Giai Bai Tap Dien

7/26/2019 Giai Bai Tap Dien

1/117

Khoa Vt L, trng H Khoa Hc, H Thi Nguyn

in hc

Chng 1: Trng tnh in

1-1. Tm lc ht gia ht nhn v electron trong nguyn t Hyr. Bit rng bn knh

nguyn t Hyr l 0,5.10-8cm, in tch ca electron e = -1,6.10-19C.

Gii:

S dng cng thc lc tng tc gia hai in tch ca nh lut Culng (vi in tch ca

electron v ht nhn hyr qe= - qp= -1,6.10-19C, khong cch r = 0,5.10-10m):

N10.23,9)10.5,0(

)10.6,1.(10.9r

qqkF 8210

2199

221

==

1-2. Lc y tnh in gia hai proton s ln hn lc hp dn gia chng bao nhiu ln,

cho bit in tch ca proton l 1,6.10-19C, khi lng ca n bng 1,67.10-27kg.

Gii:

Theo cng thc ca nh lut Culng v nh lut vn vt hp dn, ta c:

2

2

22

2

1 r

GmFv;

r

kqF ==

)ln(10.25,1)10.67,1.(10.67,6

)10.6,1.(10.9

Gm

kq

F

F 3622711

2199

2

2

2

1 ==

1-3. Hai qu cu t trong chn khng c cng bn knh v cng khi lng c treo

hai u si dy sao cho mt ngoi ca chng tip xc vi nhau. Sau khi truyn cho

cc qu cu mt in tch q0= 4.10-7C, chng y nhau v gc gia hai si dy by

gi bng 600. Tnh khi lng ca cc qu cu nu khong cch t im treo n tm

qu cu bng l= 20 cm.


2/117


Gii:

Do cc qu cu l ging nhau nn in tch mi qu cu nhn c l:

C10.22

qqq 7021

===

Hai qu cu cn bng khi:

0TFP d =++

Khi , d dng nhn thy:P

Ftg d=

vi P = mg v( )2

2

0

2

21

sin.24 l

kq

r

qkqF

d ==

tgl

kq

tgl

qP

Pl

qtg

.sin.16.sin64.sin16.4 22

2

0

22

0

2

0

22

0

2

0 ===

Thay s:

( )( ) ( )

)(157,030.30sin.2,0.16

10.4.10.9.10022

279

Ntg

P ==

)(16)(016,081,9

157,0gkg

g

Pm ====

1-4. Tnh khi lng ring ca cht lm qu cu trong bi 1-3. Bit rng khi nhng cc qu

cu ny vo du ha, gc gia hai si dy by gi ch bng 540 (= 2 i vi duha).

Gii:

F

T

2


3/117


T kt qu bi 1-3, ta c i vi qu cu t trong khng kh th:

11

22

01

2

0

.sin64 tgl

qP= (1)

Khi nhng cc qu cu vo du ho, mi qu cu s chu thm tc dng ca lc y

Acsimt P1hng ngc chiu vi trng lc. Do , bng tnh ton tng t bi trn, ta thuc:

22

22

02

2

0

1.sin64 tgl

qPP = (2)

Mt khc:

VgPVgmgP 01; === (3)

T (1), (2) v (3), ta c:

0

22

2

2

11

2

11

.sin

.sin ==

tg

tg

P

PP

)(tg.sin.tg.sin 0222

211

2

1 =

11

2

122

2

2

22

2

20

tg.sin.tg.sin.

tg.sin..

=

Thay s vi: )/(800;27;30;2;1 30

0

2

0

121

mkg=====

)/(2550800.30.30sin27.27sin

27.27sin 3002002

002

mkgtgtg

tg=

2.

2.=

1-5. Hai qu cu mang in c bn knh v khi lng bng nhau c treo hai u si

dy c chiu di bng nhau. Ngi ta nhng chng vo mt cht in mi (du) c

khi lng ring 1v hng s in mi . Hi khi lng ring ca qu cu () phi

bng bao nhiu gc gia cc si dy trong khng kh v trong in mi l nh

nhau.

Gii:


4/117


S dng cc tnh ton lm bi 1-4, v thay 1,, 1210 === , ta c:

22

2

1

21

11

2

22

2

22

2

1

.sin

.sin.sin.sin.

.sin..

tg

tgtgtg

tg

=

=

Vi iu kin gc lch gia cc si dy trong khng kh v cht in mi l nh nhau hay:

22

2

11

2

21 .sin.sin tgtg ==

biu thc trn tr thnh:

11

=

1-6. Mt electron in tch e, khi lng m chuyn ng u trn mt qu o trn bn

knh r quanh ht nhn nguyn t Hyr. Xc nh vn tc chuyn ng ca electrontrn qu o. Cho e = -1,6.10-19C, m = 9,1.10-28kg, khong cch trung bnh t electron

n ht nhn l r=10-8cm.

Gii:

lctrn chuyn ng xung quanh ht nhn theo qu o trn di tc dng ca lc hng

tm chnh l lc Culng.

Coulombht FF =

2

0

22

r4

e

r

vm

=

mr4

e

r4.m

e.rv

0

2

2

0

22

==

mr2

e

mr4

ev

00

2

==

Thay s, ta c:

)/(10.6,110.10.1,9.10.86,8.1.2

10.6,1 6103112

19

smv ==


5/117


1-7. Ti cc nh A, B, C ca mt hnh tam gic ngi ta ln lt t cc in tch im: q1

= 3.10-8C; q2= 5.10-8C; q3= -10.10

-8C. Xc nh lc tc dng tng hp ln in tch

t ti A. Cho bit AC = 3cm, AB = 4cm, BC = 5cm. Cc in tch u t trong

khng kh.

Gii:

Ta c:

+ Lc 1F

ca q2tc dng ln q1:

)(10.4,8)10.4.(10.86,8.1.4

10.5.10.3

4

3

2212

88

2

0

211 N

r

qqF

AB

===

+ Lc 2F

ca q3tc dng ln q1:

)(10.30)10.3.(10.86,8.1.4

10.10.10.3

4

3

2212

88

2

0

312 N

r

qqF

AC

===

+ D dng nhn thy: 222 ACABBC +=

Vy, tam gic ABC vung ti A. Khi :

- Lc F

c phng hp vi cnh AC mt gc xc nh bi:

'421528,010.30

10.4,8 03

3

2

1

===

F

F

tg

- Chiu ca F

nh hnh v.

- ln ca lc c tnh bng:

A

BC

F

2F

1F


6/117


)(10.11,3)10.30()10.4,8( 223232

2

2

1 NFFF =+=+=

1-8. C hai in tch bng nhau v tri du. Chng minh rng ti mi im cch u hai

in tch , phng ca lc tc dng ln in tch th q0song song vi ng thng

ni hai in tch .Gii:

Gi l ng trung trc ca on thng AB ni hai in tch q 1v q2bng nhau v tri

du. Xt in tch th q0(cng du vi in tch t ti B) t ti C nm trn . Ta c:

22

0

02

2

0

01

1)(4)(4

FAC

qq

CB

qqF ===

Xt thnh phn ca tng hp lc F

dc theo :

0cos)(coscos 2121 === FFFFF

Vy, F

ch c thnh phn hng theo phng vung gc vi , hay F

song song vi

ng thng ni hai in tch q1v q2.

2

0

3

01

2

0

01

21

sin2

sin2

sin

4

2sinsin

ABABl

qq

l

qqFFF

=

=+=

1-9. Tm lc tc dng ln mt in tch im q = (5/3).10-9C t tm na vng xuyn bn

knh r0= 5cm. tch in u vi in tch Q = 3.10-7C (t trong chn khng).

F1

F2

F

A B

C


7/117


Gii:

Ta chia na vng xuyn thnh nhng phn t dl mang in

tch dQ. Chng tc dng ln in tch q lc dF. p dng nguyn

l chng cht lc, ta c: == cos;sin dFFdFF yx

(na vng xuyn) (na vng xuyn)

Ta c:

2

004

.

r

qdQdF

=

vi

drdldlr

QdQ .; 0

0

==

dr

QqdF

2

00

24=

Do tnh i xng, ta thy ngay Fy= 0, nn

2

00

2

2

2

2

00

2 2.cos

4 r

Qqd

r

QqFF x

===

Thay s:

)(10.14,1)10.5.(10.86,8.1..2

10).3/5.(10.3 322122

97

NF

==

1-10. C hai in tch im q1= 8.10-8C v q2 = -3.10

-8C t cch nhau mt khong d =

10cm trong khng kh (hnh 1-1). Tnh:

1. Cng in trng gy bi cc in tch ti cc im A, B, C. Cho bit:

MN = d = 10cm, MA = 4cm, MB = 5cm, MC = 9cm,NC = 7cm.

2. Lc tc dng ln in tch q = -5.10 -10C t ti C.

x

y

q dFx

dFro

dl


8/117


Gii:

1. p dng nguyn l chng cht in trng:

+ in trng do q1v q2gy ra ti A cng phng cng chiu:

2

0

2

2

0

1

)(4)(421 AN

q

AM

qEEE AAA

+=+=

)/(10.5,52

)10.6(

10.3

)10.4(

10.8

10.86,8.1.4

1

4

22

8

22

8

12

mV

EA

=

+=

+ in trng do q1v q2 gy ra ti B cng phng ngc chiu:

2

0

2

2

0

1

)(4)(421 BN

q

BM

qEEE BBB

==

)/(10.6,27)10.15(

10.3

)10.5(

10.8

10.86,8.1.4

1 422

8

22

8

12 mVEB =

=

+ Phng, chiu ca EA v EBc xc nh nh trn hnh v.

Dng nh l hm s cos, ta thu c:

C

q1

B A

q2

Hnh 1-1

C

q1

B A

q2

EB

EA

EC

EC1

EC2


9/117


cos22121

22

CCCCC EEEEE +=

Ta cng c:

23,07.9.2

1079

.2coscos..2

222222222 =

+=

+=+=

NCMC

MNNCMCNCMCNCMCMN

)m/V(10.87,8)10.9.(10.86,8.4

10.8

)CM(4

qE 4

2212

8

2

0

1

C1===

)m/V(10.50,5)10.7.(10.86,8.4

10.3

)CN(4

qE 4

2212

8

2

0

2

C2===

Vy:

)/(10.34,923,0.10.50,5.10.87,8.2)10.50,5()10.87,8( 4442424 mVEC =+=

xc nh phng ca EC, ta xc nh gc l gc gia ECv CN theo nh l hm s sin:

C

CCC

E

sinEsin

sin

E

sin

E11

==

'096792,010.34,9

)23,0(1.10.87,8sin 0

4

24

==

=

2. Ta c: )(10.467,010.34,9.10.5. 4410 NEqF CC ===

Chiu ca lc FCngc vi chiu ca in trng ECtrn hnh v.

1-11. 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.

Gii:

Trn ng ni hai in tch, in trng do chng gy ra lun cng phng ngc chiu

nn ta c:

===

2

2

2

10

2

20

2

10

21

21

44

2

4 rr

q

r

q

r

qEEE


10/117


Gi s ti im M cch in tch q mt khong r, in trng trit tiu. im M cch in

tch 2q mt khong l (l-r) vi l l khong cch gia q v 2q.

0)rl(

2

r

1

4

qE

22

0

=

=

22

22 r2)rl(0

)rl(

2

r

1 ==

r2rl =

)cm(14,421

10

21

lr

+=

+=

Vy, in trng gia hai in tch q v 2q trit tiu ti im M nm trn ng ni hai

in tch ti v tr cch in tch q l 4,14 (cm).

1-12. Xc nh cng in trng tm mt lc gic u cnh a, bit rng su nh

ca n c t:

1. 6 in tch bng nhau v cng du.

2. 3 in tch m v 3 in tch dng v tr s u bng nhau.

