Copyright © 2009 Pearson Education, Inc. May Term in Guatemala GDS 3559/STS 3500: Engineering...

Copyright © 2009 Pearson Education, Inc.

May Term in Guatemala

GDS 3559/STS 3500: Engineering Public Health: An Interdisciplinary Exploration

of Community Development in Guatemala

Contact Kent Wayland ([email protected]), Eric Anderson ([email protected]), or David R. Burt, MD, at [email protected]

for more information!

Want to Learn how engineering

affects community

health?

Interested in Public Health? Global Health? Medicine?


Chapter 24Capacitance, Dielectrics, Electric Energy Storage


Recap:

12

Capacitor is any pair of conductors which

a) have a potential difference between them and

b) have on them charges of equal magnitude and opposite sign.

; 1 1 1 ; 1 10 1"puff"

Simplest version

Q CV

C Q V F C V pF F

0

/model is parallel-plate capacitor

C pp

A

d

Capacitor Capacitor CC11 is connected across is connected across

a battery of a battery of 5 V5 V. An identical . An identical

capacitor capacitor CC22 is connected across is connected across

a battery of a battery of 10 V10 V. Which one has . Which one has

more charge?more charge?

1) CC11

2) CC22

3) both have the same charge

4) it depends on other factors

ConcepTest 24.1ConcepTest 24.1 CapacitorsCapacitors

Since QQ = = CVCV and the two capacitors are

identical, the one that is connected to the

greater voltagegreater voltage has more charge charge, which

is CC22 in this case.

Capacitor Capacitor CC11 is connected across is connected across

a battery of a battery of 5 V5 V. An identical . An identical

capacitor capacitor CC22 is connected across is connected across

a battery of a battery of 10 V10 V. Which one has . Which one has

more charge?more charge?

1) CC11

2) CC22

3) both have the same charge

4) it depends on other factors

ConcepTest 24.1ConcepTest 24.1 CapacitorsCapacitors

1) increase the area of the plates increase the area of the plates

2) decrease separation between the platesdecrease separation between the plates

3) decrease the area of the plates

4) either (1) or (2)


What must be done to What must be done to

a capacitor in order to a capacitor in order to

increase the amount of increase the amount of

charge it can hold (for charge it can hold (for

a constant voltage)?a constant voltage)?

+Q –Q

ConcepTest 24.2aConcepTest 24.2a Varying Capacitance IVarying Capacitance I

Since Q = CVQ = CV, in order to increase the charge

that a capacitor can hold at constant voltage,

one has to increase its capacitanceincrease its capacitance. Since the

capacitance is given by , that can be

done by either increasing increasing AA or decreasing decreasing dd.

1) increase the area of the plates increase the area of the plates

2) decrease separation between the platesdecrease separation between the plates

3) decrease the area of the plates



dAC 0

What must be done to What must be done to

a capacitor in order to a capacitor in order to

increase the amount of increase the amount of

charge it can hold (for charge it can hold (for

a constant voltage)?a constant voltage)?

+Q –Q

ConcepTest 24.2aConcepTest 24.2a Varying Capacitance IVarying Capacitance I


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

11 2 3 1 2 3

1 2 3

2 3

[Parallel]

eqtot

eq

Q V Q Q Q CV C V C V VC C C

C

C

C C C

a bV V

a bV V

a bV V


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!


1 2 3 1 2 3

1 1 1 1 [Series]

eq eqC C C C C C C C

Q Q Q Q

1 2 3

QQ CV V

CV V V V


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.

1 2 3C C C C


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.

1 2 3

23 2 3

123 23 1

1 23123

1 23

2

1 1 1

2 2

2 3

C C C C

C C C C

C C C

C CC

C C

C CC

C C



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.

23 2C C





Q QQQ

0V

battV V

23

123

2

2

3

C C

C C





Q QQQ

0V

battV V

123

23

1

2

2 3

3

2

3

2

8

4

batt batt CQ C V CV

QV

C

Q

Q Q C

3C

1

2battV C

2 11

2 3

4

3 3

4

38

3

batt

batt

V V V

VV

V VV V V

23

123

2

2

3

C C

C C

ConcepTest 24.3aConcepTest 24.3a Capacitors ICapacitors I

o

o

C CC

Ceq

1) 1) CCeq eq = 3/2= 3/2CC

2) 2) CCeq eq = 2/3= 2/3CC

3) 3) CCeq eq = 3= 3CC

4) 4) CCeq eq = 1/3= 1/3CC

5) 5) CCeq eq = 1/2= 1/2CC

What is the equivalent capacitance, What is the equivalent capacitance,

Ceq , of the combination below? , of the combination below?

The 2 equal capacitors in seriesseries add

up as inversesinverses, giving 1/21/2CC. These

are parallelparallel to the first one, which

add up directlydirectly. Thus, the total

equivalent capacitance is 3/23/2CC.

