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H tn: 11

VT L 11L thuyt, bi tp t lun, trc nghim

* - in tch

* - nh lut Coulomb* - in trng* - Hiu in th* - Tin

TPHCM, 22/08/2013

Chng 1in tch - in trng

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BI 1. IN TCH - THUYT ELECTRON- in tch l g? Vt nh th no c gi l mang in?

- Tng tc in l g? Ti sao cc vt c th mang in?- Lm cch no nhim in cho mt vt?

I.VT MANG IN - IN TCH- Mt vt mang in (nhim in) l mt vt c kh nng ht c nhng vt nh nh khc v c

th tng tc vi cc vt mang in khc.- C 2 loi in m vt c th mang: DNG V M

Cng duy nhau ; Tri du ht nhau

- Lng in mt vt cha trong n c gi l IN TCH.- Mt vt mang in cng cn c gi l in tch.

K hiu: q, Q. n v: coulomb (C)

coulomb: mC = 10-3 C :: C = 10-6 CC : : nC = 10-9 C : : pC = 10-12 C

II.THUYT ELECTRON- Mt vt bnh thng trung ho vin c ngha l sin tch dng bng sin tch m.

. in tch m ca vt (nguyn t) l electron (qe = -1,6.10-19 C)

. in tch dng ca vt (nguyn t) l proton (qp = 1,6.10-19 C)- Trong hai loi in tch, ht no d b ht hoc tch ra hn?

- Th suy ngh: ion Ca2+, ion Cl-, ht nhn He mang in tch bao nhiu?

.............................................................................................................................................................

- Th suy ngh: mt in tch 3,2.10-10 C d hoc thiu bao nhiu electron?.............................................................................................................................................................

III.CC CCH LM NHIM IN MT VTNguyn liu Qu trnh Kt qu

C xtHai vt trung ho hocnhim in

2 vt tri du ln bng nhau

Tip xct nht 1 vt nhimin

2 vt nhim inging nhau

Hng ngt nht 1 vt nhimin

1 vt nhim intrn tng phn

Chng 1. in tch - in trng

Gc Ring Trn Bn: 3

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BI 2. NH LUT BO TON IN TCHNH LUT COULOMB

- in tch c c bo ton nhng lng, nng lng? Trong iu kin no?- Khi no xut hin lc Coulomb l g? Cc c im ca vect lc Coulomb?

I.NH LUT BO TON IN TCH

- Lp bng so snh in tch cc vt trong mi qu trnh nhim in.q1 q2 q1 + q2 q1 q2 q1 + q2

C xt 0 0 0 -1 1 0Tip xc 0 2 2 1 1 2

Hng ng 0 -1 -1 0 -1 -1

nh lut bo ton in tch: Trong mt hc lp vin, tngi sccin tchc bo ton.

q1 + q2 = q1 + q2

- Cn nh: Khi a hai qu cu kim loi ging nhau tch in q1, q2 chm nhau

q1

'= q

2

'=q1+ q

2

2

II.NH LUT COULOMB- Cc in tch tng tc vi nhau bng mt lc --> Lc tnh in (Lc Coulomb)- Lc Coulomb xut hin khi hai in tch c t gn nhau.

nh lut Coulomb: lc tng tc gia haiin tchim c phng nm trn

mtng thng ni haiin tchim, c chiu l chiu ca lc ht nu haiintchim khc du vy nu haiin tchim cng du, c lnt l thunvi tch ccin tch v tlnghch vi bnh phng khong cch gia chng.

im t Trn 2 in tch

Phng Ni tm 2 in tch

FCChiu

Theo quy lut tng tc in cng y, tri ht

ln F= kq1q2

r2

: hng sin mi (kk=ck= 1)k = 9x109(Nm2/C2)

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BI 4. CNG CA IN TRNG - HIU IN TH- Cng ca in trng c xc nh nh th no?

- Hiu in th l g? ngha ca n l g?

