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1 University of Rhode Island PHY 204: Elementary Physics II Physics Course Materials Capacitors II Gerhard Müller University of Rhode Island, Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: Abstract Part eight of course materials for Elementary Physics II (PHY 204), taught by Gerhard Müller at the University of Rhode Island. Documents will be updated periodically as more entries become presentable. Some of the slides contain figures from the textbook, Paul A. Tipler and Gene Mosca. Physics for Scientists and Engineers, 5 th /6 th editions. The copyright to these figures is owned by W.H. Freeman. We acknowledge permission from W.H. Freeman to use them on this course web page. The textbook figures are not to be used or copied for any purpose outside this class without direct permission from W.H. Freeman. Recommended Citation Müller, Gerhard, "08. Capacitors II" (2015). PHY 204: Elementary Physics II. Paper This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@URI. It has been accepted for inclusion in PHY 204: Elementary Physics II by an authorized administrator of DigitalCommons@URI. For more information, please contact digitalcommons@etal.uri.edu.
2 Capacitor with Dielectric Most capacitors have a dielectric (insulating solid or liquid material) in the space between the conductors. This has several advantages: Physical separation of the conductors. Prevention of dielectric breakdown. Enhancement of capacitance. The dielectric is polarized by the electric field between the capacitor plates. 14/9/2015 [tsl124 1/13]
3 Parallel-Plate Capacitor with Dielectric (1) The polarization produces a bound charge on the surface of the dielectric. q f q f q f q b q b q f E 0 E The bound surface charge has the effect of reducing the electric field between the plates from E 0 to E. A: area of plates d: separation between plates ±q f : free charge on plate ±q b : bound charge on surface of dielectric E 0 : electric field in vacuum E: electric field in dielectric 14/9/2015 [tsl125 2/13]
4 Parallel-Plate Capacitor with Dielectric (2) Use Gauss law to determine the electric fields E 0 and E. q f q f q f q b q b q f E 0 E Field in vacuum: E 0 A = q f ǫ 0 E 0 = q f ǫ 0 A Field in dielectric: EA = q f q b ǫ 0 E = q f q b ǫ 0 A < E 0 Voltage: V 0 = E 0 d (vacuum), V = Ed = V 0 κ < V 0 (dielectric) Dielectric constant: κ E 0 E = q f q f q b > 1. Permittivity of dielectric: ǫ = κǫ 0. 14/9/2015 [tsl126 3/13]
5 Dielectric Materials Dielectrics increase the capacitance: C/C 0 = κ. The capacitor is discharged spontaneously across the dielectric if the electric field exceeds the value quoted as dielectric strength. 14/9/2015 [tsl138 4/13]
6 Impact of Dielectric (1) What happens when a dielectric is placed into a capacitor with the charge on the capacitor kept constant? vacuum dielectric charge Q 0 Q = Q 0 electric field E 0 E = E 0 κ < E 0 voltage V 0 V = V 0 κ < V 0 capacitance C 0 = Q 0 V 0 C = Q V = κc 0 > C 0 potential energy U 0 = Q2 0 2C 0 U = Q2 2C = U 0 κ < U 0 energy density u (0) E = 1 2 ǫ 0E0 2 u E = u(0) E κ = 1 2 κǫ 0E 2 < u (0) E 14/9/2015 [tsl127 5/13]
