TỔNG HỢP BÀI TẬP VẬT LÍ 11

7/29/2019 TNG HP BI TP VT L 11

1/274

[email protected] Sutm v bin son

MC LC

Chiu lch ca tia sng..................................................................................................................................21

TNG HP BI TP VT L 11

CHNG I:IN TCH.IN TRNG

CH 1:LC TNG TC TNH IN

DNG 1: TNG TC GIA HAI IN TCH IM NG YN

A.L THUYT1.Lc tng tc gia 2 in tch im.Lc tng tc gia hai in tch im ql v q2 (nm yn, t trong chn khng) cch nhau on r c: phng l ng thng ni hai in tch. chiu l: chiu lc y nu qlq2 > 0 (cng du).

chiu lc ht nu qlq2 < 0 (tri du). ln: * t l thun vi tch cc

ln ca hai in tch,* t l nghch vi bnh

phng khong cch gia

chng.

Trong : k = 9.109N.m2/C2.q1 , q 2 : ln hai in tch (C )r: khong cch hai in tch (m)

F= 1 22

q qk

r
mailto:[email protected]:[email protected]


2/274

[email protected] Sutm v bin son : hng s in mi . Trong chn khng v khng kh =1Ch :a) in tch im : l vt m kch thc cc vt cha in tch rt nh so vi khong cch gia chng.-Cng thc trn cn p dng c cho trng hp cc qu cu ng cht , khi ta coi r l khong ccgia tm hai qu cu.

2. in tch q ca mt vt tch in: e.nq =

+ Vt thiu electron (tch in dng): q = + n.e+ Vt tha electron (tch in m): q = n.eVi: C10.6,1e 19= : l in tch nguyn t.

n : s ht electron b tha hoc thiu.3.Mt s hin tng Khi cho 2 qu cu nh nhim in tip xc sau tch nhau ra th tng in tch chia u cho mi qu cu Hin tng xy ra tng t khi ni hai qu cu bng dy dn mnh ri ct b dy ni Khi chm tay vo qu cu nh dn in tch in th qu cu mt in tch v tr v trung ha

B.BI TP T LUNBi 1. Hai in tch C10.2q 81

= , C10q 82= t cch nhau 20cm trong khng kh. Xc nh ln v v

hnh lc tng tc gia chng?S: N10.5,4 5

Bi 2. Hai intch C10.2q 61= , C10.2q 62

= t ti hai im A v B trong khng kh. Lc tng tc gichng l 0,4N. Xc nh khong cch AB, v hnh lc tng tc .

S: 30cBi 3. Hai in tch t cch nhau mt khong r trong khng kh th lc tng tc gia chng l 310.2 N. Nu

vi khong cch m t trong in mi th lc tng tc gia chng l

310

N.a/ Xc nh hng s in mi ca in mi.b/ lc tng tc gia hai in tch khi t trong in mi bng lc tng tc khi t trong khng kh th

phi t hai in tch cch nhau bao nhiu? Bit trong khng kh hai in tch cch nhau 20cm.S: 2= ; 14,14cm

Bi 4. Trong nguyn t hir (e) chuyn ng trn u quanh ht nhn theo qu o trn c bn knh 5.10 -9 cm.a. Xc nh lc ht tnh in gia (e) v ht nhn. b. Xc nh tn s ca (e)

S: F=9.10-8 N b.0,7.1016 HBi 5. Mt qu cu c khi lng ring (aKLR) = 9,8.103 kg/m3,bn knh R=1cm tch in q = -10 -6 C c trevo u mt si dy mnh c chiu di l =10cm. Ti im treo c t mt in tch m q 0 = - 10 -6 C .Tt c trong du c KLR D= 0,8 .103 kg/m3,hng s in mi =3.Tnh lc cng ca dy? Ly g=10m/s2.

S:0,614Bi 6. Hai qu cu nh, ging nhau, bng kim loi. Qu cu A mang in tch 4,50 C; qu cu B mang in tch 2,40 C. Cho chng tip xc nhau ri a chng ra cch nhau 1,56 cm. Tnh lc tng tc in gia chng.___________________________________________________________________________________________

DNG 2: LN IN TCH.

A.L THUYT

Dng 2: Xc nh ln v du cc in tch.


3/274

[email protected] Sutm v bin son

- Khi gii dng BT ny cn ch : Hai in tch c ln bng nhau th: 21 qq = Hai in tch c ln bng nhau nhng tri du th: 21 qq = Hai in tch bng nhau th: 21 qq = .

Hai in tch cng du: 212121q.qq.q0q.q =>

. Hai in tch tri du: 212121 q.qq.q0q.q = 0, t gia hai in tch cch 4q khong r/4B. Q < 0, t gia hai in tch cch 4q khong 3r/4C. Q > 0, t gia hai in tch cch q khong r/3D. Q ty t gia hai in tch cch q khong r/3

Cu 2: Hai in tch im q v 4q t cch nhau mt khong r. Cn t in tch th 3 Q c in tch dnghay m v u h 3 in tch ny cn bng:

A. Q > 0, t gia hai in tch cch 4q khong r/3B. Q < 0, t gia hai in tch cch 4q khong 2r/3C.Q tri du vi q t gia 2 in tch cch q khong r/3D.Q ty t gia 2 in tch cch q khong r/3

Cu 3: Ti bn nh ca mt hnh vung t 4 in tch im ging nhau q = + 1C v ti tm hnh vungt in tch q0, h nm in tch cn bng. Tm du v ln in tch im q0?

A. q0 = + 0,96 C B. q0 = - 0,76 C C. q0 = + 0,36 C D. q0 = - 0,96 CCu 4: Mt qu cu khi lng 10g mang in tch q1 = + 0,1C treo vo mt si ch cch in, ngi ta

a qu cu 2 mang in tch q2 li gn th qu cu th nht lch khi v tr ban u mt gc 300

, khi haiqu cu trn cng mt mt phng nm ngang cch nhau 3cm. Tm du, ln in tch q2 v sc cng casi dy:

A. q2 = + 0,087 C B. q2 = - 0,087 C C. q2 = + 0,17 C D. q2 = - 0,17 CCu 5:Ngi ta treo hai qu cu nh khi lng bng nhau m = 0,01g bng hai si dy c di nh nhau l= 50cm( khi lng khng ng k). Cho chng nhim in bng nhau chng y nhau cch nhau 6cm. Tnhin tch mi qu cu:

A. q = 12,7pC B. q = 19,5pC C. q = 15,5nC D.q = 15,5.10-10CCu 6: Treo hai qu cu nh khi lng bng nhau m bng nhng si dy cng di l( khi lng khngng k). Cho chng nhim in bng nhau chng y nhau cch nhau khong r = 6cm. Nhng c h thngvo trong ru c = 27, b qua lc y Acsimet, tnh khong cch gia chng khi tng tc trong du:

A. 2cm B. 4cm C. 6cm D. 1,6cmCu 7: Ngi ta treo hai qu cu nh khi lng bng nhau m = 0,1g bng hai si dy c di nh nhau l(khi lng khng ng k). Cho chng nhim in bng nhau chng y nhau v cn bng khi mi dy treohp vi phng thng ng mt gc 150. Tnh lc tng tc in gia hai qu cu:

A. 26.10-5N B. 52.10-5N C. 2,6.10-5N D. 5,2.10-5NCu 8: Ngi ta treo hai qu cu nh khi lng bng nhau m = 0,1g bng hai si dy c di nh nhau l= 10cm( khi lng khng ng k). Truyn mt in tch Q cho hai qu cu th chng y nhau cn bngkhi mi dy treo hp vi phng thng ng mt gc 150, ly g = 10m/s2. Tnh in tch Q:

A. 7,7nC B. 17,7nC C. 21nC D. 27nCCu 9: Ba in tch bng nhau q dng t ti 3 nh ca tam gic u ABC cnh a. Hi phi t mt intch q0 nh th no v u lc in tc dng ln cc in tch cn bng nhau:

A. q0 = +q/ 3 , gia AB B. q0 = - q/ 2 , trng tm ca tam gicC. q0 = - q/ 3 , trng tm ca tam gic D. q0 = +q/ 3 , nh A ca tam gic

Cu 10: Hai qu cu nh bng kim loi ging ht nhau tch in dng treo trn hai si dy mnh cngchiu di vo cng mt im. Khi h cn bng th gc hp bi hai dy treo l 2. Sau cho chng tip xcvi nhau ri bung ra, chng cn bng th gc lch by gi l 2 '. So snh v ':

A. > ' B. < ' C. = ' D. c th ln hoc nh hn '

Cu 1 2 3 4 5 6 7 8 9 10p n D C D B D A A B C B

in tch, Fculng - Dng 3: in tch cn bng chu td lc Culng - 2:Cu 1: Hai qu cu nh kim loi ging ht nhau mang in tch q1 v q2 t trong chn khng cch nhau20cm ht nhau mt lc 5 10- 7 N t vo gia hai qu cu mt tm thy tinh dy d = 5cm c hng s in


47/274

[email protected] tm v bin son

A. 1,2.10-7 N B. 2,2.10-7 N C. 3,2.10-7 N D.4 ,2.10-7 NCu 2: Hai qu cu ging nhau khi lng ring l D tch in nh nhau treo u ca hai si dy di nhnhau t trong du khi lng ring D0, hng s in mi = 4 th gc lch gia hai dy treo l . Khi t rangoi khng kh thy gc lch gia chng vn bng . Tnh t s D/ D0

A. 1/2 B. 2/3 C. 5/2 D. 4/3Cu 3: Bn in tch im q1, q2, q3, q4 t trong khng kh ln lt ti cc nh ABCD ca hnh vung thyhp lc tnh in tc dng ln q4 ti D bng khng. Gia 3 in tch kia quan h vi nhau:

A. q1 = q3; q2 = q1 2 B. q1 = - q3; q2 = ( 1+ 2 )q1C. q1 = q3; q2 = - 2 2 q1 D. q1 = - q3; q2 = ( 1- 2 )q1

Cu 4: Hai in tch im trong khng kh q1 v q2 = - 4q1 ti A v B, t q3 ti C th hp cc lc in tcdng ln q3 bng khng. Hi im C c v tr u:

A. trn trung trc ca AB B. Bn trong on ABC. Ngoi on AB. D. khng xc nh c v cha bit gi tr ca q3

Cu 5: Hai in tch im trong khng kh q1 v q2 = - 4q1 ti A v B vi AB = l, t q3 ti C th hp cc lcin tc dng ln q3 bng khng. Khong cch t A v B ti C ln lt c gi tr:

A. l/3; 4l/3 B. l/2; 3l/2 C. l; 2l D. khng xc nh c v cha bit gi tr ca q3

Cu 6: Hai qu cu kim loi nh ging nhau khi lng m, tch in cng loi bng nhau c treo bi haisi dy nh di lcch in nh nhau vo cng mt im. Chng y nhau khi cn bng hai qu cu cchnhau mt on r


48/274

[email protected] tm v bin sonCu hi 2: Trong cc quy tc v cc ng sc in sau y, quy tc no l sai:

A. Ti mt im bt k trong in trng c th v c mt ng sc i qua nB. Cc ng sc xut pht t cc in tch m, tn cng ti cc in tch dngC. Cc ng sc khng ct nhauD. Ni no cng in trng ln hn th cc ng sc c v dy hn

Cu hi 3: Mt in tch q c t trong in mi ng tnh, v hn. Ti im M cch q 40cm, in trngc cng 9.105V/m v hng v in tch q, bit hng s in mi ca mi trng l 2,5. Xc nh du v ln ca q:

A. - 40 C B. + 40 C C. - 36 C D. +36 CCu hi 4: Mt in tch th t ti im c cng in trng 0,16 V/m. Lc tc dng ln in tch

bng 2.10-4N. ln ca in tch l:A. 1,25.10-4C B. 8.10-2C C. 1,25.10-3C D. 8.10-4C

Cu hi 5:in tch im q = -3 C t ti im c cng in trng E = 12 000V/m, c phng thngng chiu t trn xung di. Xc nh phng chiu v ln ca lc tc dng ln in tch q:

A. F c phng thng ng, chiu t trn xung di, F = 0,36NB. F c phng nm ngang, chiu t tri sang phi, F = 0,48N

C. F c phng thng ng, chiu t di ln trn, F = 0,36ND. F c phng thng ng, chiu t di ln trn, F = 0,036NCu hi 6: Mt in tch q = 5nC t ti im A. Xc nh cng in trng ca q ti im B cch Amt khong 10cm:

A. 5000V/m B. 4500V/m C. 9000V/m D. 2500V/mCu hi 7: Mt in tch q = 10-7C t trong in trng ca mt in tch im Q, chu tc dng lc F =3mN. Tnh cng in trng ti im t in tch q. Bit rng hai in tch cch nhau mt khong r =30cm trong chn khng:

A. 2.104 V/m B. 3.104 V/m C. 4.104 V/m D. 5.104 V/mCu hi 8: in tch im q t ti O trong khng kh, Ox l mt ng sc in. Ly hai im A, B trnOx, t M l trung im ca AB. Gia EA, EB ,EM c mi lin h:

A. EM = (EA + EB)/2 B. ( )BAM EEE += 21

C.

+=

BAM EEE

112

1D.

