Chapter 24 Capacitance, Dielectrics, Electric Energy Storage · 3/06/2012 1 Copyright © 2009...

3/06/2012

1

Copyright © 2009 Pearson Education, Inc.

© 2009 Pearson Education, Inc.

This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Lecture PowerPoints

Physics for Scientists and Engineers, with Modern

Physics, 4th edition

Giancoli

Chapter 24


Chapter 24Capacitance, Dielectrics, Electric Energy Storage

3/06/2012

2


• Capacitors

• Determination of Capacitance

• Capacitors in Series and Parallel

• Electric Energy Storage

• Dielectrics

Units of Chapter 24


A capacitor consists of two conductors that are close but not touching. The space between the plates is an insulator (dielectric). A capacitor has the ability to store electric charge.

24-1 Capacitors

(a) parallel plate capacitor (b) cylindrical capacitor(rolled up parallel plate)

3/06/2012

3


(a) Parallel-plate capacitor connected to battery. (b) is a circuit diagram.

24-1 Capacitors


When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage:

The quantity C is called the capacitance.

Unit of capacitance: the farad (F):

1 F = 1 C/V

24-1 Capacitors

V

QCCVQ

3/06/2012

4


24-2 Determination of Capacitance

]capacitor plate parallel[ :eCapacitanc

shown that havebut

and between relation have we23chapter From

density charge where

:is plates parallel obetween tw field

electric the21chapter in shown As

0

0BAAB

ABBA

0

0

d

Aε

V

QC

A

QdEdV-EdV

dVVVV

V

A

QE

A

QE

l

E

E


24-2 Determination of Capacitance

Example 24-1: Capacitor calculations.

(a) Calculate the capacitance of a parallel plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0 mm air gap.

(b) What is the charge on each plate if a 12 V battery is connected across the two plates?

(c) What is the electric field between the plates?

(d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d.

3/06/2012

5



(a) Calculate the capacitance of a parallel plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0 mm air gap.

(b) What is the charge on each plate if a 12 V battery is connected across the two plates?

pC 640C 106.412105.3

pF 53F 105.3001.0

03.02.01085.8

/N.mC 108.85

1011

1112

22120

0

CVQ

d

AC



(c) What is the electric field between the plates?

(d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d.

22812

3

0

0

22120

km 110 m 101.1108.85

100.1

/N.mC 108.85

V/m 12000001.0

12

CdA

d

AC

d

VE

3/06/2012

6


24-2 Determination of CapacitanceCapacitors are now made with capacitances of 1 farad or more, but they are not parallel-plate capacitors. Instead, they are activated carbon, which acts as a capacitor on a very small scale. The capacitance of 0.1 g of activated carbon is about 1 farad.

Some computer keyboards use capacitors; depressing the key changes the capacitance, which is detected in a circuit.


Capacitors in parallel have the same voltage across each one. The equivalent capacitor is one that stores the same charge when connected to the same battery:

24-3 Capacitors in Series and Parallel

3/06/2012

7


24-3 Capacitors in Parallel

n

1iieq

321eq

321321321

:capacitors ofnumber any For

Charge Total

plates. their across voltagesame thehave capacitors All

CC

CCCV

QC

CCCVVCVCVCQQQQ


Capacitors in series have the same charge. In this case, the equivalent capacitor has the same charge across the total voltage drop. Note that the formula is for the inverse of the capacitance and not the capacitance itself!

24-3 Capacitors in Series

3/06/2012

8


24-3 Capacitors in Series

capacitor.smallest an thesmaller th always is

111 :capacitors ofnumber any For

1111

111 Voltage Total

plates.on their chargesame thehave capacitors All

eq

21eq

321eq

321321321

C

CCC

CCCQ

V

C

CCCQ

C

Q

C

Q

C

QVVVV


24-3 Capacitors in Series and ParallelExample 24-5: Equivalent capacitance.

Determine the capacitance of a single capacitor that will have the same effect as the combination shown.

3/06/2012

9


Example 24-5: Equivalent capacitance.


First replace C2 and C3 in parallel with equivalent capacitor

3223 CCC




Then calculate the equivalent capacitance of C23 and C1

in series

231

231eq

231eq

111

CC

CCC

CCC

3/06/2012

10




If all capacitors are equal: C = C1 = C2 = C3

3

2

2

)2(

231

231eq

C

CC

CC

CC

CCC

CCCC 23223


24-3 Capacitors in Series and Parallel

Example 24-6: Charge and voltage on capacitors.

Determine the charge on each capacitor and the voltage across each, assuming C = 3.0 μF and the battery voltage is V = 4.0 V.

