YF_ISM_24

YF_ISM_24 - CAPACITANCE AND DIELECTRICS 24 24.1. 24.2....

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24-1 C APACITANCE AND D IELECTRICS 24.1. IDENTIFY: ab Q C V = SET UP: 6 1 F 10 F μ = EXECUTE: 64 (7.28 10 F)(25.0 V) 1.82 10 C 182 C ab QC V −− ==× = × = EVALUATE: One plate has charge Q + and the other has charge Q . 24.2. IDENTIFY and SET UP: 0 A C d = P , Q C V = and VE d = . (a) 2 00 0.00122 m 3.29 pF 0.00328 m A C d == = PP (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 × = × EVALUATE: The electric field is uniform between the plates, at points that aren't close to the edges. 24.3. IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1. EXECUTE: (a) 6 12 0.148 10 C so 604 V 245 10 F ab ab QQ CV VC × = = × (b) 0 so A C d = P () ( ) 12 3 32 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 dE d = = × × (d) 0 so E σ = P ( ) 61 2 2 2 5 2 0 1.84 10 V/m 8.854 10 C / N m 1.63 10 C/m E × × ⋅ = × P EVALUATE: We could also calculate directly as Q/A . 6 52 1.63 10 C/m , Q A × = × × which checks. 24.4. IDENTIFY: 0 A C d = P when there is air between the plates. SET UP: 22 (3.0 10 m) A is the area of each plate. EXECUTE: 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 × = × EVALUATE: C increases when A increases and C increases when d decreases. 24.5. IDENTIFY: ab Q C V = . 0 A C d = P . SET UP: When the capacitor is connected to the battery, 12.0 V ab V = . EXECUTE: (a) (10.0 10 F)(12.0 V) 1.20 10 C 120 C ab V = × = (b) When d is doubled C is halved, so Q is halved. 60 C Q = . (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 = EVALUATE: 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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24-2 Chapter 24 24.6. IDENTIFY: ab Q C V = . 0 A C d = P . SET UP: When the capacitor is connected to the battery, enough charge flows onto the plates to make 12.0 V. ab V = EXECUTE: (a) 12.0 V (b) (i) When d is doubled, C is halved. ab Q V C = and Q is constant, so V doubles. 24.0 V V = . (ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and 3.0 V V = . EVALUATE: The electric field between the plates is 0 / EQ A = P . ab VE d = . When d is doubled E is unchanged and V doubles. When A is increased by a factor of 4, E decreases by a factor of 4 so V decreases by a factor of 4. 24.7. IDENTIFY: 0 A C d = P . Solve for d . SET UP: Estimate 1.0 cm r = . 2 Ar π = . EXECUTE: 0 A C d = P so 22 00 12 (0.010 m) 2.8 mm 1.00 10 F r d C ππ == = × PP . EVALUATE: The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it is a reasonable approximation to treat them as infinite sheets. 24.8. INCREASE: ab Q C V = . ab d = . 0 A C d = P . SET UP: We want 4 1.00 10 N/C E when 100 V V = .
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_24 - CAPACITANCE AND DIELECTRICS 24 24.1. 24.2....

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