chuong 9 - dien truong tinh

Chng 9: IN TRNG TNH

189

Chng 9

IN TRNG TNH9.1 TNG TC IN NH LUT COULOMB1 in tch nh lut bo ton in tch: T xa xa, con ngi bit hin tng mt s vt sau khi c st th chng c th ht hoc y nhau v chng ht c cc vt nh. Ngi ta gi chng l cc vt nhim in v phn bit thnh hai loi nhim in dng v m. u th k XVII, ngi ta mi nghin cu lnh vc ny nh mt ngnh khoa hc. Cc vt nhim in c cha in tch. Trong t nhin, tn ti hai loi in tch: dng v m. in tch cha trong mt vt bt k lun bng s nguyn ln in tch nguyn t in tch c gi tr nh nht trong t nhin. n v o in tch l coulomb, k hiu l C. Gi tr tuyt i ca in tch c gi l in lng. in tch ca ht electron l in tch nguyn t m: e = 1,6.10 19 C. in tch ca ht proton l in tch nguyn t dng: +e = 1,6.10 19 C. in tch dng v in tch m c th trung ho ln nhau nhng tng i s cc in tch trong mt h c lp l khng i l ni dung ca nh lut bo ton in tch. 2 nh lut Coulomb: Cc in tch cng du th y nhau, tri du th ht nhau. Tng tc gia cc in tch c gi l tng tc in. Nm 1785, bng thc nghim, Coulomb (nh Bc hc ngi Php 1736 1806) xc lp c biu thc nh lng ca lc tng tc gia hai in tch c kch thc rt nh so vi khong cch gia chng gi l in tch im, t ng yn trong chn khng. Pht biu nh lut: Lc tng tc gia hai in tch im ng yn trong chn khng c phng nm trn ng thng ni hai in tch , c chiu y nhau nu chng cng du v ht nhau nu chng tri du, c ln t l thun vi tch ln ca hai in tch v t l nghch vi bnh phng khong cch gia chng. Biu thc:

Fo = k

q 1 .q 2 r2

=

1 q 1 .q 2 . 2 4 o r

(9.1)

Trong : k =

1 = 9.10 9 (Nm2/C2) l h s t l; 4. o

190o =

Giao Trnh Vat Ly ai Cng Tap I: C Nhiet - ien

1 = 8,85.10 12 (F/m) l hng s in. 9 36.10

Trong cht in mi ng nht v ng hng, lc tng tc gia cc in tch gim i ln so vi lc tng tc trong chn khng:

F=

q .q Fo 1 q1.q 2 = k 1 22 = r 4 o r 2

(9.2)

gi l h s in mi ca mi trng . l i lng khng th nguyn, c gi tr ty theo mi trng, nhng lun ln hn 1. Bng 9.1 cho bit h s in mi ca mt s cht thng dng. Bng 9.1: H s in mi ca mt s cht Vt liu Chn khng Khng kh Du ha (20 C) Du bin th Nc (20 C) Ebnto o

1 1,0006 2,2 4,5 80 2,7 2,9

Vt liu Ru tilic (20oC) Giy S Mica Gm titan Thy tinh

25 3,5 6,5 5,5 130 5 10

q1 +F21

r12

q2 + q2 +

F12

q1 +

r21

Hnh 9.1: Lc tng tc gia 2 in tch im Nu gi r12 l vect khong cch hng t q1 n q2 th lc do q1 tc dng ln q2 c vit l:

q .q r F12 = 1 2 2 . 12 4 o r r

(9.3)

Tng t, lc do q2 tc dng ln q1 l:

q .q r F21 = 1 2 2 . 21 4 o r r

(9.4)

Chng 9: IN TRNG TNH

191. 4o r 2 r qi q j rij

Tng qut, lc do in tch qi tc dng ln in tch qj l: Fij = trong rij l vect khong cch hng t qi n qj. 3 Nguyn l tng hp cc lc tnh in:

(9.5)

Gi F1 , F2 , ..., Fn ln lt l cc lc do in tch q1, q2, , qn tc dng ln qo. Khi lc tng hp tc dng ln qo s l:

F = F1 + F2 + ... + Fn = Fii =1

n

(9.6)

Da vo nguyn l ny, ngi ta chng minh c lc tng tc gia hai qu cu tch in u ging nhng tng tc gia hai in tch im t ti tm ca chng.

9.2 IN TRNG1 Khi nim in trng: nh lut Coulomb th hin quan im tng tc xa, ngha l tng tc gia cc in tch xy ra tc thi, bt k khong cch gia chng l bao nhiu. Ni cch khc, vt tc truyn tng tc l v hn. Theo quan im tng tc gn, s d cc in tch tc dng lc ln nhau c l nh mt mi trng vt cht c bit bao quanh cc in tch l in trng. Tnh cht c bn ca in trng l tc dng lc ln cc in tch khc t trong n. Chnh nh vo tnh cht c bn ny m t bit c s cc mt ca in trng. Nh vy, theo quan im tng tc gn, hai in tch q1 v q2 khng trc tip tc dng ln nhau m in tch th nht gy ra xung quanh n mt in trng v chnh in trng mi tc dng lc ln in tch kia. Lc ny gi l lc in trng. Khoa hc hin i xc nhn s ng n ca thuyt tng tc gn v s tn ti ca in trng. in trng l mi trng vt cht c bit, tn ti xung quanh cc in tch v tc dng lc ln in tch khc t trong n. 2 Vect cng in trng: Xt im M bt k trong in trng, ln lt t ti M cc in tch im q1, q2, , qn (gi l cc in tch th), ri xc nh cc lc in trng F1 , F2 , , Fn tng ng. Kt qu thc nghim cho thy: t s gia lc tc dng ln mi in tch v tr s ca in tch l mt i lng khng ph thuc vo cc in tch th m ch ph thuc vo v tr ca im M trong in trng: F1 F2 F = = ... = n = const q1 q 2 qn

192


Hng vect c trng cho in trng ti im M c v phng chiu v ln, c gi l vect cng in trng ti im M, k hiu l E . Vy:

E=

F q

(9.7)

Vect cng in trng ti mt im l i lng c trng cho in trng ti im v phng din tc dng lc, c gi tr (phng, chiu v ln) bng lc in trng tc dng ln mt n v in tch dng t ti im . n v o cng in trng l vn/mt (V/m). Nu E khng i (c v phng chiu ln ln) ti mi im trong in trng th ta c in trng u. Nu bit vect cng in trng ti mt im, ta s xc nh c lc in trng tc dng ln in tch q t ti im :

E

+ q>0

F

F

q 0 th F E ; Nu q < 0 th F E . 3 Vect cng in trng gy bi mt in tch im: Khi mt in tch im Q xut hin, n s gy ra xung quanh n mt in trng. xc nh vect cng in trng do in tch im Q gy ra ti im M cch n mt khong r, ta t ti M in tch th q. Khi in trng ca Q s tc

Qq r dng lc ln q mt lc F xc nh theo nh lut Coulomb: F = k 2 . . So snh r r

vi (9.7), suy ra vect cng in trng ti M do in tch im Q gy ra l:

Q r Q r E=k 2. = . 2 r r 4o r r

(9.9)

Trong , r l vect bn knh hng t Q n im M. Nhn xt: Vect E c: Phng: l ng thng ni in tch Q vi im kho st M Chiu: hng xa Q, nu Q > 0 v hng gn Q, nu Q < 0.

