21F21F12Fq1.q2 >0r21F12Frq1.q2 < 0Ti liu dy thm Vt L 11
Ban c bn GV: T Hng Sn
PHN 1:IN HC. IN T HCCHNG I :IN TCH. IN TRNGCH 1: IN TCH. NH LUT
CU LNGA. TM TT L THUYT1. C hai loi in tch:in tch m (-) v in tch dng
(+)2. Tng tc tnh in: + Hai in tch cng du: y nhau;+ Hai in tch tri
du: Ht nhau;3. nh lut Cu - lng: Lc tng tc gia 2 in tch im q1; q2 t
cch nhau mt khong r trong mi trng c hng s in mi l 12 21; F Fr r c:-
im t: trn 2 in tch.- Phng: ng ni 2 in tch.- Chiu:+ Hng ra xa nhau
nuq1.q2 > 0(q1; q2 cng du)+ Hng vo nhau nuq1.q2 < 0(q1; q2
tri du) - ln:1 22..q qF kr ; Trong : k = 9.109Nm2C-2; l hng s in mi
ca mi trng, trong chn khng = 1.- Biu din:4. Nguyn l chng cht lc in:
Gi s c n in tch im q1, q2,.,qn tc dng ln in tch im q nhng lc tng tc
tnh in n n 1F ,....., F , Fth lc in tng hp do cc in tch im trn tc
dng ln in tch q tun theo nguyn l chng cht lc in. + + + i n n 1F F
..... F F FB. CC DNG BITPDng 1:Xc nh lc tng tc gia 2 in tch v cc i
lng trong cng thc nh lut Cu lng.Phng php : p dng nh lut Cu lng.-
Phng , chiu , im t ca lc ( nh hnh v) - ln : F = 22 19.| . | . 10 .
9rq q- Chiu ca lc da vo du ca hai in tch : hai in tch cng du : lc y
; hai in tch tri du : lc ht Dng 2: Tm lc tng hp tc dng ln mt in
tch.Trng THPT Ng Quyn Ph Qu Bnh Thunq1 q2 Ti liu dy thm Vt L 11 Ban
c bn GV: T Hng Sn Phng php : Dng nguyn l chng cht lc in.- Lc tng tc
ca nhiu in tch im ln mt in tch im ln mt in tch im khc : + + + nF F
F F ...2 1-Biu din cc cc lc 1Fuur,2Fuur,3FuurnFuurbng cc vecto , gc
ti im ta xt .-V cc vc t hp lc theo quy tc hnh bnh hnh . - Tnh ln ca
lc tng hp da vo phng php hnh hc hoc nh l hm s cosin. *Cc trng hp c
bit:1 2 1 21 2 1 22 21 2 1 22 21 2 1 2 1 2..(F , ) 2 osF F F F FF F
F F FE E F F FF F F F FF c + + + +r rr rr rr rC. BI TP P DNGBi 1:
Hai in tch im bng nhau c t trong khng kh cch nhau 10 cm, lc tng tc
gia hai in tch l 1N. t hai in tch vo trong du c = 2 cch nhau 10 cm.
hi lc tng tc gia chng l bao nhiu?Hng dn:- Trong khng kh:1 22| . | q
qF kr- Trong du:/1 22| . |.q qFr - Lp t s: //1 1 10,52 2 2F FFF
N.Bi 2: Hai in tch im bng nhau, t trong chn khng cch nhau mt khong
r1 = 2 cm. lc tng tc gia chng l 1,6.10-4 N. a) Tm ln hai in tch ?
b) Khong cch r2 gia chng l bao nhiu lc tc dng gia chngl 2,5.10-4
N?Hng dn: a) Ta c: 1 21 21. q qF kr ( )24 222181 191,6.10 . 2.10.
64.109 9.10F rqk Vy: q = q1= q2= 98.103C.b) Ta c: 1 22 22. q qF Kr
suy ra: 2 22 1 2 1 12 22 2 1. F r F rrF F r Vy r2 = 1,6cm.Bi 3:Hai
in tch im q1 = -10-7 C v q2 = 5.10-8 C t ti hai im A v B trong chn
khng cch nhau 5 cm. Xc nh lc in tng hp tc dng ln in tch q0 = 2.10-8
C t ti im C sao cho CA = 3 cm, CB = 4 cm.Hng dn: - Lc tng tc gia q1
v q0 l : Trng THPT Ng Quyn Ph Qu Bnh Thunq1 q2 Ti liu dy thm Vt L
11 Ban c bn GV: T Hng Sn
1 0 21 2.2.10q qF k NAC - Lc tng tc gia q2 v q0 l :
2 0 32 2.5,625.10q qF k NBC - Lc in tc dng ln q0 l :
2 2 21 21 22,08.10 F F F F F F N + + ur ur urBi 4:Hai in tch q1
= 4.10-5 C v q2 = 1.10-5 C t cch nhau 3 cm trong khng kh. a) Xc nh
v tr t in tch q3 = 1.10-5 C q3 nm cn bng ? b) Xc nh v tr t in tch
q4 = -1.10-5 C q4 nm cn bng ?Hng dn:
- Gi 13 Fur l lc do q1 tc dng ln q3 23 Fur l lc do q2 tc dng ln
q3 - q3 nm cn bng th 13 23 0 F F + ur ur r13 23 F F ur ur
13 23 , F F ur ur cng phng, ngc chiu v F13 = F23 V q1, q2, q3
>0 nn M nm gia A v B.t MA = xTa c : ( )1 3 2 32 23qq q qk
kxx
2 21243 3q x xq x x _ _ , ,x = 2 cm.b) Nhn xt : khi thay q4 =
-1.10-5 C th khng nh hng n lc tng tc nn kt qu khng thay i, vy x = 2
cm.Bi 5:Hai in tch q1 = 8.10-8 C v q2 = -8.10-8 C t ti A v B trong
khng kh cch nhau mt khong AB = 6 cm. Xc nh lc in tc dng ln q3 =
8.10-8 Ct ti C nu : a) CA = 4 cm v CB = 2 cm. b) CA =4 cm v CB =10
cm. c) CA = CB = 5 cm.Hng dn:- S dng nguyn l chng cht lc in.a) F =
F1 + F2 = 0,18 Nb) F = F1 F2 = 30,24.10-3 N c) C nm trn trung trc
AB v F = 2F1.cos = 2.F1. AHAC = 27,65.10-3 N***Trng THPT Ng Quyn Ph
Qu Bnh ThunQ2BACQ0Q1F1F2Fq1 q2 ABM qF23F13xMErrMErrTi liu dy thm Vt
L 11 Ban c bn GV: T Hng SnCH 2: IN TRNGA. TM TT L THUYT 1.Khi nim
in trng:L mi trng tn ti xung quanh in tch v tc dng lc ln in tch khc
t trong n. 2. Cng in trng: L i lng c trng cho in trng v kh nng tc
dng lc.E q FqFE . n v: E (V/m)q > 0 : F cng phng, cng chiu vi
E.q < 0 : F cng phng, ngc chiu vi E.3. ng sc in - in trng u.a.
Khi nim ng sc in:*Khi nimngscin:Lngcongdotavchra trongin trng sao
cho ti mi im trn ng cong, vector cngintrngcphngtrngvi tiptuynca ng
cong ti im , chiu ca ng sc l chiu ca vector cng in trng. *ng sc in
do in tch im gy ra:+ Xutpht t in tch dng v kt thc in tch m;+ in tch
dng ra xa v cc;+ T v cc kt thc in tch m.b. in trng unh ngha:in trng
u l in trng c vector cng in trng ti mi im bng nhau c v phng, chiu v
ln.* c im: Cc ng sc ca in trng u l nhng ng thng song song cch u.4.
