Chng VII

VT DN

Vt dn l vt c cha cc ht mang in t do; cc ht mang in ny c th chuyn ng trong ton b vt dn. C nhiu loi vt dn ( rn, lng, kh); trong chng ny ta ch nghin cu vt dn kim loi. Trong vt dn kim loi cc ht mang in t do l cc electron.

I.Vt dn cn bng tnh in

1. nh ngha: Vt dn cn bng tnh in l vt dn m trong cc in tch nm cn bng (ngha l khng chuyn ng c hng to thnh dng in)

2. iu kin cn bng tnh in

a) Vect CT bn trong vt dn bng khng.b) Ti mi im trn b mt vt dn, vect CT phi vung gc vi b mt vt dn.

Cng in trng trn b mt vt dn

l vecto n v php tuyn trn b mt VD l mt in tch mt

0

E n

n

3. Tnh cht

a) Vt dn l mt khi ng th.

b) in tch ch phn b trn b mt vt dn; bn trong vt dn, in tch bng khng (cc in tch dng v m trung ha ln nhau)

c) S phn b in tch trn mt vt dn ch ph thuc vo hnh dng ca mt , nhng ch li in tch tp trung nhiu.

II. Hin tng in hng

Khi t mt vt dn trung ho in vo trong in trng ngoi th di tc dng ca lc in trng cc electron trong vt dn s chuyn di c hng ngc chiu in trng. Kt qu l trn cc mt gii hn ca vt dn xut hin cc in tch tri du, mt c ng sc in trng ngoi i vo mang in dng, mt i din mang in m v chng c ln bng nhau. Cc in tch ny gi l in tch cm ng v hin tng ny gi l hin tng in hng.

in hng mt phn: Khi t vt dn cha mang in (B) gn qu cu mang in (A) . Gi Q v Q ln lt l in tch tng cng trn vt mang in A v ln ca cm ng xut hin trn vt dn (BC)

Q < Q

in hng ton phn: Khi vt dn (B) bao bc hon ton vt mang in (A)

Q = QB

Q > 0 --

-++

+

A B

Q > 0A

Q > 0

++

-

+

- -+

+

-

-

+

+

+

-

+

++

+-

--

III. in dung ca vt dn c lpVt dn c lp: mt vt dn c gi l c lp vin nu gn n khng c mt vt no khc cth gy nh hng n s phn b in tch can.in dung ca vt dn c lp:

V v Q l in th v in tch ca vt dn C l in dung ca vt dn, n ph thuc vohnh dng kch thc v tnh cht ca mitrng cch in bao quanh n.

VQC

IV. T in1. nh ngha: T in l mt h hai vt dn tgn nhau hoc bao bc hon ton nhau sao chogia chng xy ra hin tng in hng tonphn. Hai vt dn gi l hai bn t2. in dung ca t inGi V1 l in th ca bn tch in dng Q, V2 l in th ca bn tch in m Qin dung ca t in :

U = V1 V2 l hiu in th gia bn tch indng v m

UQ

VVQC

21

V.Tnh in dung ca t in

1. T in phng: T in phng gm hai tm kim loi phng c hnh dng v kch thc nh nhau, t song song i din nhau v cch nhau mt khong rt nh so vi kch thc ca chng.

S

-Q

V1

d

V2

V khong cch gia hai bn rt nh so vi kch thc ca mi bn nn in trng trong khong khng gian gia hai bn c th coi l in trng do hai mt phng song song v hn, tch in u bng nhau nhng tri du gy ra.

Hiu in th gia hai bn t :

in dung ca t in phng:

S

-Q

V1

d

V2

+ + + + + + + + + + + +

- - - - - - - - - - - - - - -

0 0 02 2

E E EE E E

SQdEdVVU 0

21

dS

UQC 0

2.T in cu: Trong t in cu, hai bn t l hai mt cu ng tm bn knh R1 v R2 .

Theo trn hiu in th gia hai bn t bng:

Do in dung ca t in cu l:

21

12

2121

)(11RR

RRkQRR

kQVV

)( 1221

21 RRkRR

VVQC

V. Nng lng in trng

1. Nng lng tng tc ca mt h in tch im

Trng hp h in tch im gm hai in tch im q1 v q2,cch nhau mt khong r, th nng ca in tch q2 t trong in trng ca in tch q1 l:

l in th do q2 gy ra ti v tr ca q1

l in th do q1 gy ra ti v tr ca q2

1 2 2 11 2

1 1 2 2

1 12 2

12

kq q kq kqW q qr r r

W q V q V

rkqV

21

11

kqVr

R rng W cng l th nng ca q1 trong in trng ca in tch q2. Nn W l th nng tng tc hay nng lng tng tc ca h hai in tch q1 v q2.Tng qut, nng lng tng tc in (gi tt l nng lng in) ca h n in tch im q1,q2,qn l:

Vi l in th ti v tr ca qi do cc in tch cn li gy ra.

n

iiiVqW

121

Trng hp in tch phn b lin tc

V l in th ca phn t mang in tch dq

1 .2

W dqV

2. Nng lng in ca vt dn c lp tch in

Chia vt dn thnh cc phn t VCB mang in tch dq, nng lng in ca vt dn:

3. Nng lng t in

Hay:

qVdqVVdqW21

21

21

)(21)(

21

2121 VVqVqVqW

22

21

21

21 CU

CqqUW

4.Nng lng in trng

Mt nng lng in trng l nng lng in trng trong mt n v th tch ca khng gian c in trng.