Q

+Q

r

M

-

EM

EM rM

Hnh 9.3: Cng in trng gy bi in tch im

Chng 9: IN TRNG TNH ln: E = k

193(9.10)

|Q| |Q| = 2 r 40 r 2

im t: ti im kho st M. Nu bao quanh in tch Q l mi trng in mi ng nht, ng hng, c h s in mi th cng in trng gim i ln so vi trong chn khng:

E ck Q r Q r E= . =k 2. = 2 r r 4 o r r

(9.11)

4 Nguyn l chng cht in trng: Nu cc in tch Q1, Q2, , Qn cng gy ra ti im M cc vect cng in trng E 1 , E 2 ,..., E n , th vect cng in trng tng hp ti M l:

E = E1 + E 2 + ... + E n = E ii =1

n

(9.12)

tnh cng in trng do mt h in tch phn b lin tc trn mt vt no gy ra ti im M, ta chia nh vt thnh nhiu phn t, sao cho mi phn t mang mt in tch dq coi nh mt in tch im. Khi phn t dq gy ra ti im M vect cng in trng:

dq r dq r . dE = k 2 . = 2 r r 4 o r r

(9.13)

v vect cng in trng do ton vt mang in gy ra ti M l:

E=

vat mang ien

dE

(9.14)

* Trng hp in tch ca vt phn b theo chiu di L, ta gi =

dq (9.15) d

l mt in tch di (in tch cha trn mt n v chiu di). Suy ra, in tch cha trn yu t chiu di d l dq = .d v cng in trng do vt gy ra l:

E = dE =L

1 d .r 4o r 3 L

(9.16)

* Trng hp in tch ca vt phn b trn b mt S, ta gi =

dq dS

(9.17)

l mt in tch mt (in tch cha trn mt n v din tch). Suy ra, in tch cha trn yu t din tch dS l dq = dS v cng in trng do vt gy ra l:

194


E = dE =(S)

1 4 o

dS r 3 . r (S)

(9.18)

* Trng hp in tch ca vt phn b trong min khng gian c th tch , ta gi

=

dq d

(9.19)

l mt in tch khi (in tch cha trong mt n v th tch). Suy ra, in tch cha trong yu t th tch d l dq = .d v cng in trng do vt gy ra l:

E = dE =( )

1 4 o

d ) r 3 . r (

(9.20)

T nguyn l chng cht in trng, ta chng minh c vect cng in trng do mt qu cu tch in u gy ra ti nhng im bn ngoi qu cu cng c xc nh bi (9.9), song phi coi in tch trn qu cu nh mt in tch im t ti tm ca n. 5 Mt s v d v xc nh vect cng in trng: V d 9.1: Xc nh vect cng in trng do h hai in tch im Q1 = Q2 = Q, t cch nhau mt on 2a trong khng kh gy ra ti im M trn trung trc ca on thng ni Q1, Q2 , cch on thng y mt khong x. Tm x cng in trng c gi tr ln nht. Gii Vect cng in trng ti M l E = E1 + E 2 , vi E1 , E 2 l cc vect cng in trng do Q1, Q2 gy ra ti M. Do Q1 = Q2 v M cch u Q1, Q2 nn t (9.10) suy ra: E1 = E2 = k Do :

|Q| |Q| =k . 2 r (x 2 + a 2 )k|Q| x k|Q|x . = 2 2 2 2 (x + a ) x + a (x 2 + a 2 )3/ 2

E = 2E1cos =

(9.21)

T qui tc hnh bnh hnh suy ra E nm trn trung trc ca on thng ni Q1, Q2 v hng ra xa on thng nu Q > 0 (hnh 9.4), hng li gn nu Q < 0. tm c gi tr ln nht ca E, ta c th ly o hm (9.21) theo x ri lp bng bin thin ca E(x), t suy ra gi tr ln nht. Hoc c th dng bt ng thc

1 2 1 2 a4 3 x2. Cauchy nh sau: x + a = x + a + a 3. 2 2 42 2 2

Chng 9: IN TRNG TNH

195

(x + a )2

2 3/ 2

4 2 a 27x . 4

3/ 2

a2 = 3 3 .x 2

E E2M r Q1 x a a Q2

E=Vy:

k|Q|x 2k | Q | = const 2 2 3/ 2 (x + a ) 3 3a 2 E max = 2k | Q | 3 3a 2(9.22)

E1

khi x 2 =

1 2 a a x= 2 2

V d 9.2: Xc nh vect cng in trng do mt vng dy trn, bn knh a, tch in u vi in tch tng cng Q, gy ra ti im M nm trn trc ca vng dy, cch tm vng dy mt on l x. T kt qu hy suy ra cng in trng ti tm vng dy v tm x cng in trng l ln nht. Gii Ta chia nh vng dy thnh nhng phn t rt nh sao cho in tch dq ca mi phn t y c coi l in tch im v n gy ra ti M vect cng in k.dq trng c ln: dE = . Vect d E c phn r 2

+

+

Hnh 9.4

d En

dE

M

d Et

r xdqa O

tch thnh 2 thnh phn: thnh phn php tuyn d E n song song vi trc vng dy v thnh phn tip tuyn

d E t vung gc vi trc vng dy.Cng in trng tng hp ti M l: E = d E = d E t + d E nL L L

Hnh 9.5

V ng vi mt phn t dq, ta lun tm c phn t dq i xng vi dq qua tm O ca vng dy v do lun tn ti d E' i xng vi d E qua trc ca vng dy. Tng cp d E v d E' ny c cc thnh phn tip tuyn trit tiu nhau. Do : d E t = 0 v E = d E n = n o . dE n = n o . dE.cos = n o .L L L L

r

kdq x . 2 r L(9.23)

E = no .

kx kx kQx dq = n o . 3 .Q = n o . 2 3 r L r (a + x 2 )3/ 2

196


Trong n o l php vect n v ca mt phng vng dy - qui c n o lun hng xa tm O. Vy: E lun nm trn trc vng dy v hng xa tm O nu Q > 0; hng gn O nu Q < 0 v c ln: E=

k Q .x(a 2 + x 2 ) 3 / 2

(9.24)

T (9.24) suy ra, ti tm O (x = 0) th Eo = 0. tm gi tr ln nht ca E ta p dng bt ng thc Cauchy nh v d 9.1 v thu c kt qu:

E=

k Q .x (a + x )2 2 3/ 2

k Q .x 2k Q = 2 a 3 3.a 2 .3 3.x. 2khi x2 =

Vy:

E max =

2k Q 3 3.a 2

a2 a x= 2 2

(9.25)

M rng: Nu a 0 v hng gn a x (9.27) nu < 0; c ln: E= .1 2 2 2o a +x

T (9.27) suy ra: Khi a (a tr thnh mt phng rng v hn) th E=

2 o

(9.28)

Vy in trng gy bi mt phng tch in u, rng v hn l in trng u. Khi M rt xa a, hoc a rt nh (x >> a), ta c:1/ 2

a2 = 1 + 2 a2 + x2 x x

1

a 2 kQ 1 a2 E= = 2 2 2 2x 4 o x x

(9.29)

Ton b a coi nh in tch im t ti tm O ca n.