+=

BAM EEE

11211

Cu hi 9: Cng in trng ca mt in tch im ti A bng 36V/m, ti B bng 9V/m. Hi cng in trng ti trung im C ca AB bng bao nhiu, bit hai im A, B nm trn cng mt ng sc:

A. 30V/m B. 25V/m C. 16V/m D. 12 V/mCu hi 10: Mt in tch q = 10-7C t trong in trng ca mt in tch im Q, chu tc dng lc F =3mN. Tnh ln ca in tch Q. Bit rng hai in tch cch nhau mt khong r = 30cm trong chn khng

A. 0,5 C B. 0,3 C C. 0,4 C D. 0,2 C

Cu 1 2 3 4 5 6 7 8 9 10p n C B A C D B B D C B

in trng - Dng 1: Xc nh llq Eca in tch im- 2Cu hi 1: Mt qu cu nh mang in tch q = 1nC t trong khng kh. Cng in trng ti imcch qu cu 3cm l:

A. 105V/m B. 104 V/m C. 5.103V/m D. 3.104V/mCu hi 2: Mt qu cu kim loi bn knh 4cm mang in tch q = 5.10 -8C. Tnh cng in trng trnmt qu cu:

A. 1,9.105 V/m B. 2,8.105V/m C. 3,6.105V/m D. 3,14.105V/mCu hi 3: Cho hai qu cu kim loi bn knh bng nhau, tch in cng du tip xc vi nhau. Cc in tch

phn b nh th no trn hai qu cu nu mt trong hai qu cu l rng;A. qu cu c phn b u trong c th tch, qu cu rng ch mt ngoiB. qu cu c v qu cu rng phn b u trong c th tch


49/274


D. qu cu c phn b mt ngoi, qu cu rng phn b u trong th tchCu hi 4: Mt git thy ngn hnh cu bn knh 1mm tch in q = 3,2.10 -13C t trong khng kh. Tnhcng in trng trn b mt git thy ngn :

A. E = 2880V/m B. E = 3200V/m C. 32000V/m D. 28800 V/mCu hi 5: Mt qu cu kim loi bn knh 4cm mang in tch q = 5.10 -8C. Tnh cng in trng tiim M cch tm qu cu 10cm:

A. 36.103V/m B. 45.103V/m C. 67.103V/m D. 47.103V/mCu hi 6: Mt v cu mng bng kim loi bn knh R c tch in +Q. t bn trong v cu ny mtqu cu kim loi nh hn bn knh r, ng tm O vi v cu v mang in tch +q. Xc nh cng intrng trong qu cu v ti im M vi r < OM < R:

A. EO = EM = k 2OMq

B. EO = EM = 0 C. EO = 0; EM = k 2OMq

D. EO = k 2OMq

; EM = 0

Cu hi 7: Mt qu cu kim loi bn knh R1 = 3cm mang in tch q1 = 5.10-8C. Qu cu c bao quanhbng mt v cu kim loi t ng tm O c bn knh R2 = 5cm mang in tch q2 = - 6.10-8C. Xc nhcng in trng nhng im cch tm O 2cm, 4cm, 6cm:

A. E1 = E2 = 0; E3 = 3.105 V/m

B. E1 = 1,4.105

V/m; E2 = 2,8.105

V/m ; E3 = 2,5.105

V/mC. E1 = 0; E2 = 2,8.105V/m; E3 = 2,5.105V/mD. E1 = 1,4.105 V/m; E2 = 2,5.105 V/m; E3 = 3.105 V/m

Cu hi 8: t mt in tch m, khi lng nh vo mt in trng u ri th nh. in tch s chuynng:

A. dc theo chiu ca ng sc in trng. B. ngc chiu ng sc in trng.

C. vung gc vi ng sc in trng. D. theo mt qu o bt k.

Cu hi 9: Cng thc xc nh cng in trng gy ra bi in tch im Q < 0, ti mt im trongchn khng cch in tch im mt khong r l: ( ly chiu ca vct khong cch lm chiu dng):

A. 2910.9

rQE= B. 2

910.9rQE = C.

rQE 910.9= D.

rQE 910.9=

Cu hi 10: Cng in trng gy ra bi in tch Q = 5.10-9 (C), ti mt im trong chn khng cchin tch mt khong 10 (cm) c ln l:

A. E = 0,450 (V/m). B. E = 0,225 (V/m). C. E = 4500 (V/m). D. E = 2250 (V/m).

Cu 1 2 3 4 5 6 7 8 9 10p n B B C A B C C D A A

in trng - Dng 2: Nguyn l chng cht in trng - 1Cu hi 1: Hai in tch im q1 = 5nC, q2 = - 5nC cch nhau 10cm. Xc nh vct cng in trngti im M nm trn ng thng i qua hai in tch v cch u hai in tch:

A. 18 000V/m B. 45 000V/m C. 36 000V/m D. 12 500V/m

Cu hi 2: Hai in tch im q1 = 5nC, q2 = - 5nC cch nhau 10cm. Xc nh vct cng in trngti im M nm trn ng thng i qua hai in tch v cch q1 5cm; cch q2 15cm:

A. 4 500V/m B. 36 000V/m C. 18 000V/m D. 16 000V/m

Cu hi 3: Ti ba nh ca tam gic u cnh 10cm c ba in tch bng nhau v bng 10nC. Hy xc nhcng in trng ti trung im ca cnh BC ca tam gic:


50/274

[email protected] tm v bin sonCu hi 4: Ti ba nh ca tam gic u cnh 10cm c ba in tch bng nhau v bng 10nC. Hy xc nhcng in trng ti tm ca tam gic:

A. 0 B. 1200V/m C. 2400V/m D. 3600V/mCu hi 5: Mt in tch im q = 2,5C t ti im M trong in trng u m in trng c hai thnh

phn Ex = +6000V/m, Ey = - 6 3 .103 V/m. Vct lc tc dng ln in tch q l:A. F = 0,03N, lp vi trc Oy mt gc 1500 B. F = 0,3N, lp vi trc Oy mt gc 300

C. F = 0,03N, lp vi trc Oy mt gc 1150 D.F = 0,12N, lp vi trc Oy mt gc 1200

Cu hi 6: Ba in tch im cng ln, cng du q t ti ba nh ca mt tam gic u cnh a. Xc nhcng in trng ti im t ca mi in tch do hai in tch kia gy ra:

A. E = k2

22

a

q B.E = 2k2

3

a

q C. E = k2

3

a

q D. E = ka

q 3

Cu hi 7: Hai in tch im cng ln q, tri du, t ti 2 nh ca mt tam gic u cnh a. Xc nhcng in trng ti nh cn li ca tam gic do hai in tch kia gy ra:

A. E = k 2a

q

B. E = k 23

a

q

C. E = 2k 2a

q

D. E = 2

1

k 2a

q

Cu hi 8: Bn in tch im cng ln cng du q t ti bn nh ca hnh vung cnh a. Xc nhcng in trng gy ra bi bn in tch ti tm ca hnh vung:

A. E = 2k2a

qB. E = 4k

2

2

a

q C. 0 D. E = k2

3

a

q

Cu hi 9: Bn in tch im cng ln q, hai in tch dng v hai in tch m, t ti bn nh cahnh vung cnh a, cc in tch cng du k nhau. Xc nh cng in trng gy ra bi bn in tch ti tm ca hnh vung:

A. E = 2k 23

a

q

B. E = k 23

a

q

C. E = k 223

a

q

D. E = 4k 22

a

q

Cu hi 10: Hai in tch dng q t ti A v B, AB = a. Xc nh vct cng in trng ti im Mtrn ng trung trc ca on thng AB cch trung im O ca AB mt on OM = a 3 /6:

A.E = k2a

q, hng theo trung trc ca AB i xa AB B.E = k

2

2

a

q, hng theo trung trc ca AB i vo AB

C. E = k2

3

a

q, hng theo trung trc ca AB i xa AB D. E = k

2

3

a

q, hng hng song song vi AB

Cu 1 2 3 4 5 6 7 8 9 10

p n C D D A A C B C D C

in trng - Dng 2: Nguyn l chng cht in trng - 2Cu hi 1: Hai in tch +q v - q t ln lt ti A v B, AB = a. Xc nh vct cng in trng tiim M trn ng trung trc ca on thng AB cch trung im O ca AB mt on OM = a 3 /6:

A.E = k2

2

a

q , hng song song vi AB B.E = k2

2

a

q, hng song song vi AB

C. E = k2

3

a

q, hng theo trung trc ca AB i xa AB D. E = k

2

33

a

q , hng song song vi AB


51/274

[email protected] tm v bin sonCu hi 2: Hai in tch t trong khng kh ti M v N. Ti I nm trn ng trung trc ca MN cch MNmt on IH c vct cng in trng tng hp IE nm theo ng trung trc IH v hng ra xa MNth hai in tch c c im:A. q1 > 0; q2 > 0; q1 = q2 B. q1 > 0; q2 < 0; |q1| = |q2| C. q1 < 0; q2 < 0; q1 = q2 D. q1 < 0; q2 >0; |q1| = |q2|

Cu hi 3: Hai in tch t trong khng kh ti M v N. Ti I nm trn ng trung trc ca MN cch MNmt on IH c vct cng in trng tng hp IE nm theo ng trung trc IH v hng li gnMN th hai in tch c c im:A. q1 > 0; q2 > 0; q1 = q2 B. q1 > 0; q2 < 0; |q1| = |q2| C. q1 < 0; q2 < 0; q1 = q2 D. q1 < 0; q2 >0; |q1| = |q2|

Cu hi 4: Hai in tch t trong khng kh ti M v N. Ti I nm trn ng trung trc ca MN cch MNmt on IH c vct cng in trng tng hp

IE song song vi MN th hai in tch c cim:A. q1 > 0; q2 > 0; q1 = q2 B. q1 > 0; q2 < 0; |q1| = |q2| C. q1 < 0; q2 >0; |q1| = |q2| D. B hoc CCu hi 5: Hai in tch q1 = +q v q2 = - q t ti A v B trong khng kh, bit AB = 2a. ln cng in trng ti M trn ng trung trc ca AB cch AB mt on h l:

A. 222

ha

kq

+B.

( )22222

ha

kqa

+C.

( ) 23

22

2

ha

kqa

+

D.22

22ha

kqa

+

Cu hi 6: Hai in tch q1 = +q v q2 = - q t ti A v B trong khng kh, bit AB = 2a. ti M trn ngtrung trc ca AB cch AB mt on h EM c gi tr cc i. Gi tr cc i l:

A. 22akq

B. 2akq

C. 22

a

kqD. 2

4

a

kq

Cu hi 7: Ba in tch q1, q2, q3 t trong khng kh ln lt ti cc nh A, B, C ca hnh vung ABCD.Bit vct cng in trng tng hp ti D c gi l cnh CD. Quan h gia 3 in tch trn l: A.

q1 = q2 = q3 B. q1 = - q2 = q3 C. q2 = - 2 2 q1 D. q3 = - 2 2 q2Cu hi 8: Hai in tch im q1 = 2.10-2 (C) v q2 = - 2.10-2 (C) t ti hai im A v B cch nhau mton a = 30 (cm) trong khng kh. Cng in trng ti im M cch u A v B mt khong bng a c ln l:

A. EM = 0,2 (V/m). B. EM = 1732 (V/m). C. EM = 3464 (V/m). D. EM = 2000 (V/m).Cu hi 9: Hai in tch q1 = 5.10-16 (C), q2 = - 5.10-16 (C), t ti hai nh B v C ca mt tam gic uABC cnh bng 8 (cm) trong khng kh. Cng in trng ti nh A ca tam gic ABC c ln l:

A. E = 1,2178.10-3 (V/m). B. E = 0,6089.10-3 (V/m).C. E = 0,3515.10-3 (V/m). D. E = 0,7031.10-3 (V/m).

Cu hi 10: Hai in tch im q1 = 0,5 (nC) v q2 = - 0,5 (nC) t ti hai im A, B cch nhau 6 (cm) trongkhng kh. Cng in trng ti trung im ca AB c ln l:

A. E = 0 (V/m). B. E = 5000 (V/m). C. E = 10000 (V/m). D. E = 20000 (V/m).Cu 1 2 3 4 5 6 7 8 9 10p n D A C D C A C B D C

in trng - Dng 3: q cn bng trong in trng, Etrit tiu - 1Cu hi 1: Hai in tch im q1 v q2 t ti hai im c nh A v B. Ti im M trn ng thng ni ABv gn A hn B ngi ta thy in trng ti c cng bng khng. Kt lun g v q1 , q2:

A. q1 v q2 cng du, |q1| > |q2| B. q1 v q2 tri du, |q1| > |q2|C. q1 v q2 cng du, |q1| < |q2| D. q1 v q2 tri du, |q1| < |q2|

Cu hi 2: Hai in tch im q1 = - 9C, q2 = 4 C t ln lt ti A, B cch nhau 20cm. Tm v tr im M

ti in trng bng khng:A. M nm trn on thng AB, gia AB, cch B 8cmB M t th AB i B h B 40


52/274


C. M nm trn ng thng AB, ngoi gn A cch A 40cmD. M l trung im ca AB

Cu hi 3: Hai in tch im q1 = - 4 C, q2 = 1 C t ln lt ti A v B cch nhau 8cm. Xc nh v trim M ti cng in trng bng khng:A. M nm trn AB, cch A 10cm, cch B 18cm B. M nm trn AB, cch A 8cm, cch B 16cmC. M nm trn AB, cch A 18cm, cch B 10cm D. M nm trn AB, cch A 16cm, cch B 8cmCu hi 4: Hai tm kim loi phng nm ngang nhim in tri du t trong du, in trng gia hai bn lin trng u hng t trn xung di v c cng 20 000V/m. Mt qu cu bng st bn knh 1cmmang in tch q nm l lng gia khong khng gian gia hai tm kim loi. Bit khi lng ring ca stl 7800kg/m3, ca du l 800kg/m3, ly g = 10m/s2. Tm du v ln ca q:

A. - 12,7 C B. 14,7 C C. - 14,7 C D. 12,7 CCu hi 5: Mt qu cu khi lng 1g treo u mt si dy mnh cch in. H thng nm trong intrng u c phng nm ngang, cng E = 2kV/m. Khi dy treo hp vi phng thng ng mtgc 600. Tm in tch ca qu cu, ly g = 10m/s2:

A. 5,8 C B. 6,67 C C. 7,26 C D. 8,67CCu hi 6: Mt qu cu kim loi nh c khi lng 1g c tch in q = 10 -5C treo vo u mt si dy

mnh v t trong in trng u E. Khi qu cu ng cn bng th dy treo hp vi phng thng ngmt gc 600, ly g = 10m/s2. Tm E:A. 1730V/m B. 1520V/m C. 1341V/m D. 1124V/m

Cu hi 7: Hai qu cu nh mang in tch q1 = - 2nC, q2 = +2nC, c treo u haisi dy cch in di bng nhau trong khng kh ti hai im treo M, N cch nhau2cm cng mt cao. Khi h cn bng hai dy treo lch khi phng thng ng,mun a cc dy treo v v tr phng thng ng th phi to mt in trng uEc hng no ln bao nhiu:

A. Nm ngang hng sang phi, E = 1,5.104V/mB. Nm ngang hng sang tri, E = 3.104V/mC. Nm ngang hng sang phi, E = 4,5.104V/m

D. Nm ngang hng sang tri, E = 3,5.104V/mCu hi 8: Mt vin bi nh kim loi khi lng 9.10 -5kg th tch 10mm3 c t trong du c khi lngring 800kg/m3. Chng t trong in trng u E = 4,1.105 V/m c hng thng ng t trn xung, thyvin bi nm l lng, ly g = 10m/s2. in tch ca bi l:

A. - 1nC B. 1,5nC C. - 2nC D. 2,5nCCu hi 9: Hai in tch q1 = q2 = q t trong chn khng ln lt ti hai im A v B cch nhau mtkhong l. Ti I ngi ta thy in trng ti bng khng. Hi I c v tr no sau y:

A. AI = BI = l/2 B. AI = l; BI = 2l C. BI = l; AI = 2l D. AI = l/3; BI = 2l/3Cu hi 10: Hai in tch im q1 = 36 C v q2 = 4 C t trong khng kh ln lt ti hai im A v Bcch nhau 100cm. Ti im C in trng tng hp trit tiu, C c v tr no:

A. bn trong on AB, cch A 75cm B. bn trong on AB, cch A 60cm

C. bn trong on AB, cch A 30cm D. bn trong on AB, cch A 15cmCu 1 2 3 4 5 6 7 8 9 10p n C B D C D A C C A A

in trng - Dng 3: q cn bng trong in trng, Etrit tiu - 2Cu hi 1: Ba in tch q1, q2, q3 t trong khng kh ln lt ti cc nh A, B, C ca hnh vung ABCD.Bit in trng tng hp ti D trit tiu. Quan h gia 3 in tch trn l:

A. q1 = q3; q2 = -2 2 q1 B. q1 = - q3; q2 = 2 2 q1

C. q1 = q3; q2 = 2 2 q1 D. q2 = q3 = - 2 2 q1Cu hi 2: Mt qu cu khi lng 1g treo u mt si dy mnh cch in. H thng nm trong intrng u c phng nm ngang, cng E = 2kV/m. Khi dy treo hp vi phng thng ng mtgc 600. Tm sc cng ca si dy, ly g = 10m/s2:

A 0 01N B 0 03N C 0 15N D 0 02N

M N

q1

q2


53/274

[email protected] tm v bin sonCu hi 3: Hai in tch im q v -q t ln lt ti A v B. in trng tng hp trit tiu ti:

A. Mt im trong khong ABB. Mt im ngoi khong AB, gn A hnC. Mt im ngoi khong AB, gn B hnD. in trng tng hp khng th trit tiu ti bt c im no

Cu hi 4: Hai in tch im q1 v q2 t hai nh A v B ca tam gic u ABC. in trng C bngkhng, ta c th kt lun:

A. q1 = - q2 B. q1 = q2C. q1 q2 D. Phi c thm in tch q3 nm u

Cu hi 5: Hai in tch im q1 = - q2 = 3C t ln lt ti A v B cch nhau 20cm. in trng tng hpti trung im O ca AB c:

A. ln bng khng B. Hng t O n B, E = 2,7.106V/mC. Hng t O n A, E = 5,4.106V/m D. Hng t O n B, E = 5,4.106V/m

Cu hi 6: Hai in tch im q1 = - 2,5 C v q2 = + 6 C t ln lt ti A v B cch nhau 100cm. intrng tng hp trit tiu ti:

A. trung im ca AB

B. im M trn ng thng AB, ngoi on AB, cch B mt on 1,8mC. im M trn ng thng AB, ngoi on AB, cch A mt on 1,8mD. in trng tng hp khng th trit tiu

Cu hi 7: Cc in tch q1 v q2 = q1 t ln lt ti hai nh A v C ca mt hnh vung ABCD. intrng tng hp ti nh D bng khng th phi t ti nh B mt in tch q3 c ln v du bng:

A. - q1 B. - 2 q1C. -2 2 q1 D. khng th tm c v khng bit chiu di ca cnh hnh vung

Cu hi 8: Ba in tch im bng nhau q > 0 t ti ba nh ca mt tam gic u ABC. in trng tnghp trit tiu ti:

A. mt nh ca tam gic B. tm ca tam gicC. trung im mt cnh ca tam gic D. khng th trit tiu

Cu hi 9: Ba in tch im bng nhau q < 0 t ti ba nh ca mt tam gic u ABC. in trng tnghp trit tiu ti:A. mt nh ca tam gic B. tm ca tam gicC. trung im mt cnh ca tam gic D. khng th trit tiu

Cu hi 10: Ba in tch im q1, q2 = - 12,5.10-8C, q3 t ln lt ti A, B, C ca hnh ch nht ABCD cnhAD = a = 3cm, AB = b = 4cm. in trng tng hp ti nh D bng khng. Tnh q1 v q3:

A. q1 = 2,7.10-8C; q3 = 6,4.10-8C B. q1 = - 2,7.10-8C; q3 = - 6,4.10-8CC. q1 = 5,7.10-8C; q3 = 3,4.10-8C D. q1 = - 5,7.10-8C; q3 = - 3,4.10-8C

Cu 1 2 3 4 5 6 7 8 9 10p n A D D D D C C B B A

A, U, V - Dng 1: Tnh A, U, V ca lc in trng - 1Cu hi 1: Mt in trng u cng 4000V/m, c phng song song vi cnh huyn BC ca mt tamgic vung ABC c chiu t B n C, bit AB = 6cm, AC = 8cm. Tnh hiu in th gia hai im BC:

A. 400V B. 300V C. 200V D. 100V

Cu hi 2: Mt in tch q chuyn ng t im M n Q, n N, n P trongin trng u nh hnh v. p n no l sai khi ni v mi quan h gia cngca lc in trng dch chuyn in tch trn cc on ng:

A. AMQ = - AQN B. AMN = ANP C. AQP = AQN D. AMQ = AMPCu hi 3: Hai tm kim loi phng song song cch nhau 2cm nhim in tri du. Mun lm cho in tch q

5 10 10C di h ki A 2 10 9J X h i

MQ

N

P


54/274


bn trong hai tm kim loi, bit in trng bn trong l in trng u c ng sc vung gc vi cctm, khng i theo thi gian:

A. 100V/m B. 200V/m C. 300V/m D. 400V/m

Cu hi 4: Hiu in th gia hai im M, N l UMN = 2V. Mt in tch q = -1C di chuyn t M n N thcng ca lc in trng l:

A. -2J B. 2J C. - 0,5J D. 0,5J

Cu hi 5: Mt ht bi khi lng 3,6.10 -15kg mang in tch q = 4,8.10 -18C nm l lng gia hai tm kimloi phng song song nm ngang cch nhau 2cm v nhim in tri du . Ly g = 10m/s 2, tnh hiu in thgia hai tm kim loi:

A. 25V. B. 50V C. 75V D. 100V

Cu hi 6: Mt qu cu kim loi khi lng 4,5.10 -3kg treo vo u mt si dy di 1m, qu cu nm giahai tm kim loi phng song song thng ng cch nhau 4cm, t hiu in th gia hai tm l 750V, thqu cu lch 1cm ra khi v tr ban u, ly g = 10m/s2. Tnh in tch ca qu cu:

A. 24nC B. - 24nC C. 48nC D. - 36nCCu hi 7: Gi thit rng mt tia st c in tch q = 25C c phng t m my dng xung mt t, khi hiu in th gia m my v mt t U = 1,4.108V. Tnh nng lng ca tia st :

A. 35.108J B. 45.108 J C. 55.108 J D. 65.108 J

Cu hi 8: Mt in tch im q = + 10C chuyn ng t nh B n nh C ca tam gic u ABC, nmtrong in trng u c cng 5000V/m c ng sc in trng song song vi cnh BC c chiu t Cn B. Bit cnh tam gic bng 10cm, tm cng ca lc in trng khi di chuyn in tch trn theo onthng B n C:

A. 2,5.10-4J B. - 2,5.10-4J C. - 5.10-4J D. 5.10-4J

Cu hi 9: Mt in tch im q = + 10C chuyn ng t nh B n nh C ca tam gic u ABC, nmtrong in trng u c cng 5000V/m c ng sc in trng song song vi cnh BC c chiu t Cn B. Bit cnh tam gic bng 10cm, tm cng ca lc in trng khi di chuyn in tch trn theo ongp khc BAC:

A. - 10.10-4J B. - 2,5.10-4J C. - 5.10-4J D. 10.10-4J

Cu hi 10: Mt trong ca mng t bo trong c th sng mang in tch m, mt ngoi mang in tchdng. Hiu in th gia hai mt ny bng 0,07V. Mng t bo dy 8nm. Cng in trng trongmng t bo ny l:

A. 8,75.106V/m B. 7,75.106V/m C. 6,75.106V/m D. 5,75.106V/m

Cu 1 2 3 4 5 6 7 8 9 10p n A D B B C B A C C A

A, U, V - Dng 1: Tnh A, U, V ca lc in trng - 2Cu hi 1: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V.Tnh cng in trng v cho bit c im in trng, dng ng sc in trng gia hai tm kimloi:

A. in trng bin i, ng sc l ng cong, E = 1200V/mB. in trng bin i tng dn, ng sc l ng trn, E = 800V/m

C. in trng u, ng sc l ng thng, E = 1200V/mD. in trng u, ng sc l ng thng, E = 1000V/m


55/274

[email protected] tm v bin sonCu hi 2: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V.Mt electron khng vn tc ban u chuyn ng t tm tch in m v tm tch in dng. Hi khi ntm tch in dng th electron nhn c mt nng lng bng bao nhiu:

A. 8.10-18J B. 7.10-18J C. 6.10-18J D. 5.10-18JCu hi 3: Cng ca lc in trng lm di chuyn mt in tch gia hai im c hiu in th U = 2000V

l 1J. Tnh ln in tch :A. 2mC B. 4.10-2C C. 5mC D. 5.10-4C

Cu hi 4: Gia hai im A v B c hiu in th bng bao nhiu nu mt in tch q = 1C thu c nnglng 2.10-4J khi i t A n B:

A. 100V B. 200V C. 300V D. 500VCu hi 5: Cho ba bn kim loi phng tch in 1, 2, 3 t song song ln lt nhau cch nhau nhng khongd12 = 5cm, d23 = 8cm, bn 1 v 3 tch in dng, bn 2 tch in m. E12 = 4.104V/m, E23 = 5.104V/m, tnhin th V2, V3 ca cc bn 2 v 3 nu ly gc in th bn 1:

A. V2 = 2000V; V3 = 4000V B. V2 = - 2000V; V3 = 4000VC. V2 = - 2000V; V3 = 2000V D. V2 = 2000V; V3 = - 2000V

Cu hi 6: Mt qu cu kim loi bn knh 10cm. Tnh in th gy bi qu cu ti im A cch tm qu cu40cm v ti im B trn mt qu cu, bit in tch ca qu cu l.10-9C:

A. VA = 12,5V; VB = 90V B. VA = 18,2V; VB = 36VC. VA = 22,5V; VB = 76V D.VA = 22,5V; VB = 90V

Cu hi 7: Mt qu cu kim loi bn knh 10cm. Tnh in th gy bi qu cu ti im A cch tm qu cu40cm v ti im B trn mt qu cu, bit in tch ca qu cu l - 5.10-8C:

A. VA = - 4500V; VB = 1125V B. VA = - 1125V; VB = - 4500V

C. VA = 1125,5V; VB = 2376V D. VA = 922V; VB = - 5490V

Cu hi 8: Mt git thy ngn hnh cu bn knh 1mm tch in q = 3,2.10 -13C t trong khng kh. Tnhcng in trng v in th ca git thy ngn trn b mt git thy ngn:

A. 2880V/m; 2,88V B. 3200V/m; 2,88VC. 3200V/m; 3,2V D. 2880; 3,45V

Cu hi 9: Mt ht bi kim loi tch in m khi lng 10 -10kg l lng trong khong gia hai bn t inphng nm ngang bn tch in dng trn, bn tch in m di. Hiu in th gia hai bn bng1000V, khong cch gia hai bn l 4,8mm, ly g = 10m/s2. Tnh s electron d ht bi:

A. 20 000 ht B. 25000 ht C. 30 000 ht D. 40 000 ht

Cu hi 10: Mt in trng u E = 300V/m. Tnh cng ca lc in trng trndi chuyn in tch q = 10nC trn qu o ABC vi ABC l tam gic u cnh a =10cm nh hnh v:

A. 4,5.10-7J B. 3. 10-7J C. - 1.5. 10-7J D. 1.5. 10-7J

Cu 1 2 3 4 5 6 7 8 9 10p n D A D B C D B A C D

A, U, V - Dng 1: Tnh A, U, V ca lc in trng - 3:

Cu hi 1: Xt 3 im A, B, C 3 nh ca tam gic vung nh hnh v, =600, BC = 6cm, UBC = 120V. Cc hiu in th UAC ,UBA c gi tr ln lt:A. 0; 120V B. - 120V; 0 C. 60 3 V; 60V D. - 60 3 V; 60VCu hi 2: Mt ht bi khi lng 1g mang in tch - 1C nm yn cn bng trong in trng gia hai

bn kim loi phng nm ngang tch in tri du c ln bng nhau. Khong cch gia hai bn l 2cm, ly

A

B C

E

AB

C

E


56/274


A. 20V B. 200V C. 2000V D. 20 000VCu hi 3: Mt prtn mang in tch + 1,6.10-19C chuyn ng dc theo phng ca ng sc mt intrng u. Khi n i c qung ng 2,5cm th lc in thc hin mt cng l + 1,6.10 -20J. Tnh cng in trng u ny:

A. 1V/m B. 2V/m C. 3V/m D. 4V/mCu hi 4: Gi thit rng mt tia st c in tch q = 25C c phng t m my dng xung mt t, khi hiu in th gia m my v mt t U = 1,4.108V. Nng lng ca tia st ny c th lm bao nhiukilgam nc 1000C bc thnh hi 1000C, bit nhit ha hi ca nc bng 2,3.106J/kg

A. 1120kg B. 1521kg C. 2172kg D. 2247kgCu hi 5: Mt in trng u cng 4000V/m, c phng song song vi cnh huyn BC ca mt tamgic vung ABC c chiu t B n C, bit AB = 6cm, AC = 8cm. Tnh hiu in th gia hai im AC:

A. 256V B. 180V C. 128V D. 56VCu hi 6: Mt in trng u cng 4000V/m, c phng song song vi cnh huyn BC ca mt tamgic vung ABC c chiu t B n C, bit AB = 6cm, AC = 8cm. Tnh hiu in th gia hai im BA:

A. 144V B. 120V C. 72V D. 44VCu hi 7: Cng ca lc in trng lm di chuyn mt in tch gia hai im c hiu in th U = 2000

(V) l A = 1 (J). ln ca in tch lA. q = 2.10-4 (C). B. q = 2.10-4 (C). C. q = 5.10-4 (C). D. q = 5.10-4 (C).Cu hi 8: Hai tm kim loi song song, cch nhau 2 (cm) v c nhim in tri du nhau. Mun lm choin tch q = 5.10-10 (C) di chuyn t tm ny n tm kia cn tn mt cng A = 2.10 -9 (J). Coi in trng

bn trong khong gia hai tm kim loi l in trng u v c cc ng sc in vung gc vi cc tmCng in trng bn trong tm kim loi l:

A. E = 2 (V/m). B. E = 40 (V/m). C. E = 200 (V/m). D. E = 400 (V/m).Cu hi 9: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V.Mt electron khng vn tc ban u chuyn ng t tm tch in m v tm tch in dng. Hi khi ntm tch in dng th electron c vn tc bng bao nhiu:

A. 4,2.106m/s B. 3,2.106m/s C. 2,2.106m/s D.1,2.106m/sCu hi 10: Cho hai bn kim loi phng t song song tch in tri du, th mt lectron khng vn tc banu vo in trng gia hai bn kim loi trn. B qua tc dng ca trng trng. Qu o ca lectron l:

A. ng thng song song vi cc ng sc in.

B. ng thng vung gc vi cc ng sc in.

C. mt phn ca ng hypebol.

D. mt phn ca ng parabol.

Cu 1 2 3 4 5 6 7 8 9 10

p n A B D B A A B D A B

A, U, V - Dng 2: Chuyn ng ca q trong in trng - 1:Cu hi 1: Mt electrn chuyn ng dc theo hng ng sc ca mt in trng u c cng 100V/m vi vn tc ban u l 300 km/s . Hi n chuyn ng c qung ng di bao nhiu th vn tcca n bng khng:

