BAI_TAP_TU_LUAN_DIEN_TICH_-_DIEN_TRUONG

8/6/2019 BAI_TAP_TU_LUAN_DIEN_TICH_-_DIEN_TRUONG

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NH LUT CULOMBA. PHNG PHP GII BI TP1. Hai loi in tch:

- in tch dng v in tch m- in tch dng nh nht l ca proton, in tch m nh nht l in tch c

electron

Gi tr tuyt i ca chng l e = 1,6.10-19

C2. Tng tc gia hai in tch im ng yn.

- im t: Ti in tch ang xt.- Gi: L ng thng ni hai in tch.- Chiu: l lc ynu hai in tch cng du, lc ht nu hai in tch tri du.- ln:

1 2

2

q qF k

r=

Trong k = 9.109 ( )2 2Nm / c . : l hng s in mi.3. nh lut bo ton in tch:

Trong mt h c lp v in, tng i s cc in tch l mt hng s4. Khi in tch chu tc dng ca nhiu lc:

Hp lc tc dng ln in tch L:

1 2F F F ...= + +Xt trng hp ch c hai lc:

1 2F F F= +a. Kh 1F cng hng vi 2F :

F cng hng vi 1F , 2FF = F1 + F2

b. Khi 1F ngc hng vi 2F :

1 2F F F=

Fcng hng vi1 1 2

2 1 2

F khi : F F

F khi : F F

>


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d. Khi F1 = F2 v 1 2F ,F = r

1F 2F cos 2

=

F hp vi 1F mt gc2

B. BI TP:I. BI TP V D:Bi 1: Hai in tch im cch nhau mt khong r =3cm trong chn khng ht nhau bng mlc F = 6.10-9N. in tch tng cng ca hai in tch im l Q=10-9C. Tnh in ch cmi in tch im:

Hng dn gii:p dng nh lut Culong:

1 2

2

q qF k

r=

( )2

18 2

1 2

Frq q 6.10 C

k

= = (1)

Theo :9

1 2q q 10 C+ = (2)

Gi h (1) v (2)9

1

9

2

q 3.10 C

q 2.10 C

=

= Bi 2: Hai qu cu ging nhau mang in, cng t trong chn khng, v cch nhau khonr=1m th chng ht nhau mt lc F1=7,2N. Sau cho hai qu cu tip xc vi nhau v

a tr li v tr c th chng y nhau mt lc F2=0,9N. tnh in tch mi qu cu trc vsau khi tip xc.Hng dn gii:

Trc khi tip xc

( )2

10 2

1 2

Frq q 8.10 C

k = = (1)

in tch hai qu cu sau khi tip xc:

, , 1 21 2

q qq q

2

+= =

2

1 2

5

2 1 22

q q2F k q q 2.10 Cr

+ = + =

(2)

T h (1) v (2) suy ra:5

1

5

2

q 4.10 C

q 2.10 C

=

= m


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Bi 3: Cho hai in tch bng +q (q>0) v hai in tch bng q t ti bn nh ca mt hnvung ABCD cnh a trong chn khng, nh hnh v. Xc nh lc in tng hp tc dng lmi in tch ni trn

Hng dn gii:

A B

FBDFCD

D FD CFAD F1

Cc l tc dng ln +q D nh hnh v, ta c2

1 2

AD CD 2 2

q q qF F k k

r a

= = =

( )

2 21 2

BD 22 2

q q q qF k k k

r 2aa 2= = =

D AD CD BD 1 BDF F F F F F= + + = +2

1 AD 2

qF F 2 k 2

a= =

1F hp vi CD mt gc 450.2

2 2

D 1 BD 2

qF F F 3k

2a= + =

y cng l ln lc tc dng ln cc in tch khc

Bi 4: Cho hai in tch q1= 4 C , q2=9 C t ti hai im A v B trong chn khng AB=1mXc nh v tr ca im M t ti M mt in tch q 0, lc in tng hp tc dng ln q

bng 0, chng t rng v tr ca M khng ph thuc gi tr ca q0.Hng dn gii:

q1 q0 q2

A BF20 F10

Gi s q0 > 0. Hp lc tc dng ln q0:

10 20F F 0+ =Do :

1 0 1 0

10 20 2

q q q qF F k k AM 0,4m

AM AB AM= = =

Theo php tnh ton trn ta thy AM khng ph thuc vo q0.

