Chapter S26

33
Chapter 26 Capacitance, dielectrics and electric energy storage Basic circuit devices – Resistors – Capacitors – Inductors Power supply (Battery, Generator) Use our knowledge of electric fields, potentials, and energy to describe how capacitors work.

Transcript of Chapter S26

Page 1: Chapter S26

Chapter 26Capacitance, dielectrics and

electric energy storage • Basic circuit devices

– Resistors

– Capacitors

– Inductors

– Power supply (Battery, Generator)

• Use our knowledge of electric fields, potentials, and energy to describe how capacitors work.

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Parallel Plate Capacitor

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Fig 26-4b, p.800

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Active Figure 26.4

(SLIDESHOW MODE ONLY)

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Potential difference and electric fields in a uniform electric field

+Q

-Qa

b

d

d d

bao o o0 0

ˆ ˆ ˆV j dxi dyj dy d

enclosedE

o

QE dA

o

AE A

o

E

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Potential difference and electric fields in a uniform electric field

+Q

-Qa

b

d

bao o

QdV d

A

baQ V

baQ CV

The constant of proportionality is called “capacitance.”For a parallel plate capacitor, the capacitance is:

0AC

d

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Factors affecting capacitance

• Size of the capacitor (A, d)• Geometric arrangement

– Plates– Cylinders

• Material between conductors– Air– Paper– Wax

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Units of capacitance

Coulomb CFarad F

Volt V

ba

QC

V

A Farad is a lot of capacitance. Typical capacitors are“micro, nano, pico-Farad

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Capacitance – Isolated Sphere

• Assume a spherical charged conductor

• Assume V = 0 at infinity

• Note, this is independent of the charge and the potential difference

4/ o

e e

Q Q RC πε R

V k Q R k

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Cylindrical capacitor

b a

enclosedE

o

QE dA

+Q on center conducting cylinder

-Q on outer conducting cylinder

o

QE 2 rL

o

QE

2 rL

L

b b

bao o oa a

Q Q dr Q bˆ ˆV r drr ln

2 rL 2 L r 2 L a

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Capacitance of a cylindrical capacitor

abQ CV

b a

L

o

ab

o

2 LQ QC

Q b bV ln ln2 L a a

L

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Example

• How strong is the electric field between the plates of a 0.80 F air gap capacitor if they are 2.0 mm apart and each has a charge of 72 C?

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Capacitors in Parallel

total 1 2 3Q Q Q Q

a b

abV V

1Q

3Q

2Qtotal eqQ C V

1C

2C

3C

1 1Q C V 2 2Q C V 3 3Q C V

eq 1 2 3C V C V C V C V

eq 1 2 3C C C C

1 2 3V V V V

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Capacitors in Series

1 2 3V V V V

a b

abV V

1Q 3Q2Q

eqeq

QQ C V V

C 1C 3C2C

11

QV

C

eq 1 2 3

Q Q Q Q

C C C C

eq 1 2 3

1 1 1 1

C C C C

22

QV

C 3

3

QV

C

total 1 2 3Q Q Q Q

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Capacitor circuit example

2 F

2 F3 F 4 F

What single capacitor can replace the four shown here?How much charge can the system hold?How much charge is on one of the 2 F capacitors?

3 V

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Energy Storage in Capacitors

bb

U dUV V

q dq

dU Vdq

(Like problem 23-50)

q Q

q 0

U Vdq

bao

QdV

A

bao

Q aV ln

2 L b

Parallel Plates

Concentric Cylinders

Qq Q 2 2

q 0 0

q 1 q QU dq

C C 2 2C

QV

CIn General

Change in potential energywhile charging capacitor

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Alternate Energy Expressions

2QU

2C Q

VC

2 22C V 1

U CV2C 2

21 Q 1U V QV

2 V 2

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Energy Density

Energy per unit volume: Uu

Volume

Consider a Parallel Plate Capacitor: 21U CV

2

0AC

d

V Ed

2 2 200

A1 1U E d Ad E

2 d 2

20

U U 1u E

Volume Ad 2

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Dielectrics•A dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance

•Materials with Dipoles that can align with an external electric Field. Dielectrics include rubber, plastic, and waxed paper

+ ++

--

-

+ ++

---

oE

indE

oDielectric o ind

EE E E

K

K is the Dielectric Constant

Measure of the degree of dipole alignment in the material

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Dielectrics

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Example values of dielectric constant

“Dielectric strength” is themaximum field in the dielectric before breakdown.(a spark or flow of charge)

max maxE V / d

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Effect of a dielectric on capacitance

oDielectric

EE

K o

Dielectric

EE d d

K

oDielectric

VV

K

Potential difference with a dielectric is less than the potential difference across free space

oo

Q QC K KC

V V Results in a higher capacitance.

Allows more charge to be stored before breakdown voltage.

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Effect of the dielectric constant

Parallel Plate Capacitoro o

o o

A K AC C KC

d d

oK

AC

d

Material permittivity measures degree to which the material permits induced dipolesto align with an external field

2 2 2o 0 0

1 1 1u E u K E E

2 2 2

Example modificationsusing permittivity

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d

3

What is new capacitance?

Example – Parallel Plate Capacitor

+Q

-Q

d

A

oAC

d

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DipolesThe combination of two equal charges of opposite sign, +Q and –Q, separated by a distance l

p Q2a p

2a

-Q +Q

1p

2p

1 2p p p

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Dipoles in a Uniform Electric Field

p

2a

-Q

+Q F QE

F QE

E

F a sin F a sin QEa sin QEa sin Q2aEsin pEsin

x y z

ˆ ˆ ˆi j k

r F x y z

F F F

r F sin

p E

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Work Rotating a Dipole in an Uniform Electric Field

p

2a

-Q

+Q F QE

F QE

E

dW d pEsin d dU

oLet: U( =90 ) 0 U pE cos

U p E

0U pE sin d pE cos U

0U 0

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Example P26.9

When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?

0 AQ Vd

12 2 2

09 2 4 2 2

8.85 10 C N m 150 V4.42 m

30.0 10 C cm 1.00 10 cm m

Vd

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Example P26.21

Four capacitors are connected as shown in Figure P26.21.

(a) Find the equivalent capacitance between points a and b.

(b) Calculate the charge on each capacitor if ΔVab = 15.0 V.

1

2.50 F

2.50 6.00 8.50 F

1 15.96 F

8.50 F 20.0 F

s

p

eq

C

C

C

1 1 115.0 3.00sC

5.96 F 15.0 V 89.5 CQ C V

89.5 C4.47 V

20.0 F15.0 4.47 10.53 V

6.00 F 10.53 V 63.2 C on 6.00 F

QV

C

Q C V

89.5 63.2 26.3 C

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Example P26.27

Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.

1

1

2

1

1 13.33 F

5.00 10.0

2 3.33 2.00 8.66 F

2 10.0 20.0 F

1 16.04 F

8.66 20.0

s

p

p

eq

C

C

C

C

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Example P26.35

A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?

2 12d d 2 112

C C stored energy doubles,

. Therefore, the

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Example P26.43Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.

12 4 2110

5

2.10 8.85 10 F m 1.75 10 m8.13 10 F 81.3 pF

4.00 10 m

AC

d

6 5max max 60.0 10 V m 4.00 10 m 2.40 kVV E d

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Example P26.59A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.

8 maxmax 2.00 10 V m

VE

d

60 0.250 10 FA

Cd

62max

12 80 0 max

0.250 10 40000.188 m

3.00 8.85 10 2.00 10

C VCdA

E

3.00