Gii:

1. Nu ta t ti su nh ca lc gic u cc in tch bng nhau v cng du, th cc cp

in tch cc nh i din s to ra ti tm cc in trng bng nhau nhng ngc chiu,

nn chng trit tiu ln nhau. Do vy, in trng tng cng ti tm lc gic bng khng.

E0= 0 (do tnh i xng)

2. t ba in tch dng v ba in tch m cng ln vo su nh ca lc gic u,

ta c ba cch xp nh sau:

a) Cc in tch m v dng c t xen k vi nhau:

Ta nhn thy: cc cp in trng (E1, E4), (E2, E5) v (E3, E6) cng phng cng chiu vcc in trng c cng ln.


11/117


Cc cp in tch 1-4, 2-5 v 3-6 to ra cc in trng

bng nhau v hp vi nhau cc gc bng 1200 (Hnh v).

Do tnh i xng nn in trng tng hp c gi tr bng

0.

b) Cc in tch dng v m t lin tip:Cc cp in tch 1-4, 2-5 v 3-6 to ra cc in trng bng

nhau nh hnh v:

2

0

2

0

136251424

22a

q

a

qEEEE

=====

Ta c th d dng tnh c: in trng tng cng E hng

theo phng ca in trng E14v c ln bng:

2

0

142a

qEE ==

c) Cc in tch t nh trn hnh bn:

Hai cp in tch cng du t ti cc nh i din to ra ti

O cc in trng c cng ln nhng ngc chiu. Do ,

in trng do hai cp in tch 2-5 v 3-6 to ra ti O l bng khng. Vy, in trng ti O

bng in trng do cp in tch 1-4 to ra ti O:

20

14

2 a

qEE

==

1-13. Trn hnh 1-2, AA l mt mt phng v hn tch in u vi mt in mt =

4.10-9C/cm2v B l mt qu cu tch in cng du vi in tch trn mt phng. Khi

lng ca qu cu bng m=1g, in tch ca n bng q = 10-9C. Hi si dy treo qu

cu lch i mt gc bng bao nhiu so vi phng thng ng.

1200

E14

E25

E36

1

65

4

3 2

O

E14

E25

E36

1

65

4

3 2

O

E141

65

4

3 2

O


12/117


Gii:

Ti v tr cn bng:

0=++ PFT

Trong :02

;qEqFmgP ===

T hnh v ta thy:

2309,081,9.10.10.86,8.1.2

10.10.4

2 312

95

0

====

mg

q

P

Ftg

013=

1-14. Mt a trn bn knh a = 8cm tch in u vi mt in mt = 10-8C/m2.

1. Xc nh cng in trng ti mt im trn trc ca a v cch tm a mt onb = 6cm.

2. Chng minh rng nu b 0 th biu thc thu c s chuyn thnh biu thc tnh

cng in trng gy bi mt mt phng v hn mang in u.

A

A

T

A

BA

Hnh 1-2


13/117


3. Chng minh rng nu b a th biu thc thu c chuyn thnh biu thc tnh cng

in trng gy bi mt in tch im.

Gii:

1. Chia a thnh tng di vnh khn c b rng dr. Xt di vnh khn c bn knh r (r


14/117


( ) ( )( )m/V226

10.6/10.81

11

10.86,8.2

10E

222212

8

+=

2. Nu cho b 0, ta c:

022

00 2/1

112

lim

=

+=

baE

b

in trng khi b 0 c biu thc ging vi in trng do mt phng tch in u gy ra.

3. Nu ba, p dng cng thc gn ng:

2

2

22 21

/1

1

b

a

ba

+

Vy:2

0

2

0

2

2

0

2

2

2

0 44

).(

4

.

211

2 b

q

b

a

b

a

b

aE

===

=

in trng khi ba c biu thc ging vi in trng do mt in tch im gy ra.

1-15. Mt mt hnh bn cu tch in u, mt in mt = 10-9C/m2. Xc nh cng

in trng ti tm O ca bn cu.

Gii:

Chia bn cu thnh nhng i cu c b rng dh (tnh theo phng trc ca n). i cu

c tch in tch:

( ) ..2

/

.2

cos

.2.dhR

Rr

dhrdhrdQ

h

hh

===

vi l gc gia mt i cu v trc i xng ca i cu.

dh

h

dE

O


15/117


Tnh tng t nh phn u ca bi 1-14, ta tnh c in trng dE do i cu gy ra ti

O c hng nh hnh v v c ln bng:

( ) 302/322

04

.2..

4 R

dhRhdQ

hr

hdE

h

=

+=

Ly tch phn theo h t 0 n R, ta c:

0

2

2

00

2

0 40

222

..

=

===

Rh

Rdh

R

hdEE

R

Coi 1= , ta c: )/(2,2810.86,8.1.4

1012

9

mVE ==

1-16.

Mt thanh kim loi mnh mang in tch q = 2.10

-7

C. Xc nh cng in trngti mt im nm cch hai u thanh R = 300cm v cch trung im thanh R0 =

10cm. Coi nh in tch c phn b u trn thanh.

Gii:

Chia thanh thnh nhng on nh dx. Chng c in tch l: dxRR

qdx

l

qdq

2

0

22

==

Xt in trng dEgy ra do on dx gy ra ti im ang xt. Ta c th tch dEthnh hai

thnh phn 1dE v 2dE . in trng tng cng El tng tt c cc in trng dE. Do

tnh i xng nn tng tt c cc thnh phn 1dE bng khng. Ta c:

l/2 x

0

dE2

dE1

dE

0


16/117


( )

( ) dx

xRl

qR

dxl

q

xR

R

xRr

dqdE

2/322

00

0

22

0

0

22

00

2

0

2

4

..4

1cos.

4

+=

++==

( ) = +=

+==

0

0

0

d)tgRR.(cos

Rl4

qRdxxRl4

qRdEE2/322

0

2

0

2

0

0

0

tgRx

2/l

2/l

2/322

00

02

[ ]000000

0

0

0

0000 RR4

q

R2

l.

lR2

q

lR4

sinq2sin

lR4

qd.cos

lR4

q 0

0

===

==

Thay s: )/(10.61,0.3.10.86,8.1.4

10.2 312

7

mVE =

1-17. Mt mt phng tch in u vi mt . Ti khong gia ca mt c mt l hng

bn knh a nh so vi kch thc ca mt. Tnh cng in trng ti mt im

nm trn ng thng vung gc vi mt phng v i qua tm l hng, cch tm

mt on b.

Gii:

Ta c th coi mt phng tch in c l hng khng tch in nh mt mt phng tch in

u mt v mt a bn knh a nm ti v tr l tch in u vi mt -.

+ in trng do mt phng tch in u gy ra ti im ang xt l:

0

12

=E

+ in trng do a gy ra ti im ang xt l: (xem cch tnh trong bi 1-14)

+=

220

2

/1

11

2 baE

+ in trng do mt phng v a gy ra cng phng v ngc chiu nn:

22

0

21

/12 baEEE

+==


17/117


1-18. Mt ht bi mang mt in tch q2 = -1,7.10-16C cch mt dy dn thng mt

khong 0,4 cm v gn ng trung trc ca dy dn y. on dy dn ny di

150cm, mang in tch q1= 2.10-7C. Xc nh lc tc dng ln ht bi. Gi thit rng

q1c phn b u trn si dy v s c mt ca q2khng nh hng g n s phn

b .

Gii:

Xt mt Gaox l mt tr y trn bn knh R0c trc trng vi si dy, chiu cao h (hl)

vng gia si dy v cch si dy mt khong R0l, ta c th coi in trng trn mt tr

l u. S dng nh l Otxtrgratxki-Gaox, ta c:

l

hqqhRE 1

00

00 .

1.2.

==

lR

qE

00

1

2=

Lc in tc dng ln ht bi l:

( )NlR

qqEqF 10

312

716

00

212 10

5,1.10.4.10.86,8.1.2

10.2.10.7,1

2

===

1-19. Trong in trng ca mt mt phng v hn tch in u c t hai thanh tch in

nh nhau. Hi lc tc dng ca in trng ln hai thanh c nh nhau khng nu

mt thanh nm song song vi mt phng cn thanh kia nm vung gc vi mt phng.

Gii:

Lc tc dng ln thanh nm song song l:

== ii EqFF1

v lc tc dng ln thanh nm vung gc l: == kk EqFF2

Do in trng do mt phng v hn tch in u gy ra l in trng u nn:

21 FFEE ki ==


18/117


Vy, lc tc dng ln hai thanh l nh nhau.

1-20. Mt mt phng v hn mang in u c mt in tch mt =2.10-9C/cm2. Hi

lc tc dng ln mt n v chiu di ca mt si dy di v hn mang in u. Cho

bit mt in di ca dy = 3.10-8

C/cm.

Gii:

Ta thy, lc tc dng ln dy khng ph thuc vo cch t dy trong in trng. Ta c:

+ in trng do mt phng gy ra l:02

=E

+ in tch ca dy l: Lq =

Vy, lc tc dng ln mi n v chiu di dy l:

)(4,310.86,8.1.2

1.10.3.10.2

2 12

65

0

NL

EqF ===

1-21. Xc nh v tr ca nhng im gn hai in tch im q 1v q2 ti in trng

bng khng trong hai trng hp sau y: 1) q1, q2cng du; 2) q1, q2khc du. Cho

bit khong cch gia q1v q2l l.

Gii:

Vct cng in trng ti mt im M bt k bng

21 EEE

+=

vi 1E

v 2E

l cc vct cng in trng do q1, q2gy ra.

E

= 0, th ta phi c: 21 EE

=

+ Hai in trng E1v E2cng phng, M phi nm trn ng thng i qua im t cc

in tch.

x

l

q2q1


19/117


+ Hai in trng E1v E2cng ln:

( ) 2

1

2

2

0

2

2

0

1

21

q

q

xl

x

xl4

q

x4

q

EE

=

=

=

( )xlq

qx

q

q

xl

x

2

1

2

1 ==

lqq

q

q

q1

q

ql

x

21

1

2

1

2

1

=

=

+ Hai in trng E1v E2ngc chiu:1. Nu q1, q2cng du th M phi nm gia hai in tch:

lqq

qxlx

21

10

+=<

1-22. Gia hai dy dn hnh tr song song cch nhau mt khong l= 15cm ngi ta t mt

hiu in th U = 1500V. Bn knh tit din mi dy l r = 0,1cm. Hy xc nh

cng in trng ti trung im ca khong cch gia hai si dy bit rng cc

dy dn t trong khng kh.

Gii:

Ta i xt trng hp tng qut: nu gi khong cch t im M n trc dy dn th nht

l x th cng in trng ti M l:

)(22

1

00 xlx

l

xlxE

=

+=


20/117


vi l mt in di trn dy. Mt khc:

dU = - Edx

( )[ ]

=

=

+==

r

rl

r

rlxlxdx

xlxEdxU

rl

r

lnlnln2

11

2 000

=

r

rl

U

ln

0

Th vo biu thc cng in trng v thay x = l/2, ta c:

=

=

r

rll

U

r

rl

U

ll

l

lE

ln.

2

ln

.

2.2

2

1 0

0

Thay s: ( )mVE /10.4

001,0

149,0ln.15,0

1500.2 3

=

1-23. Cho hai in tch im q1= 2.10-6C, q2= -10

-6C t cch nhau 10cm. Tnh cng ca

lc tnh in khi in tch q2dch chuyn trn ng thng ni hai in tch xa

thm mt on 90cm.

Gii:Ta c: Cng ca lc tnh in khi dch chuyn in tch q 2t im A n im B l:

A = q2.(VA VB)

Vy:)(4

.

)(44 0

21

0

2

0

12

rlr

qql

rl

q

r

qqA

+=

+=

Thay s:( )

( )JA 162,01.1,0.10.86,8.1.4

10.2.10.9,012

66

=

Du tr th hin ta cn thc hin mt cng a q2ra xa in tch q1.


