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6- Let S = 120 cm², d = 4 mm, and ER = 12 for a parallel-plate capacitor. a) Calculate the capacitance: C = EREOS/d = [12e0(120 x 10-4)]/[4 x 10-3] = 3.19 x 10-10 = 319 pF. b) After connecting a 40 V battery across the capacitor. calculate E. D. Q. and the total stored electrostatic energy: E = V/d = 40/(4 x 10-3) = 104 V/m. D EREDE = 12€0 x 10¹ = 1.06 µμC/m². Then Q = D. Blsurface x S = 1.06 x 106 x (120 x 10) = 1.27 x 10-³C = 12.7 nC. Finally W, = (1/2)CV = (1/2)(319 x 10-12) (40)2 = 255 nJ. c) The source is now removed and the dielectric is carefully withdrawn from between the plates. Again calculate E. D. Q. and the energy: With the source disconnected, the charge is constant, and thus so is D: Therefore, Q = 12.7 nC, D = 1.06 μC/m², and E D/0 10/8.85 x 10-12 1.2 x 105 V/m. The energy is then = WD ExS= -(1.06 x 10-6)(1.2 x 105)(120 × 10-4)(4 × 10-3) = 3.05 3 d) What is the voltage between the plates? V = Exd=(1.2 x 105)(4 x 10-3) = 480 V.

6- Let S = 120 cm², d = 4 mm, and ER = 12 for a parallel-plate capacitor. a) Calculate the capacitance: C = EREOS/d = [12e0(120 x 10-4)]/[4 x 10-3] = 3.19 x 10-10 = 319 pF. b) After connecting a 40 V battery across the capacitor. calculate E. D. Q. and the total stored electrostatic energy: E = V/d = 40/(4 x 10-3) = 104 V/m. D EREDE = 12€0 x 10¹ = 1.06 µμC/m². Then Q = D. Blsurface x S = 1.06 x 106 x (120 x 10) = 1.27 x 10-³C = 12.7 nC. Finally W, = (1/2)CV = (1/2)(319 x 10-12) (40)2 = 255 nJ. c) The source is now removed and the dielectric is carefully withdrawn from between the plates. Again calculate E. D. Q. and the energy: With the source disconnected, the charge is constant, and thus so is D: Therefore, Q = 12.7 nC, D = 1.06 μC/m², and E D/0 10/8.85 x 10-12 1.2 x 105 V/m. The energy is then = WD ExS= -(1.06 x 10-6)(1.2 x 105)(120 × 10-4)(4 × 10-3) = 3.05 3 d) What is the voltage between the plates? V = Exd=(1.2 x 105)(4 x 10-3) = 480 V.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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electromagnetic field)I want a detailed solution because my teacher changes the numbers. I want a detailed solution. Understand the solution
6- Let S = 120 cm², d = 4 mm, and ER 12 for a parallel-plate capacitor. a) Calculate the capacitance: C=EREOS/d = [12e0(120 x 10-4)]/[4 x 10-3] = 3.19 x 10-10 = 319 pF. EREDE b) After connecting a 40 V battery across the capacitor. calculate E. D. Q. and the total stored electrostatic energy: E = V/d = 40/(4 x 10-3) = 104 V/m. D 12€0 x 10¹ = 1.06 C/m². Then Q = D. Blsurface x S = 1.06 x 106 x (120 x 10) = 1.27 x 10-³C = 12.7 nC. Finally W, = (1/2)CV = (1/2)(319 x 10-12) (40)2 = 255 nJ. c) The source is now removed and the dielectric is carefully withdrawn from between the plates. Again calculate E. D. Q. and the energy: With the source disconnected, the charge is constant, and thus so is D: Therefore, Q = 12.7 nC, D = 1.06 μC/m², and E D/0 10/8.85 x 10-12 1.2 x 105 V/m. The energy is then = (1.00 (1.06 d) What is the voltage between the plates? V = Ex d = (1.2 x 105)(4 × 10-3) = 480 V. W₁ = D. We D.ExS: x 10-6)(1.2 x 105)(120 × 10-4) (4 × 10-³) = 3.05 J >
Transcribed Image Text:6- Let S = 120 cm², d = 4 mm, and ER 12 for a parallel-plate capacitor. a) Calculate the capacitance: C=EREOS/d = [12e0(120 x 10-4)]/[4 x 10-3] = 3.19 x 10-10 = 319 pF. EREDE b) After connecting a 40 V battery across the capacitor. calculate E. D. Q. and the total stored electrostatic energy: E = V/d = 40/(4 x 10-3) = 104 V/m. D 12€0 x 10¹ = 1.06 C/m². Then Q = D. Blsurface x S = 1.06 x 106 x (120 x 10) = 1.27 x 10-³C = 12.7 nC. Finally W, = (1/2)CV = (1/2)(319 x 10-12) (40)2 = 255 nJ. c) The source is now removed and the dielectric is carefully withdrawn from between the plates. Again calculate E. D. Q. and the energy: With the source disconnected, the charge is constant, and thus so is D: Therefore, Q = 12.7 nC, D = 1.06 μC/m², and E D/0 10/8.85 x 10-12 1.2 x 105 V/m. The energy is then = (1.00 (1.06 d) What is the voltage between the plates? V = Ex d = (1.2 x 105)(4 × 10-3) = 480 V. W₁ = D. We D.ExS: x 10-6)(1.2 x 105)(120 × 10-4) (4 × 10-³) = 3.05 J >
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