9.3 NG SC IN TRNG IN THNG1 ng sc ca in trng: a) nh ngha: ng sc ca in trng l ng m tip tuyn vi n ti mi im trng vi phng ca vect cng in trng ti im , chiu ca ng sc l chiu ca vect cng in trng. H ng sc l tp hp cc ng sc m t khng gian c in trng. Tp hp cc ng sc in trng c gi l ph ng sc in trng hay in ph. in ph m t s phn b in trng mt cch trc quan. b) Tnh cht: Qua bt k mt im no trong in trng cng v c mt ng sc. Cc ng sc khng ct nhau. V nu chng ct nhau th ti giao im s c 2 vect cng in trng iu ny l v l. ng sc ca in trng tnh khng khp kn, i ra t in tch dng, i vo in tch m.

EMM N Hnh 9.7: ng sc in trng

EN

c) Qui c v: s ng sc xuyn qua mt n v din tch dS nh, t vung gc vi ng sc bng ln ca vect cng in trng ti im M dS. T qui

198


c suy ra: ni no in trng mnh th ng sc s dy, ni no in trng yu th ng sc s tha, in trng u th cc ng sc song song v cch u nhau. Hnh 9.8 l mt s dng ng sc ca in trng. T ta thy gn cc in tch, in trng rt mnh.

+

_

a)

b)

c)

+

+

+

_

d)

e)

Hnh 9.8: Mt s dng ng sc in trng: a) in tch dng; b) in tch m; c) in trng u d) H hai in tch dng; e) H in tch dng v m

2 in thng: Trong khng gian c in trng, xt mt din tch vi cp dS nh sao cho sao cho din tch dS c coi l phng v cng in trng ti mi im trn dS l khng i. Ta nh ngha i lng v hng:

n

E dS

d E = E n .dS = EdS. cos = E .d S

(9.30) Hnh 9.9: in thng

l thng lng in trng (hay in thng) gi qua din tch vi cp dS. Trong En l hnh chiu ca vect cng in trng ln php tuyn ca dS; l gc

gia E v php vect n v n ca dS; vect din tch d S = dS. n . T suy ra in thng gi qua mt mt (S) bt k l:

E = d E = EdS cos = E d S (9.31)S S S

Chng 9: IN TRNG TNH

199

Qui c chn php vect n nh sau:

Nu mt (S) l kn th n hng t trong ra ngoi;

Nu (S) h th n chn tu .

Nh vy, in thng E gi qua mt (S) l mt s i s c th m, dng hoc bng khng. Tuy nhin | E | cho bit s ng sc in trng xuyn qua mt (S). 3 Vect in cm thng lng in cm: Thc nghim cho thy, nu in trng trong chn khng c cng Eo th trong cht in mi ng nht v ng hng, cng in trng gim ln.

=1 =2Hnh 9.10: ng sc b gin on ti mt phn cch

E E= o

(9.32)

Nh vy, khi i t mi trng ny sang mi trng khc th ng sc in trng s b gin on ti mt phn cch gia hai mi trng. iu ny i khi bt li cho cc php tnh v vi phn, tch phn.

Khc phc iu ny, ngi ta xy dng vect in cm D (cn gi l vect cm ng in, vect in dch): D = o . E

(9.33)

Trong gi l h s in mi ca mi trng. Trong chn khng = 1, trong

khng kh 1, cc mi trng khc th > 1.Thc ra cng thc (9.33) ch ng i vi cc cht in mi ng hng, cn trong cht in mi d hng, D v E c th khng cng phng. Trong chng ny, ch cp n cc cht in mi ng hng, v th D E (c thm chng 11 hiu r bn cht ca D ). Nh vy, ngoi vic m t in trng bng vect E , ngi ta cn dng vect D v tng t, ta cng c cc khi nim: ng cm ng in: l ng m tip tuyn vi n ti mi im trng vi phng ca D . Cc tnh cht v qui c v cc ng cm ng in tng t nh ng sc. Thng lng in cm (hay thng lng cm ng in, in dch thng) gi qua yu t din tch dS v gi qua mt (S) l:

200


d D = D n .dS = DdS cos = D d S D = d D = D d SS S

(9.34) (9.35)

9.4 NH L OSTROGRADSKY GAUSS (O G)1 Thit lp nh l: Xt in tch im Q > 0. Bao quanh Q mt mt cu (S), tm l Q, bn knh r. Thng lng in cm gi qua mt cu ny l: D =(S)

d

D

=

(S)

DdScos . Do tnh

i xng cu nn D = const ti mi im trn mt cu v = 0 (v php tuyn ca mt (S) lun trng vi ng cm ng in, xem hnh 9.11). Do , thng lng in cm gi qua mt kn (S) l: D = D = oE = o.(S)

DdS = D dS = DS(S)

M

Q Q = ; S = 4r2 2 2 4 o r 4r(9.36) M

Suy ra: D = Q Nhn xt: -

D

Thng lng in cm D gi qua mt cu (S) khng ph thuc vo bn knh r ca mt cu. Suy ra i vi bt k mt cu no ng tm vi (S), v d (S1), ta cng c (9.36). Nh vy, trong khong khng gian gia hai mt cu (S) v (S1), ni khng c in tch, cc ng cm ng in l lin tc, khng b mt i v cng khng thm ra. Do , nu xt mt kn (S2) bt k bao quanh Q th ta cng c (9.36).

r+ S2 S

n

S1

S3

-

Hnh 9.11: nh l O G Nu c mt kn (S3) khng bao quanh Q th c bao nhiu ng cm ng in i vo (S3) th cng c by nhiu ng cm ng in i ra khi (S3), nn thng lng in cm gi qua (S3) bng khng.

Tm li, thng lng in cm gi qua mt mt kn khng ph thuc v tr in tch bn trong n. Kt qu (9.36) cng ng cho c trng hp bn trong mt kn cha nhiu in tch, phn b bt k, khi Q l tng i s cc in tch bn trong mt kn.

Chng 9: IN TRNG TNH 2 Pht biu nh l O G:

201

Thng lng in cm gi qua mt mt kn bt k bng tng i s cc in tch cha trong mt kn .

D = Q hay

D d S = Q trongS

(S)

(9.37)

Trong chn khng th D = o E , nn ta c: E .d S =S

Q

trong (S)

o

(9.38)

v nh l O G cn c pht biu l: in thng gi qua mt mt kn bt k bng tng i s cc in tch bn trong mt kn chia cho hng s in o. 3 Dng vi phn ca nh l O G: (9.37) c gi l dng tch phn ca nh l O G. Trong trng hp in tch phn b lin tc, ta c th biu din nh l O G di dng vi phn. Mun vy, ta p dng mt nh l trong gii tch, cng c tn l nh l O G, bin mt tch phn mt thnh tch phn theo th tch. Theo , v tri ca (9.37) c vit l:

D.d S = div D.dS

(9.39)

Trong , l th tch ca khng gian gii hn bi mt kn (S) v d l yu t th tch; div l mt ton t vi phn tc ng ln mt vect v tr v mt v hng, trong h ta Descartes, ta c:

div D =

D x D y D z + + x y z

(9.40)

V in tch phn b lin tc nn v phi ca (9.37) tr thnh:

Q

trong(S)

= d

(9.41)

Thay (9.39) v (9.41) vo (9.37), ta c:

div D.d = d .