0

lT

HF

q rP Q

Bi 5: Ngi ta treo hai qu cu nh c khlng bng nhau m = 0,01g bng nhng sdy c chiu di bng nhau (khi lnkhng ng k). Khi hai qu cu nhim i

bng nhau v ln v cng du chng nhau v cch nhau mt khong R=6cm. Lg= 9,8m/s2. Tnh in tch mi qu cu

Hng dn gii:Ta c:

P F T 0+ + =


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T hnh v:

2

2

2 39

2

R R R Ftan

2.OH 2 mgR2 l

2

q Rmg R mgk q 1,533.10 C

R 2l 2kl

= = =

= = =

Bi 6: Hai in tch q1, q2 t cch nhau mt khon r=10cm th tng tc vi nhau bng lc

trong khng kh v bngF

4nu t trong du. lc tng tc vn l F th hai in tch ph

t cch nhau bao nhiu trong du?Hng dn gii:

,1 2 1 2

2 ,2

q q q q rF k k r 5cm

r r= = = =

Bi 7: Cho hai in tch im q1=16 C v q2 = -64 C ln lt t ti hai im A v B tronchn khng cch nhau AB = 100cm. Xc nh lc in tng hp tc dng ln in tch imq0=4 C t ti:

a. im M: AM = 60cm, BM = 40cm.b. im N: AV = 60cm, BN = 80cm

Hng dn gii:

A M 10F 20F F

q1 q0 q2

a. V MA + MB = AB vy 3 im M, A, thng hng M nm gia ABLc in tng hp tc dng ln q0:

10 20F F F= +

V 10F cng hng vi 20F nn:

1 0 2 010 20 2 2

q q q qF F F k k 16N

AM BM= + = + =

F cng hng vi 10F v 20F

10Fq

N F

20Fq1 q2A B

b. V 2 2 2NA NB AB NAB+ = vung tN. Hp lc tc dng ln q0 l:

10 20F F F= +2 2

10 20F F F 3,94V= + =

F hp vi NB mt gc :

tan010

20

F0,44 24

F = = =

Bi 8: Mt qu cu nh c khi lng m = 1,6g, tch in q = 2.10-7C c treo bng mt sdy t mnh.


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pha di n cn phi t mt in tch q 2 nh th no lc cng dy gim i mna.

Hng dn gii:

T

P

Lc cng ca si dy khi cha t in tch:T = P = mg

Lc cng ca si dy khi t in tch:

T = P F = P2

271 2

2

1

q qP mg mgr F k q 4.10 C

2 r 2 2kq = = = =

Vy q2 > 0 v c ln q2 = 4.10-7C

Bi 9: Hai qu cu kim loi nh hon ton ging nhau mang in tch q 1 = 1,3.10-9C vq2=6.5.10-9C, t trong khng kh cch nhau mt kh ong r th y nhau vi lc F. Chi hqu cu tip xc nhau, ri t chung trong mt lp in mi lng, cng cch nhau m

khong r th lc y gia chng cng bn Fa. Xc inh hng s in mi

b. Bit lc tc ng F = 4,6.10-6N. Tnh r.Hng dn gii:

a. Khi cho hai qu cu tip xc nhau th:

, , 1 21 2

q qq q

2

+= =

Ta c: 21 2

, 1 2

2 2

q qq .q2F F k k 1,8

r r

+ = = =

b. Khong cch r:

1 2 1 2

2

q q q qF k r k 0,13m

r F= = =

Bi 10: Hai qu cu kim loi ging nhau, mang in tch q 1, q2 t cch nhau 20cm th h

nhau bi mt lc F 1 = 5.10-7N. Ni hai qu cu bng mt dy dn, xong b dy dn i th hqu cu y nhau vi mt lc F2 = 4.10-7 N. Tnh q1, q2.