A. 2,56cm B. 25,6cm C. 2,56mm D. 2,56mCu hi 2: Trong n hnh ca my thu hnh, cc electrn c tng tc bi hiu in th 25 000V. Hi khip vo mn hnh th vn tc ca n bng bao nhiu, b qua vn tc ban u ca n:

A. 6,4.107m/s B. 7,4.107m/s C. 8,4.107m/s D. 9,4.107m/s

Cu hi 3: Mt prtn bay theo phng ca mt ng sc in trng. Lc im A n c vn tc2,5.104m/s, khi n im B vn tc ca n bng khng. Bit n c khi lng 1,67.10 -27kg v c in tch1 6 10-19C in th ti A l 500V tm in th ti B:


57/274

[email protected] tm v bin sonCu hi 4: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V.Mt electron khng vn tc ban u chuyn ng t tm tch in m v tm tch in dng. Hi khi ntm tch in dng th electron c vn tc bao nhiu:

A. 4,2.106m/s B. 3,2.106m/s C. 2,2.106m/s D. 1,2.106m/sCu hi 5: Trong Vt l ht nhn ngi ta hay dng n v nng lng l eV. eV l nng lng m mtelectrn thu c khi n i qua on ng c hiu in th 1V. Tnh eV ra Jun, v vn tc ca electrn cnng lng 0,1MeV:

A. 1eV = 1,6.1019J B. 1eV = 22,4.1024 J;C. 1eV = 9,1.10-31J D. 1eV = 1,6.10-19J

Cu hi 6: Hai bn kim loi phng nm ngang song song cch nhau 10cm c hiu in th gia hai bn l100V. Mt electrn c vn tc ban u 5.106m/s chuyn ng dc theo ng sc v bn m. Tnh gia tcca n. Bit in trng gia hai bn l in trng u v b qua tc dng ca trng lc:

A. -17,6.1013m/s2 B. 15.9.1013m/s2 C. - 27,6.1013m/s2 D. + 15,2.1013m/s2

Cu hi 7: Mt ht bi kim loi tch in m khi lng 10 -10kg l lng trong khong gia hai bn t inphng nm ngang bn tch in dng trn, bn tch in m di. Hiu in th gia hai bn bng

1000V, khong cch gia hai bn l 4,8mm, ly g = 10m/s2

. Chiu tia t ngoi lm ht bi mt mt selectrn v ri xung vi gia tc 6m/s2. Tnh s ht electrn m ht bi mt:A. 18 000 ht B. 20000 ht C. 24 000 ht D. 28 000 ht

Cu hi 8: Mt electrn chuyn ng dc theo mt ng sc ca in trng u c cng 364V/mElectrn xut pht t im M vi vn tc 3,2.106m/s i c qung ng di bao nhiu th vn tc ca n

bng khng:A. 6cm B. 8cm C. 9cm D. 11cm

Cu hi 9: Mt electrn chuyn ng dc theo mt ng sc ca in trng u c cng 364V/mElectrn xut pht t im M vi vn tc 3,2.106m/s. Thi gian k t lc xut pht n khi n quay tr vim M l:

A. 0,1s B. 0,2 s C. 2 s D. 3 s

Cu hi 10: Hai bn kim loi phng nm ngang song song cch nhau 10cm c hiu in th gia hai bn l100V. Mt electrn c vn tc ban u 5.106m/s chuyn ng dc theo ng sc v bn m. Tnh onng n i c cho n khi dng li. Bit in trng gia hai bn l in trng u v b qua tc dngca trng lc:

A. 7,1cm B. 12,2cm C. 5,1cm D. 15,2cm

Cu 1 2 3 4 5 6 7 8 9 10

p n C D C A D A D B A AA, U, V - Dng 2: Chuyn ng ca q trong in trng - 2:Cu hi 1: Mt electrn c phng i t O vi vn tc ban u v 0 vung gc vi cc ng sc ca mtin trng u cng E. Khi n im B cch O mt on h theo phng ca ng sc vn tc ca nc biu thc:A. Ehe B. Ehev +20 C. Ehev 20 D.

hm

Eev 220 +

Cu hi 2: Mt electrn c phng i t O vi vn tc ban u v 0 dc theo ng sc ca mt in trngu cng E cng hng in trng. Qung ng xa nht m n di chuyn c trong in trng choti khi vn tc ca n bng khng c biu thc:

Amv20 B

2 EeC

2

0Emve D2


58/274

[email protected] tm v bin sonCu hi 3:Electron chuyn ng khng vn tc ban u t A n B trong mt in trng u vi U AB =45,5V. Ti B vn tc ca n l:

A. 106m/s2 B. 1,5./s2 C. 4.106m /s2 D. 8.106m/s2Cu hi 4:Khi bay t M n N trong in trng u, electron tng tc ng nng tng thm 250eV. Hiuin th UMN bng:

A. -250V B. 250V C. - 125V D. 125VCu hi 5: Mt t in phng c cc bn nm ngang cch nhau khong d, chiu di cc bn l l. Gia hai

bn c hiu in th U. Mt electron bay vo in trng ca t t im O gia cch u hai bn vi vntc 0v song song vi cc bn. ln gia tc ca n trong in trng l:

A.d

UeB.md

UeC. 2

0mdv

UleD. 2

0dv

Ule

Cu hi 6: Mt t in phng c cc bn nm ngang cch nhau khong d, chiu di cc bn l l. Gia haibn c hiu in th U. Mt electron bay vo in trng ca t t im O gia cch u hai bn vi vntc 0v song song vi cc bn. lch ca n theo phng vung gc vi cc bn khi ra khi in trng c

biu thc:

A. dUe

B. mdUe

C. 20

mdv

Ule

D

2

0

2

2mdv

Ule

Cu hi 7: : Mt t in phng c cc bn nm ngang cch nhau khong d, chiu di cc bn l l. Gia haibn c hiu in th U. Mt electron bay vo in trng ca t t im O gia cch u hai bn vi vntc 0v song song vi cc bn. Gc lch gia hng vn tc ca n khi va ra khi in trng v so vi

0v c tan c tnh bi biu thc:

A.d

UeB.md

UeC. 2

0mdv

UleD. 2

0

2

2mdv

Ule

Cu hi 8: Mt electron bay vo in trng ca mt t in phng theo phng song song cng hng vicc ng sc in trng vi vn tc ban u l 8.10 6m/s. Hiu in th t phi c gi tr nh nht l baonhiu electron khng ti c bn i din

A. 182V B.91V C. 45,5V D.50VCu hi 9: Khi mt electron chuyn ng ngc hng vi vect cng in trng th:

A. th nng ca n tng, in th ca n gim B. th nng gim, in th tngC. th nng v in th u gim D. th nng v in th u tng

Cu hi 10: Mt electron c tng tc t trng thi ng yn nh hiu in th U = 200V. Vn tc cui mn t c l: A. 2000m/s B. 8,4.106m/s C. 2.105m/s D. 2,1.106m/sCu 1 2 3 4 5 6 7 8 9 10

p n D A C A B D C A B BA, U, V - Dng 2: Chuyn ng ca q trong in trng - 3:Cu hi 1: Mt prtn v mt mt electron ln lt c tng tc t trng thi ng yn trong cc intrng u c cng in trng bng nhau v i c nhng qung ng bng nhau th:

A. C hai c cng ng nng, electron c gia tc ln hnB. C hai c cng ng nng, electron c gia tc nh hnC. prtn c ng nng ln hn. electron c gia tc ln hnD. electron c ng nng ln hn. Electron c gia tc nh hn

Cu hi 2: Mt electron th cho chuyn ng khng vn tc ban u trong in trng u gia hai mtng th V1 = +10V, V2 = -5V. N s chuyn ng :

A. V pha mt ng th V1 B. V pha mt ng th V2C. Ty cng in trng m n c th v V1 hay V2. D. n ng yn

Cu hi 3: Mt electrn c phng i t O vi vn tc ban u v 0 dc theo ng sc ca mt in trng


59/274


A. Ehe B. Ehev +20 C. Ehev 20 D. hm

Eev 220 +

Cu hi 4: Trong Vt l ht nhn ngi ta hay dng n v nng lng l eV. eV l nng lng m mtelectrn thu c khi n i qua on ng c hiu in th 1V. Tnh vn tc ca electrn c nng lng0,1MeV:

A. v = 0,87.108

m/s B. v = 2,14.108

m/s C. v = 2,87.108

m/s D. v = 1,87.108

m/sCu hi 5: Hiu in th gia hai im bn ngoi v bn trong ca mt mng t bo l - 90mV, b dy camng t bo l 10nm, th in trng( gi s l u) gia mng t bo c cng l:

A. 9.106 V/m B. 9.1010 V/m C. 1010 V/m D. 106 V/mCu hi 6: Khi st nh xung mt t th c mt lng in tch - 30C di chuyn t m my xung mtt. Bit hiu in th gia mt t v m my l 2.10 7V. Nng lng m tia st ny truyn t m myxung mt t bng:

A. 1,5.10-7J B. 0,67.107J C. 6.109J D. 6.108JCu hi 7: Chn mt p n sai :A. Khi mt in tch chuyn ng trn mt mt ng th th cng ca lc in bng khngB. Lc in tc dng ln mt in tch q trong mt mt ng th c phng tip tuyn vi mt ng th

C. Vct cng in trng ti mi im trong mt ng th c phng vung gc vi mt ng thD. Khi mt in tch di chuyn t mt mt ng th ny sang mt mt ng th khc th cng ca lc inchc chn khc khngCu hi 8: Khi electron chuyn ng t bn tch in dng v pha bn m trong khong khng gian giahai bn kim loi phng tch in tri du ln bng nhau th:

A. Lc in thc hin cng dng, th nng lc in tngB. Lc in thc hin cng dng, th nng lc in gimC. Lc in thc hin cng m, th nng lc in tngD. Lc in thc hin cng m, th nng lc in gim

Cu hi 9: Hai im A v B nm trn cng mt mt ng th. Mt in tch q chuyn ng t A n B th:A. lc in thc hin cng dng nu q > 0, thc hin cng m nu q < 0

B. lc in thc hin cng dng hay m ty vo du ca q v gi tr in th ca A(B)C. phi bit chiu ca lc in mi xc nh c du ca cng lc in trngD. lc in khng thc hin cng

Cu hi 10: Mt in tch +1C chuyn ng t bn tch in dng sang bn tch in m t song song idin nhau th lc in thc hin mt cng bng 200J. Hiu in th gia hai bn c ln bng:

A. 5.10-3V. B. 200V C. 1,6.10-19V D. 2000VCu 1 2 3 4 5 6 7 8 9 10p n A A D D A C B C D B

T in - Dng 1: in dung, nng lng in trng - 1Cu hi 1: Mt t in in dung 5F c tch in n in tch bng 86C. Tnh hiu in th trn hai

bn t:

A. 17,2V B. 27,2V C.37,2V D. 47,2V

Cu hi 2: Mt t in in dung 24nF tch in n hiu in th 450V th c bao nhiu electron mi dichuyn n bn m ca t in:

A. 575.1011 electron B. 675.1011 electron

C. 775.1011 electron D. 875.1011 electron

Cu hi 3: B t in trong chic n chp nh c in dung 750 F c tch in n hiu in th 330V

Xc nh nng lng m n tiu th trong mi ln n le sng:A. 20,8J B. 30,8J C. 40,8J D. 50,8J


60/274

[email protected] tm v bin sonCu hi 4: B t in trong chic n chp nh c in dung 750 F c tch in n hiu in th 330VMi ln n le sng t in phng in trong thi gian 5ms. Tnh cng sut phng in ca t in:

A. 5,17kW B.6 ,17kW C. 8,17kW D. 8,17kW

Cu hi 5:Mt t in c in dung 500pF mc vo hai cc ca mt my pht in c hiu in th 220V.Tnh in tch ca t in:

A. 0,31C B. 0,21C C.0,11C D.0,01C

Cu hi 6: T in phng khng kh c in dung 5nF. Cng in trng ln nht m t c th chuc l 3.105V/m, khong cch gia hai bn l 2mm. in tch ln nht c th tch cho t l:

A. 2 C B. 3 C C. 2,5C D. 4C

Cu hi 7: Nng lng in trng trong t in t l vi:

A. hiu in th gia hai bn t in

B. in tch trn t in

C. bnh phng hiu in th hai bn t inD. hiu in th hai bn t v in tch trn t

Cu hi 8: Mt t in c in dung 5nF, in trng ln nht m t c th chu c l 3.10 5V/m, khongcch gia hai bn l 2mm. Hiu in th ln nht gia hai bn t l:

A. 600V B. 400V C. 500V D.800V

Cu hi 9: Mt t in c in dung 2000 pF mc vo hai cc ca ngun in hiu in th 5000V. Tnhin tch ca t in:

A. 10C B. 20 C C. 30C D. 40C

Cu hi 10: Mt t in c in dung 2000 pF mc vo hai cc ca ngun in hiu in th 5000V. Tchin cho t ri ngt khi ngun, tng in dung t ln hai ln th hiu in th ca t khi l:

A. 2500V B. 5000V C. 10 000V D. 1250VCu 1 2 3 4 5 6 7 8 9 10

p n A B C D C B C A A A

T in - Dng 1: in dung, nng lng in trng - 2Cu hi 1: Mt t in c th chu c in trng gii hn l 3.106V/m, khong cch gia hai bn t l1mm, in dung l 8,85.10-11F. Hi hiu in th ti a c th t vo hai bn t l bao nhiu:

A. 3000V B. 300V C. 30 000V D.1500VCu hi 2: Mt t in c th chu c in trng gii hn l 3.106V/m, khong cch gia hai bn t l

1mm, in dung l 8,85.10-11

F. Hi in tch cc i m t tch c:A. 26,65.10-8C B. 26,65.10-9C C. 26,65.10-7C D. 13.32. 10-8CCu hi 3: T in c in dung 2F c khong cch gia hai bn t l 1cm c tch in vi ngun inc hiu in th 24V. Cng in trng gia hai bn t bng:

A. 24V/m B. 2400V/m C. 24 000V/m D. 2,4VCu hi 4: T in c in dung 2F c khong cch gia hai bn t l 1cm c tch in vi ngun inc hiu in th 24V. Ngt t khi ngun v ni hai bn t bng dy dn th nng lng t gii phng ra l:

A. 5,76.10-4J B. 1,152.10-3J C. 2,304.10-3J D.4,217.10-3JCu hi 5: Mt t in c in dung C, in tch q, hiu in th U. Tng hiu in th hai bn t ln gpi th in tch ca t:

A. khng i B. tng gp i C. tng gp bn D. gim mt na

Cu hi 6: Mt t in c in dung C, in tch q, hiu in th U. Ngt t khi ngun, gim in dungxung cn mt na th in tch ca t:A khng i B tng gp i


61/274

[email protected] tm v bin sonCu hi 7: Mt t in c in dung C, in tch q, hiu in th U. Ngt t khi ngun, gim in dungxung cn mt na th hiu in th gia hai bn t:

A. khng i B. tng gp iC. Gim cn mt na D. gim cn mt phn t

Cu hi 8: Mt t in c in dung C, in tch q, hiu in th U. Ngt t khi ngun, gim in dungxung cn mt na th nng lng ca t:

A. khng i B. tng gp iC. Gim cn mt na D. gim cn mt phn t

Cu hi 9: Mt t in phng c in mi l khng kh c in dung l 2F, khong cch gia hai bn t l1mm. T chu c. Bit in trng gii hn i vi khng kh l 3.106V/m. Hiu in th v in tch cci ca t l:

A. 1500V; 3mC B. 3000V; 6mCC. 6000V/ 9mC D. 4500V; 9mC

Cu hi 10: Mt t in phng c in mi l khng kh c in dung l 2F, khong cch gia hai bn tl 1mm. T chu c. Bit in trng gii hn i vi khng kh l 3.106V/m. Nng lng ti a m ttch tr c l:

A. 4,5J B. 9J C. 18J D. 13,5JCu 1 2 3 4 5 6 7 8 9 10p n A A B A B A B B B B

T in - Dng 1: in dung, nng lng in trng - 3Cu hi 1: Mt t in c in dung l bao nhiu th tch ly mt nng lng 0,0015J di mt hiu inth 6V:

A. 83,3F B. 1833 F C. 833nF D. 833pFCu hi 2: Nng lng ca t in tn ti:

A. trong khong khng gian gia hai bn t B. hai mt ca bn tch in dngC. hai mt ca bn tch in m D. cc in tch tn ti trn hai bn t

Cu hi 3: Mt t in in dung 12pF mc vo ngun in mt chiu c hiu in th 4V. Tng hiu inth ny ln bng 12V th in dung ca t in ny s c gi tr:

A.36pF B. 4pF C. 12pF D. cn ph thuc vo in tchca tCu hi 4: Mt t in c in dung 20 F mc vo hiu in th ca ngun mt chiu th in tch ca t

bng 80C. Bit hai bn t cch nhau 0,8cm. in trng gia hai bn t c ln:A. 10-4V/m B. 0,16V/m C. 500V/m D. 5V/m

Cu hi 5: Khi t t in c in dung 2 F di hiu in th 5000V th cng thc hin tch in cho

t in bng:A. 2,5J B. 5J C. 25J D. 50JCu hi 6: Vi mt t in xc nh c in dung C khng i, tng nng lng in trng tch trtrong t in ln gp 4 ln ta c th lm cch no sau y:

A. tng in tch ca t ln 8 ln, gim hiu in th i 2 lnB. tng hiu in th 8 ln v gim in tch t i 2 lnC. tng hiu in th ln 2 lnD. tng in tch ca t ln 4 ln

Cu hi 7: Pht biu no sau y l khng ng?A. T in l h hai vt dn t gn nhau nhng khng tip xc vi nhau. Mi vt gi l mt bn t.

B. T in phng l t in c hai bn t l hai tm kim loi c kch thc ln t i din vi nhau.

C. in dung ca t in l i lng c trng cho kh nng tch in ca t in v c o bng thngs gia in tch ca t v hiu in th gia hai bn t


62/274

[email protected] tm v bin sonD. Hiu in th gii hn l hiu in th ln nht t vo hai bn t in m lp in mi ca t in bnh thng.Cu hi 8: Pht biu no sau y l ng?

A. Sau khi np in, t in c nng lng, nng lng tn ti di dng ho nng.

B. Sau khi np in, t in c nng lng, nng lng tn ti di dng c nng.C. Sau khi np in, t in c nng lng, nng lng tn ti di dng nhit nng.

D. Sau khi np, t in c nng lng, nng lng l nng lng ca in trng trong t in.

Cu hi 9: Mt t in c in dung C = 6 (F) c mc vo ngun in 100 (V). Sau khi ngt t inkhi ngun, do c qu trnh phng in qua lp in mi nn t in mt dn in tch. Nhit lng to ratrong lp in mi k t khi bt u ngt t in khi ngun in n khi t phng ht in l:

A. 0,3 (mJ). B. 30 (kJ). C. 30 (mJ). D. 3.104 (J).

Cu hi 10: Mt t in phng c in dung C, c mc vo mt ngun in, sau ngt khi ngun inNgi ta nhng hon ton t in vo cht in mi c hng s in mi . Khi hiu in th gia hai

bn t inA. Khng thay i. B. Tng ln ln.

C. Gim i ln. D. Tng ln hoc gim i tu thuc vo lp in mi.Cu 1 2 3 4 5 6 7 8 9 10

p n A A C C C C A D A A

T in - Dng 2: T phng - 1Cu hi 1: Mt t in phng mc vo hai cc ca mt ngun in c hiu in th 500V. Ngt t khingun ri tng khong cch ln hai ln. Hiu in th ca t in khi :

A. gim hai ln B. tng hai ln

C. tng 4 ln D. gim 4 ln

Cu hi 2: Ni hai bn t in phng vi hai cc ca acquy. Nu dch chuyn cc bn xa nhau th trong khidch chuyn c dng in i qua acquy khng:

A. Khng

B. lc u c dng in i t cc m sang cc dng ca acquy sau dng in c chiu ngc li

C. dng in i t cc m sang cc dng

D. dng in i t cc dng sang cc mCu hi 3: Ni hai bn t in phng vi hai cc ca ngun mt chiu, sau ngt t ra khi ngun ri avo gia hai bn mt cht in mi c hng s in mi th in dung C v hiu in th gia hai bn ts:

A. C tng, U tng B. C tng, U gim

C. C gim, U gim D. C gim, U tng

Cu hi 4: Ni hai bn t in phng vi hai cc ca ngun mt chiu, sau ngt t ra khi ngun ri avo gia hai bn mt cht in mi c hng s in mi th nng lng W ca t v cng in trngE gia hai bn t s:

A. W tng; E tng B. W tng; E gimC. Wgim; E gim D. Wgim; E tng


63/274

[email protected] tm v bin sonCu hi 5: Mt t in phng c in dung 7nF cha y in mi c hng s in mi , din tch mi bnl 15cm2 v khong cch gia hai bn bng 10-5m. Tnh hng s in mi :

A. 3,7 B. 3,9 C. 4,5 D. 5,3Cu hi 6: Mt t in phng hai bn c dng hnh trn bn knh 2cm t trong khng kh cch nhau 2mm.in dung ca t in l:

A. 1,2pF B. 1,8pF C. 0,87pF D. 0,56pFCu hi 7: Mt t in phng hai bn c dng hnh trn bn knh 2cm t trong khng kh cch nhau 2mm.C th t mt hiu in th ln nht l bao nhiu vo hai bn t , bit in trng nh nht c th nhthng khng kh l 3.106V/m:

A. 3000V B. 6000V C. 9000V D. 10 000V

Cu hi 8: Mt t in phng khng kh mc vo ngun in c hiu in th 200V, din tch mi bn l20cm2 , hai bn cch nhau 4mm. Tnh mt nng lng in trng trong t in:

A. 0,11J/m3 B. 0,27J/m3 C. 0,027J/m3 D. 0,011J/m3

Cu hi 9: in dung ca t in phng ph thuc vo:A. hnh dng, kch thc t v bn cht in mi

B. kch thc, v tr tng i ca 2 bn v bn cht in mi

C. hnh dng, kch thc, v tr tng i ca hai bn t

D. hnh dng, kch thc, v tr tng i ca hai bn t v bn cht in mi

Cu hi 10: Hai bn t in phng hnh trn bn knh 60cm, khong cch gia hai bn l 2mm, gia hai bnl khng kh. in dung ca t l:

A. 5nF B. 0,5nF C. 50nF D. 5F

Cu 1 2 3 4 5 6 7 8 9 10p n B D B C D D B D D A

T in - Dng 3: ghp t - 1Cu hi 1: Ba t in ging nhau cng in dung C ghp song song vi nhau th in dung ca b t l:

A. C B. 2C C. C/3 D. 3CCu hi 2: Ba t in ging nhau cng in dung C ghp ni tip vi nhau th in dung ca b t l:

A. C B. 2C C. C/3 D. 3CCu hi 3: B ba t in C1 = C2 = C3/2 ghp song song ri ni vo ngun c hiu in th 45V th in tchca b t l 18.10-4C. Tnh in dung ca cc t in:

A. C1 = C2 = 5F; C3 = 10 F B. C1 = C2 = 8F; C3 = 16 F

C. C1 = C2 = 10F; C3 = 20 F D. C1 = C2 = 15F; C3 = 30 F

Cu hi 4: Hai t in c in dung C1 = 2 F; C2 = 3 F mc ni tip nhau. Tnh in dung ca b t:

A. 1,8 F B. 1,6 F C. 1,4 F D. 1,2 F

Cu hi 5: Hai t in c in dung C1 = 2 F; C2 = 3 F mc ni tip nhau. t vo b t hiu in th mtchiu 50V th hiu in th ca cc t l:

A. U1 = 30V; U2 = 20V B. U1 = 20V; U2 = 30V

C. U1 = 10V; U2 = 40V D. U1 = 250V; U2 = 25V

Cu hi 6: Bn t in mc thnh b theo s nh hnh v, C 1 =1F; C2 = C3 = 3 F. Khi ni hai im M, N vi ngun in th C 1 cin tch q1 = 6C v c b t c in tch q = 15,6 C. Hiu in th

C1 C2

C4

C3

M N


64/274

[email protected] tm v bin sonA. 4V B. 6V C. 8V D. 10V

Cu hi 7: Bn t in mc thnh b theo s nh hnh v trn, C1 = 1F; C2 = C3 = 3 F. Khi ni haiim M, N vi ngun in th C1 c in tch q1 = 6C v c b t c in tch q = 15,6 C. in dung C4 l:

A. 1 F B. 2 F C. 3 F D. 4 F

Cu hi 8: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v.Ni b t vi hiu in th 30V. Tnh in dung ca c b t:

A. 2nF B. 3nF C. 4nF D. 5nF

Cu hi 9: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v trn. Ni b t vi hiu in th 30V. TC1 b nh thng. Tm in tch v hiu in th trn t C3:

A. U3 = 15V; q3 = 300nC B. U3 = 30V; q3 = 600nC

C.U3 = 0V; q3 = 600nC D.U3 = 25V; q3 = 500nC

Cu hi 10: Hai t in in dung C1 = 0,3nF, C2 = 0,6nF ghp ni tip, khong cch gia hai bn t ca hait nh nhau bng 2mm. in mi ca mi t ch chu c in trng c cng ln nht l 10 4V/m Hiu in th gii hn c php t vo b t bng:

A. 20V B. 30V C. 40V D. 50VCu 1 2 3 4 5 6 7 8 9 10

p n D C C D A C B C B B

T in - Dng 3: ghp t - 2Cu hi 1: Hai t in C1 = 0,4F; C2 = 0,6F ghp song song ri mc vo hiu in th U < 60V th mttrong hai t c in tch 30C. Tnh hiu in th U v in tch ca t kia:

A. 30V, 5 C B. 50V; 50 CC. 25V; 10 C D. 40V; 25 C

Cu hi 2: Ba t in ghp ni tip c C1 = 20pF, C2 = 10pF, C3 = 30pF. Tnh in dung ca b t :A. 3,45pF B. 4,45pFC.5,45pF D. 6,45pF

Cu hi 3: Mt mch in nh hnh v, C1 = 3 F , C2 = C3 = 4F. Tnh in dung ca b t:

A. 3 F B. 5 F

C. 7 F D. 12 F

Cu hi 4: Mt mch in nh hnh v trn, C1 = 3 F , C2 = C3 = 4 F. Ni hai im M, N vi hiu in th

10V. in tch trn mi t in l:A. q1 = 5 C; q2 = q3 = 20C B. q1 = 30 C; q2 = q3 = 15CC. q1 = 30 C; q2 = q3 = 20C D. q1 = 15 C; q2 = q3 = 10C

Cu hi 5: Ba t in c in dung bng nhau v bng C. c b t c in dung l C/3 ta phi ghpcc t thnh b:

A. 3 t ni tip nhau B. 3 t song song nhauC. (C1 nt C2)//C3 D. (C1//C2)ntC3

Cu hi 6: Ba t in C1 = C2 = C, C3 = 2C. c b t c in dung l C th cc t phi ghp:A. 3 t ni tip nhau B. (C1//C2)ntC3C. 3 t song song nhau D. (C1 nt C2)//C3

Cu hi 7: Hai t ging nhau c in dung C ghp ni tip nhau v ni vo ngun mt chiu hiu in th U

th nng lng ca b t l W t, khi chng ghp song song v ni vo hiu in th cng l U th nng lngca b t l Ws. ta c:A. Wt = Ws B. Ws = 4Wt

C1

C2

C3

C1

C3

C2

M N


65/274


Cu hi 8: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v.Ni b t vi hiu in th 30V. Tnh hiu in th trn t C2:

A. 12V B. 18V

C. 24V D. 30VCu hi 9: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v trn. Ni b t vi hiu in th 30V. TC1 b nh thng. Tm in tch v hiu in th trn t C1:

A. U1 = 15V; q1 = 300nC B. U1 = 30V; q1 = 600nC

C.U1 = 0V; q1 = 0nC D.U1 = 25V; q1 = 500nC

Cu hi 10: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v trn. Ni b t vi hiu in th 30V. TC1 b nh thng. Tm in tch v hiu in th trn t C2:

A. U2 = 15V; q2 = 300nC B. U2 = 30V; q2 = 600nC

C.U2 = 0V; q2 = 0nC D.U2 = 25V; q2 = 500nC

Cu 1 2 3 4 5 6 7 8 9 10p n B C B C A B B C C C

T in - Dng 3: ghp t - 3Cu hi 1: Trong phng th nghim c mt s t in loi 6F. S t phi dng t nht to thnh b t cin dung tng ng l 4,5 F l:

A. 3 B. 5 C. 4 D. 6Cu hi 2: C cc t ging nhau in dung l C, mun ghp thnh b t c in dung l 5C/3 th s t cndng t nht l:

A. 3 B. 4 C. 5 D. 6Cu hi 3: Hai t in c in dung C1 = 2 C2 mc ni tip vo ngun in c hiu in th U th hiu inth ca hai t quan h vi nhau;

A. U1 = 2U2 B. U2 = 2U1 C. U2 = 3U1 D.U1 = 3U2Cu hi 4: Hai t in c in dung C1 = 2 C2 mc ni tip vo ngun in c hiu in th U. Dm t C2vo in mi lng c hng s in mi l 2. Cng in trng gia hai bn t C1 s

A. tng 3/2 ln B. tng 2 ln C. gim cn 1/2 ln D. gim cn 2/3 lnCu hi 5: Mt t in phng t thng ng trong khng kh in dung ca n l C. Khi dm mt na ngptrong in mi c hng s in mi l 3, mt na trong khng kh in dung ca t s :

A. tng 2 ln B. tng 3/2 ln C. tng 3 ln D. gim 3 lnCu hi 6: Mt t in phng t nm ngang trong khng kh in dung ca n l C. Khi dm mt na ngptrong in mi c hng s in mi l 3, mt na trong khng kh in dung ca t s :

A. gim cn 1/2 B. gim cn 1/3 C. tng 3/2 ln D. gim cn 2/3 lnCu hi 7: B t in gm hai t in: C1 = 20 (F), C2 = 30 (F) mc song song vi nhau, ri mc vo haicc ca ngun in c hiu in th U = 60 (V). Hiu in th trn mi t in l:

A. U1 = 60 (V) v U2 = 60 (V). B. U1 = 15 (V) v U2 = 45 (V).

C. U1 = 45 (V) v U2 = 15 (V). D. U1 = 30 (V) v U2 = 30 (V).

Cu hi 8: Mt b t in gm 10 t in ging nhau (C = 8 F) ghp ni tip vi nhau. B t in cni vi hiu in th khng i U = 150 (V). bin thin nng lng ca b t in sau khi c mt t inb h th l

C1

C2

C3


66/274


67/274


S: a. I = 0,16A.6. b. 1020

Bi 2: Mt dng in khng i chy trong dy dn c cng 1,6 mA..Tnh in lng v seletron chuyn qua tit din thng ca dy dn trong thi gian 1 gi.