21/117


1-24. Tnh cng cn thit dch chuyn mt in tch q = (1/3).10 -7C t mt im M cch

qu cu tch in bn knh r = 1cm mt khong R = 10cm ra xa v cc. Bit qu cu

c mt in mt = 10-11C/cm2.

Gii:Cng ca lc tnh in khi dch chuyn in tch l:

A = q.(VA VB)

Vy: )(444

. 2102010

==

= Rdo

R

qQ

R

Q

R

QqA

)()(4

..4.

0

2

0

2

Rr

qr

Rr

rq

+=

+=

Thay s:( ) ( ) ( )JA 7

212

2277

10.42,310.11.10.86,8.1

10.10.3/1.10

=

1-25. Mt vng dy trn bn knh 4cm tch in u vi in tch Q = (1/9).10-8C. Tnh in

th ti:

1. Tm vng dy.

2. Mt im M trn trc vng dy, cch tm ca vng dy mt on h = 3cm.

Gii:

Chia vng dy thnh nhng on v cng nh dl mang in tch dq. in th do in tch

dq gy ra ti im M trn trc vng dy, cch tm ca vng dy mt on h l:

22

04 hR

dqdV

+=

in th do c vng gy ra ti M l:

+

=+

==22

0

22

0 44 hR

Q

hR

dqdVV

1. in th ti tm vng (h =0):


22/117


( )( )V

R

QVO 250

10.4.10.86,8.1.4

10.9/1

4 212

8

0

===

2. in th ti M (h = 3cm):

( )

( ) ( )

( )VhR

QVH 200

10.310.410.86,8.1.4

10.9/1

4 2

22

212

8

22

0

=

+

=

+

=

1-26. Mt in tch im q = (2/3).10-9C nm cch mt si dy di tch in u mt

khong r1 = 4cm; di tc dng ca in trng do si dy gy ra, in tch dch

chuyn theo hng ng sc in trng n khong cch r2= 2cm, khi lc in

trng thc hin mt cng A = 50.10 -7J. Tnh mt in di ca dy.

Gii:Ta c: dA = q.dV

drr2

.q)Edr.(qdA0

==

( )2

1

0

12

0

r

r0 r

rln

2

qrlnrln

2

q

r

dr

2

qdAA

2

1

====

2

1

0

r

rln.q

A2=

Vy:( )

( )mC/10.6

2

4ln.10.3/2

10.50.10.86,8.1..2 7

9

712

=

1-27. Trong chn khng liu c th c mt trng tnh in m phng ca cc vct cng

in trng trong c khong khng gian c in trng th khng i nhng gi trli thay i, v d nh thay i theo phng vung gc vi cc vct in trng

(hnh 1-3) c khng?


23/117


Gii:

Xt ng cong kn hnh ch nht nh hnh v, ta c:

dlEdV .

=

=ABCDA

AA dl.EVV

+++= DACDBCAB

dl.Edl.Edl.Edl.E

( )0CD.E0AB.E 21 ++=

( ) 0lEE 12 ==

Vy: Nu phng ca vct cng in trng khng i th gi tr ca n cng phi

khng i trong ton b khng gian. Khng c in trng no nh nu trong bi.

1-28. Tnh in th gy ra bi mt qu cu dn in mang in q bng cch coi in th ti

mt im A no bng tng cc in th do tng in tch im gy ra, trong cc

trng hp sau:

1. Ti mt im nm trn qu cu.

2. Ti mt im nm trong qu cu.

3. Ti mt im nm ngoi qu cu cch b mt ca n mt on bng a.

Gii:

E

A B

CD

E

Hnh 1-3


24/117


Chia qu cu thnh nhng vng dy tch in c chiu dy dh v cng nh bn knh

22hRr = c tch in vi mt in mt

24 R

q

= . in tch ca vng dy l:

cos

.2..

dhrdSdq ==

vi l gc gia mt vng dy v trc ca n. D thy:

R

dhqdhRR

qdqR

r

2..2.

4cos

2 ===

Tnh tng t bi 1-25, in th do vng dy gy ra ti im A cch tm O mt khong x

nh hnh v l:

( ) hxxRR

qdh

hxxhrR

dhq

xhr

dqdV

2828

.

4 22

0

222

0

22

0 ++

=+++

=++

=

Vy, in th do c mt cu gy ra l:

[ ]( )2

2

0

)(

)(0222

0

)(2.161628

.2

2

22 xR

xRtxR

q

t

dt

xR

q

hxxRR

dhqdVVxR

xRhxxRt

R

R +==

++==

+

++=

( )( )

( )

>

=+=Rx

x

q

RxR

q

xRxRxR

q

0

0

0

4

4

8

1. in th ti tm qu cu (x = 0) v trn mt cu (x = R):

R

qV

04=

2. in th ti im nm ngoi qu cu, cch mt cu mt khong l a (x = R + a):

( )aRq

V+

=04

O A

R

x

r

h


25/117


1-29. Tnh in th ti mt im trn trc ca mt a trn mang in tch u v cch tm

a mt khong h. a c bn knh R, mt in mt .

Gii:Chia a thnh nhng phn t hnh vnh khn bn knh x, b rng dx. Phn t vnh khn

mang in tch xdxdSdq 2.. == . Theo bi 1-25, in th do hnh vnh khn gy l:

22

0

22

0

22

0 24

2

4 hx

xdx

hx

xdx

hx

dqdV

+=

+=

+=

in th do c a gy ra:

[ ] 2

22

00022

0

2442

22

222 h

hRt

t

dt

hx

xdxdVV

hR

hhxt

R +==

+==

+

+=

Vy: ( )hhRV += 2202

1-30. Khong cch gia hai bn t in l d = 5cm, cng in trng gia hai bn

khng i v bng 6.104V/m. Mt electron bay dc theo ng sc ca in trng t

bn ny sang bn kia ca t in vi vn tc ban u bng khng. Tm vn tc ca

electron khi n bay ti bn th hai ca t in. Gi thit b qua nh hng ca trng

trng.

Gii:

Cng ca lc in trng gia tc cho electron l: A = eU = eEd.

Mt khc:

)0v(mv2

1mv2

1mv2

1A 1

2

2

2

1

2

2 === do

( )s/m10.26,310.1,9

10.5.10.6.10.6,1.2

m

eEd2

m

A2v 7

31

2419

2 ===


26/117


1-31. Cho hai mt phng song song v hn mang in u, mt bng nhau v tri du,

t cch nhau 5mm. Cng in trng gia chng l 104V/m. Tnh hiu in th

gia hai mt phng v mt in mt ca chng.

Gii:

Hiu in th gia hai bn:

( )VEdU 5010.5.10 34 ===

Ta li c, cng in trng gia hai mt phng song song v hn tch in u l:

( )2841200

/10.86,810.10.86,8.1 mCEE ====

1-32. Ti hai nh C, D ca mt hnh ch nht ABCD (c cc cnh AB = 4m, BC=3m)ngi ta t hai in tch im q1= -3.10

-8C (ti C) v q2= 3.10-8C (tiD). Tnh hiu

in th gia A v B.

Gii:

Trong hnh ch nht ABCD c AB = 4m, BC = 3m, nn:

( )mBCABBDAC 534 2222 =+=+==

in th ti A v B l tng in th do hai in th gy ra ti :

( )VAD

q

AC

qVA 36

3.10.86,8.4

10.3

5.10.86,8..4

10.3

.4.4 12

8

12

8

0

2

0

1 +

=+=

( )VBD

q

BC

qVB 36

5.10.86,8.4

10.3

3.10.86,8..4

10.3

.4.4 12

8

12

8

0

2

0

1 +

=+=

Vy: ( )VVVU BA 72==

1-33. Tnh cng ca lc in trng khi chuyn dch in tch q = 10-9C t im C n im

D nu a = 6cm, Q1=(10/3).10-9C, Q2= -2.10

-9C (Hnh 1-4).


27/117


Gii:

in th ti C v D bng tng in th do Q1v Q2gy ra:

( )( )V

BC

Q

AC

QV

C

20010.6

10.2

10.6

10.3/10

10.86,8.1.4

1

.4.4

2

9

2

9

12

0

2

0

1

+=

+=

( )( )V

BD

Q

AD

QVD

14110.2.6

10.2

10.2.6

10.3/10

10.86,8.1.4

1

.4.4

2

9

2

9

12

0

2

0

1

+=

+=

Cng ca lc in trng khi dch chuyn in tch q t C n D l:

( ) ( ) ( )JVVqA DC79

10.59,014120010 ===

1-34. Gia hai mt phng song song v hn mang in u mt bng nhau nhng tri

du, cch nhau mt khong d = 1cm t nm ngang, c mt ht mang in khi lng

m=5.10-14kg. Khi khng c in trng, do sc cn ca khng kh, ht ri vi vn

tc khng i v1. Khi gia hai mt phng ny c hiu in th U =600V th ht ri

chm i vi vn tc v2=2

v1 . Tm in tch ca ht.

Gii:Sc cn ca khng kh t l vi vn tc chuyn ng ca ht trong khng kh: Fc= kv.

+ Khi khng c in trng:

mg = kv1

Q2

D

Q1

Cq

a a

a

Hnh 1-4

A B


28/117


+ Khi c in trng c cng E hng ln trn:

mg Eq = kv2

T , ta rt ra:

2

1

v

v

Eqmg

mg=

112 Eqvmgvmgv =

=

=

1

2

1

21

v

v1

U

mgd

Ev

)vv(mgq

( ) ( )C10.1,45,01600

10.81,9.10.5q 18

214

=

1-35. C mt in tch im q t ti tm O ca hai ng trn ng tm bn knh r v R.Qua tm O ta v mt ng thng ct hai ng trn ti cc im A, B, C, D.

1.Tnh cng ca lc in trng khi dch chuyn mt in tch q0t B n C v t A n

D.

2. So snh cng ca lc tnh in khi dch chuyn t A n C v t D n B.

Gii:

Ta d dng nhn thy:

R

qVV

r

qVV DACB

00 4;

4 ====

1. Cng ca lc in trng khi dch chuyn in tch q0t B n C v t A n D l bng

khng: ( ) 0;0)( 00 ==== DAADCBBC VVqAVVqA

A B C DOq

Hnh 1-5


29/117


2. Cng ca lc tnh in khi dch chuyn t A n C v t D n B c cng ln:

( ) ( ) DBBDCAAC AVVqVVqA === 00

1-36. Mt ht bi ri t mt v tr cch u hai bn ca mt t in phng. T in c t

thng ng. Do sc cn ca khng kh, vn tc ca ht bi khng i v bng v1=2cm/s. Hi trong thi gian bao lu, sau khi t mt hiu in th U = 300V vo hai

bn ca t in, th ht bi p vo mt trong hai bn . Cho bit khong cch gia

hai bn l d = 2cm, khi lng ht bi m = 2.10-9g, in tch ca ht bi q = 6,5.10-

17C.

Gii:

Lc cn ca khng kh t l vi vn tc ca ht bi: Fc= kv.

+ Theo phng thng ng, ht bi c vn tc n nh v1:

1

1v

mgkkvmg ==

+ Gi s theo phng ngang, ht bi c vn tc n nh v2:

mgd

Uqv

kd

Uqvkv

d

UqEq 122 ====

+ Coi khong thi gian ht bi c gia tc n vn tc n nh v 2l rt ngn. Khi thi

gian ht bi ti c mt bn t l:

( )( )s

Uqv

mgd

v

dt 10

10.2.10.5,6.300.2

10.2.81,9.10.2

22 217

2212

1

2

2

===

1-37. Cho hai mt tr ng trc mang in u bng nhau v tri du c bn knh ln lt l

3cm v 10 cm, hiu in th gia chng l 50V. Tnh mt in di trn mi mt tr

v cng in trng ti im bng trung bnh cng ca hai bn knh.