Vct cng in trng Er do 1 in tch im Q gy ra ti mt im M cch Q mt on r
c:- im t: Ti M.- Phng:ng ni M v Q- Chiu:Hng ra xa Q nu Q > 0Hng
vo Q nu Q 0 0q < 0MMq1 q2 q1 q2 q1 q2 q1 q2 Ti liu dy thm Vt L
11 Ban c bn GV: T Hng SnDng 1: Xc nh cng in trng do in tch gy ra ti
mt imPhng php:Cng in trng do in tch im Q gy ra c: + im t:Ti im ang
xt;+ Phng:Trng vi ng thng ni in tch Q v im ang xt;+ Chiu:Hng ra xa
Q nu Q > 0v hng v Q nu Q < 0;+ ln: E = k2rQ, trong k =
9.109Nm2C-2.Dng 2: Xc nh lc in trng tc dng ln mt in tch trong in
trngPhng php:Lc tnh in tc dng ln in tch q t trong in trng:E q F F
c:+ im t: ti im t in tch q;+ Phng: trng phng vi vector cng in trngE
;+Chiu:Cng chiu viEnu q > 0 v ngc chiu viE nu q |q2| nn C nm gn
q2t CB = x 40 AC x +, c : ( )1 2 / /1 2 2 22124040 402 96,6q qE E K
kxxq x xx cmq x x + _ + + ,Bi 3:Hai in tch im q1 = 1.10-8 C v q2 =
-1.10-8 C t ti hai im A v B cch nhau mt khong 2d = 6cm. im M nm trn
ng trung trc AB, cch AB mt khong 3 cm.a) Tnh cng in trng tng hp ti
M. b) Tnh lc in trng tc dng ln in tch q = 2.10-9 C t ti M.Trng THPT
Ng Quyn Ph Qu Bnh Thun11 222 2qE kIAqE kIBq1 q2 q1 q2 ABI
E1EE2/1rE/2rEq1 q2 A B C xTi liu dy thm Vt L 11 Ban c bn GV: T Hng
SnHng dn: a) Gi12, E Eur ur l vecto cddt do q1 v q2 gy ra ti M Eur
l vecto cddt tng hp ti MTa c :1 2 E E E +ur ur ur, do q1 = | -q2 |
v MA = MB nn E1 = E2 , Vy E =2.E1.cosTrong : cos = dMA, MA = 2 2 23
3 3 2.10m+ Vy:E = 7.104 V/m.b) Lc in tc dng ln in tch q t ti Mc:-
im t: ti M- Phng, chiu: cng phng chiu vi Eur (nh hnh v)- ln: F =
|q|.E = 9 4 42.10 .7.10 1,4.10 N Bi 4: Ti 3 nh hnh vung cnh a =
30cm, ta t 3 in tch dng q1 = q2 = q3 = 5.10-9 C.Hy xc nh: a) Cng in
trng ti nh th t ca hnh vung? b) Lc tc dng ln in tch q = 2.10-6 C t
ti nh th t ny?Hng dn:a) Gi1 2 3 , , E EEur ur ur l vecto cng in
trng do q1, q2, q3 gy ra ti nh th t hnh vungV Eur l vecto cng in
trng ti .Ta c: 1 2 3 E E E E + +ur ur ur urGi 13 Eurl vecto cng in
trng tng hp ca13, E Eur urVy : Eur= 13 Eur + 2 Eur E = E13 +E2E = (
)22 22 9,5.102q qk kaa+ V/m.b) Lc in tc dng ln in tch q l : F =
|q|.E = 2.10-6.9,5.102 = 19.10-4 NBi 5:Ti 3 nh hnh vung cnh a = 20
cm, ta t 3 in tch cng ln q1 = q2 = q3 = 3.10 -6 C. Tnh cng in trng
tng hp ti tm hnh vung ?S : E = 1,35.106 V/m.Trng THPT Ng Quyn Ph Qu
Bnh Thun2rEq1 q2 1rErEABMddq1E13E3E2EE1q2q3Ti liu dy thm Vt L 11
Ban c bn GV: T Hng SnBi 6: Mt qu cu nh khi lng m = 1g, mang in tch
q = 10-5C, treo bng si dy mnh v t trong in trng u E. Khi qu cu nm
cn bng th dy treo hp vi phng thng ng mt gc60o . Xc nh cng in trng
E, bit g =10m/s2.S : E = 1730 V/m.Bi 7:Mt in tch im q = 2.106 C t c
nh trong chn khng. a) Xc nh cng in trng ti im cch n 30 cm ? b) Tnh
ln lc in tc dng ln in tch 1C t ti im ? c) Trong in trng gy bi q, ti
mt im nu t in tch q1 = 10-4 C th chu tc dng lc l 0,1 N. Hi nu t in
tch q2 = 4.10-5 C th lc in tc dng l bao nhiu ?S : a) 2.105 V/m, b)
0,2 N, c) 0,25 NTrng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm Vt L
11 Ban c bn GV: T Hng SnCH 3: CNG CA LC IN. IN TH. HIU IN TH.A. TM
T T L THUYT 1. Cng ca lc in trng:* c im: Cng ca lc in tc dng ln tc
dng ln mt in tch khng ph thuc vo dng qu o m ch ph thuc vo im u v im
cui ca qu o (v lc in trng l lc th).* Biu thc: AMN = qEdTrong , d l
hnh chiu ca qu o ln phng ca ng sc in.Ch : - d > 0 khi hnh chiu
cng chiu ng sc. - d < 0 khi hnh chiu ngc chiu ng sc.2. Lin h gia
cng ca lc in v hiu th nng ca in tch AMN = WM - WN3. in th. Hiu in
th- ien the tai mot iem M trong ien trng la ai lng ac trng cho ien
trng ve phng dien tao ra the nang khi at tai o mot ien tch q.Cng
thc: VM = qAM- Hiu in th gia 2 im trong in trng l i lng c trng cho
kh nng thc hin cng ca in trng khi c 1 in tch di chuyn gia 2 im .UMN
= VM VN = qAMN Ch : - in th, hiu in thl mt i lng v hng c gi tr dng
hoc m;- Hiu in th gia hai im M, N trong in trng c gi tr xc nh cn in
th ti mt im trong in trng c gi tr ph thuc vo v tr ta chn lm gc in
th.-Neumotientchdngbanaungyen, chchutac dung cua lc ien th no se co
xu hng di chuyen ve ni co ien the thap (chuyen ong cung chieu ien
trng).Ngc lai, lc ien co tac dung lam cho ien tch am di chuyen ve
ni co ien the cao (chuyen ong ngc chieu ien trng).- Trong in trng,
vector cng in trng c hng t ni c in th cao sang ni c in th thp;4.Lin
h gia cng in trng v hiu in thE = dUB. CC DNG BI TPDng 1:Tnh cng ca
cc lc khi in tch di chuynPhng php: s dng cc cng thc sau1. AMN =
qEdCh : - d >0 khi hnh chiu cng chiu ng sc. - d < 0 khi hnh
chiu ngc chiu ng sc.Trng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm
Vt L 11 Ban c bn GV: T Hng Sn2. AMN = WtM - WtN = WN - WM3. AMN =
UMN .q = (VM VN ).qC h :Du ca cng ph thuc vo du ca q v U v gc hp bi
chiu chuyn di v chiu ng sc.Dng 2: Tm in th v hiu in thPhng php: s
dng cc cng thc sau1. Cng thc tnh in th : MMAVqCh : Ngi ta lun chn
mc in th ti mt t v v cng ( bng 0 ) 2. Cng thc hiu in th:qAUMNMN =
VM VN 3. Cng thc lin h gia cng in trng v hiu in th trong in trng u
E = dUCh :Trong in trng, vector cng in trng c hng t ni c in th cao