7 Impact of Dielectric (2) What happens when a dielectric is placed into a capacitor with the voltage across the capacitor kept constant? vacuum dielectric charge Q 0 Q = κq 0 electric field E 0 E = E 0 voltage V 0 V = V 0 capacitance C 0 = Q 0 V 0 C = Q V = κc 0 > C 0 potential energy U 0 = 1 2 C 0V 2 0 U = 1 2 CV 2 = κu 0 > U 0 energy density u (0) E = 1 2 ǫ 0E0 2 u E = κu (0) E = 1 2 κǫ 0E 2 > u (0) E 14/9/2015 [tsl128 6/13]
8 Stacked Dielectrics Consider a parallel-plate capacitor with area A of each plate and spacing d. Capacitance without dielectric: C 0 = ǫ 0A d. Dielectrics stacked in parallel: C = C 1 C 2 with C 1 = κ 1 ǫ 0 A/2 d, C 2 = κ 2 ǫ 0 A/2 d. C = 1 2 (κ 1 κ 2 )C 0. Dielectrics stacked in series: 1 C = 1 C 1 1 C 2 with C 1 = κ 1 ǫ 0 A d/2, C 2 = κ 2 ǫ 0 A d/2 C = 2κ 1κ 2 κ 1 κ 2 C 0. 14/9/2015 [tsl129 7/13]
9 Lateral Force on Dielectric Consider two charged capacitors with dielectrics only halfway between the plates. In configuration (a) any lateral motion of the dielectric takes place at constant voltage across the plates. In configuration (b) any lateral motion of the dielectric takes place at constant charge on the plates. Determine in each case the direction (left/zero/right) of the lateral force experienced by the dielectric. (a) (b) 14/9/2015 [tsl130 8/13]
10 Geiger Counter Radioactive atomic nuclei produce high-energy particles of three different kinds: α β γ α-particles are 4 He nuclei. β-particles are electrons or positrons. γ-particles are high-energy photons. lowpressure inert gas amplifier loudspeaker Free electrons produced by ionizing radiation are strongly accelerated toward the central wire. Collisions with gas atoms produce further free electrons, which are accelerated in the same direction. An avalanche of electrons reaching the wire produces a current pulse in the circuit. 14/9/2015 [tsl123 9/13]
11 Intermediate Exam II: Problem #1 (Spring 05) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq. (b) Find the voltage V 3 across capacitor C 3. (c) Find the the charge Q 2 on capacitor C 2. C 3 = 3 µf 8V C 1 = 1µF C 2 = 2 µ F 14/9/2015 [tsl336 10/13]
12 Intermediate Exam II: Problem #1 (Spring 05) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq. (b) Find the voltage V 3 across capacitor C 3. (c) Find the the charge Q 2 on capacitor C 2. C 3 = 3 µf 8V C 1 = 1µF C 2 = 2 µ F Solution: (a) C 12 = C 1 C 2 = 3µF, C eq = 1 C 12 1 C 3 «1 = 1.5µF. 14/9/2015 [tsl336 10/13]
13 Intermediate Exam II: Problem #1 (Spring 05) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq. (b) Find the voltage V 3 across capacitor C 3. (c) Find the the charge Q 2 on capacitor C 2. C 3 = 3 µf 8V C 1 = 1µF C 2 = 2 µ F Solution: (a) C 12 = C 1 C 2 = 3µF, C eq = 1 (b) Q 3 = Q 12 = Q eq = C eq (8V) = 12µC V 3 = Q 3 = 12µC C 3 3µF = 4V. C 12 1 C 3 «1 = 1.5µF. 14/9/2015 [tsl336 10/13]
14 Intermediate Exam II: Problem #1 (Spring 05) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq. (b) Find the voltage V 3 across capacitor C 3. (c) Find the the charge Q 2 on capacitor C 2. C 3 = 3 µf 8V C 1 = 1µF C 2 = 2 µ F Solution: (a) C 12 = C 1 C 2 = 3µF, C eq = 1 (b) Q 3 = Q 12 = Q eq = C eq (8V) = 12µC V 3 = Q 3 = 12µC C 3 3µF = 4V. (c) Q 2 = V 2 C 2 = 8µC. C 12 1 C 3 «1 = 1.5µF. 14/9/2015 [tsl336 10/13]
