Study the properties and laws of electric field, magnetic field and electromagnetic field that they...

Study the properties and laws of electric field, magnetic field and electromagnetic field that they are stimulated by charges and currents.

Volume 2Volume 2

Electromagnetism Electromagnetism

Chapter 8

Electrostatic Field in Vacuum

stimulated by static charges with respect to observer.

Inertial reference frame

§8-1 Coulomb’s Law 库仑定律

§8-2 The Electric Field 电场 电场强度

§8-4 Gauss’ Law 高斯定理

§8-5 Electric Potential 电势

§8-3 Electric Field Line and Flux 电力线 电通量

§8-6 Equipotential Surface and Potential Gradient 等势面 电势梯度§8-7 The Electric Force Exerted on a Moving Particle 运动带电粒子所受电场力

§8-1 Coulomb’s Law

1. Two kinds of electric charges

positive charge

Like charges repel each other

Negative charge

Unlike charges attract each other

2. quantization of charge

Electron is the smallest negative charge in nature.

experiments show ：

e=1.60217733×10-19 C(Coulomb)

q= Ne

Proton is the smallest positive charge in nature.

The magnitude of electric charge possessed by a body is not continuous.

integer

Electrification by rubbing

3. Conservation of charge

Positive charges and negative

charges have same magnitude.

A B A B

A B

B

A

Electrification by induction ：

inducible charges have same magnitude.

The conservation of charge ： in any interaction the net algebraic amount of electric charge remains constant.

1q

2q21r

21F

12F

21321

2121 r

r

qqkF

21221

2121 r̂

r

qqkF

2112 FF

or

Coulomb’s Law ：

4. Coulomb’s Law

Point charge ：The size of charged bodies

<< their distance

In SI ：

04

1

k

0=8.8510-12 C2/Nm2

----permittivity of vacuum （真空介电系数）

f1

f2

fn

f

nffff

21

5. Superposition principle of electrostatic forces

Assume there are many point

charges in space:q0 、 q1 、q2 、 q3 … qn ，the resultant force

acting on q0 :

---vector addition

q1

q2

qn

q0

§8-2 The Electric Field

Viewpoint of action-at-a distance:

Viewpoint of field:

1.Viewpoints of the interaction about electric charges

charge charge

charge field charge

Field is a kind of matter.

The behavior of electric field as a kind of matter:

force ： E-field exerts a force on the charges in

it.

work ： E-field does work on charges during

the charges move in it.

induction and polarization ： in the field,

conductor and dielectric produce induction

and polarization.

2. Electric field 　Test charge:

small size--point charge small charge magnitude—no influence for ori

ginal field.

Test results:

D0qC

0q

A0q

B0q

same q0 is put on different points in space,

0q

P 02q

P

F

203q

P

F

3 Put different test charges on same point,

the electric forces that the test charges suffer change.

the direction and the magnitude of the force that q0

suffers is different at different points---E-field is different at different points

the ratio

000 3

3

2

2

q

F

q

F

q

F

=constant vector at same point.

the electric field is defined ：

0q

FE

SI unit ：牛顿 / 库仑 (N/C) or 伏特 / 米 (V/m)

nFFFF

21

3. the superposition principle of electric field

There are q1、 q2 、 q3 … qn in space,

q0 is put on the point P, the force acting on q0 :

00

2

0

1

0 q

F

q

F

q

F

q

FE n

At point P, the E-field is set up by q1、 q2 、 q3 … qn :

n

iin EEEEE

121

P

r

q

.point charge:

Put qoon point P ， using Coulomb Law ， qo suffe

rs

r

r

r

qqF

20

04

1

r

r

r

q

q

FE

200 4

1

--the E-field of

a point charge

4. The distributions of electric field about several different charged bodies

The distribution is spherical symmetry

1

12

1

1

01 4

1

r

r

r

qE

There areq1 ， q2 ，… , qn in space

Each charge set up its field at point P ：

.the point charge system

n

n

n

nn r

r

r

qE

2

04

1

q1

q2

q3

r1

r3

r2

P

the total field at P ：

nEEEE

21

n

i i

i

i

i

r

r

r

q

12

04

.A continuously distributed charged body

rr

rdqEd

2

041

r

r

r

dqEdE

q

204

1

At point P, element ch

arge dq produces :

The total field at P produced by entire charged

body:

Pdq

r

q

kdEjdEidEE zyx

According to the distribution of charge, dq is written as follow:

for Cartesian coordinate system ：

dv

ds

dl

dq

line distribution

area distribution

volume distribution

xr

dqdEx 3

04

1

y

r

dqdE y 3

04

1

5. Examples of calculating E-field

Steps ： divide charged body into many small charge e

lements . write out produced by dq at point PEd

jdEidEEd yx

rr

rdqEd

2

041

?dq

set up a coordinate system, write components of , such as Ed

total E-field

xx dEE yy dEE

jEiEE yx

calculate the components of , such as E

[Example] Calculate the E-field at point P produced by a charged line.