Gii:

Hiu in th gia hai mt tr ng trc c tnh theo cng thc:


30/117


1

2

0

R

R21

R

Rln

2EdrVV

2

1

==

( )

1

2

210

ln

2

R

R

VV=

( )mC/10.23,0

3

10

ln

50.10.86,8.1..2 812

=

Cng in trng gia hai mt tr ch do mt tr trong gy ra:

( )m/V63510.5,6.10.86,8.1.2

10.23,0

r2E

212

8

tb0

==

1-38. Cho mt qu cu tch in u vi mt in khi ,bn knh a. Tnh hiu in th

gia hai im cch tm ln lt l a/2 v a.

Gii:

Xt mt Gaox ng tm vi khi cu bn knh r (r < a). Do tnh i xng nn in trng

trn mt ny l nh nhau v vung gc vi mt cu. Theo nh l Otstrogratxki-Gaox:

00

3

0

2

3

3

4.

4.

rE

rqrE ===

T , ta c:

0

22

0

2

02/ 02/

2/88

3.

32/233

aa

a

ardr

rEdrVV

a

a

a

a

aa ==

===

1-39. Ngi ta t mt hiu in th U = 450V gia hai hnh tr di ng trc bng kim

loi mng bn knh r1= 3cm, r2= 10cm. Tnh:

1. in tch trn mt n v chiu di ca hnh tr.

2. Mt in mt trn mi hnh tr.

3. Cng in trng gn st mt hnh tr trong, trung im ca khong cch gia

hai hnh tr v gn st mt hnh tr ngoi.


31/117


Gii:

1. Hiu in th gia hai hnh tr c tnh theo cng thc:

( ) ( ) ( )mC

RR

U

R

RU /10.207,0

3/10ln

450.10.86,8.1..2

/ln

2ln

2

712

12

0

1

2

0

===

2. in tch trn cc mt tr:r

LrSLq2

.2.. ====

( ) ( )2`82

7

2

2

27

2

7

1

1 /10.3,310.10.2

10.207,0

2;/10.1,1

10.3.2

10.207,0

2mC

rmC

r

====

3. Cng in trng gia hai bn ch do hnh tr bn trong gy ra:

( )120 /ln2 RRrU

rE ==

+ gn st mt tr trong:( )

( )mVE /125003/10ln.10.3

4502

=

+ chnh gia hai mt tr:( )

( )mVE /57503/10ln.10.5,6

4502

=

+ gn st mt tr ngoi:( )

( )mVE /37503/10ln.1,0

450=


32/117


Chng 2: Vt dn T in

2-1.Cho hai mt cu kim loi ng tm bn knh R 1 = 4cm, R2 = 2cm mang in tch

Q1=-(2/3).10-9C, Q2= 3.10

-9C. Tnh cng in trng v in th ti nhng im

cch tm mt cu nhng khong bng 1cm, 2cm, 3cm, 4cm, 5cm.

Gii:

Cng in trng bn trong mt cu kim loi tch in q bng khng cn bn ngoi

ging nh cng in trng do mt in tch im q t ti tm cu gy ra:

2

04;0

r

qEE ngoaitrong

==

in th bn trong mt cu bng nhau ti mi im cn bn ngoi c in th ging nhin th do mt in tch im q t ti tm cu gy ra (xem bi 1-28):

r

qV

R

qV ngoaitrong

00 4;

4 ==

S dng nh l chng cht in trng v in th, chng ta tnh c cng in

trng ti cc im cn xt:

r 1cm 2cm 3cm 4cm 5cm

E1 (V/m) 0 0 0 -3742 -2395

E2 (V/m) 0 67362 29938 16841 10778

E (V/m) 0 67362 29938 13099 8383

r 1cm 2cm 3cm 4cm 5cm

V1(V) -150 -150 -150 -150 -120

V2 (V) 1350 1350 900 675 540

V (V) 1200 1200 750 525 420

E1, V1; E2, V2th t l in trng v in th gy ra do cc in tch th nht v th 2.


33/117


2-2.Mt qu cu kim loi bn knh 10cm, in th 300V. Tnh mt in mt ca qu cu.

Gii:

in th ca qu cu kim loi bn knh R c tnh theo cng thc:

R

qV

04=

vi: 24. RSq ==

00

2R

R4

R4.V

=

=

( )2812

0 m/C10.66,21,0

300.10.86,8.1

R

V

=

=

2-3.Hai qu cu kim loi bn knh r bng nhau v bng 2,5cm t cch nhau 1m, in th

ca mt qu cu l 1200V, ca qu cu kia l -1200V. Tnh in tch ca mi qu cu.

Gii:

p dng nguyn l cng in th, ta c:

)(44 0

2

0

121111

ra

q

r

q

VVV +=+=

r

q

ra

qVVV

0

2

0

122212

4)(4 +

=+=

Gii h phng trnh vi cc gi tr a = 0,025 m, r = 1 m, 9

o

10.94

1

ta nhn c:

q1=3,42.10-9C; q2=-3,42.10

-9C

2-4.Hai qu cu kim loi c bn knh v khi lng nh nhau: R = 1cm, m = 4.10 -5kg c

treo u hai si dy c chiu di bng nhau sao cho mt ngoi ca chng tip xc vi

nhau. Sau khi truyn in tch cho cc qu cu, chng y nhau v cc dy treo lch mt

gc no so vi phng thng ng. Sc cng ca si dy khi l T = 4,9.10 -4N.


34/117


Tnh in th ca cc qu cu mang in ny bit rng khong cch t im treo n

tm qu cu l l = 10cm. Cc qu cu t trong khng kh.

Gii:

Sau khi truyn in tch cho hai qu cu, chng s nhn c in tch q nh nhau no .

T hnh v, ta c:

0

4

5

9,368,010.9,4

8,9.10.4cos ====

T

mg

T

P

Khong cch gia hai qu cu l:

( ) ( ) ( )cmmlx 1212,09,36sin.1,0.2sin2 0 ====

Mt khc:

sin4

4sin. 2

020

2

Txqx

qTF

=

==

( ) ( ) ( )C10.1,29,36sin.12,0.10.9,4.10.86,8.1..4q 802412 =

Gi s q > 0 ( )C10.1,2q 8= . Khi :

( ) ( )RxR4

x.q

Rx4

q

R4

qVVV

000

12111

=

+=+=

V1 ( )V2130010.11.10.10.86,8.1..4

12,0.10.1,22212

8

=

Tng t, ta cng c: V2= 21300 (V).

T

2


35/117


2-5.Hai qu cu kim loi bn knh 8cm v 5cm ni vi nhau bng mt si dy dn c in

dung khng ng k, v c tch mt in lng Q = 13.10 -8C. Tnh in th v in

tch ca mi qu cu.

Gii:

V hai qu cu c ni vi nhau bng mt si dy dn in nn chng c cng in th V:

VrVCqVrVCq 20221011 4;4 ====

Mt khc: VrrqqQ )(4 21021 +=+=

( ) ( ) ( )V

rr

QV 9000

10.58.10.86,8.1.4

10.13

4 212

8

210

+

=+

=

( )

( )Crr

Qrq

Crr

rQ

q

88

21

22

88

21

1

1

10.585

5.10.13

;10.885

8.10.13.

=+

=+

=

=+=+=

2-6.Ti tm ca qu cu rng c lp bng kim loi c t mt in tch q. Hi khi treo mt

in tch q ngoi qu cu th n c b lch i khng? Cng cu hi trong trng

hp ta ni qu cu vi t.

Gii:

Do hin tng hng ng in, nn trn qu cu xut hin cc in tch: in tch q1cng

du vi q xut hin trn phn mt cu gn in tch q v in tch q 2tri du q xut hin trn

phn mt cu bn kia. Do qu cu trung ho in nn ln ca cc in tch ny l nh

nhau. Nhng do khong cch t q n q1nh hn ti q2nn lc ht c ln mnh hn lc

y. V vy, q b ht li gn qu cu.

Nu qu cu c ni vi t, in th trn mt cu tr thnh bng 0. Do q gy ra mt

in th V trn mt cu nn trn mt cu phi c mt in tch q3tri du vi q in thtng cng trn mt cu bng 0. Do , khi qu cu c ni t, q cng b ht li gn qu

cu.


36/117


2-7.Trc mt tm kim loi ni vi t, ngi ta t mt in tch q cch tm kim loi mt

on a. Tnh mt in tch mt trn tm kim loi ti im:

1. Cch q mt on bng a.

2. Cch q mt on bng r (r > a).

Gii:

Do tnh cht ca kim loi, khi t trc tm kim loi mt in tch q, trn mt tm s xut

hin cc in tch cm ng sao cho in trng bn trong tm kim loi bng 0. Xt ti

mt im cch q mt khong r nm trn mt tm.

+ in trng E1do q gy ra ti A:

2

0

14 r

qE

=

+ trit tiu thnh phn vung gc vi tm ca E1, cc in tch cm ng ti A to ra

in trng E2:

r

a

r

qEE .

4sin.

2

0

12

==

Dng mt Gaox dng hnh tr thit din S c trc vung gc vi mt tm xc nh E2:

0

2

00

22

ESq

S2.E

===

Vy:3r2

aq

=

Vi im cch in tch q mt khong a:

qa

r

E1

E2A


37/117


2max a2

q

=

2-8.Mt qu cu kim loi bn knh R = 1m mang in tch q = 10-6C. Tnh:

1. in dung ca qu cu.2. in th ca qu cu.

3. Nng lng trng tnh in ca qu cu.

Gii:

in dung ca qu cu:

( )FRC 10120 10.1,11.10.86,8.1..44 ==

in th ca qu cu:

( )VC

qVCVq 3

10

6

10.910.1,1

10===

Nng lng tnh in ca qu cu:

( )( )J

CVW 3

23102

10.5,42

10.9.10.1,1

2

==

2-9.Tnh in dung ca Tri t, bit bn knh Tri t l R = 6400km. Tnh bin thin

in th ca Tri t nu tch thm cho n 1C.

Gii:

in dung ca Tri t l:

( )FRC 46120 10.1,710.4,6.10.86,8.1.44 ==

Ta c:

( )VC

QV

C

QVCVQ 1400

10.1,7

14=

===


38/117


2-10. Cho mt t in hnh tr bn knh hai bn l r = 1,5cm, R = 3,5cm. Hiu in th gia

hai bn l U0= 2300V. Tnh vn tc ca mt electron chuyn ng theo ng sc in

trng t khong cch 2,5cm n 3cm nu vn tc ban u ca n bng khng.

Gii:

Cng ca lc in trng chuyn thnh ng nng ca electron:

2

2mvA =

Ta c: qEdxqdUdA == vi( )rRx

UE

/ln.

0= (Xem bi 1-39)

( )( )

( ) 2/ln/ln.

/ln

2

1200

2

1

mv

rR

llqUdx

rRx

qUA

l

l

===

( )( )

( )( )

( )smrRm

llqUv /10.3,1

5,1/5,3ln.10.1,9

5,2/3ln.2300.10.6,1.2

/ln

/ln2 731

19

120 ==

2-11. Cho mt t in cu bn knh hai bn l r = 1cm v R = 4cm, hiu in th gia hai

bn l 3000V. Tnh cng in trng mt im cch tm t in 3cm.

Gii:

in trng sinh ra gia hai bn t ch do bn t trong gy ra:2

04 x

qE

=

Mt khc:( )

UrR

rRCUq

== 04

( ) ( ) ( ) ( )mV

rRx

UrRU

rR

rR

xE /10.45,4

10.3.10.3

10.4.10.30004.

4

1 4222

22

2

0

2

0

=

=

=

2-12. Cho mt t in cu bn knh hai bn l R1= 1cm, R2= 3cm, hiu in th gia hai

bn l U = 2300V. Tnh vn tc ca mt electron chuyn ng theo ng sc in

trng t im cch tm mt khong r1= 3cm n im cch tm mt khong r2= 2cm.


