ass11 - Solutions, Assignment 11 24.1.I DENTIFY : ab Q C V...

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Unformatted text preview: Solutions, Assignment 11 24.1.I DENTIFY : ab Q C V = S ET U P : 6 1 F 10 F μ- = E XECUTE : 6 4 (7.28 10 F)(25.0 V) 1.82 10 C 182 C ab Q CV μ-- = = × = × = E VALUATE : One plate has charge Q + and the other has charge Q- . 24.2. I DENTIFY and S ET U P : A C d = P , Q C V = and V Ed = . (a) 2 0.00122 m 3.29 pF 0.00328 m A C d = = = P P (b) 8 12 4.35 10 C 13.2 kV 3.29 10 F Q V C-- × = = = × (c) 3 6 13.2 10 V 4.02 10 V/m 0.00328 m V E d × = = = × E VALUATE : The electric field is uniform between the plates, at points that aren't close to the edges. 24.3. I DENTIFY and S ET U P : It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1. E XECUTE : (a) 6 12 0.148 10 C so 604 V 245 10 F ab ab Q Q C V V C-- × = = = = × (b) so A C d = P ( 29 ( 29 12 3 3 2 2 12 2 2 245 10 F 0.328 10 m 9.08 10 m 90.8 cm 8.854 10 C / N m Cd A---- × × = = = × = × ⋅ P (c) 6 3 604 V so 1.84 10 V/m 0.328 10 m ab ab V V Ed E d- = = = = × × (d) so E σ = P ( 29 ( 29 6 12 2 2...
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This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.

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ass11 - Solutions, Assignment 11 24.1.I DENTIFY : ab Q C V...

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