Suy ra :

(div D )d = 0

(9.42)

V (9.37) ng vi mt kn (S) bt k, nn (9.42) ng vi th tch bt k. iu ny chng t :

div D = 0 hay

div D = div E =

(9.43) (9.44)

Trong mi trng ng hng, ta c:

0

202


(9.43), (9.44) l dng vi phn ca nh l O G. N din t mi quan h gia vect in cm D , vect cng in trng E vi mt in tch tng im trong in trng. 4 Vn dng nh l O G tnh cng in trng: nh l O G thng c s dng tnh cng in trng ca mt s h in tch phn b i xng khng gian, c th l i xng cu, i xng tr v i xng phng. Cc bc thc hin: Bc 1: Chn mt kn S (gi l mt Gauss) i qua im kho st, sao cho vic tnh thng lng in cm D (hoc in thng E ) c n gin nht. Mun vy, phi cn c vo dng i xng ca h ng sc suy ra qi tch nhng im c cng ln ca vect in cm (hoc vect cng in trng) vi im kho st. Bc 2: Tnh thng lng in cm D (hoc in thng E ) gi qua mt Gauss v tnh tng in tch cha trong (S). Bc 3: Thay vo (9.37) hoc (9.38) suy ra i lng cn tnh.

V d 9.4: Xc nh cng in trng gy bi khi cu tm O, bn knh a, tch in u vi mt in tch khi > 0 ti nhng im bn trong v bn ngoi khi cu. Gii Do tnh i xng cu nn h ng sc l mhng ng thng xuyn tm v hng xa tm O, v > 0. Suy ra, cc im c D = const nm trn mt cu tm O. a) Xt im M nm ngoi khi cu: Bc 1: Chn mt (S) l mt cu tm O, i qua M. Bc 2: Thng lng in cm gi qua mt Gauss (S): D = M r O a

E

D d S = D.dS = D dS = DSS S S

n

Gauss

Vi D = oE ; SGauss =4r D = 0 E.4r2

2

Tng in tch cha trong mt Gauss: Q=

Q

trong (S)

4 = d = . = . .a 3 3

Hnh 9.12: CT bn ngoi khi cu

vi l th tch khi cu Bc 3: V D =

Q

trong (S)

nn o.E.4r2 =

4 a 3 3

Chng 9: IN TRNG TNH

203

a 3 kQ E= = 2 2 3 o r r

kQ r hay dng vect: E = 2 . r r

(9.45)

M rng: i vi mt cu tch in u vi in tch tng cng Q th (9.45) vn ng. Vy, mt khi cu hoc mt mt cu tch in u vi in tch Q th in trng m n gy ra xung quanh n ging nh in trng gy bi in tch im Q t ti tm khi cu hoc mt cu. b) Xt im M bn trong khi cu: Tng t ta cng chn mt kn Gauss l mt cu, tm O, bn knh r (r < a). in thng gi qua mt Gauss l: D = 4 o E.r 2 Tng in tch cha trong mt Gauss l Q = . = . r 3 ; vi l th tch khng gian cha trong mt Gauss.

4 3

r Suy ra: E = 3 o

hay E trong

r = 3 o

(9.46)

M rng: Nu in tch ch phn b trn mt cu (v d v cu hoc qu cu kim loi) th = 0 nn trong lng qu cu E = 0, ngha l khng c in trng. Nhn xt: Cng in trng bn trong v bn ngoi khi cu bin thin theo hai qui lut khc nhau: Bn trong khi cu, cng in trng t l bc nht vi khong cch r. Bn ngoi khi cu, cng in trng t l nghch vi r2. Ngay ti mt cu, cng in trng t gi tr ln nht: O r

EM a

n

Hnh 9.13: CT bn trong khi cu

E max =

kQ a = 2 3 o a

(9.47)

Cc kt qu (9.45) v (9.46) vn ng trong trng hp qu cu tch in m, khi vect cng in trng hng vo tm O.

V d 9.5: Xc nh phn b cng in trng gy bi mt phng rng v hn, tch in u vi mt in tch mt > 0 . Gii

204


Do in tch phn b u trn mt phng nn cc ng sc vung gc vi mt phng, hng ra xa mt phng . Qi tch ca nhng im c D = const l hai mt phng i xng nhau qua mt phng . Bc 1: Chn mt Gauss (S) l mt tr c hai y song song, cch u mt phng v cha im kho st M, c ng sinh vung gc vi mt phng (hnh 9.14). Bc 2: Thng lng in cm gi qua mt Gauss l:

D = D.d S =( S)

xung quanh

D .d S +

ay tren

D.d S +

ay di

D.d S

V mt y, ta c D = const v D n ; cn mt xung quanh th D n , nn ta c: D = 0 +ay tren

DdS +

ay di

DdS = 2D dS = 2DSay = 2oESyay

Mt khc, tng in tch cha trong mt Gauss chnh l tng in tch nn trn tit din S do mt () ct khi tr. Ta c Q = .S = .SyBc 3: V D = Q nn E =

2 o(9.48)

D

n

Hay

E=

.n0 2o

n

S

Trong , n 0 l php vect n v ca mt phng . Qui c, n 0 hng ra xa mt phng ().Nhn xt: E khng ph thuc vo v tr im kho st, vy in trng do mt phng tch in u gy ra l in trng u.

Hnh 9.14: CT do mt phng tch in, rng v hn, gy ra.

Trng hp mt phng tch in m ( < 0) th (9.48) vn ng. Lc E hng li gn (). Kt qu (9.48) ph hp vi (9.28), tuy nhin phng php

vn dng nh l O G th n gin hn nhiu.

Chng 9: IN TRNG TNH

205

9.5 CNG CA LC IN TRNG IN TH, HIU IN TH1 Cng ca lc in trng:

Gi s in tch im q di chuyn dc theo ng cong (L) bt k t M n N trong in trng ca in tch im Q. Cng ca lc in trng trn qung ng ny l (xem li cch tnh cng 4.1):

F

q M

+

kQ AMN = F.d s = qE.d s = q 3 r.d r r (L) (L) (L)

+

dr

r

=k

qQ rdr dr r3 = r r2 (L) M kQ kQ A MN = q r r N M(9.49)

rN

Q

+

r+d r

N

Hnh 9.15: Tnh cng ca lc in trng

Ta thy cng AMN khng ph thuc vo ng i. Trong trng hp tng qut, khi in tch q di chuyn trong in trng tnh bt k, ta cng chng minh c cng ca lc in trng khng ph thuc vo hnh dng ng i m ch ph thuc vo v tr im u v im cui. Nu (L) l ng cong kn th AMN = 0. Vy lc in trng tnh l lc th.2 Lu thng ca vect cng in trng:

Nu k hiu ds l vi phn ca ng i dc theo ng cong (L) th cng ca lc in trng c vit l:(L)

Ed s =

A q

(9.50)

Ta gi tch phn

(L)

E d s l lu thng ca vect cng in trng dc theo

ng cong (L). Nu (L) l ng cong kn th:

(L)

Ed s = 0

(9.51)

Vy: lu thng ca vect cng in trng dc theo ng cong (L) bng cng ca lc in trng lm di chuyn mt n v in tch dng dc theo ng cong . V lu thng ca vect cng in trng dc theo ng cong kn bt k th bng khng. (9.49) v (9.50) th hin tnh cht th ca in trng tnh.

