Chapter 9 Conductors and Dielectrics in Electrostatic Field

Chapter 9 Conductors and Dielectrics in Electrostatic Field

§9-3 Dielectrics §9-3 Dielectrics 电介质电介质

§9-4 Gauss’ Law in Dielectric §9-4 Gauss’ Law in Dielectric 有电介质时的高斯有电介质时的高斯定律定律 Electric Displacement Electric Displacement 电位移电位移

§9-5 Energy in Electric Field §9-5 Energy in Electric Field 电电场的能量场的能量

§9-2 Capacitance §9-2 Capacitance 电容器电容器

§9-1 Conductors §9-1 Conductors 导体导体 Elecrostatic Induction Elecrostatic Induction 静电感应静电感应

Conductor:

§ 9-1 Conductors and § 9-1 Conductors and Electrostatic Induction Electrostatic Induction

There are many free electrons in it.

can move in the conductor randomly

ee e

e eee

e

A

B

1. The phenomena of the electrostatic induction

The charges of an insulated

conductor are redistributed

because of external E-field.

No external E-field

The process of electrostatic induction of a conductor

0E

external E-field is supplied--the electrons start to move.



0E

external

induced 'E

0E


0E

external

induced 'E

0E


0E

external

induced 'E

0E


0E

externalinduced 'E

0E


0E

external

induced 'E

0E


0E

externalinduced 'E

0E


0E

external

induced 'E

0E


0E

external

induced 'E

0E


0E

external

induced 'E

0E


0E

external

induced 'E

0E


E

静电平衡状态

0E

'E

0'0 EEE

(2). Electrostatic equilibrium conditions ：

(1). Electrostatic equilibrium state ： There is no any charge moving along a definite direction macroscopically inside the conductor or on the surface of the conductor.

2. Electrostatic equilibrium

The distribution of charges does not change with time.

The E-field equals zero everywhere inside the conductor.

The conductor is an equipotential body.The conductor surface is an equipotential surface.

The E-field at the surface of the conductor is perpendicular to the surface.

E=0

E

The behavior of E-field

The behavior of E-potential

Electrostatic field influences conductor: --electrostatic induction --make the charges in conductor redistribution.

Conductor influences electrostatic field : -- make the electrostatic field redistribution.

Example

The field is uniform before the metal sphere is put in.

E

The field is no longer uniform after the metal sphere is put in.

+++

++++

E

S--any Gaussian surface inside conductor P

S

S

SdE 0 0

内Siq

P--any point inside conductor, S--infinite small ，No excess charge inside conductor.

(1). Entire conductor ：

3. The charge distribution on conductor at the electrostatic equilibrium state.

Prove ： i

i

S

qSdE0

1

Use

No excess charges inside the conductor. They are found only on the conductor surface.

Q

(2). A conductor with a cavity ： Assume charged Q

No charge in the cavity:

Prove:

Draw a Gaussian surface S surrounding the cavity tightly.

No charge inside and internal surface of the conductor.

All the excess charges distribute on the outside surface of conductor.

Q

SSdE 0

0内S

iqi.e.,

--No net charge inside S

S

Question ？ Are there any equal magnitude and opposite sign charges on the internal surface of the conductor ？

0E

Inside the conductor

Not at all.

qQ

q

q

There are charges q in the cavity ：

On the inner surface of the conductor ：

--induction charges -q

On the external surface of the conductor ：The original charges Q of the conductor

+ induction charges +q

3.The relation about the charge distribution on the conductor surface and the its radius of curvature.

Rr

Q qR

r

Example:

Two conducting spheres of different radii connected by a long conducting wire.

They are equipotential.

R

QU R

04

1

00

2

0 4

4

4

1

rr

r

r

r

r

r

qU

R

R R

0

2

4

4

0 RR

rR

q

Q rR UU

In a qualitative way, for a conductor of arbitrary shape, the charge density distribution on its surface is inverse proportional with its radius of curvature.

rR

R

r

---

---

- -

- -

4. The relation about the E-field just outside the conductor surface and the charge density on the conductor surface.

.pE

sdE

sE

s

Sqi

i 00

11

0

E

is set up by all charges in the space (on and outside the conductor).

5. Application of electrostatic induction.

(1).Tip discharge Lighting rod( 避雷针 )

(2).Electrostatic shielding

++++ ++++

++--

Electrostatic generator

candle

Electrostatic generator and electric wind

A conductor shell can shield the external field

0E

A conductor shell that is connected with the ground can shield the influence of the fields between inside and outside the conductor.