I.CNG CA IN TRNG

- Mt in tch t trong in trng s chu tc dng ca lc in trng FE. Cng lc , nuin tch di chuyn, ta ni lc in trng thc hin cng A.- Cng thc tnh cng ca lc in trng

AMN = qEdMN

dMN: hnh chiu ca MN trn phng ca vectcngin trng

- Cng ca in trng khng ph thuc vo dng ng i m ch ph thuc vo v tr im uv im cui.--> Lc in trng l mt lc th.

II.THNNG - IN TH- HIU IN TH- in trng l mt trng th --> mt in tch t trong in trng s c th nng. Th nng

ca in tch ti im M c gi tr bng cng in trng thc hin khi dch chuyn in tch t

im M ra xa v cng. Khi gc th nng c chn xa v cng.

WtM = AM- IN TH l i lng c trng cho kh nng thc hin cng ca in trng ti mt im. Khiu: VM (V). Gc in th thng chn xa v cng.

VM = WtM/q

- HIU IN THl i lng th hin s chnh lch in th gia hai im trong in trng.

K hiu: U. n v: volt (V).

UMN = VM - VN

UMN = dMNE

dMN= MN: hnh chiu ca MN trnng scin E

- Cng thc tnh cng ca in trng theo hiu in th

AMN = qUMN

A > 0: cng phtngA < 0: cng cn

VD: in tch q > 0 chuyn ng tin thcaon in ththp --> cng pht ng

- 1 V/m l cng ca mt in trng u m trong hiu in th gia hai im cch nhau 1m trn phng ca in trng l 1 V.

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BI 5. TIN- Tin l thit b g? i lng no c trng cho tin?

- Tin dng lm g?

I. TIN L G?- Tin l thit bc s dng lu giin.

- Cu to ca tin phng n gin gm:

* 2 bn kim loi t song song

* Mt mi trng in mi t gia hai bn- Hot ng ca tin: di tc dng ca hiu in

thU, bn ni vi cc in th cao (cc +) tch in +Q, v bn cn li tch in -Q. Khi ta ni t tchin Q.

- Hiu in th cng cao, in tch trn t cng tng. Mt tin ch c thc cha (tch) in n mt

gii hn ng vi Qmax v Ugh. Nu U > Ugh --> t b thng, Q = 0, t nh mt dy dn bnh thng.

- i lng c trng cho kh nng tch in ca t l IN DUNG. T c in dung cng ln th cng tch

nhiu in cng hiu in th. Ngi ta phn bit tin bng in dung ca n.K hiu: C. n v: farad (F).

Q = CU

- in dung ca tin ch ph thuc vo cu to ca t. i vi tin phng n gin, in dung c

tnh theo cng thc

C=S

4kd

S: din tch phni din ca hai bn (m2)

d: khong cch gia hai bn (m)

: hng sin mi cain mi gia hai bn

II.NNG LNG TIN- Tin c kh nng phng in, sinh ra cng --> tin c mang nng lng.

- V gia tin l mt in trng nn ta c th ni nng lng tin l nng lng ca in trng

W =1

2CU

2=

1

2

Q2

C=

1

2QU

III.GHP TIN

Ghp t song song Ghp t ni tip

U=U1=U

2

Q =Q1+Q

2

C//= C

1+C

2

U=U1+U

2

Q =Q1=Q

2

1

Cnt=

1

C1

+1

C2

Cnt=

C1C

2

C1+C

2

Chng 1. in tch -in trng 0 v q1 + q2 > 0.t hai qu cu ti hai im A, B cch nhau 1 mm th thy lc in tc dng ln chng l 4,32.10-3 N.Sau cho chng chm nh vo nhau ri a v li v tr c, lc in by gi c gi tr l 9.10-5 N. Xcnh gi tr q1, q2.

11 Hai in tch im q1 = -2 nC v q2 = 3 nC c t ti hai im A v B nm cch nhau 4 cm trong chnkhng. Ti im D l trung im ca AB, ngi ta t mt in tch im q3 = 1 nC. Xc nh ln lctng hp tc dng ln q3.

12 Cho 3 im A, B, D nm thng hng. Ngi ta o c AB = 10 cm, BD = 4 cm, AD = 14 cm. Ln ltt 3 in tch im q1 = q2 = 2q3 = 7.10-10 C vo cc im A, B, D. Xc nh ln lc tng hp tcdng ln in tch t ti A.

