Baigiang_LTTDT

L THUYT TRNG IN T

Bin son: TS. Cung Thnh Long

B mn K thut o v Tin hc cng nghip

Vin in

Trng i hc Bch Khoa H Ni

Bi ging

(2 n v hc trnh)

1

Ti liu tham kho

1 C s L thuyt Trng in t, Nguyn Bnh Thnh, NXB i hc v

trung hc chuyn nghip

2 Electromagnetic Fields and Waves, Magdy F. Iskander, Prentice Hall

4 Electromagnetic field theory for physicists and engineers:

Fundamentals and Applications, R. Gomz Martin

5 Electromagnetic Field Theory, Bo Thid

3 Electromagnetics, John D. Kraus, McGraw-Hill Inc.

2

Ni dung chnh

1 M u v L thuyt Trng in t

2 Nhc li v Gii tch vc t

3 in trng tnh

5 T trng tnh

4 Dng in mt chiu

3

1 M u

1.1. Gii thiu

- L thuyt trng in t : mn hc c s chuyn ngnh rt quan trng

Trng l g? Th no l trng v hng v trng vec t? Th no l

trng lin tc v trng xoy? Bn cht ca trng l g? T trng

sinh ra bi cun dy mang dng in nh th no?

- Nghin cu L thuyt trng in t : hiu cc hin tng xy ra trong k thut

in

4

1 M u

1.1. Gii thiu

Bng 1.1. Cc n v dn xut ca mt s i lng in t trng

K hiu i lng n v Biu din

Y Tng dn siemen S

Tn s gc radian/giy rad/s

C in dung farad F

Mt in tch Cu-lng/mt khi C/m3

G in dn siemen S

in dn sut siemen/mt S/m

W Nng lng joule J

F Lc niu-tn N

Tn s hc Hz

Z Tr khng m

L in cm hen-ri H

Sc t ng ampe-vng

T thm hen-ri/mt H/m

in mi farad/mt F/m

P Cng sut ot W

T tr hen-ri-1 H-1

f

A t

5

1 M u

1.1. Gii thiu

Bng 1.2. Danh sch cc i lng trng c bn

Bin nh ngha Kiu n v

vc t t th vc t Wb/m

mt t thng vc t Wb/m2 (T)

mt thng lng in vc t C/m2

cng in trng vc t V/m

lc Lorentz vc t N

dng in v hng A

mt dng in vc t A/m2

in tch t do v hng C

vc t Poynting vc t W/m2

vn tc ca in tch t do vc t m/s

V in th v hng V

A

B

D

E

F

I

J

q

S

u

6

1 M u

1.1. Gii thiu

Bng 1.3. Quan h gia cc i lng trng

Hng s in mi ( )

H s t thm ( )

in dn sut ( ), lut Ohm

Phng trnh lc Lorentz

Lut Gauss (phng trnh Maxwell)

Lut Gauss (phng trnh Maxwell)

Phng trnh lin tc

Lut Faraday (phng trnh Maxwell)

Lut Ampere (phng trnh Maxwell)

D E

B H

J E

q F E u B

. D

. 0 B

.t

J

t

BE

t

DH J

7

1 M u

1.2. Khi nim trng

- nh ngha cc th hin ca mt i lng trong mt min cho trc bi

mt tp cc gi tr m mi gi tr tng ng vi mt im ca min

cho mt trng

- in t trng lan truyn trong chn khng vi vn tc ca nh sng

0 0

1/c m s

7

0 4 10 /H m

12 9

0

18.851 10 10 /

36F m

- Trng v hng, trng vc t

8

1 M u

1.3. Gii tch vc t

- Mt cng c s dng nghin cu in t trng

+ n gin, d nh

+ Khng ph thuc h trc ta

Thng nht ha, n gin ha vic biu din cc phng trnh trng

- V d:

A B CGii tch vc t

Dng v hng, trong h ta -cc

y z z y xA B A B C

z x x z yA B A B C

x y y x zA B A B C

9

1 M u

1.4. Cc cng thc tch phn v vi phn

- Ti sao biu din cng mt khi nim no di hai dng vi phn v

tch phn?

+ Dng tch phn: tin gii thch ngha ca mt phng trnh

+ Dng vi phn: tin thc hin cc php ton

- V d cng thc v tnh lin tc ca dng in:

+ Dng vi phn .t

J C th tnh tc thay i mt

in tch ti mt im khi bit mt

dng in ti im

+ Dng tch phn .s v

d dvt

J s ngha: gi tr ca dng in xuyn qua khi b mt ca mt min no

chnh bng tc gim in tch trong

min theo thi gian

10

1 M u

1.5. Trng tnh

- in trng tnh:

+ tt c cc in tch c nh trong khng gian

+ tt c cc mt in tch l hng s theo thi gian

+ in tch l ngun ca in trng

Xc nh:

+ cng in trng ti mt im bt k

+ phn b in th

+ lc tc ng gia cc in tch

+ phn b ca nng lng in trong mt min xc nh bt k

11

1 M u

1.5. Trng tnh

- in trng tnh:

Cc phng trnh in trng tnh

Lut Coulomb

Cng in trng hoc

Lut Gauss hoc

Bo ton vc t in trng hoc

Hm in th hoc

Phng trnh Poisson

Phng trnh Laplace

Mt nng lng

Quan h gia v

Lut Ohm

qF E

24

RQ

R

aE 2

1

4

R

v

dvR

aE

. D .v

d Q D s

E

0 E . 0c

dl E

V E .b

ba

a

V dl E2V

2 0V 1

.2

ew D E

D ED E

J E

12

1 M u

1.5. Trng tnh

- T trng tnh:

+ Do dng in khng i theo thi gian to nn

+ T trng khng i theo thi gian

Xc nh:

+ Cng t trng

+ Mt t thng

+ T thng

+ Nng lng d tr trong t trng

13

1 M u

1.5. Trng tnh

- T trng tnh:

Cc phng trnh t trng tnh

Phng trnh lc hoc

Lut Biot-Savart

Lut Ampe hoc

Lut Gauss hoc

Vc t t th hoc

T thng hoc

Nng lng t

Phng trnh Poisson

Lin h gia v

q F u B d Idl F B

24

rIdldr

aB

H J .c

dl I H

. 0 B . 0A

d B s

B A4

c

Idl

r

A

.s

d B s .c

dl A

1.

2mw B H

2 A J

B H B H

14

1 M u

1.6. Cc trng bin thin theo thi gian

- nh lut cm ng ca Faraday: mt trong 4 phng trnh Maxwell

- nh lut Ampe sa i: mt trong 4 phng trnh Maxwell

- Hai nh lut Gauss (cho in trng v t trng bin thin theo

thi gian): 2 trong 4 phng trnh Maxwell

H 4 phng trnh Maxwell

- Phng trnh lc Lorentz

- Phng trnh v tnh lin tc ca dng in

Gii thch tt c cc hiu ng in t trng

15

1 M u

1.7. ng dng ca trng bin thin theo thi gian

Cc qu trnh pht, thu v lan truyn nng lng

+ ng dn sng: in trng ngang v t trng ngang, bng tn hp

+ Cc ng dy truyn ti: in trng ngang, t trng ngang v in

t trng ngang

+ An-ten: bc x sng in t qua cc ngun bin thin theo thi gian vi

kch thc hu hn

16

2 Gii tch vc t

2.1. Cc i lng v hng v vc t

+ V hng: i lng vt l c th hon ton biu din bng ln ca n

+ Vc t: i lng vt l c c ln v hng

+ Biu din mt vc t theo n v ca n AA aA

A

a

Biu din hnh hc ca

mt vc t

Cc mi tn song song cng

di v cng hng biu

din cng mt vc t

Vc t khng, Hai vc t bng nhau, So snh hai i lng vc t

17

2 Gii tch vc t

2.2. Cc ton t vc t

+ Cng vc t:

C A B

A B B A (Giao hon)

A B C A B C (Kt hp)

+ Tr vc t:

D A B

+ Tch ca mt vc t v mt v hng: B kA

0k

0k

1k

1k

18

2 Gii tch vc t

2.2. Cc ton t vc t

+ Tch v hng gia hai vc t: . cosA B AB

. .A B B A

. . .A B C A B AC

. . .k A B kA B A kB

(Giao hon)

(Phn phi)

(T l)

- Tch v hng ca hai vc t khc vc t khng m bng khng: hai vc t trc giao

.cos . A

A BB B a

A

- Chiu ca ln B A

.A A A- ln ca vc t A

. .A B AC B CNu th c lun bng khng?

19

2 Gii tch vc t

2.2. Cc ton t vc t

+ Tch c hng gia hai vc t: sin nA B AB a

n A B A BCa Aa Ba a a AB

sin

A Bn

a aa

A B C A B A C kA B k A B A kB

(Phn phi)

(T l)

20

2 Gii tch vc t

2.2. Cc ton t vc t

+ Tch ba v hng:

. sin osC A B ABC c

. . .C A B A B C B C A

+ Tch ba c hng (tch ba vc t):

A B C

A B C A B C (Khng c tnh kt hp)

21

2 Gii tch vc t

2.3. Cc h trc ta

2.3.1. H ta -cc

Ta ca mt im trong h

trc ta cc

Cng vc t trong h trc ta cc

x y zr Xa Ya Za

Vc t v tr

x x y y z zA A a A a A a

x x y y z zB B a B a B a

x x x y y y z z z x x y y z zC A B a A B a A B a C a C a C a

22

2 Gii tch vc t

2.3. Cc h trc ta

2.3.1. H ta -cc

. . . 1x x y y z za a a a a a

. . . 0x y y z z xa a a a a a

0x x y y z za a a a a a

x y za a a y z xa a a z x ya a a

Tch v hng ca hai vc t

. x x y y z zA B A B A B A B

ln ca vc t 2 2 2. x y zA A A A A A

23

2 Gii tch vc t

2.3. Cc h trc ta

2.3.1. H ta -cc

Tch cho (c hng) gia hai vc t

C A B

x x y y z z x x y y z zC A a A a A a B a B a B a y z z y x z x x z y x y y x zA B A B a A B A B a A B A B a

x y z

x y z

x y z

a a a

C A B A A A

B B B

24

2 Gii tch vc t

2.3. Cc h trc ta

2.3.2. H ta tr

, ,P z osx c

siny

Mt phng

Mt phng

Mt tr

2 2 onsx y c t

+ Hai vc t cng nm trong mt mt phng hng s

z zA A a A a A a

z zB B a B a B a

25

2 Gii tch vc t

2.3. Cc h trc ta

2.3.2. H ta tr

+ Hai vc t cng nm trong mt mt phng hng s

Mt phng

Mt phng

Mt tr

z z zA B A B a A B a A B a

. z zA B A B A B A B

z

z

z

a a a

A B A A A

B B B

. 1a a . 1a a . 1z za a

0.a a 0. za a 0.za a

+ Tch chm v tch cho gia cc vc t n v

0a a 0a a 0z za a

za a a za a a za a a

26

2 Gii tch vc t

2.3. Cc h trc ta

2.3.2. H ta tr

+ Chuyn i cc vc t n v

os sinx ya c a a

sin osx ya a c a

. osxa a c

. sinya a

. sinxa a

. osya a c

os sin 0

sin os 0

0 0 1

x

y

z z

a c a

a c a

a a

+ Chuyn i mt vc t

os sin 0

sin os 0

0 0 1

x

y

z z

A c A

A c A

A A

os sin 0

sin os 0

0 0 1

x

y

z z

A c A

A c A

A A

27

2 Gii tch vc t

2.3. Cc h trc ta

2.3.3. H ta cu

sin osx r c

sin siny r

cosz r

+ Tch chm v tch cho gia cc vc t n v

. 1r ra a . 1a a . 1a a

. 0ra a . 0a a . 0ra a

ra a a ra a a ra a a

2 2 2r x y z

1osz

cr

1tany

x

28

2 Gii tch vc t

2.3. Cc h trc ta

2.3.3. H ta cu

+ Chuyn i cc vc t n v

. sin osr xa a c

. sin sinr ya a

. osr za a c

. os osxa a c c

. os sinya a c

. sinza a

. sinxa a

. osya a c

. 0za a

sin os sin sin os

os os os sin sin

sin os 0

r x

y

z

a c c a

a c c c a

a c a

29

2 Gii tch vc t

2.3. Cc h trc ta

2.3.3. H ta cu

+ Chuyn i cc vc t

sin os os os sin

sin sin os sin os

os sin 0

x r

y

z

A c c c A

A c c A

A c A

sin os sin sin os

os os os sin sin

sin os 0

r x

y

z

A c c A

A c c c A

A c A

30

2 Gii tch vc t

2.4. Cc php tnh vc t

2.4.1. o hm

+ Hm mt bin

0lims

f s s f sdf

ds s

+ Hm hai bin df f du f dv

ds u ds v ds

0

, ,limu

f u u v f u vf

u u

+ o hm ca mt trng vc t

0

lims

F s s F sdF

ds s

0

, , , ,limx

F x x y z F x y zF

x x

tng t cho cc thnh phn y, z

Dng dnh ngha div, rot (curl) (xt sau)

