Vật Lý A1-CHUONG 7

Chng VII. Trng tnh in.

CHNG VII TRNG TNH INCc in tch ng yn to ra xung quanh chng mt mi trng vt cht c bit, c gi l trng tnh in. 1. NHNG KHI NIM M U Mc d cc hin tng trong t nhin th hin di rt nhiu v khc nhau, nhng vt l hc hin i cho rng chng u thuc bn dng tng tc c bn: tng tc hp dn, tng tc in t, tng tc yu v tng tc mnh; trong s tng tc hp dn v tng tc in t l nhng tng tc rt ph bin. i vi cc vt th thng thng th tng tc hp dn rt yu v ta c th b qua. Nhng tng tc in t ni chung l ng k, thm ch nhiu khi rt ng k. 1. S nhim in ca cc vt. Hai loi in tch Thc nghim xc nhn rng, khi c xt mt thanh thy tinh vo la hay mt thanh bnit vo lng th th thanh thy tinh v thanh bnit c kh nng ht c cc vt nh. Ta ni rng, chng b nhim in hay trn cc thanh xut hin cc in tch. Trong t nhin c hai loi in tch: in tch dng v in tch m. Ngi ta quy c, in tch xut hin trn thanh thy tinh sau khi c xt n vo la l in tch dng; cn in tch xut hin trn thanh bnit sau khi c xt vo lng th l in tch m. Thc nghim cng xc nhn rng, in tch trn mt vt bt k c cu to gin on v bng mt s nguyn ln in tch nguyn t no . Ta ni, in tch b lng t ha. in tch nguyn t l in tch nh nht c bit trong t nhin v c ln e = 1,6.10 19 C. Proton v electron l nhng ht mang in tch nguyn t: proton mang in tch dng, cn electron mang in tch m. Ch rng, ngi ta pht hin nhng ht quark mang in tch , e 3 2e . Tuy 3

nhin,

nhng ht ny khng th tn ti mt cch ring bit nn ta khng ly in tch ca chng lm in tch nguyn t. Vt cht c cu to bi cc nguyn t. Mi nguyn t gm cc proton, electron mang in v cc neutron trung ha in. Cc proton v neutron xp cht trong ht nhn nguyn t. Trong mu nguyn t n gin th cc electron chuyn ng theo cc qu o quanh ht nhn. trng thi bnh thng, s proton v electron trong nguyn t lun lun bng nhau nn tng i s cc in tch trong mt nguyn t bng khng. Ta ni, nguyn t trung ha in. Nu v l do no m nguyn t mt i (hoc nhn thm) mt hoc vi electron th n s tr thnh phn t mang in dng (hoc m) v c gi l ion dng (hoc ion m). 113

Chng VII. Trng tnh in. 2. nh lut bo ton in tch Theo thuyt in t, qu trnh nhim in ca thanh thy tinh khi c xt vo la chnh l qu trnh electron chuyn di t thy tinh sang la. iu ny lm cho thy tinh tr thnh vt mang in dng. Nh vy, bn cht s c xt khng to ra in tch m ch lm cho in tch chuyn t vt ny sang vt khc, lm mt i tnh trung ha in ca mi vt trong qu trnh y. y chnh l ni dung ca nh lut bo ton in tch, v c pht biu nh sau: Cc in tch khng t sinh ra m cng khng t mt i, chng ch c th truyn t vt ny sang vt khc hoc dch chuyn bn trong mt vt. Hay: Tng i s cc in tch trong mt h c lp l khng i 3. Phn loi vt dn Xt v tnh dn in, ta c th phn loi cc cht nh sau: Cht dn in l nhng cht trong c cc ht mang in tch c th chuyn ng t do trong ton b th tch vt. Th d: kim loi, cc dung dch mui, axit, baz, Cht cch in, hay cn gi l in mi, l nhng cht trong khng c cc in tch t do m in tch xut hin u s nh x y. Th d: thy tinh, bnit, cao su, nc nguyn cht, Cht bn dn l cc cht c tnh dn in trung gian gia cc cht dn in v cch in. nhit thp, cc cht bn dn dn in km, nhng nhit cao, tnh dn in ca n tng dn. Th d: silic, germani, Cht siu dn l cc cht m cc in tch khi chuyn ng qua chng khng gp bt c s cn tr no. Nm 1911, nh vt l ngi H Lan, Kammerlingh Onnes (1853 1926) pht hin thy ngn rn mt hon ton in tr (tc tr thnh cht siu dn) nhit di 4,2K. 2. NH LUT COULOMB 1. in tch im: in tch im l mt vt mang in tch c kch thc rt nh so vi khong cch t in tch ti nhng im hoc nhng vt mang in khc m ta ang kho st. Khi nim in tch im ch c tnh tng i. Charles Coulomb l nh vt l hc ngi Php. Ban u ng nghin cu s xon ca cc si dy nh v tm c cng thc v mi lin h gia gc xon v moment lc tc dng ln dy xon. Trn c s ny, nm 1784 ng ch to mt chic cn xon chnh xc

Hnh 7-1114

Chng VII. Trng tnh in. (hnh 7-1) kho st lc tng tc tnh in gia cc in tch. Nm 1785 Coulomb tng kt cc kt qu th nghim v pht biu thnh nh lut mang tn mnh. ghi nhn cng lao ca ng, n v in tch trong h SI c gi l Coulomb (k hiu l C). 2. nh lut Coulomb trong chn khng: Gi s c hai in tch im q1 v q 2 t cch nhau mt khong r. nh lut Coulomb c pht biu nh sau: " Lc tng tc tnh in gia hai in tch q1 v q2 t trong chn khng, c phng nm trn ng thng ni hai in tch, c chiu ph thuc vo du ca hai in tch (y nhau nu hai in tch cng du, ht nhau nu hai in tch tri du), c ln t l thun vi tch s q1, q2 v t l nghch vi bnh phng khong cch r gia hai in tch " Ta c th biu din nh lut Coulomb di dng vc t:

F12 = k

q1 q 2 r12 ; r2 r

F21 = k

q1 q 2 r21 r2 rr r 21q2

(7-1)r F12

trong F12 l lc do q1 tc dng ln q2 ; F21 l r lc do q2 tc dng ln q1; r21 l bn knh vector r hng t in tch q2 n in tch q1; r12 l bn knh vector hng t in tch q1 n in tch q2; k l h s t l ph thuc vo n v s dng, Lc do q 2 tc dng ln in tch q1 l F 21 = F 12 . Trong h SI: h s k = N.m 2 1 = 9.109 2 4 0 C

r F21q1

r F 21q1

r F12q2

r

r

Hnh 7-2 Lc tng tc gia hai in tch im.

vi 0 =

C2 1 10 9 8,86.10 12 2 36 N.m

gi l hng s in.

