0p

CH4

HH

H

H

C

1. classify

§9-3 Dielectrics

Dielectrics ： isolatorAlmost no free charge inside

Non-polar molecules dielectrics

The centers of positive and negative charge coincide in the absence of external E-field.

The charges within it are bound in the range of a molecule.

Polar molecules dielectrics

The centers of positive and negative charge do not coincide in the absence of external E-field. H2O

HHO

p

0p

0E

0E

p

The polarization of the non-polar molecules

束缚电

荷 束缚电

荷

2. The polarization of dielectrics

The polarization charges appear in dielectrics when an external E-field is supplied.

bound charges

Displacement polarization

The polarization of the polar molecules

1f

2f

0E

0E

p

EpM

束缚电

荷 束缚电

荷

Orientation polarization

The difference between free charge and bound charge

Can not move in the dielectrics freely and can not leave the dielectrics.

Can move in the conductors freely and can leave the conductors.

3. The electrostatic field in dielectrics3. The electrostatic field in dielectrics

EEE

0

Special example ：

'

'

EEE 000

0E

'E

=external field + polar field

r

E

0

Two metal plates + uniform dielectrics block

r --relative permitivity

r

EE

0 1r 0EE

The relation of area charge densities between bound charge and free charge:

r

11

S

S

qSdE内0

1

内 内S S

qq )(10

0

§9-4 Gauss’ law in dielectrics

Gauss’ Law in vacuum:

=?

'

'

SFrom above:

r

11

Special example ：Two metal plates + uniform dielectrics block

内Sr

SqSdE 0

0

1

内内 SrS

qq 0

11'

'

'

S

permittivity

内S

SqSdE 0

1

Gauss’ law of E-field in dielectrics

Free charge

S

S

qSdD内

0

-- Gauss’ Law of D

NotesNotes

flux flux depends on the free charges in the Gaussidepends on the free charges in the Gaussian surface. But the itself depends on free chargean surface. But the itself depends on free charges and bound charges inside and outside Gaussian ss and bound charges inside and outside Gaussian surface.urface.

D

D

ED

-- electric displacement vector

Definition:Definition:

Free charge

E-displacement flux ( D flux)

The difference between -line difference between -line and -line. and -line. E

D

-line::originate on entire positoriginate on entire positive charges and terminate on ive charges and terminate on entire negative charges.entire negative charges.

E

-line::originate on positive froriginate on positive free charges and terminate on ee charges and terminate on negative free charges.negative free charges.

D

--lines can describe the dislines can describe the distribution of vividly.tribution of vividly.D

D

Free and bound

E

line

D

line

　　[[Example 1Example 1]A metal sphere with radius ]A metal sphere with radius RR charges po charges positive sitive qq00. Put the sphere in the infinity uniform dielec. Put the sphere in the infinity uniform dielectrics(relative permittivity is trics(relative permittivity is rr). Calculate: ). Calculate: E- field E- field outside the sphere. outside the sphere. The area density of bound charThe area density of bound charge close to the metal spherical surface and the amouge close to the metal spherical surface and the amount of bound charge on this thin surface.nt of bound charge on this thin surface.

R

0q

r

SolutionSolution::The distribution of The distribution of charges and E-field : charges and E-field : spherical spherical symmetry.symmetry.

Choose Gaussian surface Choose Gaussian surface SS : sp : spherical surface with radius herical surface with radius rr

S

rDSdD 24

0q

On the metal spherical surface, the area density On the metal spherical surface, the area density of free charges:of free charges:

r

11 2

0

41

Rq

r

r

R

0q

rDE

r

D 0

r

rq

r

ˆ4 2

0

0

2

40

r

qD

Direction:Direction:r̂

20

4 R

q

the area density of bound charges in dielectrics the area density of bound charges in dielectrics close to sphere:close to sphere:

24 Rq

The amount of bound chargesThe amount of bound charges

011 qr

----q has opposite sign with q0 ， and q < q0

[[Example 2Example 2]Two ]Two parallelparallel conductor plates have charge conductor plates have charge area density area density ++ 、、 -- respectively. respectively. Their potential Their potential difference is difference is VV00=300V =300V asas the space between them is the space between them is

vacuum. Keep the charges of the plates are constant and vacuum. Keep the charges of the plates are constant and fill dielectrics (fill dielectrics (rr=4=4)in half of their space. See in figure. )in half of their space. See in figure.

FindFind ：： the potential difference between two platesthe potential difference between two plates== ？？ the area density of bound charge on the top and bottom the area density of bound charge on the top and bottom surface of the dielectricssurface of the dielectrics== ？？

2

2

1

1

'1'1

0V0E

r

Solution Solution ：： Assume the area of plate is Assume the area of plate is SS ，， their distance is their distance is dd

In vacuum:In vacuum:0

0 E

After fill dielectrics in half space:After fill dielectrics in half space:

dEldEVbottom

top 000

1

SSS 22 21

Assume at the area possessing dielectrics:Assume at the area possessing dielectrics:

at the vacuum area :at the vacuum area : 2thenthen

221

r

r

12

1

r

1

22

58

52

Uniting Uniting and and

r

E 0

11 andand

0

22

E

dEdE 21

21 r

dEV 1 dE052 V120

The E-field between two plates:The E-field between two plates:

0

221

EE05

2 05

2 E

The potential difference between two plates:The potential difference between two plates:

56 Top surfaceTop surface

5

6 Bottom surfaceBottom surface

the area density of bound charge the area density of bound charge

11

11

r

§9-2 Capacitors and their Capacitance

1. The capacitance of an isolated conductor

Charged spherical conductor:R

qU

04

1

RU

q04 has nothing to do with q 、 U.