13 Hai in tch im q1 = -2q2 = 5 nC c t ti hai im A v B nm cch nhau 10 cm trong chnkhng. Ti im D cch A 6 cm, cch B 8 cm, ngi ta t mt in tch q3 = -2 nC. Xc nh vect lcin tng hp tc dng ln q3.

14 Cho tam gic u ABC cnh 6 mm nm trong chn khng. Ti 3 nh ca tam gic ngi ta t 3 intch ging nhau l q = -2 pC. Xc nh lc in tng hp ti in tch t ti nh A ca tam gic.

15 Mt h hai in tch q1 = -8 nC v q2 c t ti 2 im A, B cch nhau 10 cm trong chn khng. Tiim D nm cch A 6 cm, cch B 8 cm, ngi ta t mt in tch q3 = -5 nC th thy lc in tc dngln q3 l mt vect c phng song song vi AB.a. V hnh v xc nh du ca q2.b. Tnh gi tr ca q2.

16 Ngi ta t hai in tch q1 = 3 nC v q2 = 6 nC ln lt ti hai im A, B cch nhau 10 cm. Mt intch khc q3 = -2 nC c t ti im D chu tc dng ca lc in do q1 v q2 gy ra. Bit rng lc F13

vung gc vi lc F23 v ln F13 =3.F23.a. Xc nh khong cch AD, BD.b. Tnh lc tng hp tc dng ln q3.

Chng 1. in tch -in trng 0.Cho hai qu cu chm nh vo nhau th thy in tch ca mi qu cu lc sau l 1,5 nC. Xc nh in

tch q1, q2.27 Mt qu cu bng kim loi c tch in 6.10-10 C. a n chm vo mt qu cu bng kim loi khc

ging ht ri a hai qu cu cch nhau 4 mm trong chn khng. Chm nh tay (khng ni t) vomt qu cu th lc tng tc gia hai qu cu lc ny l bao nhiu?

28 Mt in tch im q c t ti im A trong du ho (hng sin mi l 2,2). Khi t mt in tchim q = 10 pC ti im B cch A 0,2 mm th ngi ta thy q b ko ra xa khi q bng mt lc 2.10-4 N.Xc nh nh gi tr ca q.

29 Mt ht ion Cl-c t im A nm trong mi trng c hng sin mi l 103. Mt in tch imq c ln 4.10-10 C c t ti B th thy q b ko v pha q0 bng mt lc F = 4.10-13 N. Xc nhdu ca q v v tr im B.

30 Hai qu cu kim loi c tch in ging nhau l q. t chng cch nhau 0,2 cm trong chn khng th

thy lc tc dng ln chng l 0,05625 N. Xc nh in tch cu mi qu cu.31 Hai hn bi nh bng kim loi c kch thc ging nhau mang cc in tch q1 = 3.10-6 C v q2 = 7.10-6

C. Ngi ta cho chng tip xc vi nhau ri tch chng ra v t chng cch nhau 6 cm trong mt cikhay cha y nc nguyn cht c hng sin mi l 81. Tnh lc in tng tc gia chng.

32 Mt in tch im q1 = -4 C c t ti im A trong mi trng c hng sin mi l 2,2. Mt intch im khc c in tch q2 = -1,1q1c t ti B cch A 2 cm. Hy v hnh vect lc in tc dngln in tch q2 v tnh ln lc ny.

33 Hai qu cu kim loi ging nhau c tch in ln lt l q1 v q2. Bit q1 > q2 > 0. t hai qu cu tihai im A, B cch nhau 2 mm th thy lc in tc dng ln chng l 0,054 N.a. Tnh tch q1.q2.b. Sau cho chng chm nh vo nhau ri a v li v tr c. Lc tng tc gia chng by gi l

0,05625 N. Xc nh gi tr q1, q2.34 Hai qu cu tch in tri du nhng c ln bng nhau c t cch nhau 4 cm. Lc tng tcgia hai qu cu ny l 2,25.10-7 N. V hnh lc tc dng v xc nh in tch ca mi qu cu.