206


3 Th nng ca in tch trong in trng:

Ta bit rng, cng ca lc th gia hai im bt k bng gim th nng ca vt gia hai im (xem 4.5): dA = F d s = dWt . i vi lc in trng F = q E nn: Suy ra, trong chuyn di t M n N th:

dWt = q E d s

(9.52)

Wt (M) Wt (N) = q

MN

E d s = A MN

(9.53)

Nu qui c gc th nng v cng ( Wt () = 0 ) th th nng ca in tch q ti im M trong in trng l i lng bng cng ca lc in trng lm di chuyn in tch q t M ra xa v cng:

Wt (M) = A M = q

M

Ed s

(9.54)

Trong trng hp tng qut, th nng sai khc nhau mt hng s cng C. Gi tr ca C ty thuc vo im m ta chn lm gc th nng. Vy th nng ca in tch q trong in trng c dng tng qut l:

Wt (M) = q E d s + C

(9.55)

i vi in trng do in tch Q gy ra th th nng ca in tch q l:

Wt (M) = q E d s + C = q

kQ kQq +C r d s+C = 3 r r

(9.56)

vi r l khong cch t in tch Q n im M; k = 9.109 (Nm2/C2). i vi in trng do h in tch im Q1, Q2, , Qn gy ra th th nng ca in tch q l:

Wt (M) = i =1

n

kqQi +C riM

(9.57)

trong riM l khong cch t in tch Qi n im M.4 in th hiu in th: a) Khi nim:

i vi cc trng th, ngi ta xy dng cc hm th. Trong C hc, hm th ca trng lc th l th nng. Nhng trong in hc, ngi ta chn hm th ca in trng l in th .

Chng 9: IN TRNG TNH

207Wt khng ph q

T cc cng thc (9.5), (9.55), (9.56) v (9.57) suy ra, t s

thuc vo in tch th q m ch ph thuc vo cc in tch gy ra in trng v vo v tr ca im kho st nn t s c trng cho in trng ti im kho st v c gi l in th ca in trng ti im kho st: V=

Wt q

(9.58)

Cng nh th nng, in th l i lng v hng c th dng, m hoc bng khng. Gi tr ca in th ti mt im ph thuc vo vic chn im no lm gc in th. Trong l thuyt, ngi ta chn gc in th v cng, khi in th ti im M trong in trng c biu thc: VM =M

Ed s

(9.59)

Trong trng hp tng qut, in th ti im M trong in trng c biu thc:

V = E d s + C

(9.60)

vi C l hng s ph thuc vo im chn gc in th. Trong thc t, ngi ta thng chn gc in th t. Hiu hai gi tr ca in th ti hai im M, N trong in trng gi l hiu (9.61) in th gia hai im : UMN = VM VN T (9.53), (9.58) v (9.61) suy ra mi quan h gia cng ca lc in trng v hiu in th: AMN = q(VM VN) = qUMN (9.62)Vy: Cng ca lc in trng trong s dch chuyn in tch q t im M n im N trong in trng bng tch s ca in tch q vi hiu in th gia hai im .

T (9.50) v (9.62) ta c: U MN

A = VM VN = MN = E d s q M

N

(9.62a)

Vy: Lu thng ca vect cng in trng t im M n im N bng hiu in th gia hai im . b) in th do cc h in tch gy ra:

T cc phn tch trn, ta c cc cng thc tnh in th: Do mt in tch im gy ra:

V=

kQ +C r kQi +C ri

(9.63)

vi r l khong cch t in tch Q n im kho st. Do h in tch im gy ra: V =

V = i

(9.64)

208


vi ri l khong cch t in tch Qi n im kho st. tnh in th do h in tch phn b lin tc trong min ( ) gy ra, ta coi min gm v s phn t nh, sao cho in tch dq ca cc phn t l nhng in tch im. Mi in tch im dq gy ra ti im kho st in th dV =

kdq v in th do ton h gy ra l: r V = dV =

kdq +C r

(9.65)

Trong r l khong cch t yu t in tch dq n im kho st. Ty theo dng hnh hc ca min ( ) m dq c tnh t (9.15), (9.17) hoc (9.19). Nu chn gc in th v cng th hng s C trong (9.63), (9.64) v (9.65) s bng khng.c) ngha ca in th v hiu in th:

T (9.62) suy ra Mc d gi tr in th ph thuc vo im chn gc in th, nhng hiu in th gia hai im M, N bt k khng ph thuc vo vic chn gc in th. Mt khc, khi UMN cng ln th cng ca lc in trng cng ln.Vy: hiu in th gia hai im M, N trong in trng c trng cho kh nng thc hin cng ca lc in trng gia hai im .

in th l i lng c trng cho in trng v mt nng lng. Trong h SI, n v o in th v hiu in th l vn (V).V d 9.6: Mt vng dy trn bn knh a, tch in u vi in tch tng cng l Q, t trong khng kh. Tnh in th ti im M trn trc vng dy, cch tm vng dy mt on x. T suy ra in th ti tm vng dy. Xt hai trng hp: a) gc in th ti v cng; b) gc in th ti tm O ca vng dy. Ap dng s: a = 5cm; x = 12 cm; Q = 2,6.10 9 C. Giid

M r a

x

O

Xt mt yu t chiu di d trn vng dy. Gi l mt in tch di th in tch cha trong d l dq = d . Theo (9.65), in th ti M l: VM =

Hnh 9.16: Tnh in th do vng dy tch in gy ra ra

L

kdq k +C= r

L

d +C r

Trong , tch phn ly trn ton b chu vi L ca vng dy. V r = a 2 + x 2 = const nn:

Chng 9: IN TRNG TNH

2092

VM =

k r

dL

+C=

k.2a a +x2

+C=

kQ a2 + x2

+ C (9.66)

a) Chn gc in th v cng. Suy ra khi x th VM 0 . T (9.66) suy ra C = 0. Vy: VM =

kQ a2 + x2

(9.67)

Thay s: VM =

kQ a2 + x2

=

9.10 9.(2,6.10 9 ) 1. (5.10 2 ) 2 + (12.10 2 ) 2

= 180(V)

(9.67) suy ra, in th ti tm O ca vng dy l thp nht: VO = Vmin =

kQ 9.10 9.(2,6.10 9 ) = = 468 (V) a 1.5.10 2

Hiu in th gia hai im OM: UOM = VO VM = 288 (V) b) Chn gc in th tm O. Suy ra khi x = 0 th VM = Vo = 0. T (9.66) suy ra C =

kQ kQ kQ . Vy: VM = 2 2 a a a +x

(9.68)

Thay s ta c: VM = 288 (V) v UOM = Vo VM = 288 (V)5 Mt ng th: Tp hp cc im trong in trng c cng in th to thnh mt mt ng th. tm dng ca mt ng th, ta gii phng trnh:

V( r ) = const = C

(9.69)

(9.69) xc nh mt h cc mt ng th. Vi mi gi tr ca C ta c mt mt ng th trong h. V d: i vi in trng do in tch im Q gy ra th phng trnh (9.69) c dng:

kQ kQ =Cr = = const r C

(9.70)

Vy, cc mt ng th l cc mt cu, tm Q. Hnh (9.17) biu din cc mt ng th ca vi h in tch khc nhau (ng nt t l giao ca cc mt ng th vi mt phng hnh v).Qui c v mt ng th: v cc mt ng th sao cho chnh lch V gia hai mt ng th bt k l nh nhau. Suy ra: ni no in trng mnh cc mt ng th