39/117


Gii:

Cng ca lc in trng chuyn thnh ng nng ca electron:

2

2mvA =

Ta c: qEdxqdUdA == vi( )122

210

RRx

RRUE

= (Xem bi 2-11)

( ) ( ) 2

mv

r

1

r

1.

RR

RRqUdx

RRx

RRqUA

2

1212

210

r

r 12

2

210

2

1

=

=

=

( ) 21

21

12

210

rr

rr.

RRm

RRqU2v

=

( )s/m10.42,110.2.10.3.10.2.10.1,9

10.10.3.10.2300.10.6,1.2v

7

22231

22219

=

2-13. Hai qu cu mang in nh nhau, mi qu nng P = 0,2N c t cch nhau mt

khong no . Tm in tch ca cc qu cu bit rng khong cch , nng lng

tng tc tnh in ln hn nng lng tng tc hp dn mt triu ln.

Gii:

Nng lng tng tc tnh in gia hai qu cu l:r

qW

0

2

14

=

Nng lng tng tc hp dn l:2

2

212

.

.

gr

PG

r

mGmW == (m1=m2)

Theo u bi, ta c:

2

0

22

2

2

0

2

2

1

4.

4 GPgq

GPrg

rq

WWk

===

( )Cg

kGPq 9

2

11612

2

2

0 10.76,181,9

04,0.10.67,6.10.10.86,8.1.44

==


40/117


2-14. Tnh in dung tng ng ca h cc t in C1, C2, C3. Cho bit in dung ca mi

t in bng 0,5F trong hai trng hp: 1) Mc theo hnh 2-3; 2) Mc theo hnh 2-4.

Gii:

+ in dung tng ng ca h hai t in mc ni tip:21

21

CC

CCC

+=

+ in dung tng ng ca h hai t in mc song song: 21 CCC += 1. Hnh 2-3: (C1nt C2) // C3

( )FCCC

CCC 75,05,0

5,05,0

5,0.5,03

21

21 =++

=++

=

2. Hnh 2-4: (C1// C2) nt C3

( )FCCC

CCCC 33,0

5,0)5,05,0(

5,0)5,05,0(

)(

)(

321

321 ++

+=

++

+=

2-15. Hiu in th gia hai im A v B bng 6V (Hnh 2-5). in dung ca t in th

nht C1= 2F v ca t in th hai C2= 4F. Tnh hiu in th v in tch trn cc

bn t in.

Gii:

Gi q l in tch trn cc t in, ta c:

UCC

CCCUq

21

21

+==

A BD

C1 C2

Hnh 2-5

A B

D

C1 C2

C3

A

DC1

C2

BC3

Hnh 2-3 Hnh 2-4


41/117


Mt khc: 22211 ; UCqUCq ==

( )V4

42

6.4

CC

UCU

UCC

CCUC

21

2

1

21

21

11

=

+

=

+

=

+=

Tng t: ( )VCC

UCU 2

42

6.2

21

12 =

+=

+=

Khi , ta c: ( )CUCq 6611 10.84.10.2 ===

2-16. Tnh in dung tng ng ca hai h cc t in C1, C2, C3, C4mc theo hnh 2-6 v

2-7, chng minh rng iu kin hai in dung tng ng bng nhau l:

4

3

2

1

CC

CC =

Gii:

Trong cch mc th 1: (C1// C3) nt (C2// C4)

( )( )

4321

42311'

CCCC

CCCCC

+++

++=

Trong cch mc th hai: (C1nt C2) // (C3+ C4)

43

43

21

212'

CC

CC

CC

CCC

++

+=

hai in dung tng ng bng nhau:

( )( )

43

43

21

21

4321

423121 ''

CC

CC

CC

CC

CCCC

CCCCCC

++

+=

+++

++=

t4

32

2

11 ;

CCk

CCk == , ta c:

( )( )( ) ( ) ( ) ( ) 42

22

1

1

42

2

42

21

2

21

4221

424221

111111C

k

kC

k

k

Ck

Ck

Ck

Ck

CkCk

CCCkCk

++

+=

++

+=

+++

++


42/117


( ) 242422

2

11

1

22

21

2

424221

2

21 CkCCk1k

1kk1k

1kCkCkCCkkCk +

+

++

+

++=+++

2

1

211

2

2122

2

11

1

21 k

1k

kkk

1k

kkkk

1k

1kk1k

1kk

+

=

+

+

+=

+

+

( )( ) 01kkkkkkkk 212122

21

2

1 =+++=+

)0k,0kdo(kk 2121 >>=

4

3

2

1

C

C

C

C=

2-17. Mt t in c in dung C1= 20F, hiu in th gia hai bn t in U1= 100V.

Ngi ta ni song song vi n mt t in th hai c hiu in th trn hai bn l U2=40 V. Xc nh in dung ca t in th hai (C2) bit hiu in th sau khi ni l U =

80V (hai bn ni vi nhau c in tch cng du).

Gii:

in tch trn cc t in trc khi ni vi nhau l:

222111 ; UCqUCq ==

Do cc bn ni vi nhau c in tch cng du nn tng in tch trn cc t in sau khi nil: ( )UCCUCUCqqq 21221121 +=+=+=

( ) ( )2211 UUCUUC =

( )F10204080

80100C

UU

UUC 1

2

12 =

=

=

2-18. Mt t in c in dung C = 2F c tch mt in lng q = 10-3C. Sau , cc

bn ca t in c ni vi nhau bng mt dy dn. Tm nhit lng to ra trong dydn khi t in phng in v hiu in th gia hai bn ca t in trc khi phng

in.


43/117


Gii:

Hiu in th gia hai bn ca t in trc khi phng in:

( )VC

qU 500

10.2

106

3

===

Nhit lng to ra trong dy dn khi t in phng in chnh l nng lng ca t inban u:

( )( )J

C

qWQ 25,0

10.2.2

10

2 6

232

====

2-19. Xc nh nhit lng to ra khi ni cc bn pha trn (bn khng ni t) ca hai t

in bng mt dy dn (hnh 2-8). Hiu in th gia cc bn pha trn ca t in v

t ln lt bng U1= 100V v U

2= -50V, in dung ca cc t in bng C

1= 2F;

C2= 0,5F.

Gii:

Trc khi ni cc t in, in tch trn cc bn t pha trn l:

222111 ; UCqUCq ==

Sau khi ni cc t in, tng in tch trn cc bn t l:

( )UCCUCUCqqq 21221121 +=+=+=

21

2211

CC

UCUCU

+

+=

Nng lng ca cc t in trc khi ni l:

OO

1 2

C2C1

Hnh 2-8


44/117


22

2

22

2

111

UCUCW +=

v sau khi ni l:

( ) ( )

( )21

2

2211

2

212

22 CC

UCUCUCCW

+

+=

+=

Nhit lng to ra ng bng thay i nng lng cc t in:

( )

( )

( )

( )21

2

2121

21

2

2211

2

22

2

1121

CC2

UUCC

CC2

UCUC

2

UC

2

UCWWQ

+

=

+

++==

( )( )( )

( )J10.5,410.5,010.22

50100.10.5,0.10.2Q 3

66

266

=+

=


45/117


Chng 3: in mi

3-1.Xc nh mt in tch lin kt trn mt mt tm mica dy 0,02cm t vo gia v p

vo hai bn ca mt t in phng c tch n hiu in th U = 400V.

Gii:

Ta c: nP= trong nnn EDP 0= .

Trong khong khng gian gia hai bn t in phng, in trng l u v vung gc vi

hai bn t. Ta c:

EDDd

UEE nn 0; ====

( ) ( )d

U1E1ED 000 ===

( ) ( )244

12 m/C10.15,110.2

40010.86,815,7

==

3-2.Bn trong mt lp in mi ng cht hng s in mi l , c mt in trng u E.

Ngi ta khot mt l hng hnh cu bn trong lp in mi y. Hy tm cng in

trng E ti tm l hng do cc in tch cm ng trn mt lp in mi to thnh lhng gy ra.

Gii:

Mt in tch trn mt phn t din tch mt dSl:

O

E

Rr

h


46/117


( ) ( ) ( )

R

EhEEP

nn0

00

1cos.11

====

vi l gc gia php tuyn ca dS v vct phn cc in mi P

.

Chia mt cu thnh cc vng c dy dh rt nh. Vng c in tch tng cng:

dhRdhrdSdQ .2sin.

=2=.=

S dng tnh ton ca bi 1-14, in trng do vng gy ra ti O cng phng vi E

v

ln bng:

( ) dQ

hr

hdEh .

4 2/322

0 +=

( ) ( )

dhh

R2

E1dh.Eh1.2.

R4

hdE 2

303

0

h

==

( )

==

R

R

2

3h dhhR2

E1dEE

( )R

Rh

R

EE

=

32

1 3

3

( )

3

1EE

=

3-3.Mt t in phng c cha in mi (= 6) khong cch gia hai bn l 0,4cm, hiu

in th gia hai bn l 1200V. Tnh:

1. Cng in trng trong cht in mi.

2. Mt in mt trn hai bn t in.

3. Mt in mt trn cht in mi.

Gii:

1. Cng in trng trong cht in mi:

( )mVd

UE /10.3

10.4

1200 53===


47/117


2. Mt in mt trn hai bn t in

( )2551200

/10.59,110.3.10.86,8.6 mCEE ===

3. Mt in mt trn cht in mi:

( ) ( )255120 /10.33,110.3.10.86,8.51' mCE ==

3-4.Cho mt t in phng, mi trng gia hai bn ban u l khng kh (1= 1), din tch

mi bn l 0,01m2, khong cch gia hai bn l 0,5cm, hai bn c ni vi hiu in

th 300V. Sau b ngun i ri lp y khong khng gian gia hai bn bng cht

in mi c 2= 3.

1. Tnh hiu in th gia hai bn t in sau khi lp y in mi.

2. Tnh in tch trn mi bn.

Gii:

in dung ca t in c xc nh theo cng thc:

d

SC 0

=

( )CUd

SUCQ 9

2

212

101

11 10.3,5300.

10.5,0

10.10.86,8.1

===

Mt khc, in tch ny sau khi lp t khng i nn:

( )VU

S

d

d

SU

C

QUUCQ 100

3

300.1.

2

11

02

101

2

222 ======

3-5.Cho mt t in phng, khong cch gia hai bn l 0,01m. Gia hai bn y du c

hng s in mi = 4,5. Hi cn phi t vo cc bn mt hiu in th bng bao

nhiu mt in tch lin kt trn du bng 6,2.10-10

C/cm2

.

Gii:

Mt in tch lin kt:


48/117


( ) ( )d

UE 00 11' ==

( ) ( )V

dU 2000

10.86,8.5,3

10.2,6.01,0

1

'12

6

0

=

=

3-6.Gia hai bn ca mt t in phng, c mt bn thu tinh (= 6). Din tch mi bn t

in bng 100cm2. Cc bn t in ht nhau vi mt lc bng 4,9.10 -3N. Tnh mt

in tch lin kt trn mt thu tinh.

Gii:

Gi lc tng tc gia hai bn t in l F. Cng dch chuyn hai bn t in li st nhau

v tr s ng bng nng lng ca t in:

S

dS

C

QFd

0

222

.22

==

S

F02=

Mt khc, ta li c:

( ) EE 00 1'; ==

( )262

3120 /10.6

10

10.9,4.10.86,8.6.2

6

5211' mC

S

F

===

3-7.Mt t in cu c mt na cha in mi ng cht vi hng s in mi = 7, na

cn li l khng kh. Bn knh cc bn l r = 5cm v R = 6cm (hnh 3-2). Xc nh in

dung C ca t in. B qua cong ca nhng ng sc in trng ti mt gii hn

cht in mi.