S: q = 5,67C ; 3,6.1019Bi 3: S electron dch chuyn qua tit din thng ca dy dn trong khong thi gian 2 s l

6,25.1018

e. Khi dng in qua dy dn c cng bao nhiu?S: I = 0,5A.

Bi 4:Dng khng i I=4,8A chy qua dy kim loi tit din thng S=1cm2. Tnh:a.S e qua tit din thng trong 1s.

b.Vn tc trung bnh trong chuyn ng nh hng ca e, bit n=3.1028(ht/m3)S: 3.1028 v 0,01mm/s

Bi 5: Trong 10s, dng tng t 1A n 4A.Tnh cng dng trung bnh v in lng chuynqua trong thi gian trn?Bi 6:T phng bn cc hnh vung cnh a=20cm, khong cch d=2mm ni vi ngun

U=500V.a mt tm thy tinh c chiu dy d,=2mm , =9 vo t vi vkhng i bng 10cm/s.Tnh cng dng in trong thi gian a tm in mi vo t?

CH 2:CC BI TP LIN QUAN N IN TR

Dng 1: IN TR DY DN.S PH THUC VO NHIT * Tnh in tr ca mt on dy dn cho bit chiu di, tit din dy v in tr sut khi ch

cn p dng cng thc

R S=

l

- Ch : cc n v o khi tin hnh tnh ton.*in tr sut ph thuc vo nhit

0

0

(1 )

(1 )

t

R R t

= +

= +

Bi 1:Dy dn Nicrom c ng knh tit din d=0,01mm. Hi di ca dy l bao nhiu R=10. Bit =4,7.10-7 m.Bi 2:Dy dn 200C c in tr 54 v 2000 C c R=90 .Tnh h s nhit in tr ca dy dn?

Bi 3:T phng in mi l thy tinh c =9 ,=.10

9

m. Tnh in tr ca t bit C=0,1FS: 7,96.105

DNG 2:IN TR MCH MC NI TIP HOC SONG SONG

Vn dng cng thc in tr tng ng

* Ni tip : Rn = iR * Song song : =sR

1

nRRR

1...

11

21

+++

Bi 1 Ch h i h h h


68/274

[email protected] tm v bin sonBit: R1 = 5 , R2 =2 , R3 = 1Tnh in tr tng ng ca mch?

S: td7

R8

=

Bi 2:Cho on mch gm n in tr R1 = 1 , R2 = 12

, ..., Rn = 1n

mc song song. Tm in tr tng ng ca mch?

S: td2

Rn(n 1)

=+

Bi 3: Cho mch in nh hnh v: R1 = 1, R2=R3 = 2 , R4 = 0,8 . Hiu in th UAB = 6V.a. Tm in tr tng ng ca mch?

S: a. 2

Bi 4.Cho mch in nh hnh v:Cho bit R1 = 4R2 = R5 = 20R3 = R6 = 12R4 = R7 = 8Tm in tr tng ng RABca mch?

(p s: RAB = 16 )DNG 3:IN TR DY DN TRN3/ in tr vng dy dn trn* in tr t l vi s o gc tm

* Ta c :0360

RRAB =

, in tr vng dy gc ln00

,

360360

RRAB=

Trong ,ABR = R - RAB

Bi 1:Dy dn in tr R c un thnh hnh trn tm O, gc AOB=.a.Tnh RAB theo R v b.nh r=3/16. RC.Tnh RAB max. Tnh gi tr cc i y.Bi 2: Dy dn in tr R c un thnh hnh trn tm O, gc AOB=., R=25 a..nh RAB =4

b.Tnh RAB max. Tnh gi tr cc i y

Bi 3:Cc on dy ng cht cng tit din c un nh hnh v.in tr AO v OB l R.Tnh in tr RAB?

R4

R1

R2

R3

A B

D

C

R1

R2

R3

R4

R5

A B

M

N

C R7

R5

R6

R4

R3

R2

R1

DB

A


69/274

U

k2R2 R

3

k1 R1 A

[email protected] tm v bin sonDNG 4:IN TR MCH PHC TP: on mch c cu to phc tp khi tnh in tr ca mch cn v li s mc in trtrong mch

* Nu bi khng k hiu cc im nt ca mch (l im giao nhau ca t nht ba dy dn) thnh s cc im nt bng k hiu. Nu dy ni c in tr khng ng k th hai u y ni ch

ghi bng mt k hiu chung.* a mch v dng n gin c cc quy tc sau:a) Qui tc 1: Chp cc im c cng in th.

Cc im c cng in th l cc im sau y:+ Cc im c ni vi nhau bng dy dn v ampe k c in tr rt nh c th b qua.

Bi 1:V li mch khi hai K cng m, K1 ng, K2 m v ngc li.

Bi 2:V li mch

Bi 1: Tnh in tr tng ng ca on mch AB nh hnh v nu:a) K1, K2 m.

b) K1 m, K2 ng.c) K1 ng, K2 m.d) K1, K2 ng.Cho R1 = 1 , R2 = 2 ,

R3 = 3 , R4 = 6 ,in tr cc dy ni khngng k.

Bi gii:a): RAB = R4 = 6

Bi 2. Cho on mch AB c tm in trR1, R2, R3, R4, R5, R6, R7, R8

c tr s u bng R = 21 .

Mc theo s nh hnh v:

N

MK

2

K1

BA

R4

R3R2R1

R8

R7

R1

R3

R2

R4

R5

R6 K2 B

N

A

r

R1

R R2

A1

U+ -

A2


70/274


Tnh in tr tng ng ca on mch AB trong cc trng hp:a, K1 v K2 u m.

b, K1 m, K2 ng.

c, K1 ng, K2 m.d, K1 v K2 u ng.

p s: a, RAB = 42b, RAB = 25.2 c, RAB = 10.5 d, RAB = 9

Bi 6: Cho mch in nh hnh v. Cc in tr R1 = 1,4; R2 = 6;R3 = 2; R4 = 8; R5 = 6; R6 = 2; Vn k V c in tr rt ln,

ampe k A c in tr rt nh. Tnh in tr tng ng ca tonmch.

b) Quy tc 2: B in trTa c th b cc in tr (khc khng) nu hai u in tr c in th bng nhau.

Bi 2:Cho mch cu in tr nh (H1.1)

Nu qua R5 c dng I5 = 0 v U5 = 0 th cc

in tr nhnh lp thnh t l thc : 1 23 4

R RR R

= = n = const

Ngc li nu c t l thc trn th I5 = 0 v U5 = 0, ta c

mch cu cn bng.

Biu thc (*) chnh l iu kin mch cu cn bng. Khi ta b qua R5 v tnh ton bnh thng

Bi 1Cho mch in nh hnh v.

Bit R1 = R2 = R3 = R4 = R5 = 10 .in tr ampe k khng ng k.

Tm RAB?A

C

A BR4

R3

R1

R3R4

R5

U+ -

V

A

R2

R1

D

C

A B

R4

R3

R5

R2

R1


71/274


72/274


73/274


Tnh in tr tng ng ca on mch trong cc trng hp sau:a)R1 = R3 = R4 = R6 = 1 ;R7 = R8 = 2 ; R2 = 3,5 ; R5 = 3 .

b) R1 = R2 = R5 = R7 = R8 = 1 ; R3 = R4 = R6 = 2 .c) R1 = 6 ; R2 = 4 ; R4 = 3 ; R5 = 2 ; R6 = 5 ; R3 = 10 ; R7 = 8 R8 = 12 p s:a)

b) R 2,18

c)3

R20

=

DNG 5: Xc nh s in tr t nht v cch mc khi bit R0 v Rt*- Nu Rt > R0th mchgm R0 ni tip vi R1 , tnh R1

- So snh R1 vi R0 :nu R1 > R0 th R1 c cu to gm R0 ni tip vi R2 ,tnh R2 . Tip tc tc cho n khi bng Rt

nu R1 < R0 th R1 c cu to gm R0 song song vi R2 ,tnh R2 .Tip tc cho n khi bng Rt*- Nu Rt < R0th mchgm R0 song song vi R1 , tnh R1- Lm tng t nh trn.

Bi 1:Tm s in tr loi R=4 t nht v cch mc mc mch c in tr tng ng R=6Bi 2:C mt s in tr loi R=12.Tm s in tr t nht v cch mc mc mch c in trtng ng R=7,5.Bi 3:Phi dng ti thiu bao nhiu in tr loi 5 mc thnh mch c Rt=8 .V s cchmc.

Dng 6/ Dng phng trnh nghim nguyn dng xc nh s in trDa vo cch ghp , lp phng trnh ( hoc h phng trnh):- Nu cc in tr ghp ni tip: xR1 + yR2 + zR3 = a vx + y + z = N , vi x,y,z l s in tr loi R1,R2,R3 v N l tng s in tr- Kh bt n s a v phng trnh 2 n, tm nghim nguyn dng

Bi 1:C hai loi in tr 5 v 7.Tm s in tr mi loi sao cho khi ghp ni tip ta c intr tng cng 95 vi s in tr t nht.Bi 2:C 50 in tr loi 8 ,3 ,1 .Hi mi loi cn my chic th khi ghp li c R=100

Bi 3: C 24 in tr loi 5 ,1 ,0,5 .Hi mi loi cn my chic th khi ghp li c R=30 Bi 4:C hai loi in tr 2 v 3.Tm s in tr mi loi sao cho khi ghp ni tip ta c intr tng cng 15 .


74/274


Bi 1: Cho mch in nh hnh v. U = 12V; R1 = 6; R2 = 3;R3 = 6. in tr ca cc kha v ca ampe k A khng ng k.

Tm cng dng in qua cc in tr khi:a. k1 ng, k2 m.

b. k1 m, k2 ng.c. k1, k2 u ng.

Bi 2: Cho mch in nh hnh (2). U = 6V; R1 = R2 = R3 = R4 = R5 = 5; R6 = 6.Tnh hiu in th hai u in tr R4.

Bi 3: Cho mch in nh hnh (3) R1 = 8; R2 = 3; R3 = 5; R4 = 4; R5 = 6;R6 = 12; R7 = 24; cng dng in qua mch chnh l I = 1A.Tnh hiu in th hai u mch v hiu in th hai u in tr R3.

Bi 4: Cho mch in nh hnh (4). R1 = 10; R2 = 6; R3 = R7 = 2; R4 = 1; R5 = 4;R6 = 2; U = 24V. Tnh cng dng in qua in tr R6.

Bi5: Chomchinnhhnhv: U

=48V; Ro = 0,5; R1 = 5; R2 = 30 ; R3 = 15; R4 = 3; R5 =12. B qua in tr cc ampe k. Tm:

a. in tr tng ng RAB.b. S ch ca cc ampe k A1 v A2.c. Hiu in th gia hai im M v N.

Bi 6: Cho mch in nh hnh v. Cc in tr R1 = 1,4; R2 =6; R3 = 2; R4 = 8; R5 = 6; R6 = 2; U = 9V. Vn k V c intr rt ln, ampe k A c in tr rt nh.Tm s ch vn k v ampe k A.

Bi 7: Cho mch in nh hnh v: U = 60V; R1 = 10; R2 = R5 = 20;R3 = R4 = 40; V l vn k l tng. B qua in tr cc dy ni.

a. Tm s ch ca vn k.b. Nu thay vn k bng mt bng n c dng in nh mc l

I = 0,4A mc vo hai im P v Q ca mch in th bngn sng bnh thng. Hy tm in tr ca bng n.

Bi 8: Trong mt th nghim vi s mch in nh hnh v. Ngun

in U =1V; in tr R = 1 cc ampe k A1 v A2 l cc ampe k ltng (c in tr bng 0), v cc dng in qua chng c th thay ikhi t th i i t bi t Khi i h h i t bi t

A

R1

R2

R3

k1

k2

U+ -

Hnh 1

Hnh 2

R6

U+-

R3R

1

R2

R4

R5

R1

U+-

R2

R5

R6

R3

R4

R7

Hnh 3

R3

R4

R1

R5

U+ -

Hnh 4

R2

R6

R7

R3R

4

R5

U+ -

V

A

R2

R1

R1

A1

U+ -

M

N

U

Ro

R4

R5

R1

R2

A1

A2

R3

V

P

QM N

UR

1

R2

R3

R4

R5


75/274

U

k2R

2 R3

k1 R1 A

[email protected] tm v bin sonampe k A2 ch 1A th ampe k A1 ch 3,5A. Nu i v tr gia R1 v R2 v chnh li bin tr r cho A2 ch 1A th A1 ch 7/3A. .Tnh R1 v R2.Bi 9: Cho mch in nh hnh v: trong hiu in th hai u onmch khng i l U = 7V, cc in tr R1 = 7, R2 = 6; AB l mtdy dn in chiu di l = 1,5m, tit din khng i S = 0,1mm 2, in tr

sut = 4.10-7.m, in tr cc dy ni v ampe k A khng ng k.a. Tnh in tr R ca dy AB.

b. Dch chuyn con chy C ti v tr sao cho chiu di AC = CB,tnh cng dng in qua ampe k.

c. Xc nh v tr C dng in qua ampe k t D n C c cng 1/3ABi 10: Cho mch in nh hnh v. in tr ca ampe k v dy ni khng ng k. Hiu inth gia hai u mch in l U. Khi m c hai kha k 1 v k2 th cng dng in qua ampe k l Io. Khi ng k1 m k2 cng dng in quaampe k l I1. Khi ng k2, m k1 cng dng in qua ampe k l I2.Khi ng c hai kha k1 v k2 th cng dng in qua ampe k l I.

a. Lp biu thc tnh I theo Io, I1 v I2.b. Cho Io = 1A; I1 = 5A: I2 = 3A; R3 = 7, hy tnh I, R1, R2 v U.

Bi 11: Cho mch in nh hnh v. in tr RMN = R. Ban u con chyC ti trung im MN. Phi thay i v tr con chy C nh th no sch vn k V khng thay i khi tng hiu in th vo UAB ln gp i.in tr ca vn k v cng ln.Bi 12: Cho mch in c s nh hnh. Cho bit: R1 = 8;R2 = R3 = 12; R4 l mt bin tr. t vo hai u A, B ca mch in

mt hiu in th UAB = 66V.