15 Unit Exam II: Problem #1 (Spring 07) Consider the configuration of two point charges as shown. (a) Find the energy U 3 stored on capacitor C 3. (b) Find the voltage V 4 across capacitor C 4. (c) Find the voltage V 2 across capacitor C 2. (d) Find the charge Q 1 on capacitor C 1. C 1 = 2µF 6V C 2 = 2µF C 3 = 3µ F C 4 = 5µ F 14/9/2015 [tsl362 11/13]
16 Unit Exam II: Problem #1 (Spring 07) Consider the configuration of two point charges as shown. (a) Find the energy U 3 stored on capacitor C 3. (b) Find the voltage V 4 across capacitor C 4. (c) Find the voltage V 2 across capacitor C 2. (d) Find the charge Q 1 on capacitor C 1. C 1 = 2µF 6V C 2 = 2µF Solution: C 3 = 3µ F (a) U 3 = 1 2 (3µF)(6V)2 = 54µJ. C 4 = 5µ F 14/9/2015 [tsl362 11/13]
17 Unit Exam II: Problem #1 (Spring 07) Consider the configuration of two point charges as shown. (a) Find the energy U 3 stored on capacitor C 3. (b) Find the voltage V 4 across capacitor C 4. (c) Find the voltage V 2 across capacitor C 2. (d) Find the charge Q 1 on capacitor C 1. C 1 = 2µF 6V C 2 = 2µF Solution: C 3 = 3µ F (a) U 3 = 1 2 (3µF)(6V)2 = 54µJ. (b) V 4 = 6V. C 4 = 5µ F 14/9/2015 [tsl362 11/13]
18 Unit Exam II: Problem #1 (Spring 07) Consider the configuration of two point charges as shown. (a) Find the energy U 3 stored on capacitor C 3. (b) Find the voltage V 4 across capacitor C 4. (c) Find the voltage V 2 across capacitor C 2. (d) Find the charge Q 1 on capacitor C 1. C 1 = 2µF 6V C 2 = 2µF Solution: C 3 = 3µ F (a) U 3 = 1 2 (3µF)(6V)2 = 54µJ. (b) V 4 = 6V. (c) V 2 = 1 6V = 3V. 2 C 4 = 5µ F 14/9/2015 [tsl362 11/13]
19 Unit Exam II: Problem #1 (Spring 07) Consider the configuration of two point charges as shown. (a) Find the energy U 3 stored on capacitor C 3. (b) Find the voltage V 4 across capacitor C 4. (c) Find the voltage V 2 across capacitor C 2. (d) Find the charge Q 1 on capacitor C 1. C 1 = 2µF 6V C 2 = 2µF Solution: C 3 = 3µ F (a) U 3 = 1 2 (3µF)(6V)2 = 54µJ. (b) V 4 = 6V. (c) V 2 = 1 6V = 3V. 2 (d) Q 1 = (2µF)(3V) = 6µC. C 4 = 5µ F 14/9/2015 [tsl362 11/13]
20 Unit Exam II: Problem #1 (Spring 08) The circuit of capacitors is at equilibrium. (a) Find the charge Q 1 on capacitor 1 and the charge Q 2 on capacitor 2. (b) Find the voltage V 1 across capacitor 1 and the voltage V 2 across capacitor 2. (c) Find the charge Q 3 and the energy U 3 on capacitor 3. C 3 = 5µF C 1 = 6µ F C 2 = 12µ F 12V 14/9/2015 [tsl377 12/13]
21 Unit Exam II: Problem #1 (Spring 08) The circuit of capacitors is at equilibrium. (a) Find the charge Q 1 on capacitor 1 and the charge Q 2 on capacitor 2. (b) Find the voltage V 1 across capacitor 1 and the voltage V 2 across capacitor 2. (c) Find the charge Q 3 and the energy U 3 on capacitor 3. Solution: C 3 = 5µF 1 (a) C 12 = 6µF 1 «1 = 4µF, 12µF Q 1 = Q 2 = Q 12 = (4µF)(12V) = 48µC. C 1 = 6µ F C 2 12V = 12µ F 14/9/2015 [tsl377 12/13]
22 Unit Exam II: Problem #1 (Spring 08) The circuit of capacitors is at equilibrium. (a) Find the charge Q 1 on capacitor 1 and the charge Q 2 on capacitor 2. (b) Find the voltage V 1 across capacitor 1 and the voltage V 2 across capacitor 2. (c) Find the charge Q 3 and the energy U 3 on capacitor 3. Solution: C 3 = 5µF 1 (a) C 12 = 6µF 1 «1 = 4µF, 12µF Q 1 = Q 2 = Q 12 = (4µF)(12V) = 48µC. (b) V 1 = Q 1 C 1 = 48µC 6µF = 8V, V 2 = Q 2 C 2 = 48µC 12µF = 4V. C 1 = 6µ F C 2 12V = 12µ F 14/9/2015 [tsl377 12/13]
