) 2!0 A * Q2 2!0 A (d) EVALUATE:! The capacitor plates and the field between the plates are shown in Figure 24.29a. = Q E# # !0 !0 A F #...
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