sang ni c in th thp;C. BI TP P DNGBi 1:Mt e di chuyn mt on 0,6 cm t
im M n im N dc theo mt ng sc in ca 1 in trng u th lc in sinh cng
9,6.10-18J 1. Tnh cng in trng E2. Tnh cng m lc in sinh ra khi e di
chuyn tip 0,4 cm t im N n im P theo phng v chiu ni trn? 3. Tnh hiu
in th UMN; UNP 4. Tnh vn tc ca e khi n ti P. Bit vn tc ca e ti M
bng khng.Gii:1.Ta c: AMN =q.E.' 'N Mv AMN > 0; q < 0; E >
0 nn ' 'N M < 0 tc l e i ngc chiu ng sc.=> ' 'N M =- 0,006 m
Cng in trng: ( ) ( )184199, 6.1010 ( / ). ' ' 1, 6.10 . 0, 006MNAE
V mq M N 2. Ta c:' 'N P= -0,004m => ANP= q.E.' 'P N =
(-1,6.10-19).104.(-0,004) = 6,4.10-18 J3. Hiu in th:-18MNMN
-19-18NPNP -199,6.10U 60( )-1,6.106,4.10U 40( )-1,6.10AVqAVq 4. Vn
tc ca e khi n ti P l:p dng nh l ng nng: AMP = WP WN => WP = AMN
+ANP = 16.10-18 J186312 2.16.105, 9.10 ( / )9,1.10dPWv m sm Bi 2:
Hiu in th gia hai im M, N trong in trng l UMN = 100V.Trng THPT Ng
Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11 Ban c bn GV: T Hng Sn a)
Tnh cng in trng lm dch chuyn proton t M n N. b) Tnh cng in trng lm
dch chuyn electron t M n N.c) Nu ngha s khc nhau trongkt qu tnh c
theo cu a v cu b.Hng dn: a. Cng in trng thc hin proton dch chuyn t
M n N.19 171. 1, 6.10 .100 1, 6.10p M NA q U J b. Cng in trng thc
hin electron dch chuyn t M n N.19 172. 1, 6.10 .100 1, 6.10e M NA q
U J c. A1>0, cnghalintrngthcslmvicdchchuynprotontMnN.A2 < 0,
in trng chng li s dch chuyn , mun a electron t M n N th ngoi lc phi
thc hin cng ng bng 1,6.10-17 J.Bi 3: Ba im A, B, C to thnh mt tam
gic vung ti C; AC = 4cm, BC = 3cm v nm trong mt in trng u. Vecto
cng in Eur trng song song AC, hng t A n C v c ln E = 5000V/m. Hy
tnh: a) UAC, UCB,UAB. b) Cng ca in trng khi e di chuyn t A n B v
trnng gy ACBHng dn:a.Tnh cc hiu in th- UAC = E.AC = 5000.0,04 =
200V.- UBC = 0 v trn on CB lc in trng. F qE ur urvung gc CB nn ACB
= 0UCB = 0.- UAB = UAC + UCB = 200V.b. Cng ca lc in trng khi di
chuyn e- t A n B.19 171,6.10 .200 3,2.10ABA J Cng ca lc in trng khi
di chuyn e- theo ng ACB.AACB = AAC + ACB = AAC = -1,6.10-19.200 =
-3,2.10-17 J cng khng ph thuc ng i.Bi 4: Mt electron bay vi vn tc v
= 1,5.107m/s t mt im c in th V1 = 800V theo hng ca ng sc in trng u.
Hy xc nh in th V2 ca im m ti electron dng li. Bit me = 9,1.10-31
kg, Hng dn:p dng nh l ng nng0 .m.v20 = e.(V1 V2)Nn : V2 = V1 -
202mve = 162V.Bi 5:ABC l mt tam gic vung gc ti A c t trong in trng
u Eur.Bit 060 ABC , AB PEur. BC = 6cm,UBC = 120V a). Tm UAC,UBA v
ln Eur. b). t thm C mt in tch q = 9.10-10 C.Tnh cng in trng tng hp
ti A.Hng dn:a. ABC V l tam gic u, vy nu BC = 6cm.Trng THPT Ng Quyn
Ph Qu Bnh ThunEACBEBACTi liu dy thm Vt L 11 Ban c bn GV: T Hng
SnSuy ra: BA = 3cm v AC = 6 33 32UBA = UBC = 120V, UAC = 0E =4000
/BAU UV md BA .b. 2 2A CA CE E E E E E + +ur ur ur= 5000V/m.Bi 6:
Hai bn kim loi phng song song mang in tch tri du t cch nhau 2cm.
Cng in trng gia hai bn l E = 3000V/m. St bn mang in dng, ta t mt ht
mang in dng c khi lng m = 4,5.10-6 g v c intch q = 1,5.10-2 C.tnh
a) Cng ca lc in trng khi ht mang in chuyn ng t bn dng sang bn m. b)
Vn tc ca ht mang in khi n p vo bn m.Hng dn:a. Cng calc in trng l:A=
qEd = 0,9 J.b. Vn tc ca ht mang in- p dng nh l ng nng4292.
2.0,92.104,5.10Avm m/s.Bi 7: Mt in tch c khi lng m = 6,4.10-15 kg
nm l lng gia hai tm kim loi song song nm ngang v nhim in tri du. in
tch ca qu cu l 1,6.10-17C. Hai tm cch nhau 3cm. Hy tnh hiu in th t
vo hai tm . Ly g = 10m/s2.Hng dn:V qu cu nm cn bng th lc in cn bng
trong lc qu cu nn:- F = P = 6,4.10-14 N.- F = q.E = . .120U q F dU
Vd q .Trng THPT Ng Quyn Ph Qu Bnh ThunALBELT EINSTEIN(14/3/1879
18/4/1955)Tnh tng i p dng cho Vt L, ch khng phi cho o c.Ti liu dy
thm Vt L 11 Ban c bn GV: T Hng SnCH 4:T IN. GHP T INA. TM TT L
THUYT1.T in-nh ngha : H 2 vt dn t gn nhau, mi vt l 1 bn t. Khong
khng gian gia 2 bn l chn khng hay in mi. T in dng tch v phng in
trong mch in. -T in phng c 2 bn t l 2 tm kim loi phng c kch thc ln
,t i din nhau, song song vi nhau.2. in dung ca t in- L i lng c trng
cho kh nng tch in ca tQCU (n v l F, mF.)- Cng thc tnh in dung ca t
in phng: dSC. 4 . 10 . 9.9. Vi S l phn din tch i din gia 2 bn.Ghi
ch : Vi mi mt t in c 1 hiu in th gii hn nht nh, nu khi s dng m t vo
2 bn t ht ln hn ht gii hn th in mi gia 2 bn b nh thng.3. Ghp t
inGHP NI TIP GHP SONG SONGCch mc : Bnthhai cat1ni vi bn th nht ca t
2, c th tip tc Bn th nht ca t 1 ni vi bn th nht ca t 2, 3, 4 in tch
QB = Q1 = Q2 = = QnQB = Q1 + Q2 + + QnHiu in thUB = U1 + U2 + +
UnUB = U1 = U2 = = Unin dungn 2 1 BC1...C1C1C1+ + + CB = C1 + C2 +
+ CnGhi ch CB < C1, C2 CnCB > C1, C2, C34. Nng lng ca t in-
Khi t in c tch in th gia hai bn t c in trng v trong t in s d tr mt
nng lng. Gi lnng lng in trng trong t in.- Cng thc: 2 2. .2 2 2QU CU
QWC B. CC DNG BI TP C BNDng 1: Tnh in dung, in tch, hiu in th v nng
lng ca t inPhng php: s dng cc cng thc sau- Cng thc nh ngha :C(F) =
UQ => Q = CU- in dung ca t in phng : C = d k 4S- Cng thc: 2 2.