P

d

l

q

1

2

L 、 q 、 d 、 1 、 2 are known

Solution . divide q dq

. Any dq produces at PEd

204

1

r

dqdE

P

d

l

q

1

2dqr

EdThe magnitude of : Ed

The direction of shows in Fig.

Ed

x

y

o

. set up Cartesian coordinate, jdEidEEd yx

ydE

xdE

. calculate Ex 、 Ey cosdEdE x sindEdE y

P

d

l

q

1

2dqr

Ed

cosdEdEE xx

cos

4cos

4

12

02

0 r

dx

r

dq

)( dxdq

sin

dr In the figure ： dctgx

dd

dx2sin

x

y

o

ydE

xdEP

d

l

q

1

2dqr

Ed

xE 2

1

cos4 0

dd

)sin(sin4 12

0

d

Same as yE 2

1

sin4 0

dd

)cos(cos4 21

0

d

jEiEE yx

. the total E

（ 1 ） If P locates on the mid-perpendicular plane of the line, i.e.

021 180

21 sinsin 21 coscos

0xE

22

0 24

l

dd

lE y

Discussion

（ 2 ） If P is very close to the line ld --the length of the charged line tends to infinity

021 1800 、

0xE

dEE y

02

E-field distribution of the infinite line with uniform charge

（ 3 ） If P is far away from the line ld 0

21 90

0xE

44

22

0

ldd

qEE y

2

04 d

q

The charged line can be regarded as a point charge.

Question ： If P locates on the elongating line of the charged line shown as in figure,

2

04

1

r

dqEE x

How do we calculate Ex 、 Ey ?

laa

qE

1

4 0

l aP

q

la

a r

dr2

04

1

Caution ！(1) If the charged body is not a point charge, we

can not use the formula

(2) If the directions of for different are not same, we can not integrate directly.

dqEd

Ed

r

r

r

qE

2

04

1

directly,

use only for point charge.

integrating its components

[Example] Find the E-field of an uniform charged ring on its axis. （ q 、 R 、 x are known)

x x

r

P

q R·

Solution Divide q dq

204

1

r

dqdE

direction Ed

EdidEEd

//

//dE

dE

0 Ed

x x

r

P

q R·

dq

cos// dEdEE

r

x

r

dq

204

1

dqr

x3

04

23

220 )(4 Rx

qx

Direction: along x axis

Discussion Discussion （ 1 ） at x=0 ， E=0.

When x ， 01

2

xE

E has extreme values on x axis.

let 0dx

dEWe get Rx

2

2

（ 2 ） when x>>R ， 323

22 xRx

204 x

qE

can be regarded as a point charge.

E

x0

R2

2

R2

2

R

P xx0

d

rEd

[Example] thin

round plate with

uniform charge

area density ,

radius R. find its

field on the axis.

21220

12 xR

xE

（ 1 ） when x « R ,

discussiondiscussion

02122

xR

x

02

EThe E-field set up by uniform sheet charge of a infinite plane.

（ 2 ） when x » R ,

2

1

2

2

21221

x

R

xR

x

2

2

2

11

x

R

20

2

2

0 42

1

2 x

q

x

RE

as a point charge

AB

AE

BE

§8-3 Electric field line and flux

1. E-field line （ line ）E

E

dS

dS --area element perpendicular to line.E

de – the number of line crossing .dSE

dS

dE e

the tangential direction of line at any point gives the direction of at that point.

E

E

the density of line gives the magnitude of

E

E

line originate on positive charges and terminate on negative charges (or go on infinity). They never originate or terminate on a no-charge point in finite space.

E

2. Electric flux eE or

The properties of line.E

Two lines never intersect at a point.E

– the number of line crossing any area.E

ESe

cosESESe

SE

S

S

S

n

The plane S is at right angle to the uniform E-field.

The plane S is at any angle with E

:the unit vector at the normal direction of the plane.

n

nSS

n

E

dS

S

EdSd e SdE

S ee d

The total E-flux crossing S ：

Take any dS on S :

An arbitrary surface S is placed in a no-uniform E-field.

S

SdE

Ede < 0

Se sdE

If S is a closed surface ：

de >0

Stipulation ： the direction of is outward.n

n

0 Se sdE

0 Se sdE

? e relate to the charge in the surface.

If there is no any charge in the closed surface,

the number of line entering it equals the number of line going out it.

E

E

If there are charges in the closed surface,

the number of line entering it does not equal the number of line going out it.