1. Mc vo hai im E v F ca mch mt ampe k c in tr nhkhng ng k v iu chnh bin tr R4 = 28. Tm s ch ca ampe kv chiu ca dng in qua ampe k.

2. Thay ampe k bng mt vn k c in tr rt ln.a. Tm s ch ca vn k. Cho bit cc dng ca vn k mc vo im no?

b. iu chnh bin tr cho n khi vn k ch 0. Tm h thc gia cc in tr R1, R2, R3 v R4khi v tnh R4.

3. Thay vn k bng mt in k c in tr R5 = 12 v iu chnh bin tr R4 = 24. Tmdng in qua cc n tr, s ch ca in k v in tr tng ng ca mch AB. Cc dng cain k mc vo im no?

Bi 13: Cho mch in nh hnh v. Cho bit:UAB = U = 132V; R1= 42, R2 = 84; R3 = 50; R4 = 40;R5 = 40, R6 = 60; R7 = 4; Rv = .

a. Tm s ch ca vn k.b. Thay vn k bng ampe k (RA= 0). Tm hiu in

th trn cc in tr v s ch ca ampe k.Bi 14: Cho mach ien nh hnh 9. R4 = R2.

Neu noi A ,B vi nguon U = 120V th I3 = 2A, UCD=30V.

Neu noi C,D vi U= 120V th UAB= 20V. Tm : R1,R2 , R3.

R-

+V

A

B

U R

N

MC

A

E

FM N

U

R1

R3

R2 R4

R7

R6

R5

R2

R3R1

R4

V

R1 R2A

BA

U

C

D


76/274


77/274


Bi 2: Cho mch in nh hnh v: E = 6 V, r = 1 , R1 = 20 , R2 = 30 , R3 = 5 . Tnh cng dngin qua mi in tr v hiu in th 2 u mch ngoi.

Bi 21: Cho mch in: E = 6V, r = 0,5, R1 = R2 = 2 , R3 = 5 , R5 = 4 ,R4= 6 . in tr ampe k v cc dy ni khng ng k. Tnh cng dngin qua cc in tr, s ch ampe k v hiu in th gia hai cc ngun in.

Bi 3 : Cho 2 in tr R1 = R2 = 1200 c

mc ni tip vo mt ngun in co sutin ng E = 180V, in tr trong khngng k. Tm s ch ca vn k mc vo mch theo cc s bn. Bit in tr ca vn k RV = 1200 .

Bi 4: Cho : E = 48V, r = 0, R1 = 2 , R2 = 8 , R3 = 6 , R4 = 16

a) Tnh hiu in th gia hai im M, N.

b) Mun o UMN phi mc cc dng vn k vo u?

Bi 5 : Cho mch in nh hnh v bi 11 vi : E = 7,8V, r = 0,4, R1 = R2 = R3 = 3, R4 = 6 .

a) Tm UMN ?

b) Ni MN bng dy dn. Tnh cng dng in qua dy ni MN.

Bi 6 : Cho mch in: E = 12 V, r = 0,1 , R4 = 4,4 , R1 = R2 = 2 , R3 = 4. Tm in tr tng ngmch ngoi, cng dng in mch chnh v cng dng in qua mi nhnh r. Tnh UAB v UCD

Bi 7 :Cho mch in nh hnh, ngun in c sut in ng E = 6,6V, in tr trong r = 0,12; bng n1 ( 6 V 3 W ) v 2 ( 2,5 V 1,25 W ).

a) iu chnh R1 v R2 sao cho 2 n sng bnh thng. Tnh cc gi tr ca R1v R2.

b) Gi nguyn gi tr ca R1,iu chnh bin tr R2 sao cho n c gi tr R2 = 1 .Khi sng ca cc bng n thay i th no so vi cu a?


78/274


Bi 8: Dng mt ngun in thp sng ln lt hai bng n c in tr R1 = 2 , R2 = 8 , khi cngsut in tiu th ca hai bng n nh nhau. Tm in tr trong ca ngun in.

Bi 9: Hy xc nh sut in ng E v in tr trong r ca mt acquy, bit rng nu n pht dng in ccng I1 = 15 A th cng sut in mch ngoi P1 = 136 W, cn nu n pht dng in c cng I2 =6 A th cng sut in mch ngoi P2 = 64,8 W.

Bi 10: Mt ngun in c sut in ng E = 6 V, in tr trong r = 2, mch ngoi c in tr R.

a) Tnh R cng sut tiu th mch ngoi P1 = 4 W.

b) Vi gi tr no ca R th cng sut in tiu th mch ngoi ln nht? Tnh gi tr .

Bi 11: Hai ngun in c sut in ng nh nhau: E1= E2= E, cc in tr trong r1 v r2 c gi tr khcnhau. Bit cng sut in ln nht m mi ngun c th cung cp c cho mch ngoi P 1 = 20 W v P2 =30 W. Tnh cng sut in ln nht m c hai ngun c th cung cp cho mch ngoi khi chng mc nitip v khi chng mc song song.

Bi 12:Cho mch in nh hnh: Cho bit E = 12 V; r = 1,1; R1 = 0,1

a) Mun cho cng sut in tiu th mch ngoi ln nht, R2 phi c gi tr bng baonhiu?

b) Phi chn R2 bng bao nhiu cng sut in tiu th trn R2 ln nht. Tnh cng sut inln nht .

Bi 13: Cho mch in c s nh hnh. Cho bit E = 15 V; r = 1; R1 = 2 . Bit cng sutin tiu th trn R ln nht. Hy tnh R v cng sut ln nht .

Bi 14 : Cho = 12(V) ,r = 2 , R1 = R2 = 6 , n ghi (6V 3W)

a. Tnh I,U qua mi in tr?b. Nhit lng ta ra n sau 2 pht?

c. Tnh R 1 n sng bnh thng ?

Bi 15: Cho = 12(V), r = 2 , R1 = 3 , R2 = 2R3 = 6 , n ghi (6V 3W)a. Tnh I,U qua mi in tr?

b. Nhit lng ta ra n sau 1 gi v cng sut tiu th?c. Tnh R1 n sng bnh thng ?

Bi 16:Cho = 12(V), r = 3 , R1 = 4 ,R2 = 6 ,R3 = 4 , n ghi (4V 4W)

a. Tnh Rt ?

b. I,U qua mi in tr?V sang ca n?c. Thay R2 bng mt t in c in dung C = 20F.Tnh in tch ca t?

R1

R2

,r

R1

R3

,rR

1

R3

R2

,rR


79/274


Bi 17 :Cho = 12(V), r = 2 , R1 = 6 ,R2 = 3 , n ghi (6V 3W)a. Tnh R t ? Tnh I,U qua mi in tr?

b. Thay n bng mt Ampe k (RA=0) Tnh s ch ca Ampe k?c. n sng bnh thng th bng bao nhiu (cc in tr khng i)?

Bi 18 :Cho = 9(V) ,r = 1,5 , R1 = 4 ,R2 = 2 ,n ghi (6V 3W)

Bit cng dng in chy trong mch chnh l 1,5A.Tnh UAB v R3?

Bi 19 :Cho = 10(V) ,r = 1 , R1 =6,6 ,R2 = 3 , n ghi (6V 3W)a. Tnh R t ,I,U qua mi in tr?

b. sng ca n v in nng tiu th ca n sau 1h20?c. Tnh R 1 n sng bnh thng ?

Bi 20:Cho = 12(V) ,r = 3 , R1 = 18 , R2 = 8 ,R3 = 6 ,n ghi (6V 6W)

a. Tnh Rt ,I,U qua mi in tr?b. sng ca n, in nng tiu th sau 2 gi 8 pht 40 giy?c. Tnh R2 n sng bnh thng ?

Bi 21: Cho = 15(V) ,r = 1 , R1 = 12 , R2 = 21 ,R3 = 3 ,n ghi (6V 6W),C = 10F.

a. Tnh Rt ,I,U qua mi in tr?b. sng ca n ,in nng tiu th R2 sau 30 pht?c. Tnh R2 n sng bnh thng ?d. Tnh R1 bit cng dng in chy qua R2 l 0,5A?

Bi 22: Cho = 18(V), r = 2 , R1 = 3 , R2 = 4 ,R3 = 12 ,n ghi (4V 4W),

a. Tnh Rt ,IA,UV qua mi in tr?b. sng ca n ,in nng tiu th n sau 1gi 30 pht?c. Tnh R3 bit cng dng in chy qua R3 lc ny l 0,7A?

Bi 23: Cho = 24(V) ,r = 1 , R1 = 6 , R2 = 4 ,R3 = 2 ,n ghi (6V 6W),C = 4F.

a. Tnh Rt ,I,U qua mi in tr?b. sng ca n ,in nng tiu th n sau 16 pht 5 giy?c. Tnh in tch ca t?

Bi 24: Cho = 15(V) ,r = 1 , R1 = 21 , R2 = 12 ,R3 = 3 ,n ghi (6V 6W),Vn k c in tr rt ln.

a. Tnh Rt ,I,U qua mi in tr?b. sng ca n ,in nng tiu th R2 sau 2 gi 30 pht?

c. Tnh R2 bit cng dng in qua n l 0,8A ?

Bi 25: Cho = 12(V) r = 0 1 R1 = R2 = 2 R3 = 4 4

R1 ,r

R1

R2

R3

A B

,rR

1

R2

n

,r

R1

R3

R2

,r

R2

R1

R3

CA B

,r

R1

R2

R3

A B

,r

R1 R2R3

C

,r

R3

R2

R1

V

A B

,r

R


80/274


n ghi (4V 4W),Vn k c in tr rt ln.RA = 0a. Tnh Rt ,I,U qua mi in tr?

b. Mc vo 2 im CD mt Vn k .Tnh s ch ca Vn k?c. Mc vo 2 im CD mt Ampe k .Tnh s ch ca Ampe k?

Bi 26: Cho = 12(V) , R1 = 10 , R2 = 3 ,R4 = 5,25 ,

Vn k c in tr rt ln ch 6,5V.RA = 0a. Tnh cng dng in chy qua R1?b. Tnh R3 v nhit lng to ra R3 sau 16 pht?c. Tnh r ca ngun?

Bi 27: Cho = 12(V) , r = 10 ,R1 = R2 = R3 = 40 , R4 = 30 ,a. Tnh R t?

b. U,I qua mi in tr?c. Mc vo 2 im AD mt Ampe k c RA = 0.

Tnh s ch ca Ampe k?

Bi 28: Cho = 16(V) , r = 0,8 ,R1 = 12 , R2 = 0,2 ,R3 = R4 = 4 ,a. Tnh R t?U,I qua mi in tr?

b. Nhit lng to ra R4 sau 30 pht?c. Thay i R4 th I4 = 1A.Tnh R4?

Bi 29: Cho = 12(V) , r = 2 ,R1 = 5 , R2 = 6 ,R3 =1,2R4 = 6 , R5 = 8 ,

a. Tnh R t?U,I qua mi in tr?b. Nhit lng to ra R4 sau 1 gi 30 pht?c. Thay i R5 th n sng bnh thng.Tnh R5?

Bi 30: Cho mch in nh hnh: E = 12V; r = 2 ; R1 = 4 , R2 = 2 .Tm R3 :

a/ Cng sut mch ngoi ln nht, tnh gi tr nyb/ Cng sut tiu th trn R3 bng 4,5Wc/ Cng sut tiu th trn R3 ln nht. Tnh cng sut ny

Bi 31: Mt Acquy c r = 0,08. Khi dng in qua acquy l 4A, n cung cp cho mch ngoi mt cngsut bng 8W. Hi khi dng in qua acquy l 6A, n cung cp cho mch ngoi cng sut bao nhiu? S:11,04WBi 32: in tr R = 8 mc vo 2 cc mt acquy c in tr trong r = 1. Sau ngi ta mc thmin tr R song song vi in tr c. Hi cng sut mch ngoi tng hay gim bao nhiu ln? S: tng

1,62 lnBi 33: Mt Acquy (E; r) khi c dng in I1 = 15A i qua, cng sut mch ngoi l P1 = 135W; khi I2 = 6Ath P2 = 64,8W. Tm E, r S: 12V; 0,2Bi 34: Ngun E = 6V, r = 2 cung cp cho in tr mch ngoi cng sut P = 4W

a/ Tm Rb/ Gi s lc u mch ngoi l in tr R1 = 0,5. Mc thm vo mch ngoi in tr R2 th cng sut

tiu th mch ngoi khng i. Hi R2 ni tip hay song song vi R1 v c gi tr bao nhiu?S: a/ 4 hoc 1b/ R2 = 7,5 ni tip

Bi 35: a/ Khi in tr mch ngoi ca mt ngun in l R1 hoc R2 th cng sut mch ngoi c cng gitr. Tnh E; r ca ngun theo R1, R2 v cng sut P

b/ Ngun in trn c in tr mch ngoi R. Khi mc them Rx song song R th cng sut mch ngoi

khng i. Tm Rx ? S: a/ r = 21RR ; E = ( PRR )21 + b/ Rx = 222

rRRr

, iu kin R > r

A BV

,r

R1

R2

R4

R3A

,rR

4

R3 R

2R

1

AB

C

D

,r

R2

R3

R1

R4

,r R3

R1

R5

R2

R4

3R2R

1R

rE;


81/274

[email protected] tm v bin sonBi 36: a/ Mch kn gm acquy E = 2,2V cung cp in nng cho in tr mch ngoi R = 0,5. Hiu sutca acquy H = 65%. Tnh cng dng in trong mch

b/ Khi in tr mch ngoi thay i t R1 = 3 n R2 = 10,5 th hiu sut ca acquy tng gp i.Tnh in tr trong ca acquy

S: a/ 2,86A b/ 7Bi 37:Ngun in E = 16V, r = 2 ni vi mch ngoi gm R1 = 2 v R2 mc song song. Tnh R2 :

a/ Cng sut ca ngun cc i b/ Cng sut tiu hao trong ngun cc ic/ Cng sut mch ngoi cc i d/ Cng sut tiu th trn R1 cc ie/ Cng sut tiu th trn R2 cc i V tnh cc cng sut cc i trn

Bi 38:Ngun E = 12V, r = 4 c dung thp sang n (6V 6W)a/ Chng minh rng n khng sng bnh thngb/ n sang bnh thng, phi mc them vo mch mt in tr Rx.

Tnh Rx v cng sut tiu th ca RxS: a/ b/ 2, 2W ( ni tip) hoc 12, 3W ( song song)

Bi 39: Cho mch in nh hnh: cc in tr thun u c gi tr bng Ra/ Tm h thc lin h gia R v r cng sut tiu th mch ngoi khng i

khi K m v ngb/ E = 24V. tnh UAB khi K m v ngBi 40: Cho mch in nh hnh: E = 20V; r = 1,6R1 = R2 = 1, hai n ging nhau.Bit cng sut tiu th mch ngoi bng 60WTnh cng sut tiu th ca mi n v hiu sut ca ngun?