Hng dn gii:

Khi cho hai qu cu tip xc nhau th: , , 1 21 2q q

q q2

+= =

p dng nh lut Culong:2

161 2 11 1 22

q .q Fr 0,2F k q .q .10

r k 9= = =


6/22

( )2

81 221 2

1 1 2

q qF 4q q .10 C

F 4 q q 15+= + =

Vy q1, q2 l nghim ca phng trnh:8

2 19

8

10C

4 0,2 3q q .10 0 q15 9 1

10 C15

= =

Bi 11: Hai qu cu nh ging nhau, cng khi lng m = 0,2kg, c treo ti cng mim bng hai si t mnh di l = 0,5m. Khi mi qu cu tch in q nh nhau, chng tcnhau ra mt khong a = 5cm. Xc inh q.

Hng dn gii:

0

lT

HF

q rP Q

Qu cu chu tc dng ca ba lc nh hnv. iu kin cn bng:

P F T 0+ + =Ta c:

22

aF 2tanP a

l4

= =

2

2

22

q ak

a 2mg a

l 4

=

9

2 2

amgq a. 5,3.10 C

k 4l a

= =

Bi 12: Hai in tch im bng nhau t trong chn khng, cch nhau khong r = 4cm. Ly tnh in gia chng l F = -10-5N

a. Tnh ln mi in tch.b. Tm khong cch r1 gia chng lc y tnh in l F1 = 2,5.10-6N.

Hng dn gii:a. ln mi in tch:

2 291 1

1 2

1

q FrF k q 1,3.10 C

r k= = =

Khong cch r1:2 2

2

2 22

2 2

q qF k r k 8.10 m

r F= = =


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Bi 13:

A

O

2F 3F

B C

1F

F

Ngi ta t ba in tch q1 = 8.10-9Cq2=q3=-8.10-C ti ba nh ca mt tam giu ABC cnh a = = 6cm trong khng khXc nh lc tc dng ln in tch q0=610-9

t ti tm O ca tam gic.Hng dn gii:Lc tng hp tc dng ln q0:

1 2 3 1 23F F F F F F= + + = +

51 0 1 021 2

q .q q .qF k 3k 36.10 N

a2 3a

3 2

= = =

52 0 1 02 3 2 2q q q .qF F k 3k 36.10 Na2 3

a3 2

= = = =

0

23 2 2F 2F cos120 F= =

Vy F = 2F1 = 72.10-5N

Aq1

O q0 03F

B C 23F

q2 1F q3

13F

Bi 14: Ti ba nh ca mt tam gic ungi ta t ba in tch ging nhaq1=q2=q3=6.10-7C. Hi phi t in tch tht q0 ti u, c gi tr bao nhiu h thnng yn cn bng.

Hng dn gii:iu kin cn bng ca in tch q3 t ti C

13 23 03 3 03F F F F F 0+ + = + =2

0

13 23 3 13 132

qF F k F 2F cos30 F 3

a= = = =

3F c phng l phn gic ca gc C

Suy ra 03F cng gi ngc chiu vi 3F .

Xt tng t vi q1, q2 suy ra q0 phi nm ti tm ca tam gic.2

70

03 3 02 2

q q qF F k k 3 q 3,46.10 C

a2 3a

3 2

= = =


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9/22

a. Tnh cng in trng tai im M nm trn ng trung trc ca AB cch 20cm

b. Tm v tr ti cng in trng bng khng . Hi phi t mt in tch q0u n nm cn bng?

S: Cch q2 40 cmBi 12: Hai bi trong khng kh cch nhau mt on R = 3cm mi ht mang in t ch = -9,6.10-13C.

a. Tnh lc tnh in gia hai in tch.b. Tnh s electron d trong mi ht bi, bit in tch ca electron l e = -16.10-19C.

S: a. 9,216.1012N. b. 6.10 6

Bi 13: Electron quay quanh ht nhn nguyn t Hiro theo qu o trn bn knR= 5.1011m.

a. Tnh ln lc hng tm t ln electron.b. Tn vn tc v tn s chuyn ng ca electron

S: a. F = 9.10-8N. b. v = 2,2.10 6m/s, f = 0,7.1016Hz

Bi 14: Hai vt nh mang in tch t trong khng kh cch nhau mt on R = 1m, nhau bng lc F = 1,8N. in tch tng cng ca hai vt l Q = 3.10 -5C. Tnh in tch mvt.