49/117


Gii:

Coi t in nh mt h hai t in mc song song m mi t in c cc bn l na mt

cu. in dung ca mi t c tnh theo cng thc:

rR

RrC

= 02

in dung ca h l:

( )

rR

Rr12

rR

Rr2

rR

Rr2CCC 00021

+=

+

=+=

( )( )F10.34,1

10

10.5.10.6.10.86,8.17.2C 10

2

2212

+

=

3-8.Trong mt t in phng c khong cch gia cc bn l d, ngi ta t mt tm in

mi dy d1< d song song vi cc bn t in. Xc nh in dung ca t in trn. Cho

bit hng s in mi ca tm in mi l , din tch ca tm bng din tch cc bn

ca t in v bng S.

Gii:

Coi t in nh ba t in mc ni tip vi cc in dung:

3

03

2

02

1

01 ;;

d

SC

d

SC

d

SC

===

vi d2v d3l khong cch gia cc mt ca tm in mi v cc bn t in.

in dung ton phn ca t in xc nh theo cng thc:

r

Hnh3-2


50/117


+=

++=++= 1

1

0

321

0321

111111dd

d

Sdd

d

SCCCC

( ) 10

1 dd

SC

+=

3-9.Hai t in phng, mi ci c in dung C = 10-6F c mc ni tip vi nhau. Tm s

thay i in dung ca h nu lp y mt trong hai t in bng mt cht in mi c

hng s in mi = 2.

Gii:

in dung ca h trc khi lp l:

2.

1C

CCCCC =

+=

in dung ca t in b lp y s tng ln ln. in dung ca h khi l:

( )( ) 1

..2

+=

+=

C

CC

CCC

thay i in dung ca h l:

( ) ( ) ( )FC

CCCCC

7612 10.7,110.

122

12

12

1

21

. +

=

+

=

+==

3-10. Mt in tch q c phn b u trong khp th tch ca mt qu cu bn knh R.

Cho hng s in mi ca mi trng bn trong cng nh bn ngoi ca qu cu u

bng . Tnh:

1. Nng lng in trng bn trong qu cu.

2. Nng lng in trng bn ngoi qu cu.

3.

Khi chia i qu cu thnh hai na qu cu bng nhau, nng lng in trng thay ith no?

Gii:

Xt mt Gaox ng tm vi qu cu c bn knh r. Do tnh i xng nn in trng trn

mt cu c ln nh nhau v vung gc vi mt cu.


51/117


+ Vi r < R:3

00

3

0

3

0

2

43.

)3/4(

.)3/4.(4.

R

qrr

R

qE

rqrE r

====

+ Vi r > R:2

00

2

44.

r

qE

qrE

==

Mt nng lng ca in trng l: 2021 Ew =

+ Nng lng bn trong qu cu l:

=

==

=

R R

R

qR

r

R

qdrr

R

qdrr

R

qrW

0 0

25

6

0

2

0

4

6

0

22

2

3

0

0140

0588

4.4

.2

1

+ Nng lng bn ngoi qu cu l:

=

==

=

R R R

q

Rr

qdrr

qdrrr

qW0

2

0

2

2

0

22

2

2

0

028

1

8

1

84.

4.

2

1

+ Khi chia i qu cu, cc bn cu s y nhau ra v chuyn v trng thi c mc nng

lng thp hn.

3-11. Vct cm ng in D

qua mt phn cch gia hai cht in mi khc nhau, s i

hng (hnh 3-3). Tm quy lut ca s i hng .

Gii:

Chia vct cm ng in thnh hai thnh phn: thnh phn hng dc theo php tuyn

nD v thnh phn hng dc theo mt ngn cch gia hai mi trng tD .

1D

2D

N

O p

1

2

Hnh 3-3


52/117


+ Xt thnh phn php tuyn nD : Do cc in tch cm ng xut hin ti mt ngn cch

gia hai mi trng, nn thnh phn php tuyn ca vect cng in trng thay i

theo biu thc:

22021011

1

2

2

1nnnn

n

n DEED

E

E====

+ Xt thnh phn tip tuyn tD : Do theo phng ngang, in trng khng b nh hng

bi cc in tch cm ng, nn:

2

1

2

1

02

2

01

121

===

t

ttttt

D

DDDEE

Khi ta c:

2

1

2

2

1

1

2

1 .

==

t

n

n

t

D

D

D

D

tg

tg


53/117


Chng4: T trng

4-1.Tnh cng t trng ca mt dng in thng di v hn ti mt im cch dng

in 2cm. Bit cng dng in I = 5A.

Gii:

S dng cng thc cng t trng cho dng in thng di v hn:

( )mAr

IH /8,39

10.2.2

5

2 2==

4-2.Hai dng in thng di v hn, c cng dng in I 1=I2= 5A, c t vung

gc vi nhau v cch nhau mt on AB = 2cm. Chiu cc dng in nh hnh v 4-7.Xc nh vect cng t trng ti im M nm trong mt phng cha I1v vung

gc vi I2, cch dng in I1mt on MA= 1cm.

Gii:

Dng I1gy ra ti M t trng H1hng t pha sau ra pha trc trang giy c ln l:

( )mAMA

IH /6,79

10.2

5

.2 2

11 ==

Dng I2gy ra ti M t trng H2hng t di ln trn c ln l:

( )mAMB

IH /5,26

10.3.2

5

.2 2

22 ==

Cng t trng tng hp c ln:

( ) ( ) ( )mAHHH /845,266,79 22222

1 +=+=

Hnh 4-7

AB

I1

I2


54/117


v hng t pha sau ra pha trc trang giy, hp vi H1gc c:

'25183

1

6,79

5,26 0

1

2 === H

Htg

4-3.

Hnh 4-8 v mt ct vung gc ca hai dng in thng song song di v hn ngcchiu nhau. Khong cch gia hai dng in AB = 10cm. Cng ca cc dng in

ln lt bng I1= 20A, I2= 30A. Xc nh vect cng t trng tng hp ti cc

im M1, M2, M3. Cho bit M1A=2cm, AM2= 4cm, BM3= 3cm.

Gii:

T trng do I1v I2gy ra cng chiu ti M2v ngc chiu ti M1v M2.

+ Ti M1: ( )mABM

I

AM

IH /120

10.12.2

30

10.2.2

20

.2.2 22

1

2

1

11 ==

H1c chiu hng t trn xung.

+ Ti M2: ( )mABM

I

AM

IH /160

10.6.2

30

10.4.2

20

.2.2 22

2

2

2

12 +=+=

H2c chiu hng t di ln.

+ Ti M3: ( )mABM

I

AM

IH /135

10.13.2

20

10.3.2

30

.2.2 223

1

3

23 ==

H3c chiu hng t di ln.

4-4.Hnh 4-9 biu din tit din ca ba dng in thng song song di v hn. Cng cc

dng in ln lt bng: I1= I2= I; I3= 2I. Bit AB = BC = 5cm. Tm trn on AC

im c cng t trng tng hp bng khng.

A

I2

1 2 3

I1

B

Hnh 4-8


55/117


Gii:

Xt im M nm trn AC. Gi 1H , 2H v 3H l cc cng t trng do I1, I2v I3gy

ra ti M. D dng nhn thy chng cng phng cng chiu trn on BC, nn im M c

cng t trng tng hp bng khng ch c th nm trn AB (do ta ch xt M nm trn

AC). t x = AM. Ta c: 2H ngc chiu vi 1H v 3H nn:

0)(2

2

)(2.2 21231 =

+

=+=

xl

I

xl

I

x

IHHHH

( ) ( ) ( )( )( )

010x5xx

xx52xx1050x15x

0x10

2x51

x1

222

=

++

=

+

( )cm3,315

50x

0x1550

=

=

Vy: im M nm trn AB v cch A mt khong x = 3,3cm.

4-5.Hai dng in thng di v hn t thng gc vi nhau v nm trong cng mt mt

phng (hnh 4-10). Xc nh vect cng t trng tng hp ti cc im M 1v M2,

bit rng:

I1= 2A; I2= 3A; AM1= AM2= 1cm; BM1= CM2= 2cm;

A

I3I1

CB

I2

Hnh 4-9


56/117


Gii:

Cc dng I1v I2gy ra ti M1v M2 cc vect cng t trng hng theo phng

vung gc vi mt phng hnh v, cng chiu ti M2v ngc chiu ti M1.

+ Ti M1:

( )mABMI

AMIH /8

10.23

102

21

.2.2 221

2

1

11

==

Do t trng do dng I1 gy ra mnh hn nn H1 hng theo phng vung gc vi mt

phng hnh v theo chiu hng ra pha sau.

+ Ti M2:

( )mACM

I

AM

IH /56

10.2

3

10

2

2

1

.2.2 222

2

2

12

+=+=

Vect cng t trng hng theo phng vung gc vi mt phng hnh v theo chiuhng v pha trc.

4-6.Tm cng t trng gy ra ti im M bi mt on dy dn thng AB c dng in

I = 20A chy qua, bit rng im M nm trn trung trc ca AB, cch AB 5cm v nhn

AB di gc 600.

Gii:

T iu kin ca u bi ta d dng c: 060,, === BMBAAMAB

S dng cng thc tnh cng t trng cho dy dn hu hn:

O BC

M2 1A

I1

I2

Hnh 4-10


57/117


( ) ( )( )mA

r

IH /8,31

10.5.4

120cos60cos20

4

coscos2

00

21

=

=

(do == 021 180; )

4-7.

Mt dy dn c un thnh hnh ch nht, c cc cnh a = 16cm, b= 30cm, c dngin cng I = 6A chy qua. Xc nh vect cng t trng ti tm ca khung

dy.

Gii:

Bn phn dy dn to nn bn canh ca hnh ch nht to ra cc t trng cng phng,

cng chiu vi nhau ti tm ca khung dy. Gi gc ABAO,= , ta c:

( )322

211 .

2.4

cos2.4coscos H

ba

b

a

Ia

I

r

IH =+

===

Tng t:2242

.ba

a

b

IHH

+==

Vy:ab

baI

a

b

b

a

ba

IHHHHH

22

224321

22 +=

+

+=+++=

Thay s: ( ) ( ) ( )mAH /1,273,0.16,0.3,016,0.6.2

22

+=

4-8.Mt dy dn c un thnh tam gic u mi cnh a = 50cm. Trong dy dn c dng

in cng I = 3,14A chy qua. Tm cng t trng ti tm ca tam gic .

Gii:

Ta nhn thy mi cnh tam gic to ra ti tm ca tam gic mt t trng cng ln,cng phng chiu. Gi khong cch t ti tm tam gic ti mt cnh l x, ta d dng c

c:


58/117


2

3

12

16.2

1

4122

42

2/coscos;

6

3

2

3

3

1222

2

21 ==

+

=

+

=====aa

a

ax

a

r

aaax

( )

( )

( )

( )mAHH

mAx

IH

/93

/3

6/3.5,0..4

2/3.2.14,3

4

coscos

1

11

==

=

=

4-9.Mt dy dn c un thnh hnh thang cn, c dng in cng I = 6,28A chy qua

(hnh 4-11). T s chiu di hai y bng 2. Tm cm ng t ti im A giao im ca

ng ko di 2 cnh bn. Cho bit: y b ca hnh thang l = 20cm, khong cch t A

ti y b b = 5cm.

Gii:

Theo nh lut Bi-Xava-Laplatx:

3

0

r

rdl.I.

4Bd

=

ta thy, in trng do phn t dng in khng gy ra ti im nm trn trc ca n (dB = 0

do 0= rdl

).