.2 2 2QU CU QWC Ch : + Ni t vo ngun: U = hng s+ Ngt t khi ngun: Q =
hng sDng 2: Ghp t inTrng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm
Vt L 11 Ban c bn GV: T Hng SnPhng php: i lng Ghp ni tip Ghp song
songin tch Q = Q1= Q2== QnQ = Q1 + Q2+.+QnHiu in th U = U1 + U2 ++
UnU = U1 = U2 == Unin dungn 2 1 bC1...C1C1C1+ + + Cb = C1 + C2 + +
CnCc trng hp c bit:a. Ghp ni tip: Cb < Ci+ Nu C1 = C2= = Cn =
C=> Cb = nC;U1 = U2 = .. = Un = nU=>U = nUi+ C1ntC2 => Cb
= 2 12 1C CC C++ C1ntC2ntC3 => Cb = 3 1 3 2 2 13 2 1C C C C C CC
C C+ +b. Ghp song song:Cb > Ci.+ Nu C1 = C2= = Cn = C=> Cb =
nC ; Q1 = Q2 = .= Qn => Qb = nQi.C. BI TP P DNGBi 1: Mt t in
phng c in mi khng kh; khong cch gia 2 bn l d = 0,5 cm; din tch mt
bn l 36 cm2. Mc t vo ngun in c hiu in th U=100 V.1. Tnh in dung ca
t in v in tch tch trn t.2. Tnh nng lng in trng trong t in.3. Nu ngi
ta ngt t in ra khi ngun ri nhng n chm hn vo mt in mi lng c hng s in
mi = 2. Tm in dung ca t v hiu in th ca t.4. Nu ngi ta khng ngt t
khi ngun v a t vo in mi lng nh phn 3. Tnh in tch v ht gia 2 bn
tGii:1.in dung ca t in: 4 29 9. 36.10 10( )9.10 .4 . 9.10 .4 .0,
005 5.SC Fd in tch tch trn t: 210 1. .100 ( )5. 5.Q CU C 2.Nng lng
in trng: 22 41 1 10 10.10 ( )2 2 5.W CU J 3.Khi nhng t vo trong
dung mi c = 2 C = 2C = 22.10( )5.FKhi ngt t ra khi ngun t in tr
thnh h c lp in tch ca t khng thay i:=>Q = Q => CU = CU =>
' 50( )' 2C UU U VC 4. Khi khng ngt t ra khi ngun hiu in th 2 bn t
khng thay i:=>U = U = 100V=> ' ' 2' 2 ( )' 5.Q Q CQ Q Q CC C
C Trng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11 Ban c bn
GV: T Hng SnBi 2 : Cho b t in mc nh hnh v.C1 = 4F,C2 = 6F ,C3 =
3,6F v C4 = 6F. Mc 2 cc AB vo hiu in th U = 100V.1. Tnh in dung ca
b t v in tch ca mi t.2. Nu hiu in th gii hn ca b t C1,2,3 (CAM) l
40V; hiu in th gii hn ca t C4 l 60V. Th hiu in th ti a t vo 2 u mch
in l bao nhiu cc t khng b nh thng?Gii:1. Cu to ca mch in: ( )1 2 3
4Cnt C Cnt C1 ]P in dung ca b t:( )( )( )1 2121 212 3446.42, 46 42,
4 3, 6 66.636 6AMAMABAMC CC FC CC C C FC CC FC C + + + + + +in tch
ca cc t:6 44412 3 66 43 3 36 412 12 12 1 2. 3.10 .100 3.10 (
)3.1050( )6.10. 3, 6.10 .50 1, 8.10 ( ). 2, 4.10 .50 1, 2.10 ( )AB
AB AB AMAMAMAMQ C U C Q QQU V U UCQ C U CQ C U C Q Q 2. in tch cc i
c th tch trn b t CAM v C4 l:QmaxAM = CAM.UmaxAM = 6.10-6.40 =
24.10-5(C)Qmax4 = C4.Umax4 = 6.10-6.60 = 36.10-5(C)M thc t ta c v
CAM; C4 mc ni tip nn khng c t no b nh thng th:QAM = Q4 ( )maxAM
max4min Q ; Q in tch ti a ca b:QAB = QAM = Q4 = QmaxAM =
24.10-5(C)Hiu in th ti a c th t vo 2 u mch in l:5624.1080(
)3.10ABABABQU VC Bi 3: Cho b t nh hnh v, bit C1 = 8F ; C2 = 6F ; C3
=3F .a) Tnh in dung tng ng ca b t.b) t vo hai u AB mt hiu in th U =
8V. Tnh hiu in th v in tch ca mi t.Hng dn:a. in dung tng ng ca b
tTa c: 2 3232 3. 6.32 .6 3C CC FC C + +- in dung tng ng: Cb = C1
+C23 = 10F .b.Hiu in th gia hai bn t C1 l: U1 = U = 8V- in tch ca t
C1: Q1 = C1.U = 6,4.10-5 C. - in tch trn mi t C2 v C3: Q2 = Q3 =
C23.U = 1,6.10-5 C.Trng THPT Ng Quyn Ph Qu Bnh ThunC3C1C2 C4AM
BC3C1C2ABTi liu dy thm Vt L 11 Ban c bn GV: T Hng Sn- Hiu in th gia
hai bn t C2: 2222,67 .QU VC - Hiu in th gia hai bn t C3 l: U3 = U
U2 = 5,33 V.Bi 4: Mt t in phng c in mi l khng kh, in dung C = 10F
gm hai bn cch nhau 2 cm.a) t tch mt in lng 0,2 mC th phi t vo hai u
t in mt hiu in th bao nhiu? b) Bit khng kh chu c cng in trng ti a l
20.105V/m. Tnh in lng cc i m t tch c.S: a) 20 V; b) 0,4 C.Bi 5: Cho
mch in nh hnh v vi:C1 = 12F ; C2 = 4F ; C3 = 3F ; C4 = 6F ; C5 = 5F
;UAB = 50 V. Tnh:a) in dung ca b t.b) in tch v hiu in th mi t.c)
Hiu in th UMN.Hng dn:a. in dung ca b tC12 = 1 21 2.3 .C CFC C +C34
= 3 43 4.2 .C CFC C +C1234 = C12 +C34 = 5F .Cb = 1234 51234 5.2,5
.C CFC C +B. in tch v hiu in th mi tTa c: C1234 nt C5 nn: q1234 =
q5 = qb = Cb.UAB = 125C Vy U5 = 55125255qVC 1234 525ABU U U V .- C1
v C2 nt nn : q12 = q1 = q2 = C12.U1234 = 3.25 = 75C .Vy :
1112226,25 .18,75 .qU VCqU VC - C3 v C4 nt nn : q3 = q4 =C34.U1234
= 50C .3334445016,7 .3508,3 .6qU VCqU VC c. Hiu in th UMN.Trng THPT
Ng Quyn Ph Qu Bnh ThunC3C1 C5A BC4C2NMOVy: UMN = UMA +UAN= - U3
+U1= - 16,7 + 6,25 = - 10,5V. +Ti liu dy thm Vt L 11 Ban c bn GV: T
Hng SnCHNG II:DNG IN KHNG ICH 1: DNG IN KHNG I. NGUN INA. TM TT L
THUYT1. Dngin khng ia. Dng in: L dng chuyn di c hng ca cc ht mang
in.-Quycchiudngin: Lchiuchuyndichngcaccht mangintch dng.Lu :+ Trong
in trng, cc ht mang in chuyn ng t ni c in th cao sang ni c in th
thp, ngha l chiu ca dng in l chiu gim ca in th trong vt dn. + Trong
kim loi, ht tham gia ti in l electron mang in tch m nn chuyn ng t
ni c in th thp sang ni c in th cao, ngha l chuyn ng ngc vi chiu ca
dng in theo quy c.b. Cng dng in:a. nh ngha:I = tq,cng dng in I c n
v l ampre (A)Trong : q l in lng, t l thi gian.+ nut l hu hn, th I l
cng dng in trung bnh;+ nut l v cng b, th i l cng dng in tc thi.c.
