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12 HMK #3 Chp 24

# 12 HMK #3 Chp 24 - 24 CAPACITANCE AND DIELECTRICS Note 24.1...

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24-1 C APACITANCE AND D IELECTRICS Note: These solutions use P 0 for the permittivity of free space. P 0 = ε 0 = 8.85E-12 F/m. 24.1. I DENTIFY : The capacitance depends on the geometry (area and plate separation) of the plates. S ET U P : For a parallel-plate capacitor, V ab = Ed, E = Q ! 0 A , and C = Q V ab . E XECUTE : (a) V ab = Ed = 4 . 00 ! 10 6 V/m ( ) 2 . 50 ! 10 " 3 m ( ) = 1 . 00 ! 10 4 V. (b) Solving for the area gives A = Q E ! 0 = 80 . 0 " 10 # 9 C (4 . 00 " 10 6 V/m )[ 8 . 854 " 10 # 12 C 2 / (N \$ m 2 )] = 2 . 26 " 10 # 3 m 2 = 22.6 cm 2 . (c) C = Q V ab = 80 . 0 ! 10 " 9 C 1 . 00 ! 10 4 V = 8 . 00 ! 10 " 12 F = 8 . 00 pF. E VALUATE : The capacitance is reasonable for laboratory capacitors, but the area is rather large. 24.2. I DENTIFY and S ET U P : 0 A C d = P , Q C V = and V Ed = . (a) C = P 0 A/d = P 0 (0.00122m 2 ) / (0.00328m) = 3.29 pF (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. 24.5. 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, ! . To produce the same ! , more charge is required when the area increases. 24

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