Thc nghim chng t, lc tng tc gia hai in tch t trong mi trng gim i ln so vi trong chn khng. Biu thc vc t ca nh lut Coulomb trong mi trng s c dng: F12 = k

q1 q 2 r12 ; r 2 r

F21 = k

q1 q 2 r21 r 2 r

mt i lng khng th nguyn c trng cho tnh cht in ca mi trng v c gi lhng s in mi. Di y l hng s in mi ca mt s cht: Cht Chn khng Khng kh bnit Thy tinh Nc nguyn cht Hng s in mi 1 1,0006 2,7 2,9 5 10 81 115

Chng VII. Trng tnh in. 3. Nguyn l chng cht cc lc in nh lut Coulomb cho php ta xc nh lc tng tc tnh in gia hai in tch im. xc nh lc tng tc gia hai vt mang in bt k ta cn phi kt hp nguyn l chng cht lc. Trng hp h in tch phn b gin on: Gi s in tch q0 t trong khng gian chu tc dng ca cc lc F1 , F2 , F3 ,...,Fn gy ra bi h cc in tch im q1 , q 2 , q 3 ,...,q n . Khi , theo nguyn l chng cht lc, lc tng hp F do h cc in tch ny tc dng ln in tch q0 s l:n r r r r r r F = F1 +F 2 +F 3 + ... + F n = Fi . i =1

r

r

r

r

r

(7-2)

Trng hp h hai vt mang in bt k phn b lin tc: xc nh lc tng tc tnh in gia hai vt , ta coi mi vt mang in nh mt h v s cc in tch im. Khi , lc tng tc tnh in tc dng ln mi vt s bng tng vector ca tt c cc lc do h in tch im ca vt ny tc dng ln mi in tch im ca vt kia. Ch : Ngi ta chng minh c rng, lc tng tc tnh in gia hai qu cu mang in u cng c xc nh bi nh lut Coulomb nu coi r l khong cch gia hai tm qu cu v in tch q ca mi qu cu nh mt in tch im t ti tm ca chng.

3. IN TRNG V VC T CNG IN TRNG 1. Khi nim in trng Thc nghim xc nhn gia hai in tch im lun c lc tng tc tnh in ngay c trong trng hp chng c t trong chn khng. Vy lc tng tc c truyn nh th no, v ch khi c mt in tch, th tnh cht vt l ca khong khng gian bao quanh in tch c g b bin i khng. Trong qu trnh pht trin ca vt l hc c hai thuyt i lp v vn trn. l thuyt tc dng xa v thuyt tc dng gn. Thuyt tc dng xa: - Tng tc gia cc in tch im c truyn i mt cch tc thi, ngha l vn tc truyn tng tc ln v hn. - Tng tc c thc hin khng cn c s tham gia ca vt cht trung gian. - Khi ch khi c mt in tch th tnh cht vt l ca khong khng gian bao quanh b bin i. Nh vy, theo thuyt ny ta phi tha nhn c s truyn tng tc m khng cn c dng vt cht no tham gia, tc l phi tha nhn c vn ng phi vt cht. Quan nim tri vi hc thuyt duy vt bin chng, do b bc b. Thuyt tc dng gn: 116

Chng VII. Trng tnh in. - Tng tc gia cc in tch im c truyn i khng tc thi, m truyn i t im ny ti im khc trong khng gian vi vn tc hu hn. - Tng tc c thc hin vi s tham gia ca mt dng vt cht c bit, l in trng. - Khi ch c mt in tch, th in tch gy ra trong khng gian bao quanh n mt in trng. in trng ny gi vai tr truyn tng tc t in tch ny n in tch khc. Tm li, thuyt tc dng gn ph hp vi quan im duy vt bin chng v c khoa hc xc nhn. Vy in trng l mt dng tn ti ca vt cht bao quanh cc in tch. c im c bn ca in trng l tc dng lc ln cc in tch t trong n. 2. Vct cng in trng a. nh ngha c trng c bn ca in trng l tc dng lc ln cc in tch nm trong . Thc nghim chng t rng: Vi mt in trng xc nh, nu ta t mt in tch dng c gi tr nh q0 ( khng lm thay i in trng ang xt, c gi l in tch th) vo mt im M no trong in trng (khi in tch chu tc dng ca lc in thuc vo q0 m ch ph thuc vo v tr ca im M. Ta c:r u r F uuuuuu r E= = const. q0r F)

r F th t s khng ph q0

(7-3)

Nh vy, vct E c th c trng cho in trng ti im M v c gi l vct cng in trng ti M, ln E c gi l cng in trng. Nu ta cho q 0 = +1 th E = F, do ta c th nh ngha vct cng in trng nh sau: nh ngha: Vct cng in trng E ti mt im l i lng c trng cho in trng ti im v phng din tc dng lc, c tr vct bng lc tc dng ca in trng ln mt n v in tch dng t ti im ". Trong h n v SI, cng in trng c n v o l Vn/mt: V/m. b. Lc in trng tc dng ln in tch im Nu bit cng in trng E ti mt im M trong in trng th khi t mt in tch q vo im , n b in trng tc dng mt lcu r r

u r

q> 0

r E

r F

r E

q 0 th F cng chiu vi E ( hnh 7.3); Nu q < 0 th F ngc chiu vi E ( hnh 7.3). 3. Vct cng in trng gy ra bi mt in tch im 117


Ta hy xc nh vct cng in trng E ti mt im M cch in tch q mt khong r. Mun vy ti im M ta t mt in tch im qo c tr s nh. Khi theo nh lut Coulomb, lc tc dng ca in tch q ln in tch qo bng:

Q M

r r

r E

Q

r E

M

kq q r F = o2 r r

Vct cng in trng do in tch im q gy ra ti im M l:E= kq r r 2 r

Hnh 7- 4 Cng in trng gy bi mt in tch im

(7-5)

trong bn knh vct r hng t in tch q n im M. Nhn xt: - Nu q > 0 th E r : E hng ra xa khi in tch q. - Nu q < 0 th E r : E hng vo in tch q. 4. Vct cng in trng gy ra bi mt h vt mang in - Nguyn l chng cht in trng a. Cng in trng gy ra bi h in tch im phn b ri rc Xt h in tch im q1, q2, ..., qn c phn b ri rc trong khng gian. Lc tng hp tc dng ln in tch q t trong in trng ca h in tch im l:

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

r

r

r

n

Vct cng in trng tng hp ti M bng:

E =

F = q

qi =1

n

Fi Fi = E i chnh l vct cng in trng do in q

Cng theo (7-5) th mi s hng tch qi gy ra ti M nn:

E=

Ei =1

n

i

(7-6)

Biu thc (7-6) l biu thc ton hc ca nguyn l chng cht in trng c pht biu nh sau: Vct cng in trng gy ra bi mt h in tch im bng tng cc vct cng in trng gy ra bi tng in tch im ca h. b. Cng in trng gy bi h in tch im phn b lin tc

