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 c) The source is now removed and the dielectric is carefully withdrawn from between the plates.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 c) The source is now removed and the dielectric is carefully withdrawn from between the plates.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.
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Publisher:Robert L. Boylestad
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Question
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
>](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F229979b8-5ee0-45a2-81bc-0e759935c064%2F1ee7e8c2-cdc9-46b8-b10d-076df4560d35%2Ffsb09jxb_processed.jpeg&w=3840&q=75)
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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