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35 Hai qu cu kim loi c tch in ging nhau l q v t cch nhau 3 cm trong chn khng. Dng taychm nh vo mt qu cu (khng ni t). Lc in tng tc gia hai qu cu lc ny l 8.10-3 N.Xc nh in tch q.

36 Hai in tch im q1 v q2 tri du (q2 < 0 < q1, |q1| > |q2|) c t cch nhau 9 mm trong chn khng.Lc tnh in gia hai in tch l 16.10-9 N. a hai in tch chm vo nhau ri a v v tr c, lctnh in gia chng by gi l 9.10-9 N. Hi hai in tch ban u c gi tr bao nhiu?

37 Hai qu cu kim loi ging nhau c tch in cng du ln lt l q1 v q2. t hai qu cu ti hai

im A, B cch nhau 2 mm th thy lc in tc dng ln chng l 0,072 N. Sau cho chng chmnh vo nhau ri a v li v tr c, lc in by gi c gi tr l 0,081 N. Xc nh gi tr q1, q2.

38 t hai in tch im q1 v q2 ti hai im A, B cch nhau 3 cm trong mi trng c hng sin mil 2.105. Xc nh lc in tc dng ln mt in tch im q3 = 0,4 nC nm cch q1 5 cm v cch q2 2cm. Bit rng q1 d 3,125.109 electron cn q2 thiu 2,5.109 electron.

39 Cho 3 im A, B, D nm thng hng theo th t trn. Bit rng AB = 10 cm, BD = 2 cm. Ln lt t 3in tch im ging nhau c gi tr q = -2.10-3C. Xc nh lc in tng hp tc dng vo in tchim t A.

40 Hai in tch im q1 = 3 C v q2 = 4 C c t ti hai im A v B nm cch nhau 10 mm trongchn khng. Ti im D cch A 4 cm, cch B 6 cm, ngi ta c mt in tch im q3 = 2.10-10 C.Xc nh ln lc tng hp tc dng ln q3.

41 Hai in tch im ging nhau q1 = q2 = 4 pC c t ti hai im A, B nm cch nhau 3 cm trong chnkhng. Khi t mt in tch im q3 = -5 pC ti im D th lc in tng hp tc dng ln q3 l baonhiu? Cho AD = 5 cm, BD = 2 cm.

42 Mt tam gic ABC vung ti A c cnh AB = 5 mm, BC = 13 mm. Ln lt t vo 3 nh A, B, C ba intch q1 = q2 = q3 = q. Lc in tng hp tc dng ln in tch q1t ti A l 0,0288 N. Xc nh gi trin tch q.

43 Mt h hai in tch q1 v q2 = 0,5 C c t ti 2 im A, B cch nhau 13 cm trong chn khng. Tiim D nm cch A 5 cm, cch B 12 cm, ngi ta t mt in tch q3 = - q2 th thy lc in tc dngln q3 l mt vect c phng vung gc vi AB. V hnh v xc nh du ca q1 v ln ca q1.

44 Mt h 3 in tch im q1, q2, q3c t ln lt ti 3 im A, B, C thng hng. Bit B l trung imAC. Xc nh ts q1/q2in tch q3 nm cn bng.

45 2 ion Na+ v ion O2-t ti 2 im A v B cch nhau 3 m. Ngi ta phi t thm mt in tch q =4.10-15 C ti mt im C th thy q cn bng. Tm v tr im C.

46 Cho ba im A, B, C nm thng hng theo th t trn (AB = 2 mm, AC = 6 mm) trong khng kh. t haiin tch im q1 = -3.10-10 C v q2 = 4.10-10 C ln lt ti hai im A, B. Hi phi t mt in tch q3 cdu v ln nh th no vo v tr C q2 cn bng?

47 Cho h gm hai in tch im m q1 v q2t trong khng kh, trong q1 = 2q2. Hai in tch cchnhau 9 mm. Hi phi t mt in tch q0u trong h ny m q0 vn cn bng?