118

Chng VII. Trng tnh in. tnh cng in trng gy bi vt ny ta tng tng chia vt thnh nhiu phn nh sao cho in tch dq trn mi phn c th xem l in tch im. Nu gi d E l vct cng in trng gy ra bi in tch dq ti im M cch dq mt khong r th vct cng in trng do vt mang in gy ra ti im M c xc nh tng t theo cng thc (7-6). r r r E = dE = k 3 dq (7-7) ca vat r ca vat Ta xt mt s trng hp c th sau y: + Nu vt l si dy (L) vi mt in tch di (C/m) th in tch trn mt vi phn di dl l dq = dl. Khi E = dE = kL L

dl r r r 3

(7-8)

+ Nu vt mang in l mt mt S vi mt in tch mt (C/m2) th in tch trn mt vi phn din tch dS l dq = dS. Khi :

E = dE = kSS

r

.dS r r r 3

(7-9)

+ Nu vt mang in l mt khi c th tch V vi mt in tch khi (C/m3) th in tch trong mt th tch vi phn dV l dq = dV. Khi :

E = dE = kV V

r

dV r r r 3

(7-10)

5. V d: a - Tnh cng in trng gy ra bi mt dy thng di v hn tch in u, mt in di > 0 ti mt im cch dy mt khong r. dq = dx Ta chia dy thnh nhiu phn c di dx, in tch dq:u r dE

in tch dq c th coi l mt in tch im v gy ra ti M vector cng in trng c phng chiu nh hnh v v c ln:dq l dE = . 2 40 r + x 2+

(

)

N+ +u r E+ + + + + + + + +

dq =dxx r H Mu r dE

Do tnh i xng nn in trng tng hp c phng vung gc vi dy tch in v hng ra xa dy ( > 0 ). Vy, nu chiu ln phng MH ta c:E = dE n = dEcos.

u r dE n

u r E

M: nn:

cos =

r r2 + x2

;

E = dEcos =

l 40

dqcos 3 r2

=

dxcos 3 r2 . 4 0

Hnh7- 5 Dy thng tch in u.

119

Chng VII. Trng tnh in. 2

Thay

d x = r tan dx = r cos 2

ta i ti kt qu: 2 0 r

E= 4 0 r

cos.d = 2 r . 2 0

Trong trng hp tng qut: E =

.

(7-11)

b- Lng cc in Lng cc in l mt h hai in tch im c ln bng nhau nhng tri du +q v q, cch nhau mt on l rt nh so vi khong cch t lng cc in ti nhng im ang xt ca trng. Vct mmen lng cc in c nh ngha l: p e = ql r trong l l vct khong cch gia hai in tch , hng t in tch (-q) n (+q). ng thng ni hai in tch gi l trc ca lng cc in. Cng in trng ti im M nm trn mt phng trung trc ca lng cc Theo nguyn l chng cht in trng th cng in trng ti M l: E M = E (+) + E ()r

(7-12)

E (+ )M

E (+ ) E ( )

r

r O pe

r+

E ( ) v E ( + ) c hng nh hnh 7- 6 v c lnbng nhau (v r- = r+) .

-q

r

+q

l Theo nh ngha lng cc in, v l 0 q 4r 2q

trong : S = 4r2;

D=

(in thng dng v i ra khi mt kn S). Khi q < 0 th D dS nn e = - D.S = 4r 2 (in thng m v ng sc i vo mt kn S).

4r2 = - q = q < 0

- Ta thy rng, in thng e khng ph thuc vo bn knh mt cu v c gi tr bng nhau i vi cc mt cu ng tm vi S ( v d S1). iu cho thy rng, khong khng gian gia hai mt cu S v S1 ni khng c cc in tch, cc ng sc l lin tc, khng mt i hoc thm ra, cng chnh v th, nn c th suy ra rng in thng qua mt S2 bt k bao quanh in tch q cng bng in thng qua S v S1 v khng ph thuc vo hnh dng ca mt S2 cng nh v tr ca in tch q bn trong n. - Nu mt kn S3 khng bao quanh q th do tnh cht lin tc ca ng sc c bao nhiu ng cm ng in i vo S3 cng c by nhiu ng cm ng in i ra khi S3, nn ta c e(vo) R. + Bc 1: V mt cu tch in u nn h ng sc c tnh cht i xng cu. + Bc 2: H ng sc trng vi cc bn knh, hng ra ngoi. Do qu tch ca

nhng im c ln D bng nhau v bng DM M. Trn mt cu S ta c D = DM = const.

l mt cu S tm O, bn knh r i qua im

126

Chng VII. Trng tnh in. + Bc 3: Mt kn S chnh l mt cu S. + Bc 4: p dng nh l O G.(S )

DdS =Dn2

qi =1

n

i

Trin khai v tri: Ti mi im trn mt S ta c

D dS(S )

r r D .dS =

v(S )

D(S )

=

=

const,

nn:

D.dS = D dS

= D.4r

M rng:

Nu khi cu tch in m ( < 0) th cc kt qu thu c vn ging nh (7-21) v (722), ch c khc l E N , E M v h ng sc in cm ngc chiu vi vct bn knh r , tc l chng hng vo tm O. Nu y l mt mt cu (rng) tch in u th: ngoi (r > R) kt qu (7-21) vn ng v

qi

i

=Q trong (r 0, ti im M cch mt phng mt khong r. T suy ra in trng gy bi hai mt phng song song v hn mang in u tri du, mt in mt l (, -).

S +

Gii: Do tnh i xng nn D ti mt im bt k trong in trng c phng vung gc vi mt mang in. Xt mt Gauss l mt mt tr kn qua M c cc ng sinh vung gc vi mt phng, c hai y song song bng nhau, cch u mt phng v tnh thng lng cm ng in qua mt tr . Khi : e = D.dS = Dn .dS =S s 2 y

uu r

u r n

+

D .dS + D .dS = 2.D.Sn n xq

vi Dn l hnh chiu ca D trn php tuyn n ca dS, S l din tch mi y. p dng nh l Ostrogradsky-Gauss: e = q = S D.2S = .S, rt ra D =

uu r

u r

-

uu r D1uu r D2

+ + + + + + + + +

2

Hnh 7-18 (7127

Chng VII. Trng tnh in. 22) hay Nhn xt: - Cc vct D (v E ) khng ph thuc vo khong cch t im M n mt phng nn in trng y l in trng u: E = const - in trng do mt phng hu hn tch in u to ra nhng v tr rt gn mt cng c xem nh l u. - Nu mt phng tch in m th kt qu thu c cng nh vy song cc vct D , E li hng vo mt phng tch in. Trng hp hai mt phng song song v hn mang in u tri du, mt in mt l (, ) th vector cm ng in D do hai mt phng mang in gy ra l: D = D1 + D2 , vi D1 v D2 ln lt l cc vector cm ng in do tng mt phng gy ra: D1 = D2 = . khong gia hai mt phng D1 v D2 nn vector cm ng in D l:D = D1 + D 2 = E = D = . 0 2 0 uu r uu r phng D1 v D2uu r uu r

