Chapter_24_Solutions - 24.1: Q = CV = (25.0 V)(7.28 F) =...

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24.1: Q = CV = (25.0 V)(7.28 µ F) = 1.82 × 10 4 C.
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24.2: a) C = ε 0 A d = 0 0.00122 m 2 0.00328 m = 3.29 pF. b) V = Q C = 4.35 × 10 8 C 3.29 × 10 12 F = 13.2 kV. c) E = V d = 13.2 × 10 3 V 0.00328 m = 4.02 × 10 6 Vm .
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24.3: a) V = Q C = 0.148 × 10 6 C 2.45 × 10 10 F = 604 V. b) A = Cd ε 0 = (2.45 × 10 10 F)(0.328 × 10 3 m) 0 = 0.0091 m 2 . c) E = V d = 604 V 0.328 × 10 3 m = 1.84 × 10 6 Vm . d) E = σ 0 = 0 E = 0 (1.84 × 10 6 ) = 1.63 × 10 5 Cm 2 .
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24.4: V = Ed = σ ε 0 d = (5.60 × 10 12 Cm 2 )(0.00180 m) 8.85 × 10 12 C 2 Nm 2 =1.14 mV
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24.5: a) Q = CV = 120 µ C b) C = ε 0 Ad d d2 means C C2 and Q Q2 = 60 c) r 2r means A 4A,C 4C, and Q 4Q = 480
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24.6: (a) 12.0 V since the plates remain charged. (b) (i) V = Q C Q does not change since the plates are disconnected from the battery. C = ε A d If d is doubled, C 1 2 C, so V 2V = 24.0V (ii) which means that A = π r 2 ,soifr 2r, then A 4A, and C 4C V 1 4 V = 3.00 V
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24.7: Estimate r = 1.0 cm C = ε 0 A d so d = 0 π r 2 C = 0 (0.010 m) 2 1.00 × 10 12 F = 2.8 mm 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.
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24.8: (a) V = Ed 100 V = (10 4 NC)d d = 10 2 m = 1.00 cm C = ε 0 A d = 0 π R 2 d R = Cd πε 0 = 4Cd 4 0 R = 4(5.00 × 10 12 F)(10 2 m)(9 × 10 9 Nm 2 C 2 ) R = 4.24 × 10 2 m = 4.24 cm (b) Q = CV = (5pF)(100 V) = 500 pC
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24.9: a) C L = 2 πε 0 ln(r b r a ) C = (0.180 m)2 0 ln(5.00 0.50 ) = 4.35 × 10 12 F b) V = QC = (10.0 × 10 12 C) (4.35 × 10 12 F) = 2.30 V
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24.10: a) C L = 2 πε 0 ln(r b r a ) ln(r b r a ) = 2 0 CL = 2 0 31.5 × 10 12 Fm = 1.77 r b r a = 5.84. b) Q L = V C L = (2.60V)(31.5 × 10 12 ) = 8.19 × 10 11 Cm .
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24.11: a) CL = 2 πε 0 ln(r b r a ) = 2 0 ln(3.5 mm 1.5 mm) = 6.56 × 10 11 Fm . b) The charge on each conductor is equal but opposite. Since the inner conductor is at a higher potential it is positively charged, and the magnitude is: Q = CV = 2 0 LV ln(r b r a ) = 2 0 (2.8 m)(0.35 V) ln(3.5 mm 1.5 mm ) = 6.43 × 10 11 C.
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24.12: a) For two concentric spherical shells, the capacitance is: C = 1 k r a r b r b r a ⎟ ⇒ kCr b kCr a = r a r b r b = kCr a kC r a r b = k(116 × 10 12 F)(0.150 m) k(116 × 10 12 F) 0.150 m = 0.175 m. b) V = 220 V, and Q = CV = (116 × 10 12 F)(220 V) = 2.55 × 10 8 C.
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24.13: a) C = 1 k r b r a r b r a ⎟ = 1 k (0.148 m)(0.125 m) 0.148 m 0.125 m ⎟ = 8.94 × 10 11 F. b) The electric field at a distance of 12.6 cm: E = kQ r 2 = kCV r 2 = k(8.94 × 10 11 F)(120 V) (0.126 m) 2 = 6082 N/C. c) The electric field at a distance of 14.7 cm: E = kQ r 2 = kCV r 2 = k(8.94 × 10 11 F)(120 V) (0.147 m) 2 = 4468 N/C. d) For a spherical capacitor, the electric field is not constant between the surfaces.
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24.14: a) 1 C eq = 1 C 1 + C 2 + 1 C 3 = 1 ((3.0 + 5.0) × 10 6 F) + 1 (6.0 × 10 6 C eq = 3.42 × 10 6 F. The magnitude of the charge for capacitors in series is equal, while the charge is distributed for capacitors in parallel. Therefore, Q 3 = Q 1 + Q 2 = VC eq = (24.0 V)(3.42 × 10 6 = 8.21 × 10 5 C. Since C and are at the same potential, 1 C 2 Q 1 C 1 = Q 2 C 2 Q 2 = C 2 C 1 Q 1 = 5 3 Q, Q 3 = 8 3 Q 1 = 8.21 × 10 5 C Q 1 = 3.08 × 10 5 C, and Q 2 = 5.13 × 10 5 b) V 2 = V 1 = Q 1 C 1 = (3.08 × 10 5 C) (3.00 × 10 6 = 10.3 V. And V 3 = 24.0 V 10.3 V = 13.7 V.
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Chapter_24_Solutions - 24.1: Q = CV = (25.0 V)(7.28 F) =...

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