Dng in khng i:'oi khong ien dong o cngoi khong ien dong cua
chieu=> I = tq,Ch :s electron chuyn qua tit din thng ca vt dn :
..I tne2. nh lut m i vi on mch ch c in tr a. nh lut m : I = URb. in
tr ca vt dn: R = S.Trong , l in tr sut ca vt dn. in tr sut ph thuc
vo nhit theo cng thc: = o[1 + (t to)]o l in tr sut ca vt dn to
(oC)thng ly gi tr 20oC. c gi l h s nhit in tr.c.Ghp in tri lng on
mch ni tip on mch song songHiu in th U = U1 + U2 + + UnU = U1 = U2
= .= UnCng dng in I = I1 = I2= = InI = I1 + I2 +.+ Inin tr tng ng
Rt = R1 + R2 ++ Rn`n 2 1 tR1....R1R1R1+ + + 3. Ngun in sut in ng
ngun ina. Ngun in+ C cu to ra v duy tr hiu in th nhm duy tr dng in
gi l ngun in.Trng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11
Ban c bn GV: T Hng Sn+ Hai cc nhim in khc nhau l nh lc l tch
electron ra khi nguyn t trung ha ri chuynelectron hay Ion dng ra
khi mi cc.b. Sut in ng ngun in- L i lng c trng cho kh nng thc hin
cng ca ngun in.Cng thc:E = Aq- in tr ca ngun in c gi l in tr trong
cu n.- Mi ngun in c c trng: (E , r)B. CC DNG BI TPDng 1:Xc nh in
lng, cng dng in theo cng thc nh ngha v tnh s elcetron chuyn qua tit
din thng ca vtdn.Phng php: s dng cc cng thc sau- Cng dng in: I = tq
hay I = tq
- S elcetron : ..I tneDng 2:Tnh in tr tng ng ca on mch.+ Nu on
mch n gin ( ch gm cc in tr mc ni tip, hoc song song) th p dng : Nu
cc in tr mc ni tip: Rt = R1 + R2 ++ Rn. Nu c n in tr ging nhauth:
Rt = n.Ri Nu cc in tr mc song song:n 2 1 tR1....R1R1R1+ + + .Nu c n
in tr ging nhau th: Rt = IRn.+ Nu on mch phc tp ta gii quyt nh sau:
* ng nht cc im c cng in th (chp mch) cc im c in th bng nhau l nhng
im ni vi nhau bng dy dn c in tr khng ng k. *V li s mch in v tnh ton
theo s .C. BI TP P DNGBi 1: Mt on dy dn c ng knh 0,4mm v in tr
200.a) Tnh chiu di on dy, bit dy c in tr sut 61,1.10 m .b) Trong
thi gian 30 giy c mt in lng 60C chuyn qua tit din ca dy. Tnh cng
dng in qua dy v s electron chuyn qua tit in trong thi gian 2
giy.Hng dn:a) in tr ca dy:ta c: R = S, vy l = 22,8m.b). Cng dng in:
I = tq = 2A.- in lng chuyn qua tit din trong thi gian 2 giy: q I.t
= 2.2 4C- S elcetron chuyn qua dy dn l: n = .2,5.10| |I te19
elcetron.Bi 2:Tnh in tr tng ng ca on mch c s sau :Chobit : R1 =
4,R2 = 2,4, R3 = 2, R4 = 5, R5 =3.Trng THPT Ng Quyn Ph Qu Bnh
ThunABR5R4R3R2R1Ti liu dy thm Vt L 11 Ban c bn GV: T Hng SnS:
0,8Bi3:Tnh in tr tng ng ca on mch c s sau:Cho bit: : R1 =6,R2 = 3,
R3 = 4, R4 = 4, Ra =0.Hng dn:V Ra =0 nn hai im M v N c cng in thVy
ta chp 2 im ny thnh mt, s c v li nh Sau:Da vo s ta tnh c: Rt = 4.Bi
4:Tnh in tr tng ng ca mch c s sau: Cho bit: : R1 =1,R2 = 2, R3 = 3,
R4 = 5, R5 =0,5. Rv = .Hng dn: - V dng in khng i khng qua t v Rv =
nn dng in khng qua vn k. Vy mch in c v li theo s sau: - Da vo s mch
in ta tnh c : Rt = 4.Bi 5: Cho mch in cs nh hnh v: Cho bit: R1
=6,R2 = R3 = 20,R4 = 2, a. Tnh in tr tng ng ca on mch khi kha k ng
vm.b. Khi kha k ng cho UAB = 24 V. tnh cng dng in qua R2.Hng dn:a.*
Khi K m mch in co s nh hnh v sau:Trng THPT Ng Quyn Ph Qu Bnh
ThunAR1R3R2ABMR4NR1R3R2R4A
BMNVRvBACR5R4R3R2R1BAR5R4R3R2R1DBAKR4R3R2R1CAC BR1R4R3R2 DTi liu dy
thm Vt L 11 Ban c bn GV: T Hng SnT s hnh v ta tnh c: Rt = 21,86. *
Khi K ng mch in c s nh hnh sau:T s mch inta tnh c: Rt = 4.b.Khi K
ng dng in qua R2 l I2:- Dng in qua R4 l:4I 23424212ABUAR .- Hiu in
th UCD l : UCD = I4.R23 = 2.10 = 20V.- Dng in qua R2 l : I2 = 2201
.20CDUAR Bi 6: Cho mch in c s nh hnh v:Cho bit: R1 =3,R2 = 6, R3 =
6, UAB = 3V. Tm:a. in tr tng ng ca on mch AC.b. Cng dng in qua
R3.c. Hiu in th gia hai im A v C.d. Cng dng in qua R1 v R2.Hng dn:
S: a) Rt = 8. b) I3 = 1,5A. c) UAC = 12V. d) I1 = 1A. I2 =
0,5A.Trng THPT Ng Quyn Ph Qu Bnh ThunACBR1R4R3R2DACBR1R3R2Angr Mari
Ampe (1775 1836)Niu tn ca in hcTi liu dy thm Vt L 11 Ban c bn GV: T
Hng SnCH 2:NH LUT M I VI TON MCHA. TM TT L THUYT 1. nh lut m ivi
ton mcha. Ton mch: l mch in kn c s nh sau: trong : ngun c Ev in tr
trongr, RN l intr tng ng ca mch ngoi. b. nh lut m i vi ton mch
NIR r+E - gim th trn on mch: UN = I.RN = E- I.r- Sut in ng ca
ngun: E= I.(RN + r).2. Trng hp c my thu in (cquy np in)
ppIRr r+ +E -ECh : + Ngun in nu dng in i ra t cc dng. + My thu
in nu dng in i vo cc dng.3. nhlut m tng qut i vi mch kn
ppIR r r+ + E - EB.DNG BI TPBi ton: Tnh ton cc i lng ca dng in
trong mch in kn.Phng php: - Da vo chiu dng in cho (hay chn) phn bit