ConcepTest 24.3aConcepTest 24.3a Capacitors ICapacitors I

o

o

C CC

Ceq

1) 1) CCeq eq = 3/2= 3/2CC

2) 2) CCeq eq = 2/3= 2/3CC

3) 3) CCeq eq = 3= 3CC

4) 4) CCeq eq = 1/3= 1/3CC

5) 5) CCeq eq = 1/2= 1/2CC

What is the equivalent capacitance, What is the equivalent capacitance,

Ceq , of the combination below? , of the combination below?

ConcepTest 24.3bConcepTest 24.3b Capacitors IICapacitors II

1) 1) VV11 == VV22

2) 2) VV11 >> VV22

3) 3) VV11 << VV22

4) all voltages are zero4) all voltages are zero

CC11 = 1.0 = 1.0 FF CC33 = 1.0 = 1.0 FF

CC22 = 1.0 = 1.0 FF

10 V10 V

How does the voltage How does the voltage VV11 across across

the first capacitor (the first capacitor (CC11) compare to ) compare to

the voltage the voltage VV22 across the second across the second

capacitor (capacitor (CC22)?)?

ConcepTest 24.3bConcepTest 24.3b Capacitors IICapacitors II

1) 1) VV11 == VV22

2) 2) VV11 >> VV22

3) 3) VV11 << VV22

4) all voltages are zero4) all voltages are zero

CC11 = 1.0 = 1.0 FF CC33 = 1.0 = 1.0 FF

CC22 = 1.0 = 1.0 FF

10 V10 V

The voltage across C1 is 10 V.

The combined capacitors C2 +

C3 are parallel to C1. The

voltage across C2 + C3 is also

10 V. Since C2 and C3 are in

series, their voltages add.

Thus the voltage across C2

and C3 each has to be 5 V,

which is less than V1.

How does the voltage How does the voltage VV11 across across

the first capacitor (the first capacitor (CC11) compare to ) compare to

the voltage the voltage VV22 across the second across the second

capacitor (capacitor (CC22)?)?

Follow-up:Follow-up: What is the current in this circuit??


A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor. Consider two sheets of charge, one positive and one negative on the surface of conductor b.

24-4 Electric Energy Storage

Work required to pull the negatively charged sheet to conductor a is:

2

2

0

2

0

2 2

2 2 2

tot ext E ext E

ex E

E

t E

W K W W W U

Q A QW U F d Q d

C

Q CVQ CV QV

UC



National Ignition Facility (NIF)

Lawrence Livermore National Laboratory


NIFLaser system driven by 4000 300 μF capacitorswhich store a total of 422 MJ. They take 60 s tocharge and are discharged in 400 μs.1)What is the potential difference across each capacitor?2)What is the power delivered during the discharge?


NIFLaser system driven by 4000 300 μF capacitorswhich store a total of 422 MJ. They take 60 s tocharge and are discharged in 400 μs.1)What is the potential difference across each capacitor?2)What is the power delivered during the discharge?

Solution:1)U = CV2/2 →V = (2U/C)1/2 →V = [2(422x106)/4000/300x10-6] ½ = 26.5 kV2)P = W/t = U/t = 422x106 /400x10-6 ~ 1012 W = 1000 GW! cf. 1.0-1.5 GW for power plant



Conceptual Example: 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. a) How has the energy stored in this capacitor changed? b) Would the answer change if the battery were not disconnected?


24-4 Electric Energy Storage2

0

2

0

2'

0

22 0

2' 0

a) ;2

2

22

2

Work done pulling plates apart

1b)

2 2

2 2 2Work done "pushing" charge back to battery

E

E E

E

EE

AQU C

C d

Q d

A

Q dU U

A

AVU CV

d

A UVU

d


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



The energy density, u, defined as the energy per unit volume, is the same no matter the origin of the electric field:

The sudden discharge of electric energy can be harmful or fatal. Capacitors can retain their charge indefinitely even when disconnected from a voltage source – be careful!


2 22

2 E EU U Q C QuVol Ad Ad A d

d2

0

0 0 0 0

2

02

1

2

E

Au


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:


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

24-5 Dielectrics


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.


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 DielectricsExample 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.


The molecules in a dielectric, when in an external electric field, tend to become oriented in a way that reduces the external field.

24-6 Molecular Description of Dielectrics


This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. This reorientation of the molecules results in an induced charge – there is no net charge on the dielectric, but the charge is asymmetrically distributed.

The magnitude of the induced charge depends on the dielectric constant:

24-6 Molecular Description of Dielectrics


• 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:


• Energy density in electric field:

• A dielectric is an insulator.

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

• For a parallel-plate capacitor:



Questions?

Copyright © 2009 Pearson Education, Inc. May Term in Guatemala GDS 3559/STS 3500: Engineering...

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Transcript of Copyright © 2009 Pearson Education, Inc. May Term in Guatemala GDS 3559/STS 3500: Engineering...