S: q1 = 2.10-5C, q2 = 10-5C hc ngc li


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IN TRNGA. PHNG PHP GII BI TP:1. Khi nim in trng:

in trng l dng vt cht:- Tn ti xung quanh in tch- Tc dng lc in ln in tch khc t vo trong n

2. Cng in trng:Vct cng in trng l i lng c trng cho in trng v mt tc dn

lc:

FE

q=

ur

3. Cng in trng ca mt in tch im Q- im t: Ti im ang xt.- Gi: L ng thng ni in tch im v im ang xt.

- Chiu: Hng vo Q nu Q < 0; hng xa Q nu Q >0- ln:

2

QE k

r=

4. Lc tc dng ln mt in tch t trong in trng:

F q.E=

q > 0 : F cng hng vi E

q < 0 : F ngc hng vi E5. Cng in trng do nhiu in tch im gy ra

1 2E E E ...= + +Xt trng hp ch c hai in trng

1 2E E E= +a. Kh 1E cng hng vi 2E :

E cng hng vi 1E , 2EE = E1 + E2

b. Khi 1E ngc hng vi 2E :

1 2E E E=

E cng hng vi1 1 2

2 1 2

E khi : E E

E khi : E E

>


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2 2

1 2E E E= +

E hp vi 1E mt gc xc nh bi:

2

1

Etan

E =

d. Khi E1 = E2 v 1 2E ,E = ur

1E 2E cos 2 =

E hp vi 1E mt gc2

S n

E

5. nh l Ostrograrski-Gouss:a. in thng:

N E.S.cos= b. nh l O-G:

1

0

1N q=

iq l tng cc in tch bn trong mt kn B. BI TP:I. BI TP V D:Bi 1: Cho hai im A v B cng nm trn mt ng sc ca in trng do mt in tcim q > 0 gy ra. Bit ln ca cng in trng ti A l 36V/m, ti B l 9V/m.

a. Xc nh cng in trng ti trung im M ca AB.b. Nu t ti M mt in tch im q0 = -10-2C th lnn lc in tc dng ln q0

bao nhiu? Xc nh phng chiu ca lc.

q A M B

EM

Hng dn gii:Ta c:

A 2

qE k 36V / m

OA= = (1)

B 2

q

E k 9V / mOB= = (2)

M 2

qE k

OM= (3)

Ly (1) chia (2)2

OB4 OB 2OA

OA = =

.

Ly (3) chia (1)2

M

A

E OA

E OM

=


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Vi:OA OB

OM 1,5OA2

+= =

2

MM

A

E OA 1E 16V

E OM 2,25

= = =

b. Lc t tc dng ln qo: M0F q E=

v q0 0) t ti hai im A v B vi AB = 2a. M l mt im nmtrn ng trung trc ca AB cch AB mt on x.

a. Xc nh vect cng in trng ti Mb. Xc nh x cng in trng ti M cc i, tnh gi tr

Hng dn gii:

E1

M E

E2x

a aA B

q H -q

a. Cng in trng ti M:

1 2E E E= +ta c:

1 2 2 2

qE E k

a x= =

+Hnh bnh hnh xc nh E l hnh thoi:

E = 2E1cos( )

3/2

2kqa

a x =

+ (1)

b. T (1) Thy Emax th x = 0:

Emax = 1 2 22kq

Ea x

=+

E

TF

P R

Bi 3: Mt qu cu nh khi lng m=0,1mang in tch q = 10-8C c treo bng sdy khng gin v t vo in trng u Ec ng sc nm ngang. Khi qu cu c

bng, dy treo hp vi phng thng n

mt gc0

45 = . Ly g = 10m/s2

. Tnh:a. ln ca cng in trng.b. Tnh lc cng dy .

Hng dn gii:aTa c:

5qE mg.tantan E 10 V / mmg q

= = =


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b. lc cng dy:

2mgT R 2.10 Ncos

= = =

Bi 4: Mt in tch im q1 = 8.10-8C t ti im O Trong chn khng.a. Xc nh cng in trng ti im M cch O mt on 30cm.

b. Nu t in tch q2 = - q1 ti M th n ps chu lc tc dng nh th no?