E=

2 o

(7-23)

uu r

uu r

uu r

uu r

uu r

uu r

uu r

2

ngoi hai mt

trc i nhau do cm ng in D l: D = D1 + D2 = 0. Xc nh in trng ca mt mt phng v hn tch in u vi

uu r

mt in mt > 0. .5. IN TH 1. Cng ca lc tnh in. Tnh cht ca trng tnh in.q0

MrM

r r u r 1 q0q r . lc tnh in: F = q 0 .E = 40 r 2 r

Gi s ta dch chuyn mt in tch im q0 trong in trng ca in tch im q. Khi , in tch q0 s chu tc dng caqr

r r+drrN

r dl

r F

N

Cng ca lc tnh in trong chuyn di v cng nh dl ur r 1 qq 0 dA = F.dl = dl.cos , (7-24) 4 0 r 2 trong = F,dl . T hnh v ta c: dl. cos dr. Do :dA = qq 0 dr . 4 0 r 2

Hnh 7-19 Cng ca lc tnh in.

(

r

r

)

(7 - 25)

128

Chng VII. Trng tnh in. Vy, cng ca lc tnh in trong s dch chuyn in tch q0 t M n N theo ng cong (C) trong in trng ca in tch im q l:AMN = dA =M N

qq 0 40

N

M

r

dr2

=

qq 0 4 0 rM

qq 0 4 0 rN

.

(7 - 26)

Kt lun: Cng ca lc tnh in trong s dch chuyn in tch im q0 trong mt in trng bt k khng ph thuc vo dng ng i m ch ph thuc vo im u v im cui ca ng cong dch chuyn. Cng ca lc tnh in trn mt ng cong kn bng khng. Nh vy, trng tnh in l mt trng th. Trong trng hp ng cong dch chuyn l mt ng cong kn th

A = F dl = qo E dl = 0 hay E dl = 0Tch phn

(7-27)u r

E dl c gi l lu s ca vct cng in trng E dc theo ng cong

kn.Vy, lu s ca cng in trng(tnh) dc theo mt ng cong kn bng khng. Pht biu ny l tnh cht th c trng ca trng tnh in.2. Th nng ca in tch trong in trng Ta bit rng, vi mi trng lc th, cng ca lc tc dng bng gim th nng trong trng lc. p dng trong trng hp trng lc th l trng tnh in. Gi WM, WN ln lt l th nng ca in tch q0 ti cc im M v N trong trng tnh in gy bi in tch q. Khi biu thc cng dch chuyn c th vit nh sau:N N r r AMN = dA = q 0 Edl = WM WN , M M

(7-28)

m:

A MN =

qq 0 4 0 rM

qq 0 4 0 rN

.

So snh hai biu thc ta thu c biu thc tnh th nng ca in tch im q0 t trong in trng ca in tch im q ti cc im M v N:WM = qq 0 qq 0 + C; WN = + C. 4 0 rM 4 0 rN

(7-29)

Nh vy, th nng ca in tch im q0 t trong in trng ca in tch im q v cch in tch im ny mt on r c xc nh nh sau:W (r) = qq 0 + C, 4 0 rWq 0 .q > 0

(7-30)

trong C l hng s ty , W cn c gi l th nng tng tc ca h in tch q v q0 . Ta nhn thy, C chnh l th nng ca q0 t ti mt im xa v cng ( r = ) i vi im t in tch

0q 0 .q < 0

r

Hnh 7-20 th th nng tng tc ca h hai 129 in tch im.

Chng VII. Trng tnh in. q. V th nng c trng cho tng tc ca cc in tch nn vi nhng khong cch v cng ln th s tng tc bng khng. Do , ngi ta qui c W ( ) = C = 0. Lc ny, biu thc th nng c th vit di dng n gin:W (r) = qq 0 . 4 0 r

(7-31)

Vi quy c W ( ) = 0, t biu thc xc nh cng dch chuyn ta suy ra biu thc th nng ca in tch im q0 trong mt in trng bt k: u r u r r r WM = q 0 .Edl = q 0 Edl.M M

(7.32)

Vy: Th nng ca in tch im q0 ti mt im trong in trng l mt i lng c gi tr bng cng ca lc tnh in trong s dch chuyn in tch t im ang xt ra xa v cng. 3. in th - Hiu in th a. in th T biu thc (7-29) v (7-30) ta suy ra rng t s W/q khng ph thuc vo ln ca in tch q m ch ph thuc vo cc in tch gy ra in trng v vo v tr ca im ang xt trong in trng. T ta nh ngha: w (7V= q 33) gi l in th ca in trng ti im ang xt. T nh ngha trn, ta suy ra biu thc tnh in th ca in trng cho mt s trng hp: - in th do mt in tch im q gy ra ti mt im cch q mt khong bng r: 1 q kq V= = 4 o r r - in th do mt h in tch im gy ra ti mt im trong in trng: q V = Vi = k i ri - in th ti mt im M trong in trng bt k: r r VM = EdlM

(7-34)

(7-35)

(7-36)

Ch : in th l i lng i s, v hng. b. Hiu in th Thay cc biu thc (7-28) v (7-30) vo (7-26), ta c: AMN = WM WN = q (VM - VN)

(7-37)

130

Chng VII. Trng tnh in. Vy: Cng ca lc tnh in trong s dch chuyn in tch im q t im M ti im N trong in trng bng tch s ca in tch q vi hiu in th gia hai im M v N . T biu thc (7-37) Nu ly q = +1 n v in tch th VM VN = AMN. C ngha l hiu in th gia hai im M v N trong in trng l mt i lng bng cng ca lc tnh in trong s dch chuyn mt n v in tch dng t im M n im N. Mt khc, nu ly q = +1 n v in tch v chn im N xa v cng th VM V = AM, m ta qui c W = 0 V = 0 nn VM = AM, tc l in th ti mt im trong in trng l mt i lng v tr s bng cng ca lc tnh in trong s dch chuyn mt n v in tch dng t im ra xa v cng. Ch : - n v o in th v hiu in th trong h SI l Vn, k hiu l V. - Trong k thut, i lng hiu in th c s dng nhiu hn i lng in th. V gi tr ca hiu in th khng ph thuc vo cch chn gc tnh in th (hoc th nng). Do vy ngi ta thng chn in th ca t hoc ca nhng vt ni t bng khng. Khi ni in th ca mt im no chnh l ni v hiu in th gia im vi t. - Mt vt tch in Q c phn b lin tc, khi mun tnh in th ti mt im no trong in trng do Q to ra th thay cho cng thc (7-33) ta s dng cng thc sau y: kdq V= (7r toan vat 38)

Mt dng khc ca cng thc (7-34) l: N r r VM - VN = EdlM

(7-

39)