Capacitance—the ability (capacity) of holding charges.

--concern with the shape and size of the conductor.

Special example ：

Definition

The capacitance of an isolated conductor:

U

qC C has nothing to do with Q 、 U.

C depends on the shape , size of the conductor and the dielectrics surrounding it.

2. The capacitance of capacitor

Capacitor ： consist of two isolated conductors that they are closed to each other.

Two plates carry equal but opposite charges.

Conductor plates

Definition

ba UU

qC

V

q

The capacitance of a capacitor:

C has nothing to do with q 、 V.C depends on the shape , size, distance of the conductors and the dielectrics surrounding it.

3. Calculation of the capacitance

(1) A parallel plate capacitor(1) A parallel plate capacitor

q

A B

r S

d

qAssume charges Assume charges qq on on conductor platesconductor plates

E

ldEV BA

EdS

qd

r 0

S

q

r 0

V

qC

d

Sr 0thenthen

Neglect fringing effect, we haveNeglect fringing effect, we have

(2) A cylindrical capacitor(2) A cylindrical capacitor

AB

ab

lr

Two coaxial cylindersTwo coaxial cylinders

aa<<rr<<bb::r

Er

02

ldEV BA

a

b

r

ln2 0

V

qC

ab

lr

ln

2 0

Assume Assume AA charges charges ++qq, , BB charges charges --qq

(3)(3) A spherical capacitorA spherical capacitor

Consist of a concentric Consist of a concentric spherical shell and a sphere.spherical shell and a sphere.

204 r

qE

r

b

aA

Br

ldEV BA

)

11(

4 0 ba

q

r

Edrba

V

qC

ab

abr

04

Assume Assume AA charges charges ++qq, , BB charges charges --qq

The common steps to calculate capacitance The common steps to calculate capacitance C C ::

Assume conductors charge Assume conductors charge q q ..

Find the E-field distribution between two plates.Find the E-field distribution between two plates.

Calculate the potential difference between two Calculate the potential difference between two

plates.plates.

Calculate Calculate C C byby using the definition of using the definition of C C ..

[[ExampleExample] Two plates separation ] Two plates separation dd are filled in air b are filled in air b

etween them to consisting of a capacitor with capetween them to consisting of a capacitor with capacitance of acitance of CC.If a sheet of paper with thickness .If a sheet of paper with thickness dd11 and relative permittivity and relative permittivity rr is inserted between th is inserted between the two plates, calculate to capacitance e two plates, calculate to capacitance CC..

1d2d

3dd

Solution:Solution:

331122 dEdEdEV 30

10

20

dddr

V

qC '

V

S 321

0

ddd

S

r

r

1d2d

3dd

Assume a charge Assume a charge q q on on plateplate ，，

S

q

E-field distribution:E-field distribution:

thenthen

r

E

01

032

EE,

The potential difference:The potential difference:

1C 2C 3C nCV

V

qC

V

qqq n

21

V

VCVCVC n

21

n

iiC

1

4.Capacitors in series and in parallel

(1) In parallel:

VV

qC

nVVV

q

21 nCqCqCq

q

/// 21

(2) In series

n

i in CCCCC 121

11111

A B

dq

q q

E

VdqdW

the work done by external forc

e to move dq from B to A ，

as the capacitor has 、 ,Eq

dqC

q

§9-5 the energy in an E-field

1. Energy of a charged capacitorA parallel plate capacitor has no charge initially.

Electrification for it,

Work changes into potential energy

So the capacitor stores potential energy when it is charged Q ：

Q

dqC

qW

0 C

Q2

2

1

2

2

1CV QV

2

1

can be used for any shape of capacitor.

This potential energy can be regarded as the energyof the E-field which is set up by Q

2. Electric field energy

2

2

1CVW 2)(

2

1Ed

d

S VE 2

2

1

Definite electric energy density in dielectrics ：

dV

dWe 2

2

1E DE

2

1

A parallel plate capacitor: S,d ,and be filled with dielectrics,

dVW eV dVEV2

2

1

volume

The space filled with E-field

[Example] A parallel plate capacitor filled with air has S ， d. A copper plate with thickness d’is inserted capacitor parallel. The capacitor is charged till potential difference U. Then cut off the battery and remove the copper plate away. Find ： How much work must we do for the copper plate removed away?

d 'd

The work for removing copper plate away

=The increment of E- field energy of capacitor

Before and after the copper plate is removed away,

Q does not change.

dd

SC

0

d 'd

UCQ 'dd

SU

0

Solution: Two methods

use C

QW

2

2

1

Beforeis removed away,

Afteris removed away,d

SC 0''

As Q does not change,

'2

1'

2

C

QW

22

02

)(2

1

''2

1'' U

dd

Sd

C

QW

20

2

1U

dd

S

22

0

)(2

1''' U

dd

dSWWW

)'''( VVe

)(2

1 20 ddSSdE

dSE 202

1 22

0

)(2

1U

dd

dS

d 'd

Before and after the copper plate is removed away, Q does not change. The E in air does not change.But the space filled with E becomes larger.

''' WWW

use dVEW V2

2

1

[Example] A spherical shell with radius R and uniform charge Q is placed in vacuum. Calculate the E-field energy of the system.

Solution

)0 RrEin （＝

)(.4

12

0

Rrr

QEout

＝

E-field distribution in space,

V edVwW

VdVE 2

02

1

Rdrr

r

Q 2

2

20

0 442

1

R

Q2

08

1