31

2 Gii tch vc t


2.4.2. Cc phn t vi phn ng, mt v khi

+ H ta -cc

dv dxdydz x xds dydza

y yds dxdza

z zds dxdyax y zdl dxa dya dza

32

2 Gii tch vc t



+ H ta tr

dv d d dz

zdl d a d a dza

ds d dza

ds d dza

z zds d d a

33

2 Gii tch vc t



+ H ta cu

2 sindv r dr d d

sinrdl dra rd a r d a

2 sinr rds r d d a

sinds rdr d a

ds rdrd a

34

2 Gii tch vc t

2.5. Tch phn ng, mt, khi

2.5.1. Tch phn ng

Phn t vi phn chiu di trn

ng c trong khng gian ba chiu

Xt hm f(x) lin tc, n tr theo x, gii hn bi x = a, x = b, ta c tch phn ng

0

1

limi

b n

i ix

ia

f x dx f x

Tng qut, tch phn ca trng v hng f trn ng c trong khng gian ba chiu:

01

limi

n

i il

ic

fdl f l

Tch phn trn ng kn, k hiu l:

35

2 Gii tch vc t


2.5.1. Tch phn ng

V d Cho vc t tnh tch phn t ti 24 9 14 8x y zA x y a yza x za .c

A dl 0,0,0P 1,1,1Qa) Theo ng: , , x t 2y t 3z t

b) Theo ng t (0,0,0) ti (1,0,0) ri ti (1,1,0) v cui cng (1,1,1)

c) Theo ng thng ni t P ti Q

Gii

2. 4 9 14 8Adl x y dx yzdy x zdz a) Ta c: nn 1

2 6 7

0

. 4 9 28 24 4c

A dl t t t t dt b) Ta tnh tch phn trn 3 on ri cng li

+ Trn c1:

1

1

0

. 4 2c

A dl xdx

+ Trn c2:

2

. 0c

A dl + Trn c3: 3

1

2

0

1. 8 4 4

0c

A dl zdz z

1 2 3

. . . . 2 0 4 6c c c c

A dl A dl A dl A dl + Do :

c) Ta c: 1

2 3

0

. 13 14 8 3.833c

A dl x x dx

36

2 Gii tch vc t


2.5.2. Tch phn mt

Tnh tch phn mt ca mt trng v hng f hoc mt trng vc t : F

10

limi

n

i in is

s

f ds f s

(l mt vc t)

10

. lim .

i

n

i in

is s

F ds F s

(T.P mt v hng ca mt trng vc t)

10

lim

i

n

i in

is s

F ds F s

(T.P mt vc t ca mt trng vc t)

V d 1

Hy ch ra rng qua mt kn ca mt hnh cu bn knh b, ta c:

0s

sd

37

2 Gii tch vc t


2.5.2. Tch phn mt

V d 1

Hy ch ra rng qua mt kn ca mt hnh cu bn knh b, ta c:

0s

sd

Gii 2

2

0 0

sinrs

ds a b d d

Ta c

sin cos sin sin cosr x y za a a a

2 2 2

2 2 2 2 2

0 0 0 0 0 0

sin cos sin sin sin cos 0x y zs

ds a b d d a b d d a b d d

M

Nn:

38

2 Gii tch vc t


2.5.2. Tch phn mt

V d 2

Tnh tch phn qua mt kn ca khi lp phng nh minh

ha trong hnh bn, trong l vc t v tr ca mt im bt k

trn mt khi lp phng?

s

sdr

.

r

Gii

x y zr xa ya za

Mt x = 1: xds dydza

1

1 1

0 0

. 1s

r ds dy dz

xds dydza Mt x = 0: 2

. 0s

r ds

Tng t cho cc mt cn li, cui cng ta c: . 3s

r ds

39

2 Gii tch vc t


2.5.3. Tch phn khi

Tch phn khi v hng: 10

lim

i

n

i in

iv v

fdv f v

Tch phn khi ca mt trng vc t 10

lim

i

n

i in

iv v

Fdv F v

V d

Phn b mt electron bn trong mt khi cu bn knh 2 mt c dng

1000 / cos / 4en r electron/m3 Tm tng in tch bn trong khi cu?

Gii

2 2

2

0 0 0

1000 1000cos / 4 sin cos / 4 16000e

v v

N n dv dv r dr d dr r

Tng s in tch bn trong khi cu

V vy tng in tch trong khi cu l:

19 1516000 1,6 10 2,56 10Q C

40

2 Gii tch vc t

2.6. Gradient ca mt hm v hng

Gi f(x,y,z) l hm kh vi gi tr thc ca x, y, z nh minh ha

trong hnh bn. Vi phn ca f t P ti Q c xc nh bng:

.x y z x y zf f f f df df

df dx dy dz a a a dxa dya dzax y z dx dy dz

x y zdl dxa dya dza Vi phn chiu di t P ti Q:

.x y zf f f

df a a a dlx y z

Nn:

. . .x y z l n ldf f f f dl

a a a N a Na adl x y z dl

x y z

f f fN a a a

x y z

Tc thay i ca hm f ln nht khi v ng hng. Do : N

l

max

dfN

dl

df/dl max th dl phi nh nht: constzyxfal ,,

V vy: constzyxfN ,,

l gradient ca hm v hng f(x,y,z), k hiu l N

f

41

2 Gii tch vc t

2.6. Gradient ca mt hm v hng

Tnh cht ca gradient ca mt hm v hng ti mt im:

Vung gc vi b mt m trn hm s cho l hng s

Ch ra hng m theo hm s thay i nhanh nht t im cho.

ln ca n th hin tc thay i ln nht ca hm s cho trn mt n v khong cch.

Vi phn c hng ca mt hm s ti mt im theo mt hng bt k th bng tch v ng

gia gradient ca hm s vi vc t n v trn hng .

x y z

f f ff a a a

x y z

Trong ta cc:

Trong ta tr:

Trong ta cu:

1z

f f ff a a a

z

1 1

sinr

f f ff a a a

r r r

42

2 Gii tch vc t

2.7. tn (div) ca mt trng vc t

.x y z x x y y z zf a a a F a F a F ax y z

Thng lng thc ca trng vc t hng ra ngoi cc b mt ca n v th tch F v

.yx z

s

FF FF ds v

x y z

yx zFF F

fx y z

Gi C:

.f F gi l tn ca trng vc t F

Nu trng vc t cho l lin tc (chng hn nh dng chy ca mt cht lng khng nn c qua mt ng dn, hay cc ng sc t ca trng bao quanh mt

nam chm) th s khng c thng lng hng ngoi: . 0F

nh l: vi mt trng vc t kh vi lin tc th thng lng thc hng ra ngoi mt mt kn s bng tch phn tn ca trng vc t ny trn min khng gian bao

bi mt kn cho . .

v s

Fdv F ds

43

2 Gii tch vc t


Trong ta cc: .yx z

FF FF

x y z

Trong ta tr: 1 1

. zF F F Fz

Trong ta cu: 221 1 1

. sinsin sin

rF r F F Fr r r r

44

2 Gii tch vc t


V d

Nghim ng li nh l tn cho trng vc t 23 3 3x y zD x a y z a z x a trong min bao bi hnh tr 2 2 9x y v cc mt phng 0, 0, 0x y z 2z v

Trc ht chng ta tnh

Gii

2. 3 3 3 6 6D x y z z x xx y z

. .v s

Fdv F ds nh l v tn:

Tch phn th tch bn v tri trong h ta tr c dng

3 /2 2 3 /2 2

2

0 0 0 0 0 0

. 6 6 6 cos 6 192.82v v

Ddv x dv d d dz d d dz

45

2 Gii tch vc t


V d (tip)

V phi:

0y 1 yds dxdza 1

3 2

1

0 0

. 3 6s x z

D ds y z dxdz

0x 2 xds dydza 2

3 2

2

2

0 0

. 3 0s y z

D ds x dydz

3 3 3ds d dza

3

/2 2

3

0 0

. 3s z

D ds D d dz

3

/2 2

2

3

0 0

. 3 cos 3 sin 3s z

D ds x y z d dz

3

3. 156.41s

D ds

2z 4 zds d d a 4

3 /2

4

0 0

. 6s

D ds x d d

4

4. 33.41s

D ds

0z 5 zds d d a

5

3 /2

5

0 0

. 9s

D ds x d d

. 6 0 156.41 33.41 9 192.82s

D ds

46

2 Gii tch vc t

2.8. Lu s (curl / rot) ca mt trng vc t

Tch phn trn mt ng khp kn ca mt trng vc t c gi l s lu

thng ca v curl (hay rot) ca l s o ca n. F

F

F

Xt mt b mt nh bao bi ng kn thnh phn lu s song song vi vc t n v vung gc vi b mt cho, trong gii hn , l:

nsa c

na 0s

cs

n lds

acurl

.1

lim.0

FF

lu s ca mt trng vc t l mt i lng vc t

+ Tnh thnh phn z ca F

curl

Trong h ta -cc:

zzyyxx aFaFaF

F

1 2 3 4

. . . . .c c c c c

F dl F dl F dl F dl F dl

Ta c:

47

2 Gii tch vc t



curl

Trong h ta -cc:

1

. .

x x

x x y y z z x x y

c x y

F dl F a F a F a dxa F x

2

. .

y y

x x y y z z y y x xc y x x

F dl F a F a F a dya F y

3

. .x x y y z z x x y yc y y

F dl F a F a F a dxa F x

4

. .

y

x x y y z z y y xc y y x

F dl F a F a F a dya F y

. x x y yy y y x x xc

F dl F x F x F y F y

48

2 Gii tch vc t


Trong h ta -cc:


curl

Trong gii hn v s dng khai

trin chui Taylor v loi b cc thnh phn bc

cao, ta c:

0x 0y

xx xy y yF

F x F x x yy

y

y yx x x

FF y F y x y

x

.y x

c

F FF dl x y

x y

0

0

1lim .