6. LIN H GIA VCT CNG IN TRNG V IN TH1. Mt ng th a. nh ngha

Mt ng th l mt m mi im trn c cng in th. Ni cch khc, mt ng th l qu tch nhng im c cng in th. Phng trnh ca mt ng th l:V ( x, y, z ) = C,

(7 - 40)

trong C l mt hng s bt k, vi mi gi tr ca C ta c mt mt ng th. Th d: 131

Chng VII. Trng tnh in. Biu thc ca in th gy ra bi in tch im q ti mt im cch in tch mt khong r l:V (r ) = W 1 q = . q 0 4 0 r

(7- 41)

T biu thc trn ta nhn thy: Tt c nhng im cch q mt khong r u c cng in th. Tp hp nhng im l cc mt cu ng tm c tm ti in tch im q, bn knh bng r. Do , phng trnh ca mt ng th trong trng hp ny l:

r = const.b. Tnh cht ca mt ng th - Tnh cht 1: Cng ca lc tnh in khi dch chuyn mt in tch q0 trn mt ng th bng khng. Tht vy, gi s ta dch chuyn in tch q0 t im M n im N bt k trn mt ng th. Khi , cng ca lc tnh in

l:

A MN = q 0 ( VM VN ) .Mt khc, do M v N u nm trn mt mt ng th nnMr dl

VM = VN . Cui cng ta c:AMN = 0.(7- 42) - Tnh cht 2: Vct cng in trng ti mt im trn mt ng th vung gc vi mt ng th ti im . th. Cng dA ca lc tnh in trong chuyn di ny l: u r r u r r dA = q 0 E.dl = 0 E.dl = 0. u r r E dl. Do :u r E

r d bt k trn mt ng l Xt mt dch chuyn v cng nh

Hnh 7-21 Vector cng in trng vung gc vi mt ng th.

r d l mt vct bt k trn mt ng th nn l Ngoi ra, do

u r E

r d ngha l l vung gc vi miu r

(7-43)

vung gc vi mt ng th ti mi im ta xt trn mt .

V cc ng sc in trng c cng phng, chiu vi E nn cc ng sc cng vung gc vi cc mt ng th ti mi im m chng i qua. - Tnh cht 3: Cc mt ng th khng ct nhau. Ti mi im ca in trng ch c mt gi tr ca u r in th. Gi s c hai mt ng th ct nhau theo mt E V V+dV giao tuyn no th tt c cc im trn giao tuyn r u s c hai gi tr in th. iu ny l v l. dl 2. H thc lin h gia in trng v in th Trong in trng E bt k xt hai im M v N rt gn nhau, ln lt c in th V v V + dV, vi dV > 0.u r

Es

M

N

u r dn

P

Hnh 7-22 Lin h gia cng 132 in trng v in th

Chng VII. Trng tnh in.uuuur r u r r t MN = dl , = E,dl . Cng ca lc tnh in khi dch chuyn mt in tch im q0 t v tr u r r dA = q 0 Edl. M ti v tr N l:

(

)

Mt khc: dA = q 0 ( VM VN ) = q 0 V ( V + dV ) = q 0 dV. u r r Edl = E.dlcos = dV. Do : suy ra: cos < 0, tc l mt gc t. iu ny c ngha l: Vct cng in trng lun lun hng theo chiu gim ca in th. Ngoi ra ta cng thu c:E.dl.cos = E l .ds = dV E l =

r l trong El = Ecos l hnh chiu ca vector cng in trng trn phng dVy: Hnh chiu ca vector cng in trng trn mt phng no v tr s bng gim in th trn mt n v di ca phng .

dV , ds

(7-44)

r d l bt k nn kt qu trn cng ng vi mi phng khc. Trong h trc ta l Do phngEx = V V V ; E y = ; E z = ; x y z

Descartes ta c: (7-45)

trong

V V V ; ; x y z

l cc o hm ring phn ca hm th V ln lt theo cc bin x, y, z.

Nh vy, ta c:u r r r u r r r V r V r V u r r u E = Ex i + Ey j + Ez k = i+ j+ k = i + j+ kV y z y z x x

hay:E = gradVu r E

(7-46)

Vy, vector cng in trng ti mt im bt k trong in trng bng v ngc du vi gradient ca in th ti im . Ta suy ra cc kt lun sau: a. Vct cng in trng E lun lun hng theo chiu gim ca in th (gc t). b. Hnh chiu ca E ln mt phng no v tr s bng gim in th trn mt n v di ca phng : dV (7-47) El = dl Trong h ta Descartes, biu thc (7-47..) c tng qut ho nh sau: r V r V V E = - gradV = - ( i + j + k ) (7-48)

x

y

z

133

Chng VII. Trng tnh in. c. Ln cn mt im trong in trng, in th bin thin nhiu (nhanh) nht theo phng php tuyn vi mt ng th (hay theo phng ca ng sc in trng v qua im ). dV dV dn ds3. ng dng a. Xc nh hiu in th gia hai mt phng song song v hn mang in u, tri du v bng nhau v ln Nh ta d bit in trng gia hai mt phng song song v hn, mang in u, tri du v bng nhau v ln l in trng u, cc ng sc in trng c phng vung gc vi hai mt phng. Gi V1 v V2 ln lt l in th ca mt phng mang in dng v mt phng mang in m, d l khong cch gia hai mt phng . Theo (7- 47) cng in trng v tr s bng gim in th trn mt n v chiu di : V V U E= 1 2 = d d b. Xc nh hiu in th gia hai im trong in trng ca mt mt cu mang in u Ga s mun xc nh hiu in th gia hai im M, N cch tm mt cu mang in u mt khong rM, rN , trong R< rM< rN, t (7 - 47) ta c :N kq dr kq dr dV = Edr = 2 r r2 VM

V

hay VM VN =

kq kq rM rN

Trong trng hp rM = R, rN = , ta s tm c biu thc in th ca mt mt cu mang in u : kq V= R u r E Vy i vi mt cu tch in u, in th ti mi im u r Er bn trong mt cu bng in th ti mi im trn mt cu v l in th ca qu cu, in th ti mi im ngoi mt cu u r E ging nh in th do mt in tch im gy ra t ti tm M cu. c. Xc nh hiu in th gia hai im trong in trng ca mt tr thng di v hn mang in u r r r2 r1 Hiu in th ti hai im M, N nm cch trc ca mt tr nhng on rM, rN c xc nh :

VM VN =

VN

VM

dV = Edr = 2 lrM rM 0

rN

rN

q

r dr q = ln N r 2 0 l rM

A -q

u O r d

B +q

134 Hnh 7-23: Lng cc in

Chng VII. Trng tnh in.d. in th v cng in trng gy bi lng cc in - in th gy bi lng cc in Ta tnh in th gy bi lng cc in ti im M cch tm ca lng cc mt khong r. Gi r1, r2 ln lt l khong cch t im M n cc in tch q, +q. Khi , in th ti M c dng:V= q r1 r2 1 q 1 q + = . 40 r1 4 0 r2 4 0 r1r2

Vi php gn ng lng cc ta c:u r r q dcos 1 pe cos 1 pe .r = = . Do , biu thc ca in th V c th vit li: V = 40 r 2 4 0 4 0 r 3 r2

r1 r2 dcos; r1 .r2 r 2 .