Hng dn gii:a. Cng in trng ti M:

M 2

qE k 8000V

r= =

b. Lc in tc dng ln q2:3

2F q E 0,64.10 N= =

V q2 0 t ti A v B trong khng kh. cho bit AB = 2a

E

2E 1E

M h

q1 a a q2A H B

a. Xc nh cng in trng ti im Mtrn ng trung trc ca AB cch Ab m

on h.b. nh h EM cc i. Tnh gi tr cc ny.

Hng dn gii:a. Cng in trng ti M:

1 2E E E= +

ta c:


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1 2 2 2

qE E k

a x= =

+Hnh bnh hnh xc nh E l hnh thoi:

E = 2E1cos ( )3/2

2 2

2kqh

a h =

+

b. nh h EM t cc i:

( ) ( )

2 2 4 22 2 2 3

3 3/22 2 4 2 2 2 2

a a a .ha h h 3.

2 2 4

27 3 3a h a h a h a h

4 2

+ = + +

+ +

Do : M 22

2kqh 4kqE

3 3 3 3aa h

2

=

EM t cc i khi: ( )2

2

M max 2

a a 4kqh h E2 2 3 3a

= = =

Aq1 q2 B

2E 3E q3 D

C13E 1E

Bi 7: Bn im A, B, C, D trong khng khto thnh hnh chc nht ABCD cnh AD a = 3cm, AB = b = 4cm. Cc in tch q1, qq3 c t ln lt ti A, B, C. Bit q2=12,5.10-8C v cng in trng tng hti D bng 0. Tnh q1, q2.

Hng dn gii:Vect cng in trng ti D:

D 1 3 2 13 2E E E E E E= + + = +V q2 < 0 nn q1, q3 phi l in tch dng. Ta c:

1 2

1 13 2 2 2

q q ADE E cos E cos k k .

AD BD BD= = =

( )

2 3

1 2 2322 2

AD ADq . q q

BDAD AB

= =

+ ( )

38

1 22 2

aq .q 2,7.10

a h

= =

+

C

Tng t:

( )

38

3 13 2 3 232 2

bE E sin E sin q q 6,4.10 C

a b

= = = =+


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II. BI TP NGHBi 1: Mt in tch t ti im c cng in trng 0,16 (V/m). Lc tc dng ln itch bng 2.10-4 (N). Tnh ln ca in tch

S: q = 8 (C).Bi 2: Cng in trng gy ra bi in tch Q = 5.10-9 (C), Tnh cng in trnti mt im trong chn khng cch in tch mt khong 10 (cm) .

S: E = 4500 (V/m).Bi 3: Ba in tch q ging ht nhau c t c nh ti ba nh ca mt tam gic u ccnh a. Tnh ln cng in trng ti tm ca tam gic

S: E = 0.Bi 4: Hai in tch q1 = 5.10-9 (C), q2 = - 5.10-9 (C) t ti hai im cch nhau 10 (cm) tronchn khng. Tnh ln cng in trng ti im nm trn ng thng i qua hai itch v cch u hai in tch .

S: E = 36000 (V/m).Bi 5: Hai in tch q1 = q2 = 5.10-16 (C), t ti hai nh B v C ca mt tam gic u ABC

cnh bng 8 (cm) trong khng kh. Tnh cng in trng ti nh A ca tam gic ABCS: E = 1,2178.10-3 (V/m).

Bi 6: Hai in tch q1 = 5.10-9 (C), q2 = - 5.10-9 (C) t ti hai im cch nhau 10 (cm) tronchn khng. Tnh ln cng in trng ti im nm trn ng thng i qua hai itch v cch q1 5 (cm), cch q2 15 (cm).

S: E = 16000 (V/m).Bi 7: Hai in tch q1 = 5.10-16 (C), q2 = - 5.10-16 (C), t ti hai nh B v C ca mt tamgic u ABC cnh bng 8 (cm) trong khng kh. Xc nh cng in trng ti nh ca tam gic ABC

S: E = 0,7031.10-3 (V/m).