y x

sc

F FF dl

s x y

yF

x

Fcurl x

yz

F

zxy

yzx

xyz a

y

F

x

Fa

x

F

z

Fa

z

F

y

Fcurl

F

+ Ta c:

FF

zzyyxxzyx aFaFaF

za

ya

xacurl

49

2 Gii tch vc t


Cng thc d nh trong cc h trc ta :

+ H ta -cc:

zyx

zyx

FFF

zyx

aaa

F

z

z

FFFz

aaa

1F+ H ta tr:

FrrFF

r

arara

r

r

r

sin

sin

sin

12

F+ H ta cu:

50

2 Gii tch vc t


ngha vt l:

Biu din s lu thng qua mt n v din tch ca trng vc t cho trn mt din tch nh c hnh dng bt k

C hng vung gc vi cc vi phn mt phng ca b mt cho

Nu lu s ca trng vc t khc khng, ta ni rng trng vc t c tnh cht xoay trn hay xoy (rotational)

Nu lu s ca mt trng vc t bng khng, th trng vc t c gi l khng c tnh cht xoay trn (tnh xoy) hay trng c tnh cht bo

ton (conservative)

51

2 Gii tch vc t

2.9. nh l Stoke

Pht biu:

cs

ldd

.. FsF

Tch phn ca thnh phn vung gc ca lu

s ca mt trng vc t trn mt b mt bng

tch phn ng ca trng vc t ly dc

theo ng bao b mt cho

52

2 Gii tch vc t

2.10. Ton t Laplace

nh ngha bng tn ca gradient ca mt hm v hng, k hiu 2

2 .f f

+ Trong h ta -cc:

. .x y z x y zf f f

f a a a a a ax y z x y z

2 2 2

2

2 2 2.

f f ff f

x y z

+ Trong h ta tr: 2 2

2

2 2 2

1 1f f ff

z

+ Trong h ta cu: 2

2 2

2 2 2 2 2

1 1 1sin

sin sin

f f ff r

r r r r r

53

2 Gii tch vc t

2.11. Phn loi trng

+ Trng loi mt:

0

0.

F

F

fF

2. 0f f

V d: in trng tnh trong mi trng khng c in tch

hay t trng tnh trong mi trng khng c dng in

PT Laplace

+ Trng loi hai:

0

0.

F

F

fF F

.

2 f

PT Poisson

V d: in trng tnh trong mt min c in tch

54

2 Gii tch vc t

2.11. Phn loi trng

+ Trng loi ba:

0

0.

F

F

JF

AF

PT vc-t Poisson

JA

JAA

2).(

Gi s: 0. A

(Tiu chun Cu-lng) JA

2

V d: T trng bn trong mt vt dn mang dng in l

trng loi ba

55

2 Gii tch vc t

2.11. Phn loi trng

+ Trng loi bn:

0

0.

F

F

HGF

0

0.

G

G

0

0.

H

H

AG

fH

f AF

V d: Cc trng kh ng trong cc mi trng c th nn c

56

2 Gii tch vc t

2.12. Mt s ng nht thc vc-t

0f

0. A

2 .f f

AAA

.2

f g f g

BABA

...

BABA

fg f g g f

fff ... AAA

AAA

fff

BAABBA

...

BAABABBABA

....

57

3 in trng tnh

3.1. Gii thiu

Nghin cu v in trng tnh gy bi cc in tch ng yn

Cc dng phn b in tch c gi tr khng i theo thi gian

nh lut Coulomb v lc tnh in gia hai in tch im ng yn trong khng gian

nh ngha cng in trng, in th, mt in tch,

Mt vi phng php (lut Gauss, phng trnh Laplace, phng trnh Poisson, phng php soi nh) gii cc bi ton in trng tnh

Khi nim in dung v qua xy dng phng trnh nng lng tch tr bn trong mt t in

58

3 in trng tnh

3.2. nh lut Coulomb (Cu-lng)

Charles Augustin de Coulomb, t thc nghim, nhn nh rng lc in gia hai in tch im:

T l thun vi tich hai in tch

T l nghch vi bnh phng khong cch gia chng

Hng dc theo ng ni gia hai in tch

y nhau (ht nhau) nu hai in tch cng (khc) du

1 212 122

12

q qF K a

R

12 12 12 1 2R R a r r

0

1

4K

(trong h n v quc t SI)

12 9

0 8.85 10 10 / 36 (hng s in mi trong chn khng)

21 12F F

F/m

59

3 in trng tnh


1 212 122

0 124

q qF a

R

1 2 1 212 3

0 1 24

q q r rF

r r

V d 1

Cho hai in tch im q1 = 0,49 C t ti im (0,0,0); q2 = 0,7 mC t ti im (2,3,6)

trong khng gian. Tnh lc tc dng ln in tch q2 ?

21 2 1 2 3 6x y zR r r a a a

Gii

1/22 2 2

21 2 3 6 7R m

9

0

19 10

4

9 3 6

0,7 3

9 10 0,7 10 4,9 102 3 6 0,18 0,27 0,54

7mC x y z x y zF a a a a a a N

ln ca lc ny bng 0,63 N

60

3 in trng tnh


+ inh lut Coulomb tha mn nguyn l xp chng :

3

1 04

ni i

t

i i

q r rF q

r r

V d 2

Cho ba in tch im q1 = q2 = q3 = 200 C t ti cc im (2,0,0), (0,2,0), (0,0,0) trong chn

khng. Tnh lc tc dng ln in tch q t ti im (2,2,0)?

Gii

1 1 12 2yR r r a R m

2 2 22 2xR r r a R m

3 3 32 2 2.828x yR r r a a R m

9 9 9

1 3

9 10 200 10 500 102 225

2y yF a a N

2 225 xF a N 3 79.6 x yF a a N

1 2 3 304.6t x yF F F F a a N

61

3 in trng tnh

3.3. Cng in trng

+ nh lut Coulomb ch ra rng: mt in tch lun tc ng mt lc ln in tch khc ngay c

khi chng cch xa nhau (tc ng t xa)

Khi cc in tch ng yn: quan nim tc ng t xa tha mn mi yu cu

Khi mt in tch chuyn ng li gn mt in tch khc: lc tng tc phi thay i lp tc (theo lut Coulomb) tri vi thuyt tng i.

Thc t: nng lng v m men ca h thng cc in tch s mt cn bng tm thi

Ph hp thuyt tng i: vi cc i tng tng tc th m men v nng lng khng c bo ton bi chnh cc i tng ny

Phi tn ti mt thc th khc, di dng mt s bin ng trong mi trng cha cc vt th tng tc, gii thch cho s tn hao nng lng v m men ca cc i tng.

Thc th c bit y c gi l trng

tn ti mt in trng hay mt cng in trng ti mi im trong khng gian bao quanh in tch

Khi mt in tch khc c t trong in trng: s c mt lc tc ng ln n (tc ng qua tip xc)

62

3 in trng tnh

3.3. Cng in trng

+ Cng in trng: lc trn mt n v in tch

tq qt

FE

0lim

F

l tng cc lc tc ng ln tq(V/m)

+ Nu l cng in trng ti im P trong khng gian th lc tc ng

ln mt in tch q t ti im l: E

EF

q

+ Cng in trng ti mt im P bt k gy bi mt in tch n v q t

ti im S:

R

sp

sp

R

q

rr

rrqaE

20

30 44

63

3 in trng tnh

3.3. Cng in trng

+ Cng in trng gy bi n in tch im

n

i i

ii

rr

rrq

13

04

E

l vc t khong cch hng t im t in tch qi ti im

cn o cng in trng irr

+ Cng in trng gy bi cc phn b in tch ng, mt v khi

- Mt in tch ng 0

limll

q

l

- Mt in tch mt 0

limss

q

s

- Mt in tch khi 0

limvv

q

v

64

3 in trng tnh

3.3. Cng in trng

Cng in trng gy bi phn b in tch ng:

'3

,

,

04

1dl

rr

rr

c

l

E

Cng in trng gy bi phn b in tch mt:

,3

,

,

04

1ds

rr

rr

s

s

E

65

3 in trng tnh

3.3. Cng in trng

Cng in trng gy bi phn b in tch khi:

,3

,

,

04

1dv

rr

rr

v

v

E

66

3 in trng tnh

3.3. Cng in trng

V d

Gii

in tch phn b u di dng mt vng trn c bn knh b nh m t trong hnh v pha

di. Xc nh cng in trng ti mt im bt k trn trc ca vng trn ny?

- Phn t vi phn chiu di theo hng phn b in tch trong ta

tr l 'bd

- Vc t khong cch t mt phn t vi phn in tch ti im

quan st P(0,0,z):

zR ba za

2 2 2'

' '

3/2 3/22 2 2 2

0 00 0 0

1

4 4

l lz z

bbdb z b d z d

b z b z

E a a a a

- Do :

' 'os sinx ya a c a 2 2 2

' ' ' ' '

0 0 0

os sin 0x yd c d d

a a a

67

3 in trng tnh

3.3. Cng in trng

V d 1 (tip)

Cng in trng ti im P trn trc ca vng xuyn bng

3/22 2

02

lz

bz

b z

E a

ti tm vng trn th cng in trng bng khng

68

3 in trng tnh

3.3. Cng in trng

V d 2

Mt a mng hnh vnh khuyn vi bn knh trong a v bn knh ngoi

b, c mt in tch mt u s

Xc nh cng in trng ti mt im bt k trn trc z vi

z 0.

p s

2 ' ' '

3/2 1/2 1/2'2 2 2 2 2 2

0 00

1 1

4 2

b

s sz z

a

zd dz

z a z b z

E a a

2 ' ' ''

3/2'2 2

0 04

b

sz

a

d dz

z

E a a

69

3 in trng tnh

3.3. Cng in trng

V d 2 (tip)

T kt qu va tnh suy ra cng in trng trong cc trng hp sau

1/2

2 20

1

2

sz

z

a z

E a

1/2

2 20

1 1

2

sz

z

z b z

E a02

sz

E a

70

3 in trng tnh

3.4. in thng v mt in thng

+ Gi s t mt in tch th ti mt im trong in trng v cho php n dch chuyn n di chuyn dc theo mt ng no di tc dng ca lc in. ng ny gi l mt

ng lc (ng sc) hay mt ng thng lng

+ S lng cc ng sc gy bi mt in tch bng vi ln ca in tch theo n v

Cu-lng. Cc ng (trng) ny biu din thng lng in hay in thng.

! Cc ng thng lng in khng tn ti trong thc t nhng chng l mt khi nim hu ch cho vic biu din, quan st v m t cc in trng

+ Cng in trng ti mt im bt k lun c phng tip tuyn vi ng in thng

i qua im

71

3 in trng tnh


+ Cc tnh cht ca in thng:

- c lp vi mi trng

- ln ph thuc duy nht vo in tch to ra n

- Nu mt in tch im c bao bi mt hnh cu gi tng vi bn knh R, th in

thng xuyn qua mt cu theo phng vung gc v c gi tr nh nhau ti mi im

- Mt in thng, tc thng lng trn mt n v din tch, t l nghch vi R2

Mt in thng c th nh ngha theo cng in trng D E

0D E

+ Khi in trng gy bi in tch im q, mt in thng ti bn knh r l:

24r

q

rD a 2/C m

72

3 in trng tnh


3.4.1. in thng:

s

ds D

: phn t vi phn din tch trn b mt s nh hnh minh ha ds

Thng lng in qua mt s s ln nht khi v cng hng D ds

V d

Bit mt in thng 10 5 3r D a a a mC/m2

Xc nh in thng xuyn qua b mt ca min xt xc nh bi 0z 2 2 2 36x y z

Gii

36sin rds d d a/2 2

0 0

. 360 sin 720s

ds d d

D mC

73

3 in trng tnh


3.4.2. nh lut Gauss

Tng thng lng in hng ra bn ngoi khi xuyn qua mt mt kn th bng tng in tch

b gii hn trong b mt

s

Qsd

.D

+ nh lut Gauss cng c th biu din thng qua cng in trng

s

Qsd

0

.