(7- 49)

- Cng in trng gy bi lng cc in

tnh cng in trng gy ra lng cc in ti M, ta phi tnh gradient ca in th V. Mun vy, ta s dng h ta cc ( r, ) nh sau: H ta cc c gc ti in tch q, trc cc hng theo vector moment lng cc in pe .u ru r

Khi , hnh chiu Er ca vector E theo phng ca bn knh vector r v hnh chiu E theo phng vung gc vi vector r l:Er = p sin V 2pe cos V . = ; E = = e 3 r r 4 0 r 3 4 0 rr r 1 2pe cos.e r + pe sin.e . 4 0 r3r

r

Vy, biu thc ca vector cng in trng ti M ( r, ) l:u r r r E = Er er + E e =

Gi tr cng in trng ti M ( r, ) :2 2 E = Er + E =

pe 40 r 3

. 3cos 2 + 1.

Ta xt hai trng hp c bit: + Nu = , tc im M nm trn mt phng trung trc ca trc lng cc, th biu thcr u r 1 pe .e , tc l vct E c phng ca vector cng in trng 3 40 r u r trng vi phng ca trc lng cc v ngc chiu vi pe .u r E

2

u r c th vit li: E =

135

Chng VII. Trng tnh in.+ Nu = 0, tc im M nm trc lng cc, th biu thc ca vector cng in trngu r E

u r c vit li: E =u r

r u r 1 pe .e r , tc l vector E c phng trng vi phng ca trc lng cc 2 0 r 3

v cng chiu vi pe .Ch : T biu thc ca cng in trng gy bi lng cc in ta nhn thy: Nu o in trng ca mt lng cc in ti nhng im cch xa n th khng th xc nh c q v d mt cch ring bit m ch xc nh c tch pe = qd ca chng. Nh vy, moment lng cc in l

mt tnh cht c bn ca lng cc in.in trng gy bi lng cc in gim rt nhanh theo khong cch, lun bin thin theo1 . r3

V0

Hnh 7-24. Th gy ra bi lng cc in.

1 9 10 C 9 t ti (0,1; 0; 0) v in tch m q t ti (-0,1; 0; 0) trong mt phng (Oxy). b) Hnh v biu din cc ng ng th trong mt phng (Oxy).a) th biu din th ca lng cc in c in tch dng q =

136


HNG DN HC CHNG VIII. MC CH, YU CU 1. Hiu v vn dng c thuyt electron v nh lut bo ton in tch gii thch cc hin tng in. 2. Hiu v vn dng c nh lut Coulomb tnh lc tng tc gia cc in tch im. 3. Hiu c ngha v bn cht ca in trng, in th, in thng. 4. Vn dng c nguyn l chng cht in trng tm in trng gy bi h cc in tch im, hay bi cc vt mang in c hnh dng bt k. 5.Vn dng nh l Oxtrogradxki - Gaux tm in tr ng gy bi cc vt mang in c hnh dng i xng nh mt cu, mt phng, mt tr v.v 6. Tnh c cng ca lc in trng khi dch chuyn mt in tch im trong in trng. Biu din cng qua th nng ca in tch im trong trng. 7. Tm c mi lin h gia in trng v in th. II. TM TT NI DUNG 1. C hai loi in tch: in tch dng v in tch m. in tch c cu to gin on. N gm nhng phn t mang in nh nht, gi l in tch nguyn t e = 1,6.10-19 C. in tch q ca mt vt c th biu din bng q= ne, trong n l mt s nguyn dng. Lc tng tc gia cc in tch im c xc nh bng nh lut Coulomb:

F0 =

q1 . q 24 0 r2

=k

q1 . q 2 r2

; trong k =9.10-9

Nu hai in tch cng du lc F l lc y, nu hai in tch khc du nhau th lc F l lc ht. 2. Mi in tch u gy trong khng gian bao quanh mt in trng c cng E. in trng l mt dng c bit ca vt cht, n gi vai tr truyn tng tc gia cc in tch. Biu hin ca in trng l, khi c mt in tch qo t vo trong in trng th qo s chu tc dng ca mt lc in F =q E . Cng in trng gy bi in tch im q ti mt im cch xa mt on r:

kq r r 2 r in trng tun theo nguyn l chng cht. p dng nguyn l ny, ta tm c in trng gy bi h in tch im hay gy bi mt vt mang in bt k. Mun vy, ta ch cn chia vt ny thnh v s cc phn t mang in tch dq v cng nh, c th coi dq l in tch E=im. Phn t in tch ny gy ra in trng dE . T , ta tm c in trng E gy bi ton b vt mang in tch q

137


r E = dE =ca vat

r r k 3 dq ca vat r

Trong dq = dV; dq = dS haydq = dl ty vo trng hp vt l khi mang in u, mt mang in u hay l mt dy dn thng mang in u. 3. tm in trng ca cc vt mang in i xng, ngi ta thng dng nh l Oxtrogradxki- Gaux. Theo nh l ny, in thng gi qua mt mt kn bng tng i s cc in tch q nm trong mt kn

e =

(S )

DdS = qi

i

T y, ta tm in trng E gy bi mt dy tch in u di v hn, mt mt phng v hn tch in , mt mt tr di v hn tch in u... 4. Vct cm ng in ( in cm): D = o E 5. Cng in trng gy bi mt si dy thng di v hn mang in u ti mt im cch dy mt khong r: E = ( l mt in di ca dy) 2 o r 6. Cng in trng gy bi mt mt phng mang in u ti mt im cch dy mt khong r : ( l mt in mt) E= 2 o 7.Cng ca lc in trng khi dch chuyn in tch qo trong in trng

AMN

q .q = dA = F .dS = q o E.dS = o 4 o M M M

N

N

N

rN

rM

r

dr2

=

qo q qo q = WM W N 4 o rM 4 o rN

Th nng ca in tch qo t trong in trng gy bi in tch q ti khong r l: qo q W= 4 o r Cng ca lc tnh in khi dch chuyn in tch qo theo ng cong kn bng khng. Trng tnh in l mt trng th. Biu thc ton hc biu din tnh cht th:

E.dS = 0S

8. in th ti mt im M trong in trng E: VM

AM = = Edr qo M

9. Cng ca lc tnh in khi dch chuyn in tch im qo t im A n im B trong in trng: A = qo(VA- VB) 10. Hiu in th gia hai im A v B : VA- VB = E.dlA B

138

Chng VII. Trng tnh in.U vi U = V1- V2 l hiu in th, d l khong d

Trong trng hp in trng u E =

cch gia hai mt ng th tng ng. 11. Hiu in th gia hai mt cu ng tm mang in u, bng nhau, tri du: q( R2 R1 ) V1 V2 = 4 0 R1 R2 12. Lin h gia in trng v in th: V E= = gradV rIII. CU HI NTP