210


s st nhau; ni no in trng yu cc mt ng th s xa nhau; in trng u, cc mt ng th l nhng mt phng song song cch u nhau.Tnh cht ca mt ng th:

Cc mt ng th khng ct nhau. V nu chng ct nhau th ti giao im s c hai gi tr khc nhau ca in th (v l). Khi in tch di chuyn trn mt ng th th lc in trng khng thc hin cng. Tht vy, nu in tch q di chuyn t M n N trn mt ng th th cng ca lc in trng l AMN = q(VM VN). M VM = VN , vy AMN = 0. Vect cng in trng E ti mi im trn mt ng th lun vung gc vi mt ng th . Tht vy, gi s in tch q di chuyn trn mt ng th

theo mt on d s bt k, ta lun c dA = F d s = q E d s = 0 E d s . M d s l vi phn ng i theo mt hng bt, nn E phi vung gc vi mi ng d s trn mt ng th ngha l E phi vung gc vi mt ng th. Vy, ng sc in trng phi vung gc vi mt ng th.

+

_

a)

b)

c)

+

+

+

_

d)

e)

Hnh 9.17: Mt s dng mt ng th (nt t) gy bi: a) in tch dng; b) in tch m; c) in trng u d) H hai in tch dng; e) H in tch dng v m

Chng 9: IN TRNG TNH

211

9.6 LIN H GIA CNG IN TRNG V IN THTa bit cng in trng c trng cho in trng v phng din tc dng lc; cn in th c trng cho in trng v mt nng lng. Nh vy gia cng in trng v in th phi c mi quan h vi nhau. Sau y chng ta s tm mi quan h . Trong khng gian c in trng, ly hai mt ng th st nhau (I) v (II), m in th c gi tr ln lt l V v (V + dV). Gi s in tch q di chuyn t im M (I) n im N (II) theo cung ds bt k. Ta c cng ca lc in trng l:V V + dV

M

dn

ds

(I)

N (II)

dA = q E d s (*)Mt khc:

Hnh 9.18: Quan h gia CT v in th.

dA = q(VM VN) = q[V (V + dV)] = qdV (**) So snh (*) v (**) suy ra:

E d s = Eds cos = dV

(9.71)

vi l gc hp bi vect cng in trng E v vect ng i d s .Trng hp 1: Nu d s hng v ni c in th cao, ngha l dV > 0, th t (9.71)

suy ra, gc > 900 , ngha l E hng v ni c in th thp.Trng hp 2: Nu d s hng v ni c in th thp, ngha l dV < 0, th t (9.71)

suy ra, gc < 900 , ngha l E cng hng v ni c in th thp.Kt lun 1: Vect cng in trng lun hng theo chiu gim ca in th.

Gi E s = Ecos l hnh chiu ca E ln phng ca d s th theo (9.71) ta c: E s .ds = E.ds.cos = dV, hay:

Es =

dV ds

(9.72)

Kt lun 2: Hnh chiu ca vect cng in trng ln mt phng no bng gim in th trn mt n v chiu di theo phng .

212


Nu chiu vect cng in trng E ln ba trc Ox, Oy, Oz ca h ta Descartes th ta c: E x =

V V V ; Ey = ; Ez = x y z

(9.73)

Trong ,

V V V l o hm ring phn ca hm th V i vi cc bin x, y, , , x y z

z. Trong gii tch vect, (9.73) c vit di dng:

E = E x . i + E y . j + E z .k = (Hay:

V V V .i+ . j+ .k) x y z

(9.74)

E = gradV

(9.75)

trong vect gradV gi l gradien ca in th V.Kt lun 3: Vect cng in trng ti mt im bt k trong in trng bng v ngc du vi gradien ca in th ti im .

Nu xt theo phng ng sc ca in trng (M v N nm cng mt ng sc) th E s = E v MN nm trn php tuyn ca cc mt ng th. Do ta vit ds = dn v ta c:

E=

dV dn dV dV ds dn

(9.76)

V E s E nn t (9.72) v (9.76) suy ra:

(9.77)

Kt lun 4: ln cn mt im trong in trng th in th s bin thin nhanh nht theo phng php tuyn ca mt ng th (hay phng ca ng sc in trng v qua im ).

Nu gi n o l vect n v hng dc theo chiu ca ng sc in trng th ta c th biu din mi quan h gia cng in trng v in th bng cng thc:

E=

dV .no dn( 2)

(9.78)

i vi in trng u, nhn hai v ca (9.76) vi dn, ri ly tch phn ta( 2)

c: Hay

V2 V1 =

(1)

dV = E dn = E.d(1)

U12 = V1 V2 = E.d

(9.79)

Chng 9: IN TRNG TNH

213

trong d l khong cch gia hai mt ng th i qua im (1) v im (2) (hay khong cch gia hai im tnh dc theo mt ng sc in trng). Vn dng mi quan h gia cng in trng v in th ta s tnh c cng in trng nu bit in th v ngc li.V d 9.7: Xc nh in th gy bi khi cu tm O, bn knh a, tch in u vi mt in tch khi > 0 ti nhng im bn trong v bn ngoi khi cu. Cho bit h s in mi bn trong v bn ngoi khi cu u bng 1. Xt 2 trng hp: a) Chn gc in th v cng; b) chn gc in th ti tm O. Gii

Xt im M bn trong khi cu. Cng in trng ti M, theo (9.46) l:

E trong

r dV r = . n o (*) . Thay vo (9.78), ta c = 30 dn 3 o

V ng sc hng theo bn knh, nn r v

n o cng phng vi phng bn knh. Do : dV dV r = = dV = rdr dr dn 3 o 3 oVM

rO

A M a

N

E

n

dV = 3 o VO

rM

rdr0

VM VO =

2 rM 6 o

(9.80)

Hnh 9.19: S phn b in th bn trong v bn ngoi khi cu tch in

Tng t, xt im N bn ngoi khi cu,

dV kQ thay (9.45) vo (9.78) ta suy ra: = 2 dr r VN VA = kQ(

VN

VA

dV = kQ ra

rM

dr2

1 1 ) rN a

(9.81)

trong VA l in th ti im trn b mt khi cu.a) Trng hp 1: chn gc in th ti v cng th khi rN ; VN 0

(9.81)

VA =

kQ a

(9.82)

214


Thay (9.82) vo (9.81) ta tnh c in th ti im N bn ngoi khi cu:

VN =

kQ rN

hay Vngoai =

kQ r

(9.83)

T (9.80) suy ra, khi M trng vi A th ta c:

4 1 1 kQ a 2 = a 3 .. = . VA VO = 6 0 3 4 o 2 a 2aKt hp vi (9.82) suy ra: VO =

3kQ 2a

(9.84)

Thay (9.84) vo (9.80) ta c in th bn trong khi cu l: Vtrong =

3kQ r 2 2a 6 o

(9.85)

b) Trng hp 2: chn gc in th ti tm O th VO = 0. T (9.80) suy ra: Vtrong =

r 2 6 o

(9.86)

Do , in th ti mt cu l: VA =

kQ a 2 = 6 o 2a

(9.87)

Thay (9.87) vo (9.81) ta c: Vngoi =

kQ 3kQ r 2a+

(9.88)