E

+ Nu cc in tch phn b trong mt th tch bao bi mt mt kn th

s v

vdvsd

.D (dng tch phn ca nh lut Gauss)

nh lut Gauss c th c dng xc nh tng in tch b gii hn trong mt mt kn nu bit cng in trng hoc mt in thng ti mi im trn mt

74

3 in trng tnh


3.4.2. nh lut Gauss

+ T phng trnh dng tch phn ca nh lut Gauss, p dng nh l v tn, ta c

. vv v

dv dv D

. v D (dng im hay dng vi phn ca nh lut Gauss)

Cc ng in thng pht ra t bt k im no trong khng gian m ti tn ti mt mt in tch dng

Mt in thng l mt s o ca cc in tch t do c mt trong mt min xt

75

3 in trng tnh


3.4.2. nh lut Gauss

V d

Mt in tch phn b u trn mt cu c bn knh a nh minh ha trong hnh v di.

Xc nh cng in trng ti mi im trong khng gian?

Gii

Chn mt mt cu Gauss bn knh r sao cho trn

cng in trng l hng s

+ Khi r < a : cng in trng bng khng

+ Khi r > a : tng in tch b gii hn trong mt cu l

24 sQ a

Mt khc, ta c: s

rErsd24.

E

Theo nh lut Gauss

2

2 2

0 04

sr

aQE

r r

r a

76

3 in trng tnh

3.5. in th

+ Nu t mt in tch th dng q trong in trng th s c mt lc in

tc ng ln in tch

.qF E

Di tc dng ca lc ny, in tch dch chuyn mt on dl

Cng thc hin bi trng bng E

. .edW dl q dl F E

+ Nu mt in tch th chuyn ng ngc vi hng ca vc t cng in

trng di tc dng ca ngoi lc th cng thc hin bi ngoi lc bng

.extdW dl F

+ Gi thit rng ngoi lc cn bng vi lc in, Ton b cng do ngoi lc thc

hin dch chuyn in tch th t b ti a l:

.

a

ab

b

W q dl E

77

3 in trng tnh

3.5. in th

+ Nu dch chuyn in tch theo mt ng kn, th cng thc hin phi bng

khng

c

ld 0.

E

Trng iu kin tnh th khng xoy hay c tnh bo ton E 0 E

+ Khi rot bng khng, c th biu din in trng theo mt trng v hng V nh

sau:

V E

Do

. .

a a

ab

b b

W q dl q V dl E .V dl dV

.a

b

Va

ab a b ab

b V

W q dl q dV q V V qV E

v ln lt l in th cc im a v b tng ng vi im gc no aV bV

78

3 in trng tnh

3.5. in th

Cng dch chuyn mt in tch dng ngc hng tc dng ca in trng th bng tng th nng ca in tch

Hiu in th l s thay i th nng trn mt n v in tch trong gii hn gi tr in tch ny tin ti khng

0lim .

a

abab

qb

WV dl

q E

+Xt trng hp in tch im q t ti gc ta :

- Cng in trng ti mt im cch in tch q mt khong r 2

04r

q

rE a

- Hiu in th gia hai im bt k 1

2

2

0 0 1 2

1 1

4 4

r

ab

r

q qV dr

r r r

- Th tuyt i ca mt im P cch in tch im ti gc ta mt khong R l:

04a

qV

R

79

3 in trng tnh

3.5. in th

in th ti mt im bt k do cc dng phn b in tch gy nn

' ''

0

1

4

v

v

dvV

r

r r

' ''

0

1

4

s

s

dsV

r

r r

' ''

0

1

4

l

c

dlV

r

r r

Cc mt ng th

80

3 in trng tnh

3.6. Lng cc in

Mt lng cc in : mt cp in tch bng nhau v gi tr, ngc du, nm rt

gn nhau

Xc nh in th v cng in trng ti mt im bt k trong khng gian gy bi lng cc in

Mt lng cc in Cc khong cch xp x khi P xa lng cc

81

3 in trng tnh

3.6. Lng cc in

Khi khong cch gia cc in tch rt nh so

vi khong cch t lng cc ti im quan st

th tng in th ti im P khi s bng:

2 1

0 1 2 0 1 2

1 1

4 4

r rq qV

r r r r

Nu cc in tch c t i xng trn trc z

v im quan st rt xa chng

1 0,5 cosr r d 2 0,5 cosr r d

22 2

1 2 0,5 cosr r r d r

2

0

cos

4

q dV

r

V = 0 khi 090

82

3 in trng tnh

3.6. Lng cc in

+ nh ngha vc t m men lng cc p

ln p qd Hng thng t in tch m ti in tch dng

zqdp a

+ Khi , in th ti P c th vit

2

0

cos

4

q dV

r

2 2 3

0 0 0

.cos .

4 4 4

rpVr r r

p a p r

+ Lu

- Vi lng cc: in th ti mt im bt k t l nghch vi bnh phng

khong cch t im ti lng cc

- Vi in tch ring l: in th ti mt im bt k t ln nghch vi

khong cch t im xt ti in tch

83

3 in trng tnh

3.6. Lng cc in

+ Cng in trng

25

0

3 .

4

r

r

p r r pE

t l nghch vi lp phng khong cch t im xt ti lng cc

cc ng sc in trng nm trc tip trn trc z trong mt phng chia i lng cc

/ 2, .z a a

3

04 r

pE

Cc ng

ng th

Cc ng

E Khi nim lng cc in rt hu ch khi gii thch th hin ca mt vt liu cch in (in

mi) khi t n trong in trng

84

3 in trng tnh

3.7. Cc loi vt liu trong in trng

3.7.1. Vt liu dn in trong in trng

L loi vt liu cha mt s lng ln cc electron t do

Electron t do: - lin kt lng lo vi ht nhn ca n

- c th t do di chuyn trong vt dn

- p ng vi hu ht cc in trng v cng nh

- di chuyn lin tc khi chu tc dng lc

Cc tinh th kim loi: - 1, 2 hoc 3 electron ha tr trn mt nguyn t

- t do i vi ht nhn

- di chuyn ngu nhin di tc dng nhit

Vt dn c cch li: khng c hin tng tri cc electron ha tr theo hng xc nh

Khi c in trng bn trong vt dn di tc dng ca mt ngun nng lng bn ngoi: cc electron t do to nn dng in trong vt dn

85

3 in trng tnh



+ M t vt liu: theo in dn sut (thay v s electron t do)

s electron ha tr tng: in dn sut ca vt liu gim

+ Khng tn ti in tch d tha bn trong vt dn cch ly: b y ra b mt cho n

khi lc y cn bng vi lc cn b mt

mt in tch khi bn trong vt dn bng khng

0v

+ t vt dn c cch ly vo bn trong mt

in trng ngoi:

0E Cng in trng bn trong vt dn:

Mt in tch khi v cng in trng KHNG duy tr bn trong vt dn di cc iu kin tnh. Mi vt

dn to thnh mt min ng th trong khng gian

86

3 in trng tnh



in tch phn b u bn trong mt khi cu bn knh a. Mt v cu lm bng cht dn in,

bn knh trong b, bn knh ngoi c, c t ng tm vi khi cu in tch bn trong, nh

m t trong hnh di. Xc nh cng in trng ti mi im trong khng gian?

V d

Gii

+ Min I: Vi bn knh r a

Tng in tch trong min 34 .3

vQ r

Do phn b in tch u: vc t cng in trng hng dc theo bn knh, c gi tr bng nhau

trn mt cu Gauss

s

rErd24. sE

Theo nh lut Gauss s

Qd

0

.

sE

03

v r

r

E a 0 r a

87

3 in trng tnh



V d (tip)

+ Min II: Vi bn knh .a r b

Tng in tch trong min 34

3vQ a

Theo nh lut Gauss 3

2

03v r

a

r

E a a r b

+ Min III: Vi bn knh .b r c

bn trong vt dn bng khng E

B mt ti r b mang in tch m vi gi tr bng tng in tch dng

trong khi cu bn trong

Mt in tch mt trn b mt ny 3

23sb v

a

b

sbbQ 24

88

3 in trng tnh



V d (tip)

+ Min IV: Vi bn knh .r c

Ti phi mang in tch dng c gi tr

bng lng in tch m mt trong ,r c

Mt in tch mt trn b mt ny 3

2.

3sc v

a

c

24 ,scc Q Tng in tch mt trong b mt ny

Theo nh lut Gauss 3

2

03v r

a

r

E a r c

89

3 in trng tnh


3.7.2. Cc cht in mi trong in trng

+ in mi l tng (cht cch in): vt liu khng c electron t do trong cu trc

li ca n

+ Cc electron chu tc ng ca cc lc ko bn trong rt mnh, ngc vi hng di

chuyn ngu nhin ca chng

khi mt in trng c duy tr bn trong mt cht in mi di tc dng ca mt ngun nng lng bn ngoi, th s khng c dng in trong cht in

mi

+ in mi b phn cc di nh hng ca lc in:

tm ca in tch dng ca mt phn t khng cn trng vi tm ca in tch m na

90

3 in trng tnh



Mt in mi trng thi bnh thng,

vi tm ca cc cp in tch tri du

trng nhau

S phn tch ca cc cp in tch trong mt

in mi b phn cc

91

3 in trng tnh



+ in th ti mt im P bn ngoi mt in mi b phn cc

0limv v

pP

l m men lng cc ca th tch p v

'd dvp P

- Vc t phn cc : s m men lng cc

trn mt n v th tch P

- in th ti im P gy bi vi phn m men lng cc dp

'

2

0

.

4

RdV dvR

P a

- Thu c cng thc ' '

' '

0 0

.1 1 .