ngun in v my thu in.- Tnh in tr tng ng ca mch ngoi bng cc phng php
bit.- p dng nh lut m ca mch kn:ppIRr r+ +E -ECh :+ Nu tm c I > 0
th l chiu thc ca dng in trong mch. + Nu I < 0 ch chiu dng in
trong mch l chiu ngc li.+ Nu mch c t in th khng c dng in chy qua t
in.C. BI TP P DNGBi 1: Cho mch in c s nh hnh v: E= 6V, r = 1, R1 =
0,8, R2 = 2, R3 = 3.Tnh hiu in th hai cc ca ngun in v cng dng in
chy quacc in tr.Hng dn:- in tr tng ng mch ngoi: Rt = 2. - Cng dng
in qua mch chnh I = I1:
d tIR r+E= 2A. - Hiu in th hai u R1: U1 =I1.R1 = 1,6 V.Trng THPT
Ng Quyn Ph Qu Bnh Thun-+E,rRNIEp,rpE, rIRE,rR1R2R3Ti liu dy thm Vt
L 11 Ban c bn GV: T Hng Sn- Hiu in th hai u R1 v R3: U2 = U3 = U U1
= 4 1,6 = 2,4 V. - Cng dng in qua R2 : I2 = 221,2UAR.- Cng dng in
qua R3: I3 = 33UR = 0,8 A.Bi 2: Cho mch in c s nh hnh v:Trong : E=
1,2 V, r = 0,1, R1 = R3 = 2.R2 = R4 = 4. Tnh hiu in th gia hai im
A, B.Hng dn: - in tr on MN l: RMN = 1,5 V. - Dng in qua mch chnh: I
= 0,2 A. - Hiu in th gia M, N : UMN = I.RMN = 0,3A. - Cng dng in
qua R2: I2 = 1 20,05 .MNUAR R+ - Hiu in th gia A,N: UAN = I2.R2 =
0,2V. - Hiu in th gia N v B: UNB = I.R4 = 0,88V. - Hiu in th gia A
v B : UAB = UAN + UNB = 1,08 V.Bi 3: Cho mch in c s nh hnh v:E=
7,8V, r = 0,4, R1 = R3 = R3 =3,R4 = 6. a.Tnhcng dng in qua mch chnh
v mi in tr.b.Tnh hiu in th UMN.Hng dn:- in tr tng ng ca mch: Rt =
3,6. - Cng dng in qua mch chnh: d tIR r+E= 1,95A.- Hiu in th hai du
A v B: UAB = I.RAB = 7,02 V.- Cng dng in qua R1v R3: I13 = 131,17
.ABUAR- Cng dng in qua R2 v R4: I= = 240,78 .ABUAR- Hiu in th : U1
= UAM = I1.R1 = 3,51V- Hiu in th : U2 = UAN = I2.R2 = 2,34 V.Vy:
UMN = UMA + UAN = UAN UAM = -1,17 V.Bi 4: Mt ngun in c mc vi mt bin
tr, khi in tr ca bin tr l 14 th hiu in th gia hain cc ca ngun in l
10,5V v khi in tr ca bin tr l 18 th hiu in th gia hai cc ca ngun in
l 10,8V. Tnh in tr trong v sut in ng ca b ngun ny.Hng dn:- T cng
thc: UN = E- I.r v NIR r+E NU E- NR r +E.r UN(RN+r) = E .RN.- Khi
RN = 14 10,5(14+r) = 14E . (1),- Khi RN = 18 10,8 (18+r) = 18E.
(2).Gii h phng trnh ta c r = 2, th vo ta c E = 12V.Trng THPT Ng
Quyn Ph Qu Bnh ThunABR1R4R3R2NME,rA BR1R4R3R2NE,rMTi liu dy thm Vt
L 11 Ban c bn GV: T Hng SnBi 5: Cho mch in c s nh hnh v, b qua cc
on dy ni, cho bit E = 3V; R1 = 5, Ra = 0, ampe k ch 0,3A, vn k ch
1,2V. Tnh in tr trong ca ngun in.Hng dn:- Ta c: U1 = I.R1 = 1,5 V.-
Hiu in th mh ngoi: UN = U1 + U2 = 2,7V.- C:UN = E- I.r r = 1.Bi 6:
Cho mch in c s nh hnh v:Bit R2 = 2,R3 = 3. Khi K m, vn k ch 6V,Khi
K ng vn k ch 5,6V v ampe k ch 2A.a. Tnh sut in ng v in tr trong ca
ngun in.b. Tnh R1 v cng dng in qua R2 v R3.Hng dn:a. Khi k m, vn k
ch gi tr ca sut in ng ca ngun: V UV = E- I.r c I = 0, vy E= 6V.Khi
k ng, vn k ch hiu in th hai u ngun in: UV = E- I.r r = 0,2.b. Theo
nhlut m, ta c: I = 2,8V VtdtdU URR I .Mt khc, R1 = Rt R12 = 1,6.-
Cng dng in quaR2 v R3 l: U23 = I.R23 = 2,4V. 23223 21,2 .0,8 .UI
ARI I I A Trng THPT Ng Quyn Ph Qu Bnh
ThunAVR1R2E,rAVR3E,rR1R2KGheeooc Ximn M (1789 1854)Ti liu dy thm Vt
L 11 Ban c bn GV: T Hng SnCH 3: NH LUT OHM I VI CC LOI MCH INMC
NGUN IN THNH BA.TM TT L THUYT1. nh lut Ohm cha ngunUAB = -E+ I. (R
+r) .i vi ngun in, dng in i vo cc m v i ra t cc dng.2. nh lut Ohm
cho on mch cha my thu in UAB =E+ I. (R +r) .i vi my thu, dng in i
vo cc dng v i ra t cc m.3. Cng thc nh lut m tng qut cho on mch cha
ngun v my thu.UAB = t E tI.(RAB+r).Trong :+ Ly (+ I) khi dng in i t
A n B. + Ly (- I) khi dng in i t B n A. + Ly (+ E ) khi A ni vi cc
dng. + Ly (-E ) khi A ni vi cc m.4. Ghp ngun in thnh ba. Mc ni
tip:- Sut in ng b ngun: Eb = E1 + E2 + E3 +. + En - in tr trong b
ngun:rb = r1 + r2 + r3 +. + rn ch : Nu c n ngun ging nhau.Eb = nErb
= n.rb. Mc xung i:2 12 1r r rE E Ebb+ - Nu E1 > E2 th E1 l ngun
pht v ngc li.c. Mc song song ( cc ngun ging nhau). - Sut in ng b
ngun: Eb = E.- in tr trong b ngun: rb = rn.d. Mc hn hp i xng (cc
ngun ging nhau).Gi: m l s ngun trong mt dy.n l s dy.- Sut in ng b
ngun : Eb =m.E.- in tr trong b ngun :rb = . mrn.* Tng s ngun trong
b ngun:N = n.m.* Cng dng in trong mch s l: I = .NEmr nR +B. PHNG
PHP GII BI TPTrng THPT Ng Quyn Ph Qu Bnh ThunA BE,r RA
BEp,rRE1,r1E2, r2E1,r1E2, r2E1,r1E2,r2E3,r3En,rnEb,rbE,rE,rE,rnE,r
E,rE,r E,rE,r E,rnmTi liu dy thm Vt L 11 Ban c bn GV: T Hng Sn1.