3/06/2012

11




plateson their charge same thehave seriesin Capacitors

C 108.00.4100.2

isit across th battery wi theleaves that charge the

example previous from backwards Working

F 102.0μF 2.0 3

3.02

3

2

example previous From

66eq

eq

6eq

VCQ

C

CC




ltageBattery vo V 4.01.32.7 that Note

V 3.1100.32

100.8

2

V 3.1100.32

100.8

2

C 104.0 charge theshare

plates. their across same thehave and parallelin are &

V 2.7100.3

100.8

ab

6

6

233

6

6

232

63223

32

6

6

11

V

C

QV

C

QV

QQQC

VCC

C

QV

3/06/2012

12


24-3 Capacitors in Series and ParallelExample 24-7: Capacitors reconnected.

Two capacitors, C1 = 2.2 μF and C2 = 1.2 μF, are connected in parallel to a 24 V source as shown. After they are charged they are disconnected from the source and from each other and then reconnected directly to each other, with plates of opposite sign connected together. Find the charge on each capacitor and the potential across each after equilibrium is established.


Example 24-7: Capacitors reconnected.C1 = 2.2 μF and C2 = 1.2 μF, V = 24 V

withstarted they charge total theequals

capacators twoon the charges of sum The 3.

'&'

by given iscapacator each on charge The 2.

capacatoreach across same theis V' Voltage 1.

onreconnectiAfter

μC .882242.1 & μC 52.8242.2

:bygiven is charge

source voltage the toconnected are capacators When the

2211

2211

VCqVCq

CVq

VCQVCQ

CVQ

3/06/2012

13


Example 24-7: Capacitors reconnected.

C1 = 2.2 μF and C2 = 1.2 μF, V = 24 V

μC 5.81.72.1'

μC 161.72.2'

V 7.1104.3

100.24'

gives equations Combining

μC .0428.288.52

same thebemust figuresboth in platesupper on the charge totalThe

22

11

6

21

21

2121

VCq

VCq

CC

qqV

QQqq


A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor:

24-4 Electric Energy Storage

3/06/2012

14


A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor:


:can write we using

1 2

21

00

CVQC

Qdqq

CdqVWU

QQ

QVCVC

QU 2

1221

2

21


24-4 Electric Energy StorageExample 24-8: Energy stored in a capacitor.

A camera flash unit stores energy in a 150 μFcapacitor at 200 V.(a) How much electric energy can be stored?(b) What is the power output if nearly all this energy is released in 1.0 ms?

3/06/2012

15


24-4 Electric Energy StorageExample 24-8: Energy stored in a capacitor.

A camera flash unit stores energy in a 150 μFcapacitor at 200 V.

(a) How much electric energy can be stored?

(b) What is the power output if nearly all this energy is released in 1.0 ms?

J 3.02

)200(10150 262

21

CVU

W300010

0.3 :Power

3 t

UP


24-4 Electric Energy StorageConceptual Example 24-9: Capacitor plate separation increased.

A parallel plate capacitor carries charge Q and is then disconnected from a battery. The two plates are initially separated by a distance d. Suppose the plates are pulled apart until the separation is 2d. How has the energy stored in this capacitor changed?

doubled. isEnergy 2/

same theis charge but the

2 then 2 If

22

21

2

21

0

C

Q

C

Q

C

QU

C/Cddd

AC

3/06/2012

16



Example 24-10: Moving parallel capacitor plates.

The plates of a parallel-plate capacitor have area A, separation x, and are connected to a battery with voltage V. While connected to the battery, the plates are pulled apart until they are separated by 3x. (a) What are the initial and final energies stored in the capacitor? (b) How much work is required to pull the plates apart (assume constant speed)? (c) How much energy is exchanged with the battery?




Area = A, Separation x 3x, Battery voltage = V.

(a) What are the initial and final energies stored in the capacitor?

x

AV

x

AV

x

AVUUU

x

AV

x

AVVCU

x

AVVCU

VVd

AC

326

energyin Change632

2

)constant is that Note(

20

20

20

12cap

20

202

221

2

202

121

1

batt0

3/06/2012

17





(b) How much work is required to pull the plates apart (assume constant speed)?

x

AVd

x

AVdQEdFW

dQEdFW

x

x

x

x

x

x

x

x

x

x

3

V/E is hissymmetry t(by V/2E use

only E todue plate oneon force theneed We

20

3

2

20

33

21

33

ll

ll

ll

ll





(c) How much energy is exchanged with the battery?

battery. theintoback flows Charge 3

2

33

ΔΔ Systemon Work 2

02

02

0capbatt

battcap

x

AV

x

AV

x

AVUWU

UUW

3/06/2012

18


Heart defibrillators use electric discharge to “jump-start” the heart, and can save lives.