4 4

n

s v

V ds dvR R

P a P

92

3 in trng tnh



+ in th ti im P gy bi mt in mi b phn cc l tng i s ca hai thnh

phn: mt thnh phn b mt v mt thnh phn th tch

- nh ngha mt in tch mt gii hn .sb n P a

- nh ngha mt in tch khi gii hn .vb P

' '

' '

0

1

4

sb vb

s v

V ds dvR R

- Do :

S phn cc ca mt vt liu in mi dn ti hnh thnh cc phn b in tch gii hn (khng ging cc in tch t do)

93

3 in trng tnh



+ Nu mt min in mi cha mt in tch t do ph thm, th cng in

trng trong min in mi c tnh l

0 0

.. v vb v

PE

0. v E P l mt in tch t do v

+ Trong chn khng 0P

D

.v

+ Mt in thng trong bt k mi trng no nh sau

0 D E P

lun biu din mt in tch t do trong bt k mi trng no .D

94

3 in trng tnh



+ Trong mi trng in mi b cm ng bi mt in trng ngoi:

- in mi tuyn tnh EPEp

~~ (vc t phn cc t l thun vi in trng ngoi)

- in mi ng hng: cc c tnh in c lp v hng

- in mi ng nht: cc phn ca vt liu ging nhau hon ton

+ Ch xt vt liu loi A (tuyn tnh, ng hng, ng nht), v th:

0 P E c gi l nhy in

0 1 D E gi l h s in mi tng i, k hiu l 1 .r

0 r D E E

95

3 in trng tnh



V th, trong bt k mi trng no, cc in trng tnh cng lun tha mn cc phng trnh sau

0 E

. v D

D E

Tng cng in trng ko ton b electron ra khi cc phn t: hin tng nh thng in mi (sau in mi s ging nh mt cht dn in)

96

3 in trng tnh


3.7.3. Cc cht bn dn trong in trng

+ Cht bn dn: lng nh cc electron ha tr c th t do chuyn ng mt

cch ngu nhin bn trong cu trc tinh th

L cht dn in km

Nu t in tch d vo bn trong cht bn dn: dch chuyn ra pha b mt ca cht bn dn do tc ng ca cc lc y, chm hn so vi cht

dn in

Trng thi cn bng: khng c in tch d bn trong cht bn dn

+ Cht bn dn c t trong in trng: electron t do dch chuyn, to nn

mt in trng trit tiu in trng ngoi

Trong in trng tnh: bn dn ging nh dn in

Trong in trng tnh, c th nhm tt c cc loi vt liu thnh hai dng: cht dn in v cht in mi

97

3 in trng tnh

3.8. iu kin bin

3.8.1. Phn t php tuyn ca mt in thng

p dng nh lut Gauss: tng in tch trong hp dt l .s s

Mi trng 1

Mi trng 2

Giao din

1 2. .n n ss s s D a D a 1 2.n s a D D 1 2n n sD D hay hay

Cc phn t php tuyn ca mt in thng khng lin tc nu tn ti mt mt in tch mt t do trn giao din gia hai mi trng

98

3 in trng tnh

3.8. iu kin bin

3.8.1. Phn t php tuyn ca mt in thng

D E

1 1 2 2.n s a E E 1 1 2 2n n sE E + Giao din gia hai cht in mi khc nhau: khng c phn b in tch mt t do

trn giao din

Cc thnh phn php tuyn ca mt in thng l lin tc trn bin ca mt cht in mi

1 2n nD D

1 1 2 2n nE E

+ Nu mi trng 2 l vt dn, mi trng 1 l in mi:

02 D

1 1.n n sD a D 1 1n sE

Thnh phn php tuyn ca mt in thng trong mt mi trng in mi ngay trn b mt ca mt vt dn phi bng mt in tch mt ca vt dn

99

3 in trng tnh

3.8. iu kin bin

3.8.2. Thnh phn tip tuyn ca cng in trng

+ in trng c bo ton trong t nhin c

dl 0.E

+ Xt ng kn abcda nm vt ngang giao din nh hnh v

ab v cd c di w bc v da c chiu di 0h

Mi trng 1

Mi trng 2

Giao din Giao din

1 2. . 0 E w E w 1 2 . 0 E E whay

tw w a 1 2. 0t a E E hay 1 2t tE E

100

3 in trng tnh

3.8. iu kin bin

3.8.2. Thnh phn tip tuyn ca cng in trng

Cc thnh phn tip tuyn ca cng in trng l lin tc ti giao din gia hai mi trng khc nhau

+ Nu mi trng 1 l in mi v mi trng 2 l dn in: thnh phn tip

tuyn ca in trng trong mi trng 1 ngay st mi trng 2 phi bng khng

Trng tnh in bn trn mt vt dn lun vung gc vi b mt vt dn

1 2 0n a E E

101

3 in trng tnh

3.9. T in v in dung

+ Hai vt dn cch ly t cnh nhau nh hnh minh ha to thnh mt t in

+ Cung cp mt ngun nng lng bn ngoi: c th truyn cc in tch t vt dn

ny sang vt dn kia (np t)

+ Trong qu trnh np: hai vt dn c lng in tch bng nhau nhng tri du

to nn mt in trng trong min in mi v do to nn mt hiu in th gia hai vt dn

+ Hiu in th gia hai vt dn t l thun vi lng in tch c chuyn qua

min in mi

Vt dn a

Vt dn b

102

3 in trng tnh

3.9. T in v in dung

+ in dung l t s gia in tch ca mt vt dn trn in th ca n so vi vt dn

cn li

a

ab

QC

V

V d:

Hai phin dn in t song song, mi phin c din tch A v cch nhau mt khong

d, nh hnh minh ha, to thnh mt t in hai bn phng song song (t in phng).

in tch trn mt phin l +Q v trn phin cn li l Q. in dung ca t c biu

thc ra sao? Biu din nng lng tch tr trong mi trng ny (t in) theo in

dung ca t?

103

3 in trng tnh

3.9. T in v in dung

Gii

Gi s khong cch gia hai tm dn in l rt nh so vi kch thc ca chng, v

in tch phn b u trn cc tm kim loi. Khi , cng in trng gia hai

tm kim loi l:

sz

E a s

Q

A

Q l in tch trn tm a nm pha trn ti z = d, A l din tch b mt mi tm kim

loi, in tch trn tm b ti z = 0 l Q.

in th ca tm a so vi tm b l

0

.

a d

s sab

b

d QdV dl dz

A

E

Do , in dung ca t in l ab

Q AC

V d

Nng lng tch tr trong h thng bng

2 2 2 2 21 1 1 1 1

2 2 2 2 2s ab

v

Ad dW E dv Q Q CV

A C

104

3 in trng tnh

3.10. Cc phng trnh Laplace v Poisson

+ Trc: xc nh cc trng tnh in trong mi trng m phn b in tch l xc

nh ti mi ni

+ Nhiu trng hp:

- Trn cc mt bin, hoc mt in tch mt hoc in th l xc nh

- Phi xc nh cng in trng trc khi chng ta c th tnh phn b in tch

+ Lut Guass trong mt mi trng tuyn tnh:

D E . v E

V E . vV

. . vV V 2 . vV V

105

3 in trng tnh


+ Trong mi trng ng nht : 0

2 vV

(phng trnh Poisson)

Phn b in th trong mt min khng gian ph thuc vo phn b in tch cc b trong min

+ Vi vt dn ng nht: mt in tch khi t do bng khng

2 0V (phng trnh Laplace)

Trong min khng cha in tch: tm hm in th V tha mn phng trnh Laplace theo cc iu kin bin

Cng in trng .V E

106

3 in trng tnh


V d

Hai tm kim loi trong hnh minh ha c din tch A v cch nhau mt khong d to

thnh mt t in phng. Tm pha trn c in th V0, tm pha di c in th bng

khng. Hy xc dnh (a) phn b in th, (b) cng in trng, (c) phn b in

tch trn mi tm kim loi v (d) in dung ca t in phng.

Gii

Hai tm kim loi to thnh hai mt ng th trong

mt phng xy ti z = 0 v z = d : in th V s ch

l hm ca z

Trong min khng gian khng cha in tch gia

hai tm kim loi, phng trnh Laplace suy bin

thnh: 2

20

V

z

V az b z = 0, V = 0 b =0 z = d, V = V0 a = V0/d

0

zV V

d

107

3 in trng tnh


V d (tip)

+ Cng in trng 0z z

VVV

z d

E a a

+ Mt in thng 0z

V

d

D E a

+ Mt in tch b mt ca tm kim loi pha di 00s zV

d

+ Mt in tch mt trn tm kim loi pha trn 0s z d

V

d

+ Tng in tch trn tm kim loi pha trn 0s

V AQ A

d

+ in dung ca t in phng cho 0

Q AC

V d

108

3 in trng tnh

3.11. Phng php soi nh

gi thit: cc in tch t thn chng tn ti v khng c bt c th g trong min

xt c th gy nh hng ln cc trng ca chng

Tuy nhin, cc in tch (hoc phn b in tch) thng kh gn cc b mt dn

in, v nh hng ca nhng mt dn ny phi c k n khi tnh tng cc trng

trong min xt

Vi cc trng tnh: - mt vt dn to nn mt mt ng th

- khng c cc trng bn trong mt vt dn cch ly

- cc trng lun vung gc vi b mt vt dn

gip lng ha phn b in tch trn b mt ca mt vt dn v nh hng ca n ln cc trng trong min xt

Vi lng cc: - in th ti bt k im no trn mt phng ct i lng cc cng

bng khng

- cng in trng vung gc vi mt phng ny

mt ct i ny tha mn cc yu cu ca mt mt dn in

109

3 in trng tnh


Mt mt dn t trng vi mt ct i ny: cc thnh phn trng ca lng cc khng thay i

+ Nu cc in tch m bn di mt phng dn in ny b loi b:

- Phn b trng pha trn mt phng vn gi nguyn

- Tng in tch cm ng trn b mt vt dn l q

+ Nu c mt in tch im q nm cch mt khong h pha trn mt mt phng dn

in rng v hn:

xc nh c in th v in trng ti bt k im no trn mt phng ny

Bng cch: - B qua mt phng dn in cho

- Tng tng c mt in tch im q nm i xng vi in

tch q cho pha bn kia ca mt phng dn in

q c gi l nh ca in tch thc q

110

3 in trng tnh


Trong phng php nh:

- Mt phng dn in b loi b tm thi

- Mt in tch o c t pha sau n

- in tch o c cng ln nhng ngc du vi in tch thc

- Khong cch gia in tch thc v o bng hai ln khong cch gia

in tch thc v mt phng dn in cho

Mt phng dn in

in tch im t gia hai mt phng dn in song song: s lng nh bng v cng

111

3 in trng tnh


V d

Mt in tch im q nm pha trn mt mt phng dn in rng v dy v hn. Tnh

in th v cng in trng ti mt im P bt k. Ch ra rng tng in tch trn

b mt ca mt phng dn in bng q.

Gii

Hnh di minh ha in tch im q ti im (0,0,d) pha trn mt mt phng dn

in. xc nh cc trng, chng ta t mt in tch o q vo im (0,0,-d) v tm

thi b qua s tn ti ca mt phng dn in ni trn

Mt phng dn in

112

3 in trng tnh


V d (tip)

+ in th ti im P(x,y,z) vi 0z

1 2

1 1

4

qV

R R

Mt phng dn in

1/2

22 2

1R x y z d

1/2

22 2

2 .R x y z d

+ Trn b mt ca mt phng dn in: V = 0

+ Cng in trng ti im P

3 3 3 3 3 3

2 1 2 1 2 14x y z

q x x y y z d z dV

R R R R R R

E a a a

113

+ Trn b mt vt dn, cng in trng gim xung cn

3 in trng tnh


V d (tip)

3

2

4z

qd

R E a

1/22 2 2R x y d

Mt phng dn in

+ Mt in tch mt trn b mt vt dn ti z = 0

3

2

4s

qd

R

+ Tng in tch cm ng trn b mt ca mt phng

dn rng v hn l

2

3/22 2

0 0

2

4s

s

qd dQ ds d q

d

114

4 Dng in mt chiu

4.1. Gii thiu

+ Xt chuyn ng ca cc in tch trong mt vt dn in, khi c in trng tnh

duy tr bn trong vt dn

+ t mt vt dn trong in trng:

+ Gi s t vo hai pha i din ca vt dn mt lng in tch ht bng nhau nhng

tri du:

Bn trong vt dn tn ti mt min khng c in tch

Cc in tch t phn b li trn cc b mt ca vt dn

Vt dn mt trng thi cn bng Mt in trng c thit lp bn trong vt dn

Cc in tch ht tri du dch chuyn li gn nhau, tip xc, trit tiu ln nhau Khi tt c cc in tch ht bin mt: in trng bn trong vt dn bin mt, vt dn tr li trng thi cn bng

dch chuyn ca cc in tch to nn mt dng in

115

4 Dng in mt chiu

4.1. Gii thiu

+ Dng in l tc in tch c vn chuyn qua mt im cho trc, bn trong

mt mi trng dn in

dqi

dt

+ Trong phn ny, ch tho lun v dng in khng i (dng in c gi tr hng s

theo thi gian)