V d 9.8: Xc nh cng in trng v in th gy bi hai mt phng song song, rng v hn, cch nhau mt khong d, tch in u vi mt in tch mt l + v . Cho bit h s in mi ca mi trng bao quanh hai mt phng l . Chn gc in th mt phng . Gii

Hnh 9.20: in trng gy bi 2 mt phng rng v xc nh cng in trng gy bi hn, tch in u. hai mt phng ny, ta c th vn dng trc tip nh l O G. Tuy nhin c th lp lun n gin da vo kt qu ca v d 9.5 nh sau: Cng in trng ti im M bt k lun l tng

hp ca hai in trng do tng mt phng gy nn: E = E1 + E 2 . Trong E 1 l vect cng in trng do mt phng + gy ra, lun hng xa mt phng ny;

Chng 9: IN TRNG TNH

215

E 2 l vect cng in trng do mt phng gy ra, lun hng gn mt phng nn: ny. V E1 = E2 = 2 o i vi nhng im nm ngoi hai mt phng (vng (1) v (3)) th E = 0. i vi nhng im nm gia hai mt phng th E hng t + sang v c ln: E = E1 + E2 =

o

(1)

xM x O

Vy: in trng trong khong gia hai mt phng l in trng u, c cng :

+

E= o

(9.89)

-

(2) E (3) Hnh 9.21

tnh in th, ta chn trc Ox nh

dV dV .n o = .i ; hnh (9.21). Ta c: E = dn dx

i l vect n v hng theo trc Ox ( i n o )V

Suy ra :

VO

dV = Edx V VO = Ex0

x

V chn gc in th mt phng nn VO = 0. Do : V = Ex =

x o

(9.90)

Bn ngoi pha , E = 0 V = const = V- = 0; Bn ngoi pha +, E = 0 V = const = V+ =

d o d o(9.91)

Hiu in th gia hai mt phng l: U = V+ V- =

216


9.7 BI TON C BN CA TNH IN HCBit trc s phn b ca in tch, tm s phn b ca cng in trng v in th. V ngc li, bit trc s phn b ca cng in trng hoc in th, tm s phn b ca cc in tch. l ni dung c bn ca bi ton tnh in hc. gii bi ton ny, ta s dng nh l O G v mi quan h gia cng in trng v in th. Gi s trong mi trng ng hng c h s in mi , in tch phn b lin tc vi mt in tch khi th theo nh l O G dng vi phn, ta c :

div E =

o

(*)

Mt khc, theo mi quan h gia cng in trng v in th th :

E = gradV

(**).

Thay (**) vo (*), ta c : diV(gradV) =

0(9.92)

Hay :

V =

0V = 0

Nu khng c in tch ( = 0) th ta c :

(9.93)

(9.92) c gi l phng trnh Poisson, cn (9.93) c gi l phng trnh Laplace. l hai phng trnh c bn ca tnh in hc. Trong ton t l ton t vi phn cp hai, c gi l Laplacian hay ton t Laplace. Trong h ta Descartes, ton t c dng :

V =

2V 2V 2V + + x 2 y 2 z 2

(9.94)

Trong h ta cu, ton t c dng :

V =

1 2 V 1 V 1 2V r (sin . ) + 2 2 + 2 r 2 r r r sin r sin 2

(9.95)

Nh vy, gii bi ton c bn ca tnh in hc, thc cht l gii phng trnh Poisson hoc phng trnh Laplace. nghim ca cc phng trnh trn c ngha vt l, ta phi c nhng iu kin gii hn, gi l iu kin bin. Khi phng trnh c bn ca tnh in hc s c nghim duy nht.V d 9.9 : Trong chn khng, in th phn b theo qui lut V =

4yz (SI). Xc x2 +1

nh in th, vect cng in trng v mt in tch ti im P(1, 2, 3).

Chng 9: IN TRNG TNH Gii -

217

in th ti P : VP =

4.2.3 = 12V 12 + 1

Vect cng in trng ti P :

Ex = Ey = Ez =

V 8xyz 8.1.2.3 = 2 = 2 = 12V / m 2 x (x + 1) (1 + 1) 2 V 4z 4.3 = 2 = 2 = 6 V / m y x +1 1 +1 V 4y 4.2 = 2 = 2 = 4 V / m z x +1 1 +1

Vy : E = (12, 6, 4) v E = 122 + 62 + 42 = 14V / m-

Mt in tch ti P tinh t (9.92): = 0 .V M :

2 V 8xyz 8yz(3x 2 1) 8.2.3(3.12 1) = = = 12 = (x 2 + 1)3 (12 + 1)3 x 2 x (x 2 + 1) 2 2 V 4y = =0 z 2 z x 2 + 1 2V 2V 2V + + = 12 x 2 y 2 z 2

2 V 4z = =0 ; y 2 y x 2 + 1 Thay vo (9.94) ta c : V =

Vy : = 0 .V = 8,85.1012.12 = 1, 062.109 C / m3

9.7 LNG CC IN1 nh ngha :

Lng cc in l mt h gm hai in tch im bng nhau v ln nhng tri du, lin kt vi nhau, t cch nhau mt khong rt nh so vi nhng khong cch t n n im ta xt (hnh 9.22). Nhng vt th vi m thng c cu trc nh nhng lng cc in. V d phn t mui n NaCl l mt lng cc in, gm ion Na+ v Cl.

+ +q

_ q

Hnh 9.22: Lng cc in

218


c trng cho tnh cht in ca lng cc, ngi ta dng i lng mmen lng cc in hay mmen in ca lng cc, c nh ngha l :

pe = q

(9.96)

l vect hng t in tch q n +q, c mdun bng khong cch gia Trong q v +q. ng thng ni hai in tch q v +q gi l trc ca lng cc in.2 Vect cng in trng gy bi lng cc in : Xt im M nm trn mt phng trung trc ca lng cc in.Vect cng

in trng do lng cc in gy ra ti M l : E = E1 + E 2 . Trong E1 ,v E 2 l vect cng in trng do in tch q v +q gy ra ti M (hnh 9.23). D thy : E1 = E 2 = k

q r12 q nn E = 2E1 sin = 2k 2 sin r1, r nn r1 2r1 kp kq Do : E = 3 = 3e r r

E2 E r(9.97) (9.98)

M

M : sin =

E1

r

q

r1

r2

hay dng vect : E = vi k = 9.109 Nm2/C2.

k pe r 3

+

+q

pe

Vy : vect cng in trng do lng cc in gy ra ti mt im trn mt phng trung trc ca lng cc in lun ngc chiu vi vect mmen in ca lng cc. Tng t ta cng xc nh c vect cng in trng ti im N nm trn trc ca lng cc in, cc tm O ca lng cc in mt khong r (hnh 9,24) th lun cng chiu vi vect mmen lng cc in:N

Hnh 9.23: Vect cng in trng ti im M trn mt phng trung trc ca lng cc in

E

qr

pe

O

+

+q

E=

2k pe r 3

(9.99)

Hnh 9.24: Vect cng in trng ti im N trn trc ca lng cc in

Chng 9: IN TRNG TNH 3 Lng cc in t trong in trng ngoi :