A dielectric is an insulator, and is characterized by a dielectric constant K.

Capacitance of a parallel-plate capacitor filled with dielectric:

24-5 Dielectrics

Using the dielectric constant, we define the permittivity:

d

AKC 0

0 K

3/06/2012

19


Dielectric strength is the maximum field a dielectric can experience without breaking down.

24-5 Dielectrics

The higher the dielectric constant, the higher the capacitance. e.g. for the same area and plate separation a strontium titanate dielectric would produce a capacitance 300 times that if the gap was air filled.


24-5 DielectricsHere are two experiments where we insert and remove a dielectric from a capacitor. In the first, the capacitor is connected to a battery, so the voltage remains constant. The capacitance increases, and therefore the charge on the plates increases as well.

3/06/2012

20


24-5 DielectricsIn this second experiment, we charge a capacitor, disconnect it, and then insert the dielectric. In this case, the charge remains constant. Since the dielectric increases the capacitance, the potential across the capacitor drops.


24-5 Dielectrics

Example 24-11: Dielectric removal.A parallel plate capacitor, filled with a dielectric with K = 3.4,is connected to a 100 V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m2 and are separated by d = 4.0 mm.(a) Find the capacitance, the charge on the capacitor, the

electric field strength, and the energy stored in the capacitor.

(b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor.

3/06/2012

21


24-5 Dielectrics

Example 24-11: Dielectric removal.K = 3.4, V = 100, A = 4.0 m2 and d = 4.0 mm.(a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor. (0 = 8.8510-12 C2/N.m2)

C = Kε0A/d = 3.010-8 F

Q = CV = 3.010-6 C

E = V/d = 25000 V/m = 25 kV/m

U = ½ CV 2 = 1.510-4 J


24-5 Dielectrics

Example 24-11: Dielectric removal.K = 3.4, V = 100, A = 4.0 m2 and d = 4.0 mm.(b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor. (0 = 8.8510-12 C2/N.m2)

C = ε0A/d = 8.910-9 F [Q = 3.010-6 C (no change)]

V = Q/C = 340 V

E = V/d = 8.5104 V/m = 85 kV/m

U = ½ CV 2 = ½ 8.910-9(340)2 = 5.110-4 J

The increase in energy comes from work to remove the dielectric

3/06/2012

22

43

Some Sample Q & A

1. A 5 F capacitor is charged up to 5 V, what is the charge on each plate. (ans. 25 C)

2. What is the capacitance of a pair of 2 metre square parallel metal plates separated by a 10 mm thick plastic sheet having a dielectric constant of 3.2

(0=8.85x10-12 F/m)? (ans. 11.3x10-9 F)

C = 5 µF, V = 5 VC = q/V → q = CV = 5 × 10-6 × 5 = 25 × 10-6 C = 25 µC

nF 11.3F 101.130.01

4108.853.2 812

0

d

AKεC

K = 3.20 = 8.85 x 10-12 F/md = 10 mm = 0.01 mA = 2 x 2 = 4 m2

44

Since d

AKC 0and

So the voltage on a charged parallel plate capacitor is:

3. What happens if the plates of a charged capacitor are pulled apart?

A

d

K

qV

0

K and 0 are constants and q and A do not change. If d increases then V will also increase.

V

qC

3/06/2012

23

45

4. In a certain application (eg. Electronic flash) a capacitor is required to store and then deliver 60 J of energy.

What value of capacitance is required if the supply voltage is 100 V ? (ans. 12000 F)

U = ½CV2 → C = 2U/V2

= (2 × 60)/1002 = 12 × 10-3 F = 12 mF (12000 µF)

U = 60 JV = 100 V

5. What is the equivalent capacitance of the arrangement if the value of each capacitor is 20 F ?

(ans. 30 F)

20uF 20uF

20uF

10uF

20uF

30uF

Replace series capacitors:

For parallel capacitors:

μF 10

10

1

20

1

20

1111

eq

21eq

C

CCC

Ceq = C1 + C2 = 10 + 20 = 30 µF

3/06/2012

24


• Capacitor: nontouching conductors carrying equal and opposite charge.

• Capacitance:

• Capacitance of a parallel-plate capacitor:

Summary of Chapter 24



• Capacitors in parallel:

• Capacitors in series:

3/06/2012

25


• A dielectric is an insulator.

• Dielectric constant gives ratio of total field to external field.

• For a parallel-plate capacitor:


Chapter 24 Capacitance, Dielectrics, Electric Energy Storage · 3/06/2012 1 Copyright © 2009...

Documents

Transcript of Chapter 24 Capacitance, Dielectrics, Electric Energy Storage · 3/06/2012 1 Copyright © 2009...