Xt t in phng, theo trn ta c:

Kt qu ny cng ng cho in trng bt k.

Do nng lng in trng nh x trong mt th tch l:

20

20

1 1 12 2 2

1 1 .2 2e

W qU SEd E

W E E D

dwW e

V d:

1) Cho hai mt cu kim loi ng tm bn knh R1 = 4cm, R2 = 2cm mang in tch Q1 = -(2/3).10-9C, Q2 = 3.10-9C. Tnh CT v in th ti nhng im cch tm mt cu nhng khong cch 1cm, 2cm, 3cm, 4cm.

Ta c:a) Ti r = 1cm

1 2 1 2;E E E V V V

1 2

1 2

1 2

9 99 9

2 2

0, 0 0

29.10 ( .10 ) 9.10 (3.10 )34.10 2.10

1200

E E EkQ kQVR R

V

b) r = 2cm2

1 2 2 2

9 94

4

4

1 21

1 2

9 99 9

2 2

0;(2.10 )

9.10 .3.10 67,5.10 /4.10

67,5.10 /

29.10 ( .10 ) 9.10 (3.10 )3 12004.10 2.10

kQE E

V m

E V mkQ kQV VR R

V

c) r = 3cm

19 9

422 2 2 4

4

1 2

1

9 99 9

2 2

0;

9.10 .3.10 3.10 /(3.10 ) 9.10

3.10 /

29.10 .( .10 ) 9.10 .(3.10 )3 7504.10 3.10

EkQE V m

E V mkQ kQVR r

V

d) r = 4cm

V v ngc chiu nhau nn cng chiu vi (hng xa tm) v c ln

E = E2 E1 = 4375V/m

mVr

kQE

mVR

QkE

/10625)10.4(

310.10.9

/3750)10.4(

)10.32.(10.9

22

99

22

2

22

99

21

11

1E 2E

9 99 9

1 22 2

1

29.10 ( .10 ) 9.10 (3.10 )3 5254.10 4.10

kQ kQV VR r

E

2E

2) Cho ba in tch imq1 = -4.10-8C ; q2 = 5.10-8C; q3 = 3.10-8C ln lt t ti 3 nh A, B, C ca mt hnh ch nht ABCD cnh AB = 3cm; BC = 4cm. Tnh nnglng tng tc ca h 3 in tchny.

1 2 3

9 8 9 832

2 2

9 8 9 831

2 2

9 8 9 81 2

2 2

1 ( )2

9.10 .5.10 9.10 .3.103.10 5.10

9.10 .( 4).10 9.10 .3.103.10 4.10

9.10 .( 4).10 9.10 .5.105.10 4.10

A B C

A

B

C

W q V q V q V

kqkqVAB AC

kqkqVAB BCkq kqVAC BC

W

3)Hai on dy thng, mnh ging nhau, mi on c chiu di 2L c tch in u vi mt in di . Ngi ta t hai on dy cng nm trn mt ng thng, khong cch gia hai u gn nht bng a. Tm th nng tng tc ca hai on dy ny trong chn khng.

2L x

a

in th do on dy 1 gy ra ti im M c ta x:

Th nng tng tc gia on dy th nht v

phnt mang in tch ca on dy th hai:

2ln L xV kx

dx

2

2 22 2

2

2( ) ln

ln(2 ) ln

4 ln( 4 ) 2( 2 )ln( 2 ) ln

a L a L

a a

L xdW dxV k dxx

W k L x dx k xdx

k a L a L a L a L a a

4) Hai qu cu kim loi bn knhln lt l R1 v R2 vi R2 = 2R1t rt xa nhau, c ni vi nhaubng si dy dn mnh c tchin tng cng l Q = 9.10-8C. Tnhin tch trn mi qu cu.

Sau khi ni in th hai qu cu bng nhau nn:

' ' ' '1 2 1 2

1 2 1 2' '1 2 1 2

1 2 1 2 1 2

' 811

1 2

' 822

1 2

( ) ( )

6.10

3.10

kQ kQ kQ kQR R R R

k Q Q k Q Q kQR R R R R R

R QQ CR R

R QQ CR R

5)Hai qu cu kim loi tch in, tip xc vi nhau v cn bng in. Bn knh v in tch ca chng ln lt l R1; q1 v R2v q2. Nu R2 = 2R1, so snh in tch ca hai qu cu.

in th ca hai qu cu:

M: V1 = V2 nn:

1 2 2 11 2

1 1 2 2 1 2;kq kq kq kqV V

R R R R R R

1 2 2 1

1 1 2 2 1 2

1 2 2 1

1 1 1 1

2 1

3 2 34

kq kq kq kqR R R R R Rkq kq kq kqR R R R

q q

6) Mt qu cu bn knh R mang in tch Q phn b u vi mt in tch khi . Tnh:a) Nng lng in trng bn trong qu cu

b) Nng lng in trng bn ngoi qu cu

c) Nng lng in trng trong ton khng gian

p dng cng thc:

a) Bn trong qu cu:

2 20

1 ; 42

W E d d r dr

2 2 2 52

0 20 00

33

2

1 442 9 904 33 4

110

R

in

in

r RW r dr

QQ RR

QW kR

03rE

b) Bn ngoi qu cu

c)

2

QE kr

2 2 22

0 4

1 142 2ext R

k Q QW r dr kr R

235in ext

QW W W kR