Phng php gii bi tp nh lut m i vi on mch- Xc nh chiu dng in trong on
mch (hay chn chiu ).- Xc nh in tr tng ng ca on mch RAB.- Vn dng nh
lut m tng qut i vi on mch: UAB = t E tI.(RAB+r).Trong :+ Ly (+ I)
khi dng in i t A n B. + Ly (- I) khi dng in i t B n A. + Ly (+ E )
khi A ni vi cc dng. + Ly (-E ) khi A ni vi cc m.- Tm cc i lng theo
yu cu bi ton.2. Phng php gii bitp v nhlut m ton mch- Xc nh b ngun
(mc ni tip, song song hay hn hp) tm Eb, rbtheo cc phng php bit.- Xc
nh mch ngoi gm cc in tr c mc ni tip hay song song tm Rt theo cc
phng php bit.- Vn dng nh lut m i vi ton mch: I = d.bt bER r +- Tm
cc i lngtheo yu cu bi ton.C.BI TP VN DNGBi 1: Cho mch in c s nh hnh
v:Trong , E1 = 8V, r1 = 1,2, E2 = 4V, r2 = 0,4,R = 28,4. Hiu in th
UAB = 6V.a. Tnh cng dng in trong mch v chiu ca n.b. Tnh hiu in th
UAC v UCB.Hng dn:a. Gi s dng in trong mch c chiu t A n B.- p dng nh
lut m ta c: UAB = - E1 + E2 + I.(R + r1 + r2) hay I = 13A. - V I
> 0 nn dng in trong mch c chiu t A n B.b. Hiu in th UAC Ta c:
UAC = - E1 + r1.I = - 7,6 V. Hiu in th UCB Ta c: UCB = E2 + I.r2 =
13,6 V.Bi 2: Cho mch in nh hnh v:Bit, E = 1,5 V, r = 0,25, R1 = 12,
R2 = 1, R3 = 8, R4 = 4. Cng dng in qua R1 0,24.a. Tnh sut in ng v
in tr trong b ngun.b. Tnh UAB v cng dng in qua mch chnh.c. Tnh
R5.Hng dn:
S: a. 6 V, 0,5; b. 4,8 V, 1,2A; c. 0,5.Bi 3: Cho mch in c s nh
hnh v: Trng THPT Ng Quyn Ph Qu Bnh ThunCRB AE1,r1E2, r2RR5A
BR1R4R3R2Ti liu dy thm Vt L 11 Ban c bn GV: T Hng SnBit, E = 1,5 V,
r = 1, R = 6.Tnh cng dng in qua mch chnh.S: 0,75A.Bi 4: Cho mch in
c s nh hnh v: Mi ngun c: E = 2V, r = 0,4.Cc in tr,R1 = 30,R2 = 20,
R3 = 10.Xc nh chiu dng ln dng in qua R3.Hng dn: EAB = E = 2V. rab =
0,2. ECD = 3E = 6VrCD = 3r = 1,2A.V ECD > EAB nn dng in qua R3 t
C n B. Vy:I = 0,17A.
Bi 5: Cho mch in c s nh hnh v:Bit, E1 = 20V, r1 = 4, E2 = 12V,
r2 = 2.R1 = 2,R2 = 3, C = 5C .Tnh cc dng in trong mch v in tch ca t
C. Hng dn:- Gi s dng in c chiu nh hnh v:Ta c: 1 111 12 222 231 2NM
MNNM MNMNU E E UIr rU E E UIr rUIR R+ + +Ti M ta c; I3 = I1 + I2.Gi
UMN = U ta c: 1 21 2 1 2E U E U UR R r r ++Gii phng trnh ny ta c U
= 11,58V.Suy ra : I1 = 2,1A I2 = 0,2A I3 = 2,3A.- Vy chiu dng in l
ng vi chiu tht ca chn.UR2 = I3.R2 = 6,9V.- in tch ca t C l: Q =
C.UR2 = 5. 6,9 = 34,5C .Bi 6: Cho mch in c s nh hnh v: Bit, E1 = E2
= E3 = 3V, r1 = r2 = r3 = 1.R1 = R2 = R3 = 5, R4 = 10.Tnh:a. sut in
ng v in tr trong ca b ngun.Trng THPT Ng Quyn Ph Qu Bnh
ThunR3R1R2CBADE1,r1CR1 R2E2,r2N MPBE1,r1AR1R4R2R3E2,r2E3,r3QTi liu
dy thm Vt L 11 Ban c bn GV: T Hng Snb. Hiu in th gia P v Q.Hng dn:
Ta c: E1 ni tip E3 v mc xung i vi E2.Vy Eb = E1 + E3 E2 = 9V v dng
in c chiu nh mch.- in tr ca b ngun: rb = r1 + r2 + r3= 3.- in tr
tng ng ca mch ngoi:Rt = 12 3412 34.6RRR R +.- Cng dng in trong mch
chnh:I = 1 .btd bEAR r+- Hiu in th gia A v B.UAB = I.RAB =
6V.121234340,60,4 .ABABUI ARUI AR - Hiu in th gia P v Q.UPQ = UPA +
UAQ = - I12.R1 + I34.R3 = - 1V.CH 4:IN NNG. CNG SUT INTrng THPT Ng
Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11 Ban c bn GV: T Hng SnNH
LUT JUN-LEN- XA.TM TT L THUYT1. Cng v cng sut ca dng ina. Cng ca
dng in hay in nng tiu th ca on mch c tnh:A = U.q = U.I.t Trong : U
(V) l hiu in th gia hai u on mch I (A) cng dng in qua mch t (s) thi
gian dng in chy qua mchCh : 1KWh = 3600.000 J.b. Cng sut in- Cng
sut in ca mt on mch l cng sut tiu th in nng ca on mch .P = At=
U.I(W)c nh lut Jun-len-x (nhit lng ta ra trn vt dn)Q = R.I2.t2. Cng
v cng sut ca ngun ina. Cng cangun in- Cng ca ngun in l cng ca dng
in chy trong ton mch.Biu thc: Ang = q. E = E.I.t.b. Cng sutca ngun
in - Cng sut ca ngun in bng cng sut tiu th ca ton mch.Png = At=
E.I3. Cng v cng sut ca cc dng c ch ta nhit a. Cng: A = U.I.t =
RI2.t = 2.UtR b. Cng sut :P = U.I = R.I2 = 2UR.4. Hiu sut ngun inH
= cochN NNAU RA E R r +B. CC DNG BI TPDng 1:Xc nh in tr cng sut tiu
th mch ngoi t gi tr ln nht.- Cng sut mch ngoi : P = RN.I2 = RN.