Dng in hng cn c gi l dng in mt chiu

116

4 Dng in mt chiu

4.2. Dng in dn

+ Electron t do: dch chuyn xuyn qua ton b cu trc tinh th

+ Cc electron t do: tham gia vo qu trnh dn in trong kim loi

+ Cc ion dng: nng, lun nm c nh ti v tr thng l ca chng trong li tinh

th khng tham gia vo dng in trong kim loi

Dng in trong mt vt dn kim loi: dng in dn (dng chuyn dch ca cc electron)

duy tr c dng in khng i bn trong mt vt dn: cn cung cp lin tc cc electron ti mt u v loi b chng u cn li

+ Xt mt vt dn m hai u ca n c ni ti mt qu pin

Tn ti chnh lch in th gia hai u dy dn tn ti mt in trng bn trong vt dn

+ S thay i vn tc ca cc electron di tc ng ca lc

in theo trc z: mt phn rt nh trong vn tc chuyn ng

ngu nhin ca n lc in to ra vn tc tri

117

4 Dng in mt chiu

4.2. Dng in dn

+ S tri ca cc electron dc theo phng tc ng ca in trng to nn mt

dng in chy qua vt dn

Tri

+ Hng quy c ca dng in c chn trng vi

hng ca in trng

+ Dng in chy qua vt dn c gi tr nh nhau ti

mi tit din ngang

duy tr dng in hng: mi im trn vt dn khng

th ng vai tr nh mt "ngun" hay mt "kho" in tch

4.3. Dng in i lu

+ Chuyn ng ca cc ht in tch trong chn

khng to nn dng in i lu

+ Xt v d bn: Khi vn tc ca cc electron

tng th mt in tch gim i

dng in i lu khng trng thi trung ha tnh in Khng cn vt dn duy tr lung in tch, khng tun theo nh lut Ohm

118

4 Dng in mt chiu

4.4. Mt dng in i lu (dch)

+ Xt mt min vi phn b in tch khi,

cc in tch di chuyn di tc dng ca

mt in trng:

tld U

+ Xt mt ca s o vi din tch b mt

vung gc vi vn tc tri

nas

s

ldvdq vv

.s

+ Lng in tch dch chuyn qua ca s

+ Dng in xuyn qua b mt s Uss

.. vvdt

ldI

+ nh ngha dng in ny theo mt dng in dch J

sJ

.I

UJ

v

+ Dng in chy qua mt mt s c tnh bng s

sdI

.J

119

4 Dng in mt chiu

4.4. Mt dng in i lu (dch)

cng thng c gi l mt dng in khi J

+ Nu c cc in tch dng v in tch m, vi cc mt in tch khi v v

di chuyn vi vn tc trung bnh U

U

Tng mt dng in s l UUJ

vv

Dng in l khng i trong mt min no , nu mt dng in l hng s ti mi im trong min

4.5. Mt dng in dn

eU

vn tc trung bnh (hay vn tc tri) ca mt electron bn trong mt vt dn di

tc dng ca in trng E

em khi lng ca electron thi gian trung bnh xy ra mt va chm

Tn hao ng lng trong khong thi gian eem U

Tc trung bnh tn hao ng nng trong va chm /eem U

Tc electron thu ng nng t lc in E

e

120

4 Dng in mt chiu

4.5. Mt dng in dn

trng thi xc lp hng EU

em ee

ee

m

e EU

hay EU

ee ue

em

eu

vi - linh ng electron

Vn tc tri ca mt electron trong vt liu dn in t l thun vi in trng tc dng ln vt dn

+ Nu c N electron trong mt n v th tch th mt electron l Nev

+ Do , mt dng in dn, trong mi trng dn in, l

ev UJ

EEJ

eNeu

eNeu - in dn sut ca vt liu

Mt dng in mt im bt k trong vt dn t l thun vi cng in trng

Vi vt liu tuyn tnh, v cng hng J

E

121

4 Dng in mt chiu

4.5. Mt dng in dn

V d. Mt in p 10V c duy tr trn hai u ca mt si dy ng di 2m. Bit

thi gian trung bnh gia cc va chm l 2,7.10-14 s, hy xc nh vn tc tri ca cc

electron t do.

Gii. Gi s si dy ko di theo trc z (phng thng ng, chiu dng hng t

di ln trn) v u trn ca dy c in th cao hn. Khi , cng in trng

trong si dy s l

zz aaE

52

10

(V/m)

linh ng electron xc nh theo cng thc

3

31

1419

10747,4101,9

107,2106,1

ee

m

eu

Vn tc tri ca cc electron t do l

zzee u aaEU 33 1074,23510747,4 (m/s)

122

4 Dng in mt chiu

4.6. in tr ca vt dn

ssd

ld

I

dVdR

.

.

J

E

dV l hiu in th gia hai u ca vt dn

l cng in trng bn trong vt dn E

l mt dng in dn EJ

+ in tr ca ton b vt dn

a

b

s

sd

ldR

.

.

J

E

I

V

sd

ld

R ab

s

a

b

.

.

J

E

I l dng in chy qua mi mt s

123

V d. Mt hiu in th c gi tr V0 c duy tr gia hai u ca mt on dy

ng vi chiu di l. Nu A l din tch tit din ngang ca dy, hy vit biu thc

in tr ca dy dn ny. Gi tr ca in tr bng bao nhiu, nu V0 = 2 kV, l = 200

km, A = 40 mm2 ?

4 Dng in mt chiu

4.6. in tr ca vt dn

Gii. Gi s rng dy dn dng ng theo phng ca trc z, u dy trn ni vi cc

c in th cao hn ca ngun V0. Cng in trng trong dy dn trong trng

hp ny s l:

zl

VaE 0

+ Mt dng in khi ti mt tit din ngang bt k ca dy dn s l:

zl

VaEJ 0

+ Dng in chy qua dy dn l

AVds

l

VsdI

ss

00.

J

+ Do , in tr ca dy dn A

l

A

l

I

VR

0

+ Thay s

8510.40

10.20010.7,16

38

R

124

4 Dng in mt chiu

4.7. Phng trnh v tnh lin tc ca dng in

+ Cho mt min dn in c bao bi mt s nh hnh v

+ Dng in tng hng ra ngoi mt kn

s

sdti

.J

+ Khi c lung in tch i ra ngoi: mt in

tch bn trong mt s gim i mt lng ng bng

nh vy

dt

dQti

v

vdvQ v

v

s

dvdt

dsd

.J

(dng tch phn ca phng trnh v tnh lin tc)

+ C th vit thnh

v

v

v

dvt

dv

J

. 0.

dvt

v

vJ

125

4 Dng in mt chiu


0.

t

vJ

(dng vi phn ca phng trnh v tnh lin tc)

t

v

J

.+ C th vit di dng

Cc im m mt in tch khi thay i: ngun ca mt dng in khi

+ Vi mt min dn in c dng in hng chy qua

0. s

sd

J 0. J

hay

dng in i vo v i ra khi mt mt kn bt k l bng khng

+ Ti mt im 0I (nh lut Kirchoff v dng in)

126

4 Dng in mt chiu


+ Thay EJ

ta c

0. E

0.. EE

+ Mi trng ng nht:

0. E

+ Thay VE

(V l in th ti mt im bt k bn trong min dn)

02 V

Phn b in th bn trong mt mi trng dn in nghim ng phng trnh Laplace khi mi trng dn in l ng nht v mt dng in khng

i theo thi gian

127

4 Dng in mt chiu


V d. Mt vt liu vi in dn sut , vi m v k l cc hng s, lp y

khng gian gia hai vt dn in hnh tr, ng tm vi bn knh ln lt l a v b

nh minh ha trong hnh v. Nu V0 l hiu in th gia hai vt dn, v L l chiu

di ca mi vt dn, hy vit biu thc in tr ca vt liu, mt dng in v

cng in trng bn trong vt liu cho?

km /

Gii. Do dng in tng I chy qua bt k mt

tit din ngang no u phi bng nhau, nn

mt dng in trong min dn in l

aJ

L

I

2

Mt khc, ta c cng in trng trong min

dn l:

a

JE

kmL

I

2

128

+ Hiu in th gia hai vt dn in c tnh t cng thc

4 Dng in mt chiu


V d (tip)

LkIM

kam

kbm

Lk

I

kmL

IdldV

a

bc

2ln

22.0

E

+ in tr bn trong min dn in xc nh c l

Lk

M

I

VR

20

+ Dng in chy qua min dn l 00 2 V

M

Lk

R

VI

+ Cng in trng, mt dng in:

aE

0VMkm

k

aEJ

0V

M

k

129

4 Dng in mt chiu

4.8. Thi gian hi phc

+ Xt mt mi trng ng hng, ng nht, tuyn tnh v b cch ly, vi hng s in

mi v in dn sut

+ Gi s c mt in tch vi mt khi c t thm vo mi trng v

in tch t thm b y ln b mt ca mi trng t ti trng thi cn bng tnh in

Trong qu trnh di chuyn ca in tch ny, phng trnh v tnh lin tc phi c tha mn

0.

t

vJ

EJ

0.

t

v E

vE

. 0

v

v

t

Nghim:

t

v e

0

trng thi cn bng tnh in trong mi trng k trn s dn tr li theo dng hm m

130

4 Dng in mt chiu

4.8. Thi gian hi phc

gi l thi gian phc hi

+ L khong thi gian cn thit in tch trong mt mi trng dn in

bt k gim v cn 36,8% gi tr ban u ca n

+ Mi trng dn in t ti trng thi cn bng sau qung thi gian

bng 5 ln thi gian phc hi

131

4 Dng in mt chiu

4.9. nh lut Joule

+ Lc tc ng ln in tch bn trong mt th tch dv bt k, t trong in trng

EF

dvd v l mt in tch khi ca min xt v

+ Nu cc in tch di chuyn c mt qung ng dtld U

th cng thc hin bi in trng bng:

dvdtdvdtldddW v EJEUF

...

+ Cng sut do in trng thc hin l:

dvdt

dWdp EJ

.

+ nh ngha mt cng sut p l cng sut trn mt n v th tch

pdvdp EJ

.p

dng vi phn ca nh lut Joule

132

4 Dng in mt chiu

4.9. nh lut Joule

+ Cng sut tng ng vi mt th tch v

vv

dvpdvP EJ

.

dng tch phn ca nh lut Joule

+ Xt cc in tch t do chuyn ng bn trong mt mi trng dn in:

lc tc ng bi in trng cn bng vi tn hao ng nng do va chm

cng sut cp bi in trng b tiu tn di dng nhit

mt cng sut p th hin tc ta nhit theo thi gian trn mt n v th tch

133

4 Dng in mt chiu

4.9. nh lut Joule

- Xt dy dn c chiu di L, tit din u A, in th V gia hai u dy, th mt

cng sut bng:

2

L

Vp W/m3

Tng cng sut tn hao do nhit trn ton b dy dn:

R

V

L

AVP

22

A

LR

- Vi mt vt dn in tuyn tnh 2. Ep EE

- Tng cng sut tiu tn bng v

dvEP 2

W

134

4 Dng in mt chiu

4.10. Cc iu kin bin cho mt dng in

Min 1

Min 2

0. s

sd

J

0.0.. 2121 JJaJaJa

nnn ss

21 nn JJ

Thnh phn php tuyn ca mt dng in lin tc khi qua mt gii hn gia hai mi trng

135

02

2

1

1

JJa

n

4 Dng in mt chiu


Min 1

Min 2

+ Do thnh phn tip tuyn ca cng in trng

lin tc ti giao din, tc

021 EEa

n

C th vit phng trnh cho thnh phn tip tuyn ca J

EJ

2

1

2

1

t

t

J

J

+ Chng ta c

2

22

1

11

t

n

t

n

J

J

J

J

2

1

2

1

tan

tan

136

4 Dng in mt chiu


+ Xt giao din gia mt min dn in km (mi trng 1) v mt min dn in tt

(mi trng 2)

Min 1

Min 2

nm trong khong t 0 ti 900 2

12

trong mi trng 1 gn nh vung gc vi

giao din J

E

12

12 nn EE

rt nh gn nh khng tn ti bn trong mi trng 2

Phi tn ti mt mt in tch b mt t do ti giao din

2

2

1

11

2

12211

12

21121 1

nnnnns JEDDD

137

4 Dng in mt chiu


V d

Gi s mi trng 1 (z 0) c hng s in mi bng 2 v in dn sut l 40 S/m.