CH 6: HAI PHNG PHP GII BI TON IN MT CHIU

PHNG PHP 1:PHNG PHP NGUN TNG NG

(C nhiu phng php gii bi ton in mt chiu, phn ny ch gii thiu 2 phng php c bn)I. L THUYT

1. Ngun in tng ng ca b ngun ni tip:b AB( ) 1 2 n

b 1 2 n

e U e e ... e

r r r ... r

= = + + +

= + + +

mch ngoi h

- c bit: Nu c in tr R ghp ni tip vi ngun (e;r) th b

ngun l: bb

e e

r r R

= = +

2. Cc trng hp b ngun ghp song song cc ngun ging nhau, ghp hn hp i xng ccngun ging nhau

3. Trng hp tng qutBi ton: Cho mch in nh hnh v, cc ngun c sut in ng v in tr trong tng ng l(e1;r1); (e2;r2);.... (en;rn). n gin, ta gi s cc ngun c cc dng ni vi A tr ngun (e 2;r2)Tm sut in ng v in tr trong ca b ngun ny nu coi A v B l hai cc ca ngun intng ng.

Gii- Gi s ngun in tng ng c cc dng A, cc m B. Khi ta c: II.N

- in tr trong ca ngun tng ng:n

1b AB 1 2 n i

1 1 1 1 1 1...

r r r r r r = = + + + =

- tnh eb, ta tnh UAB. Gi s chiu dng in qua cc nhnh nh hnh v (gi s cc ngun u l

ngun pht).

e1;r

1e

2;r

2e

n;r

n

A B

A

I1

I2

In

rE;

BA

2R1R

rE;


82/274


- p dng nh lut m cho cc on mch:

1 AB1

11 AB 1 1 1

2 AB2 AB 2 2 2 2

2n AB n n n

n ABnn

e UI

rAe B : U e I r

e UAe B : U e I r I

rAe B : U e I r

e UI r

=

= + = + =

= =

- Ti nt A: I2 = I1 + I3 + ... + In. Thay cc biu thc ca dng in tnh trn vo ta c phngtrnh xc nh UAB:

3 AB2 AB 1 AB n AB

2 1 3 n

e Ue U e U e U...

r r r r

+ = + + +

- Bin i thu c:

n1 2 n i

11 2 n iAB

1 2 n b

e e e e...

r r r r U

1 1 1 1...

r r r r

+ +

= =

+ + +

- Vy

ni

1 ib

b

e

re

1

r

=

.

* Trong quy c v du nh sau: i theo chiu t cc dng sang cc m m ta gi s cangun tng ng (tc chiu tnh hiu in th):- Nu gp cc dng ca ngun trc th e ly du dng.- Nu gp cc m ca ngun trc th e ly du m.

* Nu tnh ra eb < 0 th cc ca ngun tng ng ngc vi iu gi s.-nu tnh ra I

. Cc dng ca ngun tng ng A.

- Gi s chiu dng in qua cc nhnh nh hnh v. p dng nh lut m cho cc on mch tnh cng dng in qua cc nhnh:

e1;r

1

e2;r

2

en;r

n

A B

I1

I2

I3

R1

R2

R3


83/274


1 AB1

1 11 AB 1 1 1 1

2 AB2 AB 2 2 2 2 2

2 2

3 AB 3 3 3 3

3 AB3

3 3

e U 12 2 10I A

r R 3 3Ae B : U e I (r R )

e U 9 2 11Ae B : U e I (r R ) I A

r R 3 3Ae B : U e I (r R )

e U 3 2 1I A

r R 3 3

= = = += + + + = + + = = =

+ = + = = =

+

Chiu dng in qua cc nhnh nh iu gi s.

Bi 2: Cho mch nh hnh v: e 1 = 24V; e2 = 6V; r1 = r2 = 1; R1 = 5; R2 = 2; R l bin tr. Vigi tr no ca bin tr th cng sut trn R t cc i, tm gi tr cc i .

Gii- Ta xt ngun in tng ng gm hai nhnh cha hai ngun e1 v e2. Gi s cc dng cangun tng ng A. Bin tr R l mch ngoi.- in tr trong ca ngun in tng ng l:

bb AB 1 1 2 2

1 1 1 1 1 1 1r 2

r r r R r R 6 3 2= = + = + = =

+ +

- Sut in ng ca b ngun tng ng l:1 2

1 2

b AB

b

e e 24 6r r 6 3e 4V U 0

1 1

r 2

= = = = > .

- cng sut trn R cc i th R = rb = 2. Cng sut cc i l:2 2b

max

b

e 4P 2W

4r 4.2= = =

Bi 3: Cho mch in nh hnh v: e1 = 6V; e2 = 18V; r1 = r2 = 2; R0 = 4; n ghi: 6V - 6W; Rl bin tr.a. Khi R = 6, n sng th no?

b. Tm R n sng bnh thng?

Giia. Khi R = 4. Ta xt ngun in tng ng gm hai nhnh cha haingun e1 v e2. Gi s cc dng ca ngun tng ng A. Bin tr Rv n l mch ngoi.- in tr trong ca ngun in tng ng l:

bb 1 0 2

1 1 1 1 1 2r 1,5

r r R r 6 2 3= + = + = =

+

- Sut in ng ca ngun tng ng l:

1 2

1 0 2

b

b

e e 6 18r R r 6 2

e 12V 01 2r 3

+= = = < . Cc dng ca

ngun tng ng B.- in tr v cng dng in nh mc ca n l: mR 6 ; I 1A= =- Cng dng in qua n cng l dng in trong mch chnh

b m

b

e 12 8I A I I

R R r 4,5 6 1,5 9= = = =

+ + + ++ +

- nh lut m cho cc on mch: AR2B: I2 = UAM/(R2 + R3) = 14/12 = 7/6A => UNM = I2.R3 =7/3V.AR1M: UAM = 14V = e2 + I1(R1 + r2) = 9 + 6I1 => I1 = 5/6A => UBM = e2 + I1r2 = 9 + 5/6 = 59/6V.- Vy UBN = UBM + UMN = 59/6 - 7/3 = 7,5V > 0.

- T : PR(max) =2 2

bR (max) b

b

e 7,5P 5,625W, khi R r 2,5

4r 4.2,5= = = = =

.

.

PHNG PHP DNG NH LUT KICHOFFA.L THUYT

Vi quy c dau cua I: (+) cho dong ti nut.(-) cho dong ra khoi nut.

Nt mng:Giao ca t nht 3 nhnhPhng trnh (1) co the c viet oi vi moi mot trong tong so mnut mang trong mach ien Tuy nhien ch co (m-1) phng trnh oc

I.nh luat Kirchhoff 1 (nh luat nut)Tai mot nut mang, tong ai so cac dong ien bang khong

n: so dong ien quy tu tainut mang ang xet.

e1;r

1e

2;r

2

A B

R1

R2

R3

M

N

I2

I1


85/274


86/274


Neu suat ien ong cua nguon ien cha biet tren motoan mach tnh c co gia tr dng th v tr gia nh cua cac cccua no (bc 1) la phu hp vi thc te; con neu suat ienong co gia tr am th phai oi lai v tr cac cc cua nguon.Ket luan

Dung hai nh luat Kirchhoff, ta co the giai c hau het nhng baitap cho mach ien phc tap. ay gan nh la phng phap c ban e giaicac mach ien phc tap gom nhieu mach vong va nhanh, neu can tmbao nhieu gia tr cua bai toan yeu cau th dung hai nh luat nay chungta lap c bay nhieu phng trnh nut mang va mac mang, sau o giaihe phng trnh ta se tm c cac gia tr ma bai toan yeu cau. Tuy nhien, e giai nhng mach ien co nhieu nguon, nhieu ientr mac phc tap th giai he phng trnh nhieu an rat dai, tnh toanphc tap. V the trong nhng mach khac nhau, chung ta nen ap dung cacphng phap phu hp e giai quyet bai toan mot cach nhanh nhat.

Bai 1: Cho mot mach ien co s o nh hnh veE1=25v R1=R2=10E2=16v R3=R4=5r1=r2=2 R5=8

Tnh cng o dong ien qua moi nhanh.

Giasdong ien chay trong machcochieu nh hnh ve:*nh luat Kirchoff cho cac nut mang :Tai C, B : I=I +I =I +I (1)3 4 1 5Tai A : I =I +I1 2 3

(2)

Tai D: I =I +I (3)4 2

*nh luat Kirchoff cho mat mang:Mach BACB: E =I R +I R +Ir 10I +5I +2I=16 (4)2 1 1 3 3 2 1 3

5

Mach ADCA: 0=I R +I R -I R 10I +5I -5I =0 (5)2 2 4 4 3 3 2 4 3Mach DCBD: E +E =I R +I R +I r +Ir (6)1 2 4 4 1 25 5 5

5I +10I +2I=414 5

T(1), (2), (3), (4), (5), (6) ta cohephng trnh:

I-I -I =0 (1)1 5I -I -I =0 (2)1 2 3I -I +I =0 (3)2 4 510I +5I +2I=16 (4)1 3

10I +5I -5I =0 (5)2 4 35I +10I +2I=414 5

=>

10I +5I +2I=16 (4)1 310I +5I -5I =0 (5)2 4 3I-I +I -I =0 (7)1 2 4I -I -I =0 (2)1 2 3

12I-10I +5I =41 (8)1 4I =I-I (1)(6) 15

10I +5I +2I=16 (41 312I-10I +5I =41 (81 4I-I -I =0 (93 410I -15I +5I =0 (11 3 4

I =I -I2 1 3I =I-I15

2I+10I +5I =161 317I-10I -5I =41 3

I=3 (A)

I =0.5 (A)1 1I =-0.5 (A)5I+10I -20I =0 21 3 I =1 (A)I =I -I 32 1 3

I =2 (A)I =I-I 44 3I =2.5 (A)I =I-I 515


87/274


Bi 2:E=14Vr=1V R3=3R4=8 R1=1R2=3 R5=3

Tm I trong cac nhanh?

GiaiTa gia s chieu cua dong ien nh hnh ve.*nh luat mat mang:AMNA: 0=I

1R

1-I

5R

5-I

2R

2

0=I1-3I

5-3I

2(1)

MBNM: 0=I3R

3-I

4R

4+I

5R

5

0=3I3-8I

4+3I

5(2)

ANBA: E=Ir+I2R

2+I

4R

4

14=I+3I2+8I

4(3)

*nh l nut mang:-Tai N: I

2-I

5-I

4=0 (4)

-Tai B: I-I4-I3=0 (5)

T i A I I I 0 (6)

A

E,r

M

N


88/274


Ta chon I,I2,I4 lam an chnh va bien oi I1,I3,I5 theo bien trenT (1) ta co :

I1-3I5-3I2 =0 I-I2-3(I2-I4)-3I2=0 I-7I2+3I4=0

T (2) ta co: 3I3-8I4+3I5=0 3(I-I4)-8I4+3(I2-I4)=0 3I-14I4+3I2 =0

Ta c h pt:I+3I2+8I4=14I-7I2+3I4=0 I=3.56(A) I2=0.92(A) I4=0.96(A)3I+3I2-14I4=0

I1=I-I2=2.24(A)I3=I-I4=2.6(A)I5=I2-I4=-0.04(A). Vy dng i t m n N.B.BI TPCh :

a/. chp cc im cng in th: "Ta c th chp 2 hay nhiu im c cng inth thnh mt im khi bin i mch in tng ng."(Do VA-Vb = UAB=I RAB Khi RAB=0;I 0 hoc RAB 0,I=0 Va=VbTc A v B cng inth)Cc trng hp c th: Cc im 2 u dy ni, kha K ng, Am pe k c intr khng ng k...c coi l c cng in th. Hai im nt 2 u R5 trong mchcu cn bng...b/. B in tr: ta c th b cc in tr khc 0 ra khi s khi bin i mchin tng ng khi cng dng in qua cc in tr ny bng 0.Cc trng hp c th: cc vt dn nm trong mch h; mt in tr khc 0 mc songsong vi mt vt dn c in tr bng 0( in tr b ni tt) ; vn k c in trrt ln (l tng).

4/. Vai tr ca am pe k trong s :* Nu am pe k l tng ( Ra=0) , ngoi chc nng l dng c o n cn c vai trnh dy ni do :C th chp cc im 2 u am pe k thnh mt im khi bin i mch intng ng( khi am pe k ch l mt im trn s )Nu am pe k mc ni tip vi vt no th n o cng d/ qua vt.Khi am pe k mc song song vi vt no th in tr b ni tt( ni trn).

Khi am pe k nm ring mt mch th dng in qua n c tnh thng qua ccdng 2 ntm ta mc am pe k ( d theo nh l nt).* Nu am pe k c in tr ng k, th trong s ngoi chc nng l dng co ra am pe k cn c chc nng nh mt in tr bnh thng. Do s ch ca n

cn c tnh bng cng thc: Ia=Ua/Ra .5/. Vai tr ca vn k trong s :/ t h k i t t l ( l t )


89/274

[email protected] tm v bin son*Vn k mc song song vi on mch no th s ch ca vn k cho bit HT gia2 u on mch :UV=UAB=IAB. RAB

*TRong trng hp mch phc tp, Hiu in th gia 2 im mc vn k phi ctnh bng cng thc cng th: UAB=VA-VB=VA- VC + VC- VB=UAC+UCB....

*c th b vn k khi v s mch in tng ng .*Nhng in tr bt k mc ni tip vi vn k c coi nh l dy ni ca vn k( trong s tng ng ta c th thay in tr y bng mt im trn dy ni),theo cng thc ca nh lut m th cng qua cc in tr ny coi nh bng 0 ,( IR=IV=U/ =0).b/. Trng hp vn k c in tr hu hn ,th trong s ngoi chc nng l dngc o vn k cn c chc nng nh mi in tr khc. Do s ch ca vn k cnc tnh bng cng thc UV=Iv.Rv...

Mi bi tp c th c nhiu cch gii, vi mi bi tp phi quan st tm c cch gii hpl.I.CC BI TNH TON N THUNBi 1.Tnh hiu in th gia hai cc ca mt ngun c sut in ng l , bit in tr trong v ngoil nh nhau ?

s:2

Bi 2.Nu mc in tr 16 vi mt b pin th cng dng in trong mch bng 1 A. Nu mc intr 8 vo b pin th cng bng 1,8 A. Tnh sut in ng v in tr trong ca b pin

S: R=2;=18VBi 4: Cho mch in nh hnh v. Ngun in c sut in ng E = 7,8V,v in trtrong r = 0,4. Cc in tr mch ngoi R1 = R2 = R3 = 3, R4 = 6.a. Tnh cng dng in chy qua cc in tr v hiu in th hai u mi in tr.b. Tnh hiu in th gia hai im C v D.c. Tnh hiu in th hai u ngun in v hiu sut ca ngun in.

S:a.I1=I2=1,17A, I3=I4=0,78A, U12=3,5V; U3=2,34V; U4=4,68Vb.UCD=-1,17

Documents

TỔNG HỢP BÀI TẬP VẬT LÍ 11