Chapter 24 Capacitance, dielectrics and electric energy storage Basic circuit devices –Resistors...

Chapter 24Capacitance, 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.

Parallel Plate Capacitor

Fig 26-4b, p.800

Active Figure 26.4

(SLIDESHOW MODE ONLY)

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

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

Factors affecting capacitance

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

– Plates– Cylinders

• Material between conductors– Air– Paper– Wax

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

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

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

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

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?

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

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

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

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

Alternate Energy Expressions

2QU

2C Q

VC

2 22C V 1

U CV2C 2

21 Q 1U V QV

2 V 2

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

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

is the Dielectric Constant

Measure of the degree of dipole alignment in the material

Dielectrics

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

Effect of a dielectric on capacitance

oDielectric

EE

o

Dielectric

EE d d

oDielectric

VV

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

oo

Q QC C

V V Results in a higher capacitance.

Allows more charge to be stored before breakdown voltage.

Effect of the dielectric constant

Parallel Plate Capacitoro o

o o

A AC C C

d d

o

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 E E

2 2 2

Example modificationsusing permittivity

d

3

What is new capacitance?

Example – Parallel Plate Capacitor

+Q

-Q

d

A

oAC

d

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

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

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

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

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

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

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

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

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