24_InstructorSolutions

24_InstructorSolutions - CAPACITANCE AND DIELECTRICS 24...

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24-1 C APACITANCE AND D IELECTRICS 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 ( )( ) 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 ( )( ) 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. 24.4. I DENTIFY : 0 A C d = P when there is air between the plates. S ET U P : 2 2 (3.0 10 m) A = × is the area of each plate. E XECUTE : 12 2 2 12 3 (8.854 10 F/m)(3.0 10 m) 1.59 10 F 1.59 pF 5.0 10 m C × × = = × = × E VALUATE : C increases when A increases and C increases when d decreases. 24.5. I DENTIFY : ab Q C V = . 0 A C d = P . S ET U P : When the capacitor is connected to the battery, 12.0 V ab V = . E XECUTE : (a) 6 4 (10.0 10 F)(12.0 V) 1.20 10 C 120 C ab Q CV μ = = × = × = (b) When d is doubled C is halved, so Q is halved. 60 C Q μ = . (c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4. 480 C. Q μ = E VALUATE : When the plates are moved apart, less charge on the plates is required to produce the same potential difference. With the separation of the plates constant, the electric field must remain constant to produce the same potential difference. The electric field depends on the surface charge density, σ . To produce the same σ , more charge is required when the area increases. 24
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