1. So snh s ging nhau v khc nhau ca nh lut Coulomb gia cc in tch im q1, q2 v nh lut hp dn v tr Niutn gia cc vt c khi lng m1 v m2. C nhn xt g v ln gia hai lc . 2. Nguyn l chng cht in trng c vn dng nh th no trong vic tm in trng gy bi mt vt mang in tch q bt k ( vn dng khi vt mang in l mt mt phng v hn hay mt mt tr v hn mang in u). 3. Vc t cng in trng: nh ngha, biu thc, ngha. Lin h gia vct cng in trng v in th. Vn dng mi lin h xc nh hiu in th gia hai mt phng song song v hn mang in u. 4. ng dng nh l Oxtrogradxki- Gaux, tm cng in trng gy bi mt mt cu mang in u ti mt im nm trong v ngoi mt cu rng mang in u. 5. ng dng nh l Oxtrogradxki- Gaux, tm cng in trng gy bi mt dy dn thng di v hn mang in u. 6. ng dng nh l Oxtrogradxki- Gaux, tm cng in trng gy bi mt mt phng v hn mang in u. 7. nh ngha m men lng in, tm biu thc ca mmen lc tc dng ln lng cc in, khi lng cc in t trong mt in trng u. 8. Thit lp biu thc cng ca lc tnh in khi chuyn mt in tch im qo trong in trng gy bi in tch im q. 9. Vit biu thc th nng ca mt in tch im qo trong in trng gy bi mt in tch im q. T rt ra biu thc in th gy bi mt in tch im q ti mt im cch n mt on r. 10. Tnh cht th ca trng tnh in th hin nh th no? Vit biu thc ton hc th hin tnh cht th ca trng tnh in bt k . 11. Nu nh ngha mt ng th v tnh cht ca mt ng th.

139


IV. BI TP Th d 1: Mt vng trn lm bng mt dy dn mnh bn knh R mang in tch dng Q phn b u trn dy. Hy xc nh cng in trng ti im M nm trn trc ca vng dy, cch tm mt on h. Gii: Cng in trng do vng dy gy ra ti mt im no

dE

dE n

M r h

dEt

bng tng cc cng in trng dE do cc phn t in tch dQ nm trn vng dy gy ra. Ti im M cng in trng do phn t in tch dQ gy ra l: r r kdQ r dE = 2 r r Theo nguyn l chng cht, cng in trng ti M bng:

r E M = dE =vong

dQ r ) k r 3 r ( vongthnh hai thnh phn

Hnh 7-24in trng gy bi vng dy trn tch in u

Trc tin ta phn tch vct dE bng khng. Cn li

dE t v dE n . V cc in tch dQ phn b i xng qua im O nn tng cc thnh phn d E t

EM =

vong

r dE n

V cc vct dE n cng phng, chiu nn E M c im t ti M, c phng ca trc vng dy v chiu hng ra xa vng dy. V ln th

EM =

vong

dEn .

Theo hnh 7- 4 ta c dEn = dEcos ( l gc gia dE v OM ). in trng gy bi dQ kdQ ti M bng: dE = 2 r 3/ 2 h khdQ 2 v r2 = R2 + h2 nn dEn = V cos = ( R + h2 ) r 3/ 2 kh Vy: EM = dEn = ( R 2 + h 2 ) dQvong

vong

hay:

EM =

khQ

(R

2

+ h2 )

3/ 2

140


Th d 2: Xc nh hiu in th gia hai mt phng song song v hn mang in u tri du, mt in mt l (, -). Hng dn: Gi V1 v V2 ln lt l in th ca mt phng mang in dng v mt phng mang in m, d l khong cch gia hai mt ta c:V V1 V1 V2 dV = 2 = dl d d d . V1 V2 = Ed = 0 E=

Th d 4: Xc nh hiu in th gia hai im nm cch tm mt cu mang in u nhng khong R1 v R2 ( R 2 > R1 > R ).

Hng dn: Ta c:q dr dV = Edr = 4 0 r 2 V1 V2 =V2

V1

q dV = 4 0 .

R2

R1

r

dr2

q 1 1 R R 4 0 1 2

Trongtrnghp R1 = R v R 2 = ( V2 = 0 ) ,tatmcbiuthctnhinthVca mtmtcumanginu:V= q . 40 R

BI TP T GII 7.1 Hai qu cu kim loi ging ht nhau, kch thc khng ng k t cch nhau 60 cm th chng y nhau vi mt lc F1 = 7.10-5N. Ni hai qu cu bng mt si dy kim loi mnh ri b si dy i th chng y nhau vi lc F2= 1,6.10-4 N. Hy xc nh in tch ban u ca mi qu cu. qq HD: Lc tng tc gia hai qa cu trc khi ni F1 = k 1 22 r q1 + q 2 2 Lc tng tc gia hai qa cu sau khi ni F2 = k r 22

q1 = 14.10 8 C ; q 2 = 2.10 8 C

141

Chng VII. Trng tnh in.7.2 Hai qu cu nh ging ht nhau tng tc vi nhau trong chn khng, mt qu cu mang in tch 6.10-9C, cn qu th hai mang in tch -3.10-9C. Khong cch gia hai qu cu bng 5,0 cm. Tm lc tng tc gia chng. 1 q1 q 2 = 6,6.10 5 ( N ) HD: Trong chn khng lc ht l F = 4 o r

Khi hai qu cu tip xc vi nhau, mt phn in tch ca chng s trung ha nhau, phn q q2 ' = 1,5.10 9 C nn lc tng tc gi in tch cn li s chia u cho nhau q1' = q 2 = 1 2 chng s l F ' =

q'4 o r 2

= 0,81.10 5 N

7.3 Hai qu cu t trong chn khng c cng bn knh v cng khi lng c treo u hai si dy sao cho hai mt ngoi ca chng tip xc vi nhau. Sau khi truyn cho cc qu cu mt in tch qo = 4.10-7C, chng y nhau v gc gia hai si dy by gi bng 60o. Tnh khi lng ca cc qu cu nu khong cch t cc im treo n tm cu bng l = 20cm. HD: Gi gc hp bi hai dy l 2 . O in tch ca mi qu cu l q =qo/2. 2 l 2 q T , vi T iu kin cn bng dn ti F= P tg = q1 q2 4 or 2 I B F A r q2 2 sin = P = = 0,157 N 2l 4 o .4l 2 sin 2 .tg P R

Hnh-bt37.4 Cho hai in tch im q1 = 8.10-8 C; q2 = 3.10-8 C t trong khng kh ti hai im M , N cch nhau 10 cm. Cho MA = 9cm; NA=7cm; MB = 4cm;NB = 6cm;qo= 5.10-10C

a. Tnh cng in trng ti hai im A v B. b. Tnh in th ti A v B. c.Tnh cng dch chuyn in tch qo t A n B.HD: a. im A: E A = E A1 + E A 2 MA2 + NA2 MN 2 5 = cos = 2.MA.NA 21 o o = 76 42' ; = 67 09' / q1 / E A1 = 8,91.104V / m; 2 4 o MA / q2 / E A2 = 9,34.10 4 V / m 4 o NA2