E

E

1. The E-flux crossing a sphere surface with a point charge q in its center.

q

S

sdESe

22

0

44

rr

q

dsES

SdsE

0q

§8-4 Gauss’s Law of electrostatic field

q

S

Discussion Discussion

e does not relate with r .

S’ If S’is an arbitrary

closed surface surro

unding q,

then

sdES

e

0q

sdES

2. If the charge is outside the closed surface S,

+ qS

e = 0

3. e crossing any closed surface S with point charge system.

Inside S :

iqq 1 1iq

Outside S:

nii qq 1

Then e

iq

S

S

sdE

000

1

iqq

内S

ie q0

1

q

4. e crossing any closed surface S with any charged body

S

qe dq

0

1

Inside S

Summary above results Gauss’LawIn any electrostatic field, the electric flux crossing any closed surface equals the algebraic sum of the charges enclosed by the surface divided by 0

内S

iSe qSdE0

1

The closed surface S –Gaussian surface

or q

S

e dqdSE0

1

S 内

Notes:

① is the algebraic sum of the charges enclosed by the Gaussian surface

内S

iq

i.e., the e crossing S depends on the charges enclosed by S only, and has nothing to do with the charges outside S.

the total at any point of S is concerned with all the charges (inside and outside S).

E

0 S

e dSE

indicates that the electrostatic field is a 有源场

If the distribution of charges and its E-field has some symmetry, Gauss’s Law can be used to calculate the E-field.

Common steps: Analyze the symmetry of charges and its E-field. Choose a suitable enclosed surface as Gaussian s

urface S. Calculate e crossing S.

Calculate the algebraic sum of charges inside S. Use Gauss’s Law to calculate E 。

4. Applications of Gauss’s Law

[Example 1] Calculate the E-field distribution of a infinite line with uniform charges. (Assume the linear density of charges is )

Solution ： Analyze the symmetry

--axial symmetry

λ

E

Choose a suitable Gaussian surface

--cylinder surface

hr

P

PE1S

2S

3S

Se SdE

Calculate e crossing S ：

321 SSS

SdESdESdE

rhE 2 Calculate the algebraic sum of charges inside S

hqi Use Gauss’s Law ：

ii

s

e qSdE0

1

0

2 h

rhE

rE

02

E-field distribution of the infinite line with uniform charge

Similar problem: the charge distribution on a infinite cylinder surface with radius R. The charges per meter of length of the cylinder isλ.

axial symmetry

Gaussian surface – cylinder surface

（ 1 ） r R>

S

SdE

SdE

下 SdE

上 SdE

侧

i

iq ： l

rlE 2

use i

i

s

e qSdE0

1

rlE 20

l

G-surface

rE

02

e

0 E

（ 2 ）r R

0E

（ r > R ）

Distribution of E-field of the cylinder charges

r02

)( Rr<

[Example 2] Calculate the E-field distribution of a infinite plane with uniform charges. (Assume the area density of charges is σ)

Solution:Analyze the distribution character of E-field

P

E-field :area symmetry

σ

SE

E

σ

E

.P

Make a cylinder surface through point P as Gaussian surface.

G-surface

02

E

S

SdE

SdE

左 SdE

右 SdE

侧0 SESE SE2

Use i

i

s

E qSdE0

1

SE20

S

Sqi

i ：

e

[Example 3] Calculate the E-field distribution of a sphere surface with uniform charge q.

(1) r > R

spherical symmetry

Make a sphere as Gaussian surface through P.

R

+++++ +

+++

+ ++++

+ +

qrP

G-surface

e

24 rE

S

SdE

S

EdS 00cos

qqi

i

204

1

r

qE

use i

i

s

e qSdE0

1

0

24

qrE

(2) r < R

G-surface

E-field—spherical symmetry

Gaussian surface--sphere

P.r

E

+ +

++++ +

+

+

+ ++++

+ +

R

q

S

e SdE

S

EdS 00cos

24 rE 0

iiq

0 E

use i

i

s

e qSdE0

1

04 2 rE

204

1

r

q

The distribution of E-field for charged spherical surface:

E0 （ r>R ）

（ ） r < R

12r

r

E

0 R

R

q（ 1 ） <r R

r

G-sueface

Similar question ： uniform charged spherical body

E

S

SdE

24 rE e

3

3

4rqi

33

rR

q3

3 3

4

3

4r

R

q

24 rE 33

0

rR

q

r

R

qE

304

1

（ 2 ）r R>

G-surface

R

r

E

.P

S

SdE

24 rE 0

q

204

1

r

qE

The distribution of E-field with charged spherical body:

rR

q3

04

1

E

（ r > R ）204

1

r

q

)( Rr <

rR

E

o

204

q

R