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Unformatted text preview: Chapter 23 Practice problem solutions # 19, 20,21,26,30,34,40,44,48,52,54 19. (a) We find the electric field to be 2 12 2 2 (1.1 C) 1.99 MV/m (8.85 10 C /N m )(0.25 m) q E A  = = = (b) The potential difference is (1.99 MV/m)(5 mm) 9.94 kV V Ed = = = (c) The energy stored is 2 1 1 1 (1.1 C)(9.94 kV) 5.47 mJ 2 2 2 U CV qV = = = = A SSESS For completeness, the capacitance of the capacitor is 2 12 2 2 10 3 (8.85 10 C /N m )(0.25 m) 1.1 10 F 11 nF 5 10 m A C d  = = = = The value is typical of a capacitor. 20. The separation is much smaller than the linear dimensions of the plates, so the discussion in Section 23.2 applies. (a) From Equations 23.1 and 23.2, we have 2 2 2 2 1 1 2 2 /2 (7.2 C) (1.2 mm)/2 W CV Q C Q d A = = = = 12 2 (8.85 10 F/m)(5 cm) 1.41 J. = (b) The additional work required to double the charge on each plate is 2 (2 ) /2 W Q d A W = = 3 4.22 J. W = 21. D EVELOP Using Equations 23.13, the energy stored in a parallelplate capacitor is related to the charges on the plates Using Equations 23....
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This note was uploaded on 06/03/2010 for the course PHYS 299 taught by Professor Hoston,amahd,bakanowski during the Summer '08 term at University of Louisville.
 Summer '08
 Hoston,Amahd,Bakanowski
 Physics

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