16/22

IN TH -HIU IN THCHUYN NG CA IN TCH TRONG IN TRNG U

A. PHNG PHP GII BI TPI. IN TH - HIU IN TH

N

E

M d H

1. Cng ca lc in trng u:A = qEd

d: L hnh chiu ca di trn mt nsc bt k

2. in th:a. in th ti mt im trong in trng

MM

AV

q=

MA cng ca lc in trng lm in tch q di chuyn t M

b. in th ti mt im M gy bi in tch q:

M

qV k

r=

c. in th ti mt im do nhiu in tch gy ra:

V = V1 + V2 + + Vn3. Hiu in th:

MNMN M N AU V V q= =

AMN l cng ca lc in trng lm di chuyn in tch q t M n N3. Th nng tnh in:

Wt(M) = q.VM

M N Ed

4. Lin h gia cng in trng vhiu in th

MN

EU

d

=

Vc t cng in trng hng t ni cin th ln ti b.

II. CHUYN NG CA IN TCH TRONG IN TRNG U:1. Gia tc:

F qEa

m m= =r

- ln ca gia tc:


17/22


18/22

Bi 2: Ba im A, B, C l ba nh ca mt tam gic vung trong in trng u, cng E=5000V/m. ng sc in trng song song vi AC. Bit AC = 4cm, CB = 3cm. GACB=900.

a. Tnh hiu in th gia cc im A v B, B v C, C v Ab. Tch cng di chuyn mt electro t A n B

Hng dn gii:

A C E

B

a. Ta c:ABU E.AB.cos E.AC 200V= = =

0

BCU E.BCcos90 0= =

CA ACU U 200V= =

b. Cng dch chuyn electron:17

AB ABA e.U 3,2.10 J= =

Bi 3: Mt electron bay vi vn tc v = 1,12.107m/s t mt im c in th V1 = 600V, the

hng ca cc ng sc. Hy xc nh ddienj th V2 im m electron dng li.Hng dn gii:

p dng nh l ng nng:

2

1

1A mv

2= = -6,65.10-17J

Mt khc:A

A eU U 410Jq

= = =

1 2 2 1U V V V V U 190V= = =Bi 4: Mt electron bt u chuyn ng dc theo chiu ng sc in trng ca mt tin phng, hai bn cch nhau mt khong d = 2cm v gia chng c mt hiu in th U 120V. Electron s c vn tc l bai nhiu sau khi dch chuyn c mt qung ng 3cm.

Hng dn gii:p ng nh l ng nng:

2

2

1A mv

2=

Mt khc:

A =F.s =q.E.s=qU

.sd

Do :

6

2

2.q.U.sv 7,9.10 m / s

m.d= =

Bi 5: Mt electron bay t bn m sang bn dng ca mt t in phng. in trng tronkhong hai bn t c cng E=6.104V/m. Khong cch giac hai bn t d =5cm.


19/22

a. Tnh gia tc ca electron.b. tnh thi gian bay ca electron bit vn tc ban u bng 0.c. Tnh vn tc tc thi ca electron khi chm bn dng.

Hng dn gii:a. Gia tc ca electron:

16 2e EFa 1.05.10 m / sm m

= = =

b. thi gian bay ca electron:

2 91 2dd x at t 3,1.10 s2 a

= = = =

c. Vn tc ca electron khi chm bn dng:v = at = 3,2.107m/v

Bi 6: Gia hai bn kim loi t song song nm ngang tch in tri du c mt hiu in thU1=1000V khong cch gia hai bn l d=1cm. ng gi hai bn c mt git thy ng

nh tch in dng nm l lng. t nhin hiu in th gim xung ch cn U2 = 995VHi sau bao lu git thy ngn ri xung bn dng?Hng dn gii:

-

F

P+

Khi git thy ngn cn bng:

1 11

U UP F mg q m q

d gd= = =

Khi git thy ngn ri:

2 2P F qU

a gm md

= = Do :

22 1 2

1 1

U U Ua g g g 0,05m / s

U U

= = =

Thi gian ri ca git thy ngn:

21 1 dx at d t 0,45s2 2 a

= = = =

Bi 7: Mt electron bay vo trong mt in trng theo hng ngc vi hng ng svi vn tc 2000km/s. Vn tc ca electron cui on ng s l bao nhiu nu hiu ith cui on ng l 15V.