Mi trng 2 (z 0) c hng s in mi bng 5 v in dn sut l 50 nS/m. Nu

c ln l 2 A/m, v so vi phng vung gc vi giao din gia hai mi

trng, hy tnh v ? Mt in tch b mt trn giao din bng bao nhiu?

2J

602

1J

1

Gii

Ta c 160cos22

nJ A/m2 732,160sin22

tJ A/m

2

T iu kin bin ca mt dng in: 121 nn JJ

6,1385732,11050

10409

6

22

11

tt JJ

A/m2

A/m2

Do , mt dng in trong mi trng 1 l

6,1385)6,1385(1 2/1222/121211 tn JJJ A/m2

138

4 Dng in mt chiu


V d (tip)

Gc lch ca so vi phng php tuyn ca giao din bng 1J

96,891

6,1385tantan

1

11

a

J

Ja

n

t

Mt in tch trn b mt giao din

88,036

10.

1050

5

1040

2.1

9

962

2

1

11

ns J mC/m

2

139

4 Dng in mt chiu

4.11. S tng ng gia v D

J

+ Vi dng in hng ta c 0. J

+ Trong mt min khng c in tch ta c 0. D

EJ

ED

0 E

0 J

0 D ( , l hng s)

+ Ti giao din gia hai mi trng dn in: 21 nn JJ

+ Gia hai mi trng khng dn in: 21 nn DD

+ Cc iu kin bin:

2

1

2

1

t

t

J

J

2

1

2

1

t

t

D

D

140

4 Dng in mt chiu


J

V d

Cho hiu in th gia hai bn phng song song l V0. Cho din tch ca mi bn l A,

khong cch gia chng l d, mi trng dn in gia hai bn cc c c trng bi

hai thng s, hng s in mi v in dn sut , hy xc nh dng in chy qua

min ny bng cch s dng tnh tng ng gia cc trng v ?

D

J

Gii

+ Cng in trng trong mi trng gia hai bn phng song song

zd

VaE 0

+ Mt in thng trong mi trng gia hai bn cc bng

zVd

aD

0

+ S dng tnh tng ng gia v chng ta nhn c mt in tch khi bn

trong min D

J

141

4 Dng in mt chiu


J

V d (tip)

zVd

aJ

0

+ V vy, dng in chy trong mi trng tnh theo cng thc

R

VV

d

AsdI

s

00.

J

A

dR

a

b

s

n

a

b

s

s

abld

dsE

ld

ds

V

QC

.. EE

+ Ngoi ra:

a

b

s

n

a

b

s

abld

dsE

ld

sd

V

IG

..

.

EE

J

CG

142

4 Dng in mt chiu

4.12. Sc in ng

+ Vi in trng tnh ta c: 0. c

ld

E

+ Mt dng in khi trong mt mi trng dn in EJ

+ Dng in chy qua vt dn ss

sdsdI

.. EJ

in trng tnh tuyt i khng th gy nn v duy tr dng in chy trn mt ng kn

Cn phi c mt ngun nng lng duy tr mt dng in khng i bn trong mt vng kn

Ngun nng lng ny l phn t khng bo ton trong mch in chng to nn mt in trng khng bo ton 'E

143

4 Dng in mt chiu

4.12. Sc in ng

+ in trng tng trong mt vng kn 'EE

+ Cng sut tiu tn trn tt c cc phn t trong vng kn:

v

dvP JEE

.'

+ Gi s rng dng in khng i trong vng kn c phn b u

dvJ

thay bng lId

cc

ldIldIP

'..' EEE

+ nh ngha sc in ng trong mt vng kn nh sau

c

ld

'.E

144

4 Dng in mt chiu

4.12. Sc in ng

+ Vi mt nhnh ca mch in ni gia hai nt a v b

ababb

a

b

a

b

a

b

a

VVldldldld

'...'.1

EEEEJ

0ab nhnh cha ngun

+ Vt dn hnh tr trn vi tit din A, chiu di L ni gia hai im a v b

Vt dn

in

Vt dn

in

AIJ /

IRA

IL

IRVV abab

+ Xt vng kn Va = Vb

n

j

j

m

i

i IR11

(lut Kirchoff v in p)

145

5 T trng tnh

5.1. Gii thiu

+ Pht hin qung st t ha vnh cu lnh vc nghin cu t tnh hc

+ Kh nng t nh hng ca nam chm theo hng bc nam: lc t

- Vt liu c th b nh hng (t ha) bi lc t: vt liu t

- Vt liu b t ha: nam chm

+ T trng c mi lin h vi nam chm: ging

mi lin h gia in trng v mt in tch

+ Thc nghim: cc cc cng du y nhau,

cc cc khc du ht nhau

+ Cc th nghim vi nam chm ch ra rng: hai cc (bc v nam) ca mt nam

chm khng th tch ri cc t cch ly (c lp) khng phi l mt thc th vt l

146

5 T trng tnh

5.1. Gii thiu

cy cu qua h ngn cch in v t: dng in cng chnh l cc ngun t trng

+ Khm ph quan trng:

(1820), khi Hans Christian Oersted (c-x-tt) pht hin mt thanh nam

chm nh b lch i bi dng in chy trong mt dy dn

+ Bio v Savart: cng thc xc nh mt t thng ti mt im bt k gy bi

mt vt dn c dng in chy qua

+ Andr Marie Ampre: pht hin s tn ti ca lc t gia hai vt dn mang

dng in

+ Chng ny tp trung nghin cu t trng tnh (t trng c to ra bi cc

dng in khng i)

147

5 T trng tnh

5.2. nh lut Bio - Savart

+ Mt t thng ti im P gy bi mt phn t dy dn mang dng in khng

i c chiu di ld

2R

lIdkd R

aB

B

d vi phn mt t thng (T) (Wb/m2)

4

0k7

0 10.4 H/m

30

4 R

lIdd

RB

cR

lId3

0

4

RB

148

5 T trng tnh


+ Biu din phn t dng in lId

theo mt dng in khi

dvlId vJ

thu c cng thc biu din mt t thng

v

v dvR3

0

4

RJB

+ Biu din mt t thng theo mt dng in mt:

s

s dsR3

0

4

RJB

+ Biu din 5.2 theo mt in tch q chuyn ng vi vn tc trung bnh U

Adldq v

UJ

dqdvv

30

4 R

q RUB

149

5 T trng tnh


V d 1

Cho mt dy dn nh (tit din c th b qua) vi chiu di hu hn t z = a ti z = b,

nh hnh minh ha. Hy xc nh mt t thng ti mt im P trong mt phng xy?

Mt t thng ti P bng bao nhiu nu chiu di dy l v cng?

Gii

+ V chng ta c zIdzlId a

zzaaR

aR

dzIlId

+ V vy mt t thng ti P bng

aaB

22220

2/322

0

44 a

a

b

bI

z

dzIb

a

150

5 T trng tnh


V d 1

+ Mt t thng mt im bt k trong mt phng xy gy bi mt dy dn di v hn

mang dng in a b

aB

2

0I

Trn mt mt phng bt k vung gc vi dy, cc ng sc t lun c dng cc ng trn bao

quanh dy dn

Lu : Bin thin ln ca theo ging ca theo B

E

Di mt s iu kin nht nh, c th c s tng ng gia in trng tnh v t trng tnh (hng ca chng khc nhau)

151

5 T trng tnh


V d 2

Cho mt vng dy hnh trn, bn knh b, mang dng in khng i I nm trong mt

phng xy, nh hnh minh ha. Hy vit biu thc mt t thng ti mt im trn

hng dng ca trc z? Tm cng thc xp x mt t thng ti mt im trn trc z

cch xa vng dy?

Gii

+ Do a

bdld zzb aaR

dbzbld z aaR

2

+ Ta c biu thc mt t thng

zz

zb

Ib

zb

dIbz

zb

daIba

aB

2/322

20

2

02/322

02

02/322

20

244

152

5 T trng tnh


V d 2

Trn trc ca vng dy mang dng in, mt t thng ch c duy nht thnh phn theo hng z

+ Cho z = 0, thu c mt t thng ti tm vng dy z

b

IaB

2

0

+ Khi im quan st trn trc v cch xa vng dy bz 32/322 zzb

zz

IbaB

3

20

2

+ Khi im quan st cch xa vng dy, kch thc vng dy s rt nh so vi

khong cch z

coi vng dy mang dng in nh mt phn t lng cc t

153

5 T trng tnh


V d 2

+ nh ngha m men lng cc t nh sau zz IAbI aam

2

mt t thng 3

0

2 z

mB

+ Hnh nh cc ng sc t gy bi mt vng dy mang dng in:

- min pha trn vng dy ging nh cc bc ca mt nam chm

- min pha di vng dy ging nh cc nam ca mt nam chm

khi mt vng dy mang dng in, n to nn mt nam chm in

154

5 T trng tnh

5.3. nh lut Ampe v lc t

Xc nh lc tc ng ln mt vt dn mang dng in do mt vt dn mang dng in khc gy ra

+ Khi hai phn t mang dng in v tng tc, lc t do phn t 1 tc ng

ln phn t 2 l 11 ldI

22 ldI

321

21112202

4 R

ldIldId

RF

+ Lc t do dy dn mang dng in th nht gy nn trn dy dn mang dng in th

hai l

12

321

211122

02

4cc

R

ldIldI

RF

1222

2

BF

c

ldI

(phng trnh nh lut Ampre v lc t)

155

5 T trng tnh


1

321

211101

4c

R

ldI RB

+ mt t thng gy bi dy dn mang dng in th nht ln v tr ca phn t

mang dng in th hai 1B

+ Tng qut: khi mt vt dn mang dng in t trong mt t trng ngoi th dy

dn chu tc dng ca mt lc t bng: B

BF

c

lId

+ Biu din theo mt dng in khi dv

v

v BJF

(dng tng qut ca nh lut Ampre v lc t)

156

5 T trng tnh


+ Nu

l mt in tch khi 1v l vn tc trung bnh ca in tch 1U

A1 l tit din ca vt dn mang dng in 1

1111 ldAdq v

1111 UJ

dqdvv

l mt t thng trong min t vt dn B

Lc t tc ng ln in tch q1 bng BUF

111 q

+ Lc t tc ng ln in tch q1 trong t trng do q2 chuyn ng vi vn tc trung

bnh gy ra l 2U

122211312

01

4RUUF

qqR

(c s ca nh lut Am-pe v nh lut Bio-Savart)

157

5 T trng tnh


V d 1

Cho mt dy dn un cong nh trong hnh v, nm trn mt phng xy v mang mt

dng in khng i I. Nu mt t thng trong min dt dy dn l ,

hy xc nh lc t tc dng ln dy dn ny? zBaB

Gii

+ Lc t tc dng ln on dy dn c di t

Lax ax

y

a

La

zx

a

La

zx

c

IBLdxIBBIdxlId aaaaaBF

1

+ Lc t tc dng ln on dy dn c di t

ax Lax

y

La

a

zx IBLdxIBF aaa

2

158

5 T trng tnh


+ Lc t tc dng ln on dy c dng na ng trn bn knh a l:

y

yxzz

IBa

dIBadIBaadIBBIad

a

aaaaaaaF

2

sincos

0000

3

+ Lc t tc dng ln c si dy bng:

yLaIB aFFFF

2321

( bng tng lc t tc dng ln si dy thng c chiu di ) La 2

+ Trong trng hp hai dy dn mang dng in, cng thc nh lut Am-pe c dng:

2 1

1212321

2102

1

4c c

ldldR

IIRF

159

5 T trng tnh


+ S dng hng ng thc vc t ta c BACCABCBA

..

