ass11 - Solutions Assignment 11 24.1.IDENTIFY C = Q Vab SET...

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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 : 0 A C d = P , Q C V = and V Ed = . (a) 2 0 0 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) 0 so A C d = P ( 29 ( 29 12 3 3 2 2 12 2 2 0 245 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) 0 so E σ = P ( 29 ( 29 6 12 2 2 5 2 0 1.84 10 V/m 8.854 10 C / N m 1.63 10 C/m E σ - - = = × × = × P E VALUATE : We could also calculate σ directly as Q/A . 6 5 2 3 2 0.148 10 C 1.63 10 C/m , 9.08 10 m Q A σ - - - × = = = × × which checks.
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