Hng dn gii:p dng nh l ng nng:

2 22 62 1

2 1

2 e Umv mve U v v 3.10 m / s

2 2 m = = + =

Bi 8: Mt electron bay trong in trng gia hai bn ca mt t in tch in v


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cch nhau 2cm vi vn tc 3.107m/s theo ngsong song vi cc bn ca t in. Hiu in thgia hai bn phi l bao nhiu electron lch i 2,5mm khi i c on ng 5cm tronin trng.

Hng dn gii:Ta c

e E e UF amda U

m m md e

= = = = (1)

Mt khc:2

2

22 2

1 2h 2h 2hvh at a

2 t ss

v

= = = =

(2)

T (1) v (2):2

2

2mhvU 200V

e s= =


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II. BI TP NGHBi 1: Hai tm kim loi song song, cch nhau 2 (cm) v c nhim in tri du nhauMun lm cho in tch q = 5.10-10 (C) di chuyn t tm ny n tm kia cn tn mt cnA=2.10-9 (J). Coi in trng bn trong khong gia hai tm kim loi l in trng u vc cc ng sc in vung gc vi cc tm. Tnh cng in trng bn trong tm kimloi .

S: E = 200 (V/m).

Bi 2: Mt lectron chuyn ng dc theo ng sc ca mt in trng u. Cng in trng E = 100 (V/m). Vn tc ban u ca lectron bng 300 (km/s). Khi lng clectron l m = 9,1.10-31 (kg). T lc bt u chuyn ng n lc vn tc ca lectron bnkhng th lectron chuyn ng c qung ng l bao nhiu.

S: S = 2,56 (mm).Bi 3: Hiu in th gia hai im M v N l UMN = 1 (V). Cng ca in trng lm dcchuyn in tch q = - 1 (C) t M n N l bao nhiu

S: A = - 1 ( J).

Bi 4: Mt qu cu nh khi lng 3,06.10

-15

(kg), mang in tch 4,8.10

-18

(C), nm l lngia hai tm kim loi song song nm ngang nhim in tri du, cch nhau mt khon2(cm). Ly g = 10 (m/s2). Tnh Hiu in th t vo hai tm kim loi

S: U = 127,5 (V).Bi 5: Cng ca lc in trng lm di chuyn mt in tch gia hai im c hiu in thU = 2000 (V) l A = 1 (J). ln ca in tch l bao nhiu.

S: q = 5.10-4 (C).Bi 6: Mt in tch q = 1 (C) di chuyn t im A n im B trong in trng, n thc mt nng lng W = 0,2 (mJ). Tnh hiu in th gia hai im A, B.

S: U = 200 (V).Bi 7: Hai in tch im q1 = 0,5 (nC) v q2 = - 0,5 (nC) t ti hai im A, B cch nha6(cm) trong khng kh. Tnh cng in trng ti trung im ca AB.

S: E = 10000 (V/m).Bi 8: Hai in tch im q1 = 0,5 (nC) v q2 = - 0,5 (nC) t ti hai im A, B cch nha6(cm) trong khng kh. Tnh in trng ti im M nm trn trung trc ca AB, cctrung im ca AB mt khong l = 4 (cm).

S: E = 2160 (V/m).Bi 9: Mt in tch q = 10-7 (C) t ti im M trong in trng ca mt in tch im Qchu tc dng ca lc F = 3.10-3 (N). Cng in trng do in tch im Q gy ra tim M c ln bng bao nhiu.

S: EM = 3.104 (V/m).Bi 10: Mt in tch im dng Q trong chn khng gy ra ti im M cch in tch mkhong r = 30 (cm), mt in trng c cng E = 30000 (V/m). ln in tch Q l:

S: Q = 3.10-7 (C).


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Bi 11: Hai in tch im q1 = 2.10-2 (C) v q2 = - 2.10-2 ( C) t ti hai im A v B ccnhau mt on a = 30 (cm) trong khng kh. Tnh cng in trng ti im M cch A v B mt khong bng a

S: EM = 2000 (V/m).

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BAI_TAP_TU_LUAN_DIEN_TICH_-_DIEN_TRUONG