48 Hai in tch im q1 = 2 pC v q2 = -8 pC c t ti hai im A, B nm cch nhau 6 mm trong chnkhng. Khi t mt in tch im q3 ti im D th lc in tng hp tc dng ln q3 = 2 nC bng 0.a. Xc nh v tr im D.b. Xc nh lc in tng hp tc dng ln q1 lc .

49 Treo 2 qu cu kim loi tch in bng nhau ln 2 si dy rt di ging nhau gn chung mt u.Khong cch ca chng lc ny l 5 cm. Dng tay chm nh vo mt qu cu. Sau , khong cchca hai qu cu l bao nhiu? Cho g = 10 m/s2.

50 Hai qu cu nh ging nhau bng kim loi, c khi lng 5 g, c treo vo cng mt im O bng haisi chkhng dn, di 10 cm. Hai qu cu tip xc vi nhau. Tch in cho mt qu cu th thy hai qucu y nhau cho n khi dy treo hp vi nhau mt gc 600. Tnh in tch truyn cho qu cu.Cho g = 10 m/s2.

51 Hai qu cu nh ging nhau bng kim loi, c tch in ln lt l q1 v q2, c treo vo cng mtim O bng hai si chkhng dn, di bng nhau th thy hai si dy hp vi nhau mt gc 600. Chohai qu cu tip xc th thy gc ca hai si dy tng ln thnh 900. Tnh ts q1/q2 Cho g = 10 m/s2.

52 Mt qu cu kim loi tch in q1 = 4.10-10

C nng 0,2 g c treo bng mt si dy nh, khng co gin.Khi a mt in tch q2 = -2.10-4 C nm di v cch q1 1 mm th lc cng dy s l bao nhiu? Chog = 10 m/s2.

Chng 1. in tch -in trng 0 ti hai im A, B cch nhau 4 mm. Ti im D cch A 6 mm,cch B 2 mm, ngi ta o c cng in trng c ln l 875000 V/m. Xc nh ln caq2.

30 Xc nh cng in trng ti im M nm cch u hai u on thng AB 5 cm. Bit rng ti A,B c hai in tch ging nhau l -2 nC v AB = 8 cm.

31 Cho ba im A, B, C nm thng hng theo th t trn. Ln lt t vo hai im A, B hai in tch q1 vq2 = 1 pC th thy cng in trng ti C b trit tiu.a. C th kt lun c g v du ca q1?b. Nu bit B l trung im ca AC th gi tr ca q1 l bao nhiu?

32 a mt in tch im q vo mt in trng u c ln 2000 V/m hng t trn xung di ththy q chu tc dng ca mt lc hng ln trn c ln 3.10-5 N. Xc nh gi tr q.

33 Hai bn kim loi tch in tri du c t nm ngang vi bn dng trn ti thnh in trng uc cng 50000 V/m. Mt ht bi c khi lng 0,1 mg nm l lng gia hai bn ny. Xc nh intch ca ht bi ny.

34 Ti im A trong mi trng c hng sin mi l 2,2, ngi ta t mt in tch im q = -2,31 nC.a. Xc nh cng in trng ti im B nm cch A 1 mm.b. Ti B, t thm mt in tch im q = 2 nC. Xc nh ln lc in tc dng ln q.

35 Mt qu cu tch in dng c t trong mt in trng u c ln 4500 V/m. Lc in tcdng ln qu cu c ln 2,7.10-5 N. Hi qu cu ny b thiu bao nhiu electron?

36 Hai bn kim loi tch in tri du c t song song nm ngang. a mt electron vo gia hai bnth thy electron b ko ln trn bng mt lc 2.10-10 N. Xc nh du ca mi bn kim loi v ln cacng in trng gia hai bn. Cho qe = -1,6.10-19 C.

37 Ngi ta t vo gia hai bn kim loi tch in tri du mt qu cu tch in dng th thy qu cunm yn.a. Cho bit hai bn ny c t nh th no?

b. Bit khi lng qu cu ny l 1 g, cng in trng gia hai bn l 3,2.108 V/m. Xc nh sht electron b mt trn qu cu ny.

Chng 1. in tch -in trng