23 Unit Exam II: Problem #1 (Spring 08) The circuit of capacitors is at equilibrium. (a) Find the charge Q 1 on capacitor 1 and the charge Q 2 on capacitor 2. (b) Find the voltage V 1 across capacitor 1 and the voltage V 2 across capacitor 2. (c) Find the charge Q 3 and the energy U 3 on capacitor 3. Solution: C 3 = 5µF 1 (a) C 12 = 6µF 1 «1 = 4µF, 12µF Q 1 = Q 2 = Q 12 = (4µF)(12V) = 48µC. (b) V 1 = Q 1 C 1 = 48µC 6µF = 8V, V 2 = Q 2 C 2 = 48µC 12µF = 4V. (c) Q 3 = (5µF)(12V) = 60µC, U 3 = 1 2 (5µF)(12V)2 = 360µJ. C 1 = 6µ F C 2 12V = 12µ F 14/9/2015 [tsl377 12/13]
24 Unit Exam II: Problem #2 (Spring 12) Consider a parallel-plate capacitor of capacitance C = 6pF with plates separated a distance d = 1mm and a potential difference V = V V = 3V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. (d) Find the area A of each plate. Q Q E V V d 14/9/2015 [tsl428 13/13]
25 Unit Exam II: Problem #2 (Spring 12) Consider a parallel-plate capacitor of capacitance C = 6pF with plates separated a distance d = 1mm and a potential difference V = V V = 3V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. (d) Find the area A of each plate. Q Q E Solution: (a) E = V d = 3V 1mm = 3000V/m. V V d 14/9/2015 [tsl428 13/13]
26 Unit Exam II: Problem #2 (Spring 12) Consider a parallel-plate capacitor of capacitance C = 6pF with plates separated a distance d = 1mm and a potential difference V = V V = 3V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. (d) Find the area A of each plate. Q Q E Solution: (a) E = V d = 3V 1mm = 3000V/m. (b) Q = CV = (6pF)(3V) = 18pC. V V d 14/9/2015 [tsl428 13/13]
27 Unit Exam II: Problem #2 (Spring 12) Consider a parallel-plate capacitor of capacitance C = 6pF with plates separated a distance d = 1mm and a potential difference V = V V = 3V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. (d) Find the area A of each plate. Q Q E Solution: (a) E = V d = 3V 1mm = 3000V/m. (b) Q = CV = (6pF)(3V) = 18pC. V V d (c) U = 1 QV = 0.5(18pC)(3V) = 27pJ. 2 14/9/2015 [tsl428 13/13]
28 Unit Exam II: Problem #2 (Spring 12) Consider a parallel-plate capacitor of capacitance C = 6pF with plates separated a distance d = 1mm and a potential difference V = V V = 3V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. (d) Find the area A of each plate. Q Q E Solution: (a) E = V d = 3V 1mm = 3000V/m. (b) Q = CV = (6pF)(3V) = 18pC. V V d (c) U = 1 QV = 0.5(18pC)(3V) = 27pJ. 2 (d) A = Cd ǫ 0 = (6pF)(1mm) C 2 N 1 m 2 = m 2. 14/9/2015 [tsl428 13/13]
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University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 18. RL Circuits Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
05. Electric Potential I
University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 05. Electric Potential I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons
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University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 12. Magnetic Field I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons
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