A

A

[Example 1]A neutral conductor sphere with radius R is put on the side of a point charge +q . Assume the distance between the spherical center and the point charge is d. Calculate: The E-field and potential at point 0 set up by the induced charges on the sphere. If the sphere is connected with the ground, how much is the net charge on the sphere?

qd

R0

'q 'q

The examples about electrostatic induction

qd

R0

'q 'q

x

Solution

The total field at 0 = the field set up by q + the field set up by ±q

Assume the induction charges are±q

=0 ！！

EEE q

0

qEE

'

0

)(

4 20

id

q

id

q 2

04

'q 'q

qd

R0 x

the potential at 0 set up by ±q :

'

04

''

q R

dqU

0

d

qUq

04

'0 UUU q d

q

04

the potential at 0 set up by q :

the sphere is connected with ground.

R

q

d

qU

0

1

00 44

0q

d

Rq 1

Assume net charge q1 is left on the sphere.

the potential U0 at 0 = Uq + Uq1

qd

R0

1q

[Example 2] Two large parallel plates with the area S carry charge Qa and QB respectively.

QAQB

Find ： The charge and field distribution.

solutionAssume the charge surface density areσ1,σ2,σ3 andσ4 on the four surfaces .

σ1 σ2

σ3 σ4

Draw a Gaussian surface1 2 3 4

S

SSsdES

320

1

0P

4321 EEEEE p

0

41

Ⅰ Ⅱ Ⅲ

32

at point P :

and AQS )( 21

BQS )( 43

S

QQ BA

241

S

QQ BA

232

1 2 3 4

Ⅰ Ⅱ Ⅲ

The field distribution:

E1 E1E2 E2

E3 E3

E4 E4

4321 EEEEE I

0

4321

2

E=0 inside the plates

outside the plates:

4321 EEEEE II

direction: point to left

point to right

4321 EEEEE III point to right

2 3

Ⅰ Ⅱ Ⅲ

0

3

0

2

22

ES

q

0

EⅠ = E Ⅲ= 0

0

E

discussion Two plates carry equal magnitude and opposite sign charge

QQQ AB

041

S

Q 32

the charges distribute on the inner surface only.

S

qE

ES

qE

III

I

00

4

0

1

00

4

0

1

22

022

1 4

Ⅰ Ⅱ Ⅲ

Charges distribute on the exterior surface only.

S

Q 41

032

Two plates carry equal magnitude and same sign charge

QQQ AB

1r 2r

3r[Example 3] conductor sphere with radius r1 carries ＋ q and conductor spherical shell with inner and exterior radii r2 and r3 carries ＋ Q.

Calculate the E distribution, the potential of sphere and shell U1 and U2, potential difference U△

Connect sphere and shell with a wire, find E, U1 and U2 , U △ =?.

If the shell is connected with ground, find E, U1 , U

2, U =△ ? If the sphere is connected with ground, find the ch

arge distribution. U2=?

solution ： the field distribution:

0: 11 Err

rr

qE

rrr

ˆ4

:

20

2

21

0332 Errr

rr

QqE

rr

ˆ4

:

20

4

3

1r 2r

3rq

q

qQ

sphere potential:

1

1 rrdEU

3

2

12

02

0 44 r

r

rdr

r

Qqdr

r

q

32104

1

r

Qq

r

q

r

q

3

2

142 r

r

rrdErdE

3

2 rrdEU

3

4rrdE

3

204r

drr

Qq

304 r

Qq

21 UUU

210

11

4 rr

q

The shell potential:

Potential difference:

0321 EEE

Connect sphere and shell with a wire,

rr

QqE ˆ

4 20

4

3

421 rrdEUU

drr

Qqr

3

204 304 r

Qq

All charges are on the exterior surface of the shell.

The shell is connected with ground,

rr

qE ˆ

4 20

2

0431 EEE

qq

2

11

r

rrdEU 2

12

r

rrdE

210

11

4 rr

q

121 UUUU

U2 ＝ 0 ， no any charge on the exterior surface of the shell.

The sphere is connected with ground,

3020101 4

'

4

'

4

'

r

Qq

r

q

r

qU

0

Qrrrrrr

rrq

122313

12'

'q

'q

'qQ

U1=0. Assume sphere charges q' ， then the inner surface of shell charges -q' ， its exterior surface charges （ Q ＋ q' ）

As r3 r1 < r3 r2 , q'<0

The potential of shell:

1323210

12

302 44

'

rrrrrr

rrQ

r

QqU

Documents

Chapter 9 Conductors and Dielectrics in Electrostatic Field