Ch 23 Practice Pr soln

# Ch 23 Practice Pr soln - Chapter 23 Practice problem...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 parallel-plate capacitor is related to the charges on the plates Using Equations 23....
View Full Document

## 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.

### Page1 / 2

Ch 23 Practice Pr soln - Chapter 23 Practice problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online