2 1 2 1

21321

1213

21

2122102

..

4c c c c

R

ldldld

R

ldIIR

RF

- Tch phn u tin bn v phi vit thnh:

2 1

1221

.1

c c

ldldR

- Nu dy dn mang dng in I2 to thnh mt vng kn

1 2

1221

.1

c s

ldsdR

+ Lc t tc dng ln mt vng dy c hnh dng bt k mang dng in:

1 2

21321

212102

.

4c c

R

ldldIIRF

160

5 T trng tnh


V d 2

Cho mt khung dy hnh ch nht mang dng in khng i I2 t gn mt dy

dn thng mang dng in khng i I1, nh minh ha trong hnh 5.12. Hy vit

biu thc lc t tc dng ln khung dy?

Gii

Vi phn chiu di trn cc cnh AB hay CD ca

khung dy

zdzld a

22

Vi phn chiu di trn cc cnh BC hay DA ca

khung dy

ydyld a

22

Phn t vi phn chiu di ca si dy thng mang

dng in I1

zdzld a

11

Trn cc cnh BC v DA ca khung dy . = 0 1ld

2ld

161

5 T trng tnh


V d 2 (tip)

+ Xc nh lc t tc dng ln cnh AB:

zy zzb aaR

1221

y

a

a

zyL

L

AB

baLbaLb

II

dz

zzb

zzbdz

II

a

aaF

2222210

22/3212

2

121

210

2

4

+ Lc t tc dng ln cnh CD ca khung dy:

yCD caLcaLc

IIaF

2222210

2

ycaLcaLc

baLbaLb

IIaF

22222222210 11

2

+ Lc t tc dng ln khung dy:

162

5 T trng tnh

5.4. M men xon t

+ Vt dn mang dng in nm trong t trng ngoi: chu tc dng ca lc t c

xu hng di chuyn vt dn theo hng vung gc vi c t trng v vt dn

+ Cun dy mang dng in t trong t trng: lc t tc dng ln cun dy c th

khin cun dy xoay trn

+Xt mt vng dy hnh ch nht mang dng in mt chiu I t trong t trng

nh hnh minh ha a)

163

5 T trng tnh

5.4. M men xon t

+ Lc t tc dng ln cnh ab ca cun dy l: yab BILaF

+ Lc t tc dng ln cnh cd l ycd BILaF

ng tc ng ca hai lc khng trng nhau: cc lc ny to ra mt m men xon, c xu hng khin cun dy quay xung quanh trc ca n

- Cnh tay n bn pha cnh ab x

Wa

2- Bn pha cnh cd

x

Wa

2

Bn pha cnh ab chu mt m men xon:

zabxab BILWW

aFaT

2

1

2

Bn pha cnh cd chu mt m men xon:

zcdxcd BILWW

aFaT

2

1

2

164

5 T trng tnh

5.4. M men xon t

+ Tng m men xon tc dng ln cun dy: zcdab BILWaTTT

+ Vit li theo m men lng cc t nh sau BmT

yy IAILW aam

165

5 T trng tnh

5.4. M men xon t

+ Gi s di tc dng ca m men xon, cun dy b xoay quanh trc ca n v to

mt gc so vi trc y nh minh ha trong hnh minh ha c)

Lc t tc dng ln cnh bc by gi khc khng v bng

zc

xyxbc BIWBdydxI aaaaF

cos

2

Lc t tc dng ln cnh da l zc

xyxda BIWBdydxI aaaaF

cos

2

Tng lc t theo phng trc z bng khng

166

5 T trng tnh

5.4. M men xon t

+ Cc m men xon tc ng ln hai cnh ab v cd trong trng hp ny ln lt l:

zyyxab BILWBILW

aaaaT

sin2

1cossin

2

zyyxcd BILWBILW

aaaaT

sin2

1cossin

2

+ Tng m men xon bng

BmaTTT

zcdab BILW sin

lun c xu hng thng hng vi m

B

Khi mt phng cun dy vung gc vi t trng: cun dy b kt ti v tr , khng th xoay tip

(*)

167

5 T trng tnh

5.4. M men xon t

V d

Cho mt cun dy hnh trn gm 200 vng dy, vi tit din trung bnh l 10cm2. B

mt cun dy to vi t trng u c mt t thng 1,2 T mt gc 30 nh minh

ha trong hnh v di. Hy xc nh m men xon tc ng ln cun dy nu n

mang mt dng in vi cng 50A?

Gii

Hnh chiu cnh ca cun dy qua

thy hng ca m men lng

cc th hin trong hnh b). M

men lng cc t nm trn mt

phng xy v c ln bng:

10101050200 4 NIAm At.m2

Do , m men xon tc ng ln cun dy l:

zzyyx aaaaaBmT 39,102,1.60sin102,160cos60sin10 N.m

168

5 T trng tnh

5.5. T thng v nh lut Gauss cho t trng

+ Hnh a): cc ng sc t xuyn qua mt s

gii hn bi ng bin c

+ Mt t thng c th u hoc khng B

+ Chia mt s thnh n phn t vi phn v cng

nh, u trn mi phn t (hnh b) B

+ Thnh phn t thng xuyn qua mt phn t

vi phn b mt c nh ngha nh sau:

iii sB

.

+ Tng t thng xuyn qua mt s l

n

i

ii

1

. sB

+ Tng qut, t thng xuyn qua mt mt h s bng

s

dsB

.

169

5 T trng tnh


+ T thng i vo mt mt kn lun bng t thng ri khi mt kn

+ Vi mt mt kn s ta c

0. s

dsB

(dng tch phn ca nh lut Gauss cho t trng)

+ p dng trc tip nh l v tn, ta c 0. v

dvB

Thc cht l 0. B

(dng vi phn ca nh lut Gauss cho t trng)

Mt t thng c tnh lin tc

Hai cng thc trn hon ton tng qut, chng vn ng ngay c vi cc dng in bin thin theo thi gian di bt c quy lut no

170

5 T trng tnh


V d

Nu , tnh t thng xuyn qua na mt cu bn knh R, c tm ti gc ta

v y nm trn mt phng z = 0 ? zBaB

Gii

BRdBddBd

R R

s

2

0

2

0 0

2.

sB

Wb

171

5 T trng tnh

5.6. T th vc-t

+ Mt t thng c tnh cht lin tc (solenoidal) do tn ca n bng khng

c th c biu din di dng xoy (rot) ca mt i lng vc t khc

AB

vc t t th, n v l v-be trn mt (Wb/m) A

+ T nh lut Bio-Savart

cR

ldI3

0

4

RB

3

1

RR

R

Do

c

ldR

I 1

4

0

B

ldRR

ldld

R

110 ld

C

cR

ldI

4

0B ,

172

5 T trng tnh

5.6. T th vc-t

+ Do php tch phn v php vi phn tng ng vi hai tp bin khc nhau, nn

cR

ldI

4

0B

+ Ta nhn c biu din ca t th vc t: c

R

lId

4

0A

+ Vi vng dy mang dng in: c

R

lId

4

0A

+ Tng qut ha theo mt dng in khi:

v

v

R

dvJA

4

0

173

5 T trng tnh

5.6. T th vc-t

+ l vc t t th m xoy (rot) ca n to ra mt t thng A

B

+ Mt i lng vc t c nh ngha hon ton khi v ch khi c xoy (rot) v

tn (div) ca n c nh ngha tng minh phi nh ngha tn ca A

+ Trong t trng tnh, chng ta nh ngha:

0. A

(tiu chun Coulomb)

+ C th biu din t thng theo t th vc t:

ss

dd sAsB

).(.

+ p dng nh l Stoke ta c

c

ld

.A

174

5 T trng tnh

5.6. T th vc-t

V d

Cho mt dy dn thng rt di, nm dc theo trc z, mang dng in mt chiu I c

hng theo chiu dng trc z. Hy tm biu thc ca vc t t th ti mt im bt

k nm trn mt phng vung gc v ti v tr ct i dy dn? Tm mt t thng

ti im ?

Gii

zzaaR

z

L

L

z

c

LLLLI

z

dzI

R

lId

a

aA

22220

22

00

lnln4

44

+ Vi mt dy dn rt di:

LL

LLLL 22

1

222

175

5 T trng tnh

5.6. T th vc-t

V d (tip)

LLLLLL

221

2222

+ Thay vo cng thc, ta c: zLI

aA

2ln

2

0

+ Mt t thng ti im P nh sau:

aaAB

22

0 1

2 L

ILAz

LL 22 Ldo ta thu c

aB

2

0I

176

5 T trng tnh

5.7. Cng t trng v nh lut Am-pe v dng in

5.7.1 Cng t trng

+ Chng ta nh ngha mt in thng theo cng in trng

ED

+ nh ngha cng t trng trong chn khng

0

BH

HB

0

cng t trng khng ph thuc vo h s t thm

trong chn khng v cng hng B

H

177

5 T trng tnh


5.7.2 nh lut Am-pe v dng in

+ nh lut Ampe v dng in: tch phn cng t trng theo mt ng kn

chnh bng dng in qua min bao bi ng kn

Ild

c

.H (dng tch phn ca nh lut Ampe)

nh lut Ampe rt thun tin xc nh t trng m khng phi tnh cc tch phn phc tp nh khi s dng nh lut Bio-Savart

+ Do: s

v dI sJ

. s

v

c

dld sJH

..nn

s

v

s

dd sJsH

..+ p dng nh l Stoke:

+ Mt cch tng qut: vJH

(dng vi phn ca nh lut Ampe)

178

5 T trng tnh



V d 1

Mt dy dn hnh tr rng rt di vi bn knh trong l a, bn knh ngoi l b c t

dc theo trc z v mang dng in I c chiu trng vi chiu dng trc z nh hnh

minh ha. Gi thit phn b dng in trong dy dn l u, hy xc nh cng

in trng ti mt im bt k trong khng gian?

Gii

+ Do dng in c phn b u nn chng ta c th biu

din n theo mt in tch khi:

zv abI

aJ

22

Cc ng sc t c dng cc ng trn ng tm

C 3 min khng gian m chng ta s xt cng t trng trn mi min ring bit

179

5 T trng tnh



V d 1 (tip)

a - Min 1: a

Vi bt k ng kn no bn trong l rng ca dy dn cho, ta u c dng in b

bao bi chng bng khng. Do :

0H

b - Min 2: ba

+ Tng dng in b bao trong min ny l:

22

222

0222

.ab

aIdd

ab

IdI

as

vm

sJ

+ Mt

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