Cc cnh CD v BE khng sinh ra t trng ti A. Cc cnh BC v DE sinh ra ti A cc t

trng hng theo phng vung gc vi mt phng hnh v nhng ngc chiu.

b2.4)cos.(cosI

b.4)cos.(cosIBBB 210210DEBC

==

22

2

0

22

0

b4lb4

l.I

b4

l

2/l.

b4

I

b2.4

cos2.IB

+=

+

==

E

D

C

B

l A

b

Hnh 4-11


59/117


( )( )T10.24,2

10.5.42,010.5.4

2,0.28,6.10.4.1B 6

2222

27

+

=

4-10. Mt dy dn di v hn c un thnh mt gc vung, trn c dng in 20A chy

qua. Tm:

a) Cng t trng ti im A nm trn mt cnh gc vung v cch nh O mt on

OA = 2cm (hnh 4-12);

b) Cng t trng ti im B nm trn phn gic ca gc vung v cch nh O mt

on OB = 10cm.

Gii:

a) T trng trn trc dy dn bng 0, nn t trng ti A ch do mt cnh gc vung gyra:

( )( )mA

R

I

HA /8,7910.2.4

01.20

4

2cos0cos.

2

=

=

b) T trng do hai cnh gc vung gy ra ti cng phng, cng chiu:

21

BR4

cos4

cosI

R4

4

3cos0cosI

H

+

=

( )m/A3,77

2

1,04

12

2.20

2

1,04

2

21.20

HB

+

+

+

=

4-11. Mt dy dn di v hn c un thnh mt gc 560. Cng dng in chy qua

dy dn I = 30A. Tm cng t trng ti im A nm trn phn gic ca gc v cchnh gc mt on a = 5cm (hnh 4-13).


60/117


Gii:

T trng do hai cnh ca gc nhn gy ra ti A cng phng v cng chiu:

( ) ( )0

000

28sin.a2

)28cos1(I

R4

180cos28cosI

R4

152cos0cosIH

+=

+

=

( )m/A10.8,328sin.05,0.2

)28cos1(30H 2

0

0

+

=

4-12. Trn mt dy dn c un thnh mt a gic n cnh u ni tip trong vng trn bn

knh R c mt dng in cng I chy qua. Tm cm ng t B ti tm ca a gic.

T kt qu thu c, xt trng hp n .

Gii:

Gi H0l cng t trng do mt cnh a gic c dng in cng I chy qua gy ra

ti tm a gic. Do tnh i xng, nn t trng ti tm a gic s bng:

0nHH= , vi n l s cnh ca a gic.

p dng cng thc tnh cng t trng cho on dy dn thng hu hn, ta thu c:

a

IH

4

)cos(cos 210

=

trong : a l di cnh a gic.

D thy: nnnRa

+=== 2;2;cos 21

Vy:n

tgR

I

nR

nI

nR

nnI

H

.2

cos4

sin2.

cos4

2cos

2cos

0 ==

+

=

Aa

O

Hnh 4-13


61/117


ntg

R

nIHB

20

0 ==

Khi cho n , ta c:

( )

( )R

I

n

ntg

R

IB

n 2/

/.

2lim 00

0/

==

chnh l cm ng t do dng in trn bn knh R gy ra ti tm O ca vng trn.

4-13. Trn mt vng dy dn bn knh R = 10cm c dng in cng I = 1A chy qua.

Tm cm ng t B:

a) ti tm O ca vng dy;

b) ti mt im trn trc ca vng dy v cch tm O mt on h = 10cm.

Gii:

Chia nh vng dy thnh cc on dy dn rt ngn dl. on dy gy ra ti A cm ng t

Bd

c th phn tch thnh hai thnh phn 1Bd

v 2Bd

. Do tnh i xng nn tng tt c cc

vct thnh phn 1Bd

bng khng. Ta c:

( ) ( ) 2/322

2

0

2/322

0

3

0

2

02

22.4

4.

.

4cos.

hR

IR

RhR

IR

dlr

IR

r

R

r

dlIdBdBB

+=+=

====

+ Cm ng t ti tm O (h = 0):

( )TR

I

R

IRBO

67

0

3

2

0 10.3,61,0.2

1.10..4

22

===

dB1

dB2dB

R

h

dl

A

I


62/117


+ Cm ng t ti im trn trc ca vng dy cch tm O mt on h = 10cm:

( ) ( ) ( )T

hR

IRB

A

6

2/322

27

2/322

2

0 10.3,21,01,02

1,0.1.10..4

2

+

=+

=

4-14.

Ngi ta ni hai im A, B ca mt vng dy dn kn hnh trn vi hai cc ca ngunin. Phng ca cc dy ni i qua tm ca vng dy, chiu di ca chng coi nh ln

v cng (hnh 4-14). Xc nh cng t trng ti tm ca vng dy.

Gii:

Ta thy, do cc dy ni hoc l rt xa hoc l nm theo phng i qua tm O nn t

trng tng hp do cc dy ni gy ra ti O l bng khng. Gi H1v H2ln lt l t trng

do hai on dy AMB v ANB gy ra ti O. Hai t trng ny cng phng ngc chiu. Do

:

2

2

2

1

2

1221121

RU

r2l

RU

r2l

r2l.

rI

r2l.

rIHHH

===

( ) ( ) 0

r2

US

r2

US

S/l.r2

Ul

S/l.r2

Ul22

2

2

2

1

2

1 ===

trong : + I1, I2: cng dng in trong AMB v ANB

+ l1, l2: chiu di cc cung AMB v ANB

+ R1, R2: in tr ca cc on dy AMB v ANB

+ r, , S: bn knh, in tr sut v tit din ca vng dy

+ U: hiu in th gia hai im AB.

B

A

EMNO

Hnh 4-14


63/117


4-15. Cng t trng ti tm ca mt vng dy dn hnh trn l H khi hiu in th

gia hai u dy l U. Hi nu bn knh vng dy tng gp i m mun gi cho cng

t trng ti tm vng dy khng i th hiu in th gia hai u dy phi thay i

nh th no?

Gii:

Ta c:

( ) 242..2/.2.

2

1

2 r

US

rr

US

Slr

U

R

U

rr

IH

=====

vi: r, , S l bn knh, in tr sut v tit din ca vng dy.

Vy: Mun cng t trng H khng i khi bn knh vng dy r tng ln 2 ln th hiu

in th gia hai u dy phi tng ln 22= 4 ln.

4-16. Hai vng dy dn trn c tm trng nhau v c t sao cho trc ca chng vung

gc vi nhau. Bn knh mi vng dy bng R = 2cm. Dng in chy qua chng c

cng I1= I2= 5A. Tm cng t trng ti tm ca chng.

Gii:

Do hai vng dy c cng bn knh vng dy, cng cng dng in nn chng gy ra

ti tm O cc t trng c ln nh nhau:

( )mAR

IHH /125

10.2.2

5

2 221 ====

Do cc vng c t trng tm v vung gc vi nhau nn 1H

v 2H

c phng vung gc

vi nhau:

( )mAHHHHHHH /177125.22 12

2

2

121 ==+=+=

4-17. Hai vng dy dn ging nhau bn knh R = 10cm c t song song, trc trng nhau

v mt phng ca chng cch nhau mt on a = 20cm. Tm cm ng t ti tm ca mi

vng dy v ti im gia ca on thng ni tm ca chng trong hai trng hp:

a) Cc dng in chy trn cc vng dy bng nhau (I1= I2= 3A) v cng chiu.


64/117


b) Cc dng in chy trn cc vng dy bng nhau (I1= I2= 3A) nhng ngc chiu.

Gii:

S dng kt qu ca bi 4-13, ta c, cm ng t do vng dy gy ra ti im nm trn trc

ca vng dy bn knh R cch tm vng mt on h c ln l:

( ) 2/3222

0

2 hR

IRB

+=

a) Nu cc dng in chy trn cc dy l cng chiu, th cc vect cm ng t do cc vng

to ra cng chiu ti mi im trn trc ca cc vng dy:

21 BBB +=

+ Ti tm vng 1 (h1= 0, h2= a) v ti tm vng 2 (h1= a, h2= 0):

( ) ( ) ( )T

aR

R

R

IBB OO

5

2/322

27

2/322

2

0 10.1,22,01,0

1,0

1,0

1

2

3.10.41

221

++=

++==

+ Ti im chnh gia hai vng dy (h1= h2= a/2):

( )Ta

R

IRBM

5

2/32

2

27

2/32

2

2

0 10.35,1

4

2,01,0

1,0.3.10.4

42

.2

+

=

+

=

b) Nu cc dng in chy trn cc dy ngc chiu, th cc vect cm ng t do hai vng

to ra ngc chiu nhau ti mi im trn trc vng dy:

21 BBB =

+ Ti tm vng 1 (h1= 0, h2= a) v ti tm vng 2 (h1= a, h2= 0):

( ) ( ) ( )T

aR

R

R

IBB OO

5

2/322

27

2/322

2

0 10.7,12,01,0

1,0

1,0

1

2

3.10.41

221

+=

+==

nhng cc vect1O

B

v2O

B

ngc chiu nhau:1O

B

cng chiu vi 1B

;2O

B

cng chiu vi 2B

+ Ti im chnh gia hai vng dy (h1= h2= a/2):

0

42

42

2/32

2

2

0

2/32

2

2

0 =

+

+

=a

R

IR

aR

IRBM


65/117


4-18. Mt si dy c v cch in ng knh (k c v) bng d = 0,3mm c un thnh

mt ng xon c phng gm N = 100 vng. Bn knh ca vng ngoi cng R = 30

mm. Cho dng in I = 10mA chy qua dy. Tnh:

a)

Mmen t ca ng xon c .b) Cng t trng ti tm ca ng xon c.

Gii:

Chia ng xon c thnh nhng on dy rt nh. Xt mt on dl rt ngn trn vng dy

nm cch tm ng xon c mt on r v c ni vi tm ng xon c bng hai on

dy thng. Khi :

rdI

rdlIdHrdIdSIdpm

4.

4.;

2... 2

2

====

Mt khc:

dN

Rdkdrkr

.2.. ===

drRr

NIdHdr

R

NIrdp

m2

;2

==

Do tt c cc vect mmen t v cng t trng u cng phng cng chiu nn ta c:

( )23

2223R

2/d

2

mm m.A103

03,0.10.100.3NIR

2/d

R

3r

RNIdr

RNIrdpP

=

===

[ ]

====

2/d

Rln

R2

NI

2/d

Rrln

R2

NIdr

Rr2

NIdHH

R

2/d

( )m/A9010.5,1

03,0ln

03,0.2

10.100H

4

2

=

4-19. Mt qu cu ng cht khi lng m, bn knh R, mang mt in tch q. in tch q

c phn b u trong th tch qu cu. Ngi ta cho qu cu quay xung quanh trc

ca n vi vn tc gc . Tm mmen ng lng L, mmen t Pmca qu cu ; t

suy ra t s Pm/L?


66/117


Gii:

+ Mmen ng lng ca qu cu l:

2

5

2mRIL ==

+ Mt phn t in tch dq quay xung quanh mt trc vi tn s

s tng ng vi mt dng in cng dqI .= .p

Dng in ny c mmen t: ISdpm=

vi S l din tch ca vng trn qu o ca phn t in tch. Dng h to cu nh hnh

v:

SdVSdqdpm ...2

..

==

trong : + 33

4 R

q

= l mt in tch ca qu cu

+ dddrrdV ...cos2=

+ 222 cos' rrS ==

Do cc vect mmen t ca cc phn t khc nhau u nm trn trc quay nn mmen t

ca ton b qu cu bng:

===

2

0

2/

2/

3

R

0

4

3

22

3mm

dd.cosdrr.R8

q3dV.cosr.

R34

q.

2dpP

( ) ( )

=

2

0

2/

2/

2

R

0

4

3 dsind.sin1drr.

R8

q3

dV

O

x

y

z

r

r


67/117


[ ]0

2.

2/

2/

3

sinsin.

0

R

5

r.

R8

q3 35

3

=

5

qR2.

3

4.

5

R.

R8

q3

5

qR2.

3

4.

5

R.

R8

q3 25

3

25

3

====

Suy ra t s:m

q

mR

qR

L

Pm

22

5.

