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Unformatted text preview: apacitor are in parallel and this combination is in series with the 9.0 $ F capacitor. EXECUTE:! 24.24. % 1 1 1 1& # #' 0 ( . (15 0 x ) $ F # 72 $ F and x # 57 $ F . Ceq 8.0 $ F ) (11 0 4.0 0 x ) $ F 9.0 $ F * EVALUATE:! Increasing the capacitance of the one capacitor by a large amount makes a small increase in the equivalent capacitance of the network. !A IDENTIFY:! Apply C # Q / V . C # 0 . The work done to double the separation equals the change in the stored d energy. 1 Q2 SET UP:! U # CV 2 # . 2 2C EXECUTE:! (a) V # Q / C # (2.55 $ C) (920 " 10!12 F) # 2770 V !0 A says that since the charge is kept constant while the separation doubles, that means that the capacitance d halves and the voltage doubles to 5540 V. Q 2 (2.55 " 10!6 C) 2 (c) U # # # 3.53 " 10!3 J . When if the separation is doubled while Q stays the same, the 2C 2(920 " 10!12 F) capacitance halves, and the energy stored doubles. So the amount of work done to move the plates equals the difference in energy stored in the capacitor, which is 3.53 " 10!3 J. EVALUATE:! The oppositely charged plates attract each other and positive work must be done by an external force to pull them farther apart. IDENTIFY and SET UP:! The energy density is given by Eq.(24.11): u # 1 !0 E 2 . Use V # Ed to solve for E. 2 (b) C # 24.25. EXECUTE:! Calculate E : E # Then u # 1 !0 E 2 # 2 24.26. 1 2 .8.854 " 10 V 400 V # # 8.00 " 104 V/m. d 5.00 " 10!3 m !12 C2 / N + m 2 /. 8.00 " 104 V/m / # 0.0283 J/m3 2 EVALUATE:! E is smaller than the value in Example 24.8 by about a factor of 6 so u is smaller by about a factor of 62 # 36. Q !A IDENTIFY:! C # . C # 0 . Vab # Ed . The stored energy is 1 QV . 2 d Vab SET UP:! d # 1.50 " 10!3 m . 1 $ C # 10!6 C 0.0180 " 10!6 C # 9.00 " 10!11 F # 90.0 pF 200 V !A Cd (9.00 " 10!11 F)(1.50 " 10!3 m) (b) C # 0 so A # # # 0.0152 m 2 . d 8.854 " 10!12 C2 /(N + m 2 ) !0 EXECUTE:! (a) C # (c) V # Ed # (3.0 " 106 V/m)...
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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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