219

Gi s t lng cc in vo in trng u, sao cho vect mmen in p e ca lng cc to vi vect cng in trng E 0 mt gc . Khi in trng tc dng ln lng cc in hai lc ngc chiu: F+ = q E 0 v F = q E 0 (hnh 9.25). Tng ca hai lc ny bng khng nn lng cc in khng tnh tin trong in trng. Tuy nhin, hai lc F+ v F to thnh mt ngu lc lm lng cc in quay trong in trng. Mmen ca ngu lc l :

M = F+ d = qE 0 sin = pe E 0 sin Hay dng vect : M = pe x E 0

(9.100) (9.101)

Vect M c phng ca vung gc vi mt phng cha p e v E 0 , chiu xc nh theo qui tc inh c thun (xem chng 0). Di tc dng ca mmen ngu lc, lng cc in s quay theo chiu sao cho vect+q

pe ti trng vi hng ca vect E 0 . Nulng cc l cng ( khng i), n s nm cn bng v tr ny. Nu lng cc l n hi, n s b bin dng hoc phn li nu km bn. Trong trng hp lng cc in t trong in trng khng u, n s b xoay n v tr sao cho vect p e ti trng

+pe

F+

d

F

q

E0

Hnh 9.25: Lng cc in t trong in trng ngoi

vi hng ca vect E 0 , sau lc in trng s ko lng cc in tnh tin v pha in trng mnh. Cc kt qu trn y c ng dng gii thch hin tng phn cc in mi, hin tng cc vt nh nh mu giy, bi vi, ... b ht vo cc vt nhim in v l nguyn l hot ng ca l nu, nng bng sng viba (xem C s vt l tp 4 David Halliday, dch gi m Trung n).

220

Giao Trnh Vat Ly ai Cng Tap I: C Nhiet - ien BI TP CHNG 9

9.1 So snh lc hp dn v lc tnh in gia cc cp ht s cp sau y rt ra nhng kt lun cn thit : a) electron v electron ; b) electron v proton ; c) proton v proton. Bit khi lng proton gp 1840 ln khi lng electron. 9.2 Theo gi thuyt ca N.Bohr, electron trong nguyn t Hydro chuyn ng quang ht nhn theo qi o trn c bn knh r = 5,3.10 9 cm. Tnh vn tc gc, vn tc di v tn s vng ca electron. 9.3 Hn bi st mang in tch +2C, vy n tha hay thiu bao nhiu electron? 9.4 Hai qu cu kim loi nh ging ht nhau, tch in q1, q2, th tng tc nhau mt lc F. Nu cho chng chm nhau ri a v v tr c th lc tng tc by gi l bao nhiu? p dng s : q1 = +2C, q2 = 4C, F = 0,8N. 9.5 Hai qu cu kim loi nh ging nhau, c treo bi hai si dy mnh khng dn in vo mt cng mt im. Tch cho mt trong hai qu cu th chng lch nhau mt gc 2 = 10014. Gii thch hin tng v tnh in tch ca mi qu cu, bit chiu di dy treo l = 40cm. Bit khi lng mi qa cu l 100g. 9.6 Hai in tch im q1 = 3.10 8 C v q2 = 1,2.10 7 C, t cch nhau mt khong AB = 20cm trong khng kh. Xc nh vect cng in trng ti im M, bit: a) MA = MB = 10cm; b) MA = MB = AB c) MA = 12 cm; MB =16cm d) MA = 10cm; MB = 30cm

e) Tm im N m ti cng in trng trit tiu.9.7 Trong mt min (), in tch phn b vi mt = ( r ), Hy vit biu thc

xc nh vect cng in trng E v in th V ti v tr c vect bn knh

r . Cho hng s in mi trong v ngoi min () u bng 1.

9.8 in th ca in trng gy bi mt h in tch c dng: V = a(x2 + y2) + bz2 trong a, b l cc hng s dng. a) Xc nh vect cng in trng ti im M(x,y,z). b) Nhng mt ng th c dng nh th no? 9.9 Mt khng gian mang in vi mt in tch bin i theo qui lut = o/r, trong o l hng s v r l khong cch tnh t gc to n im kho st.

Tnh cng in trng E v in th V theo r (khng xt min gn gc to ).9.10 Si dy mnh, thng, di 2a, tch in u vi mt in di > 0. Xc nh vect cng in trng v in th ti im M nm trn mt phng trung trc ca si dy, cch si dy mt on h. Chn gc in th v cng.

Chng 9: IN TRNG TNH

221

9.11 So snh cng in trng v in th ti hai im A, B, C trong in trng m t hnh 9.26. 9.12 Mt mt phng A B B thng ng, rng v B hn, tch in u C vi mt in mt A A 6 2 = 8,85.10 C/m . a) b) c) Mt qu cu nh khi lng m = 1g, Hnh 9.26 tch in q = 2.10 8 C, c treo vo im A mp() bng si dy rt mnh, khng dn in. Tnh gc lch ca dy treo so vi phng thng ng. (ly g = 10m/s2). Q C B D A 9.13 Mt in tch Q t ti tm ca hai ng r trn ng tm, bn knh r v R. Xt mt ng R thng qua tm O ct c hai ng trn ti cc im A, B, C, D nh hnh (9.27). a) Tnh cng ca lc in trng thc Hnh 9.27 hin khi in tch q di chuyn t B n C v t A n D. b) So snh cng ca lc in trng khi in tch q di chuyn t A n C v t D n C. c) Cc kt qu trn c thay i khng nu q di chuyn gia cc im nhng theo cc cung trn? 9.14 t in tch m (-Q) ti gc ta trong mt phng (Oxy). So snh cng in trng v in th ti A(5,0) v B(0, - 5). Suy ra cng ca lc in trng khi in tch +q di chuyn t A n B mang du m hay dng ? 9.15 Si dy mnh tch in u vi mt in di c un thnh cung trn AB bn knh R, chn gc tm 2 . Xc nh vect cng in trng v in th ti tm O ca cung AB, chn gc in th v cng. 9.16 Hai si dy mnh, rt di, song song, cch nhau mt khong 2a, tch in tri du vi mt in di l + v . Xc nh vect cng in trng v in th V ti (Chn gc in th mt phng trung trc ca hai dy):

a) M nm trn on thng ni hai dy, vung gc vi hai dy, cch dy tch in dng mt on x b) N cch u hai dy, cch mt phng cha hai dy mt khong h.9.17 Chm cu c bn knh R, gc m 2, tch in u vi mt in mt +. Xc nh vect cng in trng v in th ti tm O ca chm cu. Chn gc in th v cng.

222


9.18 Hai vng trn tch in u, cng bn knh R = 6cm , ng trc, hai tm O1 v O2 cch nhau mt khong a = 8cm. Vng th nht tch in q1 = +4C. Tnh in tch ca vng th hai, bit rng, khi in tch th q0 = 1C di chuyn t O1 n O2 th ng nng ca n tng 0,6J. 9.19 t nh nhng mt in tch im q = +2nC vo in trng gy bi si dy mnh di, tch in u th thy in tch ny di chuyn vo gn dy. Khi n qua v tr cch dy 4cm th c ng nng 0,015mJ. Xc nh du v mt in di trn dy. 9.20

t mt lng cc in c mmen lng cc pe = 6,24.10 30 Cm vo in

trng u c cng E = 30kV/m sao cho p e v E to vi nhau mt gc 300. Tnh mmen lm quay lng cc in.

chuong 9 - dien truong tinh

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