E A1A

E A2N q2

EA

M q1 B

Hnh - bt4

- i vi im B: E B = E B1 + E B 2 142

Chng VII. Trng tnh in./ q1 / / q2 / 45.10 4V / m ; EB 2 = 7,5.10 4V / m 2 2 4 o MB 4 o NB

EB1 =

- Vy EB= EB1+ EB2 = 52,5.104 V/m v hng t M n N b.Tnh in th ti im ti im A v B: / q1 / / q2 / VA= VA1+VA2 = + 4,14.103V 4MA 4 o NA VB=VB1+VB2=/ q1 / / q1 / + 13,5.103V 4 o MB 4 o NB

c. Cng dch chuyn qo t A n B: A = qoUAB= qo(VA- VB) = - 46,80.10-7J5 7.5 Tm lc tc dng ln mt in tch im q = .10 9 C t tm O ca na vng dy 3 trn bn knh R= 5cm tch in u mang in tch Q =3.10-7C t trong chn khng.

HD: Chn trc Ox nh hnh v. Dy tch in u c mt Q . Phn t dl mang in dq= dl = R.d . Lc tc dng = R q.dq ln phn t in tch dq l dF = k 2 R Do i xng nn cc dF nm trn trc Ox+

dl

2

dFxR qdFHnh - bt.5

F =

nuaVD +

dF cos =2

q R cos k d R2

x

2

= kqQ

cos qQ d = 1,14 .10 3 N 2 2 2 R 2 o R

2

7.6 Mt si dy kim loi mnh di 8cm t trong khng kh tch in u, in lng ca dy l q1 = 35.10-5C. in tch im q2 t trn phng ca si dy cch im gia dy mt on r = 6cm. Dy tc dng ln q2 mt lc l F2 = 12.10-5 N. Hy xc nh in tch q2.

HD: Chn trc Ox nh hnh v. Chia dy thnh cc q phn t dx v mang in 2 dq= dx = 1 dx s tc l dng mt lc dF ln in tch q2

lO

x dx r

q2

x

Hnh - bt.6

143

Chng VII. Trng tnh in.l 2

q dx qq dx qq 1 dF = k 2 2 F = k 1 2 F =k 1 2 2 l (r x ) l (r x ) (r x ) l 2

+

+l / 2 l / 2

q2 =

o F (4r l )2 2

q1

7,62.1011 C

7.7 Treo mt qu cu nh c khi lng m = 1g mang in tch q=10-9C gn mt mt phng v hn thng ng mang mt in mt =4.10-9 C/m2. Xc nh gc lch ca si dy so vi phng thng ng. + T + P + Fc = 0 + FC q + o T tg = = 13 + P 2 o mg Fc + + + P

Hnh - bt.97.8 Ti ba nh A,B,C ca mt hnh ch nht trong khng kh t ba in tch q1,q2,q3. Cho AB = a = 3cm; BC = b = 4cm; q2 = -2,5.10-6 C. a. Xc nh cc in tch q1 v q3 in trng ti D bng khng. b. Xc nh in th gy ra ti im D ca h in tch im.

HD: a.

Do q2 < 0 nn E B chiu

hng t D ti B v hnh chi E B1 , E B 2 . Vy in trng ti D bng 0 th u ca n xung DA v DC ln lt l: E A v E C phi c chiu nh hnh v q1 > 0 v q3 > 0. T hnh v ta c: EA= EB2=EBcos q1 q2 b = 2 2 2 4 o b 4 o (b + a ) b 2 + a 2B

B, q2 a

bEB

C, q3 E B2 D EC EA E AC

A, q1

E B1

Hnh - bt.11

q1 = 2,7.10 6 C

7.9 Cho hai in tch q v 2q t cch nhau 10 cm. Hi ti im no trn ng ni hai in tch y in trng trit tiu. HD: Gi s ti im M trn ng thng ni hai in tch q v 2q, gi x l khong cch t

q n M, in trng do h hai in tch gy ra trit tiu th E = E1 + E 2 = 0 ; E1 v E2 ln lt l cng in trng do q1 v q2 gy ra ti M E1=E2

144

Chng VII. Trng tnh in.q4 ox2

Ta c

=

2q x = 4,14.10 2 m 2 4 o (l x)

7.10 Xc nh cng in trng tm mt lc gic u cnh a, bit rng su nh ca n c t: a. Su in tch bng nhau v cng du. b. 3 in tch m v 3 in tch dng v tr s u bng nhau, t xen k . c. 3 in tch m v 3 in tch dng v tr s u bng nhau, t lin tip . HD: p dng nguyn l cht in trng, c hai trng hp a,b in trng ti tm u q bng 0. Trng hp t 3 in tch dng v 3 in tch m t lin tip E = 4 o a 2 7.11 Cho hai in tch im q1= 2.10-6 C, q2= -10-6C t cch nhau 10 cm. Tnh cng ca lc tnh in khi in tch q2 dch chuyn trn ng thng ni hai in ra xa thm mt on 90 cm. S: A = - 0,162J 7.12 Mt a trn bn knh R = 8cm tch in u vi mt in mt = 10 3 C / m 2 .

a. Xc nh cng in trng ti mt im nm trn trc ca a v cch tm a mt on h= 6 cm. b. Chng minh rng nu h0 th biu thc thu c s chuyn thnh biu thc tnh cng in trng gy bi mt mt phng v hn mang in u. c. Chng minh rng nu h>>R th biu thc thu c chuyn thnh biu thc tnh cng in trng gy bi mt in tch im. HD: Chia a thnh thnh nhng hnh vnh khn tm O. Din tch dS ca hnh vnh khn gii hn bi cc vng trn tm bn knh x v x + dx l: dS = 2 xdx. Vy, in tch dq trn dS l: dq = dS = 2xdx. Do tnh i xng ca bi ton nn in trng gy bi cc phn t din tch dS ti M u c phng Ox, c chiu hng ra xa a. C ln l h.dq ; dE = 4 o ( R 2 + h 2 ) 23

u r dE

A r x O B x+dx h

M

uur dE 2 uu r dE1

u r dE

x

Hnh - bt.12 145

p dng nguyn l chng cht in trng, in trng tng hp do c a trn gy ra ti 2hxdx 1 1 M l: E = dE = = 3 2 o R2 0 caVD 4 o ( x 2 + h 2 ) 2 1+ 2 h R


Nu h>>R th

R2 R2 R2 1 2 1 1+ 2 2 2h h 2h R2 1+ 2 h 12

R q Nn E = = 2 4 oh 4 oh 2+ h

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