5 2

2

==

.

4-20. Mt khung dy hnh vung abcd mi cnh l = 2cm, c t gn dng in thng di

v hn AB cng I = 30A. Khung abcd v dy AB cng nm trong mt mt phng,

cnh ab song song vi dy AB v cch dy mt on r =1cm (hnh 4-15). Tnh t thng

gi qua khung dy.

Gii:

Chia khung thnh cc di nh song song vi dng in thng. Xt di cch dng in mt

on x c din tch dS = l.dx. T ta tnh c t thng do dng in gi qua khung dy:

==abcdabcd

dS.BSd.B

[ ]

+=

+==

+

r

lrln

2

Il

r

lrlln

2

Ildx.l.

x2

I 00lr

r

0

( )Wb10.32,11

21ln

2

02,0.30.10.4 77

+=

A

B

I

a b

cd l

r

Hnh 4-15


68/117


4-21. Cho mt khung dy phng din tch 16cm2quay trong mt t trng u vi vn tc

2vng/s. Trc quay nm trong mt phng ca khung v vung gc vi cc ng sc t

trng. Cng t trng bng 7,96.104A/m. Tm:

a) S ph thuc ca t thng gi qua khung dy theo thi gian.

b)

Gi tr ln nht ca t thng .

Gii:

Ta c: cos.BS=

vi l gc gia vect cm ng t v php tuyn ca khung.

Mt khc: 0 += t

Vy: ( ) ( )0000 coscos +=+= ttHS

vi tn s gc ( )sradn /42 ==

Gi tr ln nht ca t thng:

( )WbHS 444700 10.6,110.16.10.96,7.10.4 ==

( ) ( )Wbt 04 4cos10.6,1 +=

4-22. Mt thanh kim loi di l = 1m quay trong mt t trng u c cm ng t B = 0,05T.

Trc quay vung gc vi thanh, i qua mt u ca thanh v song song vi ng sc ttrng. Tm t thng qut bi thanh sau mt vng quay.

Gii:

Ta c t thng qut bi thanh sau mt vng quay l t thng gi qua din tch hnh trn

tm ti trc quay, bn knh lv vung gc vi ng sc t:

( )Wb

lBBS

16,00cos.1..05,0

cos..cos.

2

2

=

==


69/117


4-23. Cho mt dng in I = 5A chy qua mt dy dn c hnh tr, bn knh tit din

thng gc R = 2cm. Tnh cng t trng ti hai im M1v M2cch trc ca dy

ln lt l r1=1cm, r2= 5cm.

Gii:

Chn ng cong kn l ng trn c tm nm trn trc dy dn, bn knh r. p dng

nh l v lu s ca t trng (nh l Ampe):

=

=n

i

i

C

IldH1

.

Do tnh i xng nn cc vect cng t trng bng nhau ti mi im trn C v lun

tip tuyn vi C. Do : =

=n

i

iIrH1

2.

a) Gi s dng in phn b u trn thit din dy dn , th vi cc im nm trong dy dn:

2

22

2.2.

R

Irr

R

IrH ==

22 R

IrH

=

b) Vi cc im nm bn ngoi dy dn:

r

IHIrH

22. ==

+ Vi r1= 1cm:( )

( )mAH /2010.2.2

10.522

2

1 =

Vi r2= 5cm: ( )mAH /1610.5.2

522

=

4-24. Mt dng in I = 10A chy dc theo thnh ca mt ng mng hnh tr bn knh

R2=5cm, sau chy ngc li qua mt dy dn c, bn knh R1= 1mm, t trng

vi trc ca ng. Tm:a) Cm ng t ti cc im cch trc ca ng r1= 6cm v r2= 2cm;

b) T thng gy ra bi mt n v chiu di ca h thng. Coi ton b h thng l di v

hn v b qua t trng bn trong kim loi.


70/117


Gii:

L lun tng t bi 4-23, ta c: =

=n

i

iIr

B1

0

2

Vi r > R2 : ( ) 0.20 == IIr

B

Vi R2> r > R1: ( )TIr

B 42

7

0 1010.2.2

10.10.4

2

===

Do t trng bn ngoi dy dn v trong kim loi bng khng nn t thng do mi n v

chiu di h thng gy ra l:

===

1

20

R

R

0

R

Rln.

2

Ildr.l.

r2

IdS.B

2

1

( )Wb10.8,71

50ln

2

1.10.10.4 67

=

4-25. Cho mt ng dy in thng di 30cm, gm 1000 vng dy. Tm cng t trng

bn trong ng dy nu cng dng in chy qua ng dy bng 2A. Coi ng knh

ca ng dy rt nh so vi chiu di ca ng.

Gii:

Ta c th coi ng dy l di v hn, nn t trng bn trong ng dy l u v c tnh

theo cng thc:

( )mAl

NInIH /10.7,6

3,0

2.1000 3===

4-26. Dy dn ca ng dy tit din thng c ng knh bng 0,8mm, cc vng dy c

qun st nhau, coi ng dy kh di. Tm cng t trng bn trong ng dy nu

cng dng in chy qua ng dy bng 1A.


71/117


Gii:

Do cc vng dy c qun st nhau, nn chiu di ng dy c th tnh bng:

Ndl=

Cng t trng bn trong ng dy l:

( )mAd

I

Nd

NI

l

NInIH /1250

10.8

14======

4-27. Hi t s gia chiu di l v ng knh D ca mt ng dy in thng phi bng bao

nhiu c th tnh cng t trng ti tm ca ng dy theo cng thc ca ng dy

di v hn m khng sai qu 1%.

Gii:

Ta i tnh cm ng t ti im O trn trc ca ng dy. Vng dy cch O mt on x gy

ra ti O cm ng t:

( ) 2/3220

2 xR

ISB

+=

Cm ng t do ndx vng dy cch O mt on x gy ra bng:

( ) dx

xR

ISnB

2/322

0

2 +=

vi n l s vng dy trn mt n v chiu di ca ng dy. T , cm ng t tng hp do c

ng dy gy ra ti O l:

( )2

1

222

0

d

d

2/322

0

d

d

xRR2

ISnxdx

xR2

ISnB

1

2

+=

+=

d2 d1

R

x

O 12


72/117


( )2

1

222

0

d

d

2/322

0

d

d

xRR2

ISnxdx

xR2

ISnB

1

2

+=

+=

++

+=

22

2

2

21

2

1

2

0

dR

d

dR

d

R2

ISnB

Do 222

2

212

1

2

12 cos;cos; =+

=+

=dR

d

dR

dRS , ta c:

)cos(cos2

1210 += nIB

Ti tm ng dy (d1= d2= l/2):

+=

2

2

0220 21/1

1

l

D

nIlDnIB do 12

2

l

D

+ Vi ng dy di v hn, cm ng t bn trong ng dy l:

nIB 0=

Sai s mc phi l:2

2

2l

D

B

BBB =

=

B khng vt qu 1% th: 1,7

02,0

1%1

2

2

2

D

l

l

D

Vy, chiu di ca ng dy cn ln hn ng knh t nht 7,1 ln.

4-28. Mt dy dn thng di 70cm c t trong mt t trng u c cm ng t B =

0,1T. Dy dn hp vi ng sc t trng mt gc = 300. Tm t lc tc dng ln dy

dn khi cho dng in I = 70A chy qua dy dn.

Gii:Theo cng thc ca lc t:

( )NBIlFBlIF 45,230sin.7,0.70.1,0sin. 0 ====


73/117


4-29. Trong mt t trng u cm ng t B = 0,1T v trong mt phng vung gc vi cc

ng sc t, ngi ta t mt dy dn un thnh na vng trn. Dy dn di s = 63cm,

c dng in I = 20A chy qua. Tm lc tc dng ca t trng ln dy dn.

Gii:

Chia dy dn thnh cc on dy c chiu di dl rt nh. Ta c:

BldIFd

= .

Lc tng hp tc dng ln dy dn l:

( ) BlIBldIBldIFdF

==== ..

vi l

l vct ni t im u n im cui ca dy dn.

Lc tng hp khng ph thuc vo hnh dng ca dy dn m ch ph thuc vo ng ni

im u v im cui dy, y l ng knh ca na ng trn:

( )Ns

BIBIlF 8,090sin.63,0

.2.20.1,0sin2

sin 0 ===

4-30. Mt ng dy thng trn c dng in I = 10mA, c t trong mt t trng u sao

cho trc ca ng dy trng vi phng ca ng sc t trng. Cc vng dy c

qun bng dy ng c ng knh d = 0,1mm. Bn knh ca mi vng dy R = 10mm.

Hi vi gi tr no ca cm ng t B ca t trng ngoi, vng dy s b ko t? Chobit ng sut ca dy ng khi b t P= 2,3.10

8 N/m2.

Gii:

Theo kt qu bi 4-29, lc tc dng ln na vng dy l:

BIRBIs

F 22

==

(vi R l bn knh vng dy)

Lc ny phn b trn hai tit din thng ca dy dn. Gi FPv BP ln lt l lc ko v cmng t khi dy ng b t. Ta c:

TF 2= (do hai phn tit din thng ca dy dn song song)


74/117


( )( )T

IR

dB

dIRB P

PP

P

4

33

23822

10.8,110.10.10.10.4

10.1,0..10.3,2

4422 ===

4-31. Cho mt ng dy in thng di, c trn mi mt c n = 5000 vng dy. Ti tm ng

dy, ngi ta t mt cun dy nh gm N = 200 vng. Cc vng dy ca cun nh cng knh d = 10mm. Cun dy c gn u mt n cn sao cho trc ca n

vung gc vi trc ca ng dy (hnh 4-16). Lc u cun dy c cn bng bi mt

s qu nng (n cn nm trng vi trc ca ng dy). Khi cho qua ng dy v cun

dy cng mt dng in I = 20mA th cn bng b ph hu. Hi phi t thm qu nng

c trng lng bng bao nhiu cn bng c thit lp li? Bit rng cnh tay n

ca cn c chiu di l = 300mm.

Gii:

Khi cho dng in chy qua cun dy, cun dy tng ng vi mt nam chm c m

men t: 24

1NIdNISpm ==

Khi , cun dy s chu tc dng ca t trng do ng dy gy ra. Mmen lc tc dngln cun dy:

( ) 22002

4

1.

4

1dNnInINIdBpM m =

==

cn bng c thit lp li, cn t thm qu nng c trng lng P sao cho

22

04

1dNnIMPlMP ===

( )Nl

dNnIP 73

227220 10.3,1

10.300.4

01,0.02,0.5000.200.10.4.1.

4

==

AO

l

B

Hnh 4-16


75/117


4-32. t mt a bng ng bn knh R = 5cm trong mt t trng u c cm ng t

B=0,2T sao cho mt phng ca a vung gc vi ng sc t trng. Cho mt dng

in I= 5A chy dc theo bn knh ab ca a (hnh 4-17). Hi:

a) Chiu quay ca a nu chiu ca t trng i t pha sau ra pha trc mt phng

hnh v;

b) Mmen lc tc dng ln a.

Gii:

Phn a nm dc theo bn knh ab c dng in chy qua s b t trng tc dng lc v

lm cho a quay. p dng quy tc bn tay tri, ta tm c chiu quay ca a hng theo

chiu kim ng h.

Xt mmen lc t tc dng ln mt on dl dc theo ab:

dlBIlrdFdM .. ==

( ) ( )mNBIRdlBIldMMR

.10.5,12210.5.5.2,0

2.

4

222

0

=====

4-33. Hai cun dy nh ging nhau c t sao cho trc ca chng nm trn cng mt

ng thng. Khong cch gia hai cun dy l = 200mm rt ln so vi kch thc di

ca cc cun dy. S vng trn mi cun dy u bng N = 200 vng, bn knh cc

vng dy R=

Documents

Giai Bai Tap Dien