222NNNE ER rrRR _ + _ , + , P = PMax th NNrRR _ + , nh nht. Theo BT
C-si th : NNrRR _ + , 2.rTrng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy
thm Vt L 11 Ban c bn GV: T Hng Sn Du = xy ra khiN NNrR R rR Khi : P
= PMax = 24.ErDng 2: Bi ton v mch in c bng n. - Trn bng n thng ghi
HT nh mc v cng sut nh mc ca bng n. - Tnh cng nh mc ca n: PIU - in
tr nh mc ca n: 2URP+ Nu I < I: n sng yu hn bnh thng(U < U).+
Nu I > I: n sng hn bnh thng (U > U).* Trng hp n sng bnh thng
th ta thm gi thuyt: thc thc I I vaU U C. BI TP P DNGBi 1: 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 l 4W.b. Vi gi tr no ca R th cng sut
tiu th mch ngoi ln nht. Tnh gi tr .Hng dn:a. Cng sut tiu th mch
ngoi: P = R.I2 = R.( )22ERr + khi P = 4W th4 = R.( )2262 R+R = 1 v
R = 4.b. Ta c: : P = R.I2 = R. 222E ER rrRR _ + _ , + , P = PMax th
rRR _ + , nh nht.Theo BT C-si th : rRR _+ , 2.rDu = xy ra khi 2
.NrR R rR Khi : P = PMax = 24.Er=264,54.2 W.Bi 2: Cho mch in c s nh
hnh v: Bit, E = 15V, r = 1,, R1 = 2, R l bin tr.Tm R cng sut tiu th
trn R l cc i.Tnh gi tr cc i khi .Trng THPT Ng Quyn Ph Qu Bnh
ThunE,rR1RTi liu dy thm Vt L 11 Ban c bn GV: T Hng SnHng dn:Ta c:
PR = 2URMt khc: UR = I.RN = 11 11. 30.. 3 2R R E RR R R R RrR R+
+++.Vy: PR = ( )22 2900 9003 2 . 23RR RRR _ ++ ,Theo BT C-si, ta c
: 23 2 6 RR _+ ,, du = xy ra khi : 23 RR _ , hay R = 23.Vy : PRMax
= ( )290037,5W.2 6Bi 3: Cho mch in c s nh hnh v:Bit. E = 16 V, r =
2, R1 = 3, R2 = 9.1 v 2 l2 n ging nhau. Vn k ch 3V, in trVn k rt
ln.a. Tm in tr mi n.b. Hai n sng nh th no bit cng sut nh mc ca mi n
l 6W.c. Thay vn k bng 1 ampe kc Ra = 0. tnh cng dng in qua ampe
k.Hng dn: a. Sut in ng v in tr trong ca b ngun :Eb = E = 16V v rb
=12r- Cng dng in qua mch chnh :1 12 216132bD D bEIR R R R r + +
++Mt khc, ta c : 1232VD DUIR R R = 6.b. Hiu in th nh mc ca mi n :Um
=. 6.6 6dm DP R V .M UV = 3V < Um nn n sng m hn.c. Khi thay vn k
bng ampe k th dng in khng qua 2 n m ch qua ampe k, s ch ampe k lc
ny l :1 21,23 .bbEI AR R r + +Trng THPT Ng Quyn Ph Qu Bnh
ThunR2VR1E,r12E,rTi liu dy thm Vt L 11 Ban c bn GV: T Hng SnBi
4:Cho mch in gm ngun c sut in ng Ev in tr trong r = 2, mch ngoi gm
in tr R1 = 9 v R2 = 18 mc song song, bit cng sut ca in tr R1 =
9W.a. Tnh cng dng in qua R2.b. Tnh sut in ng E.c. Tnh hiu sut ca
ngun.S : a) 0,5A ; b) 12V ; c) 75%.Bi 5:Mt ngun in c E =12V, r = 4,
thp sng bng n (6V 6W).a. Chng minh n sng khng bnh thng.b. Phi mc
thm Rxvo mch nh th no n sng bnh thng. Tm Rxv cng sut ta nhit trn Rx
trong mi trng hp tng ng.Hng dn:a. Cng nh mc ca n: I = 1 .PAU- in tr
ca n l:26URP - Cng dng in thc t qua n l:1,2 .EI AR r+Vy:I > I nn
n sng khng bnh thng.b. C 2 cch mc:* Khi Rx mc ni tip vo mchTa c: I
= 1 2 x xEI A RR R r + +- Cng sut trn Rx l: Px = I2.Rx = 2W.* Khi
Rx mc song song vo mchTa c: n sng bnh thng th U = U = 6V- Cng dng
in qua mch chnh l: I =1,5 1,5 1 0,5 .xEUA I Ar Khi : Rx = 12xUI
.Cng sut trn Rx l: Px = Ix2. Rx = 0,52.12 = 3W.Bi 6:Cho mch in c s
nh hnh v :Bit,E = 6V, r = 2, R1 = 6, R2 = 12, R3 = 4.a. Tnh cng dng
in chy qua R1.b. Tnh cng sut tiu th in nng trn R3.c. Tnh cng ca
ngun sn ra trong 5 pht.Hng dn:a. in tr tng ng ca mch ngoi l: R = 8.
- Cng dng in qua mch chnh: I = 0,6A. - Cng dng in chy qua R1 l: I1
= 0,4A.b. Cng sut tiu th in nng trn R3 l: P3 = 1,44W.c.Cng ca ngun
in sn ra ttrong 5 pht: A = 1080 J.Trng THPT Ng Quyn Ph Qu Bnh
ThunR1E,rR3R2Ti liu dy thm Vt L 11 Ban c bn GV: T Hng SnCHNG
III:DNG IN TRONGCC MI TRNGA. TM TT L THUYT1.Dng in trong kim loi-
Bn cht dng in trong kim loi l dng chuyn di c hng ca cc electron ngc
chiu in trng.Trng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11
Ban c bn GV: T Hng Sn- in tr sut ca kim loi ph thuc vo nhit : = 0[1
+ (t t0)].: h s nhit in tr (K-1).0 : in tr sut ca vt liu ti nhit
t0.- Sut in ng ca cp nhit in: E = T(T1 T2).Trong T1 T2 l hiu nhit
gia u nng v u lnh;T l h s nhit in ng.- Hin tng siu dn: L hin tng in
tr sut ca vt liu gim t ngt xung bng 0 khi khi nhit ca vt liu gim
xung thp hn mt gi tr Tc nht nh. Gi tr ny ph thuc vo bn thn vt
liu.2. Dng in trong cht in phn- Trong dung dch, cc axit, ba z, mui
b phn li thnh ion.- Dng in trong cht in phn l dng chuyn di c hng ca
cc ion trong in trng theo hai hng ngc nhau.- Hin tng gc axit trong
dung dch in phn tc dng vi cc dng to thnh cht
inphntantrongdungdchvccdngb mni gi lhintng dng cc tan. - Cc nh lut
Faraday: (ch ng trong trng hp in phn dng cc tan).+ nh lut Faraday
th nht: Khi lng vt cht c gii phng in cc ca bnh in phn t l thun vi
in lng chy qua bnh .m = kqTrong , k l ng lng in ho ca cht gii phng
in cc.+ nh lut Faraday th hai:ng lng in ho k ca mt nguyn t t l vi
ng lng gam nAca nguyn t . H s t l l F1, trong F c gi l s Faraday.k=
F1.nAKt hp hai nh lut Faraday ta thit lp c cng thc tnh khi lng cht
in phn gii phng in cc: m = F1.nAItLu : + m(kg) =710 . 65 , 91.nAIt
+m(g) =410 . 65 , 91.nAIt F = 96.500C/mol.3. Dng in trong cht kh-
Trong iu kinthng thchtkhkhng dn in. Chtkh ch dn in khi trong lng n
c s ion ha cc phn t.- Dng in trong cht kh l dng chuyn di c hng ca
cc ion dng, ion m v cc electron do cht kh b ion ha sinh ra.- Khi
dng ngun in gy hiu in th ln th xut hin hin tng nhn ht ti in trong
lng cht kh.- Qu trnh phng in vn tip tc c quy tr khi khng cn tc nhn
ion ha cht kh t bn ngoi gi l qu trnh phng in t lc.- H quang in l qu
trnh phng in t lc hnh thnh dng in qua cht kh c th gi c nhit cao ca
catod n pht c eletron bng hin tng pht x nhit in t. 4. Dng in trong
chn khngTrng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11 Ban c
bn GV: T Hng Sn- L dng chuyn ng ngc chiu in trng ca cc electron bt
ra t in cc.- Diot chn khng ch cho dng in i qua theo mt chiu, n gi l
c tnh chnh lu.- Dng electron c tng tc v i hng bng in trng v t trng
v n c ng dng n hnh tia catot (CRT).5. Dng in trong cht bn dn- Mt s
cht phn nhm chnh nhm 4 nh Si, Ge trong nhng iu kin khc nhau c th dn
in hoc khng dn in, gi l bn dn.- Bn dn dn in hng hai loi ht ti l
electron v l trng. - bn dn tinh khit, mt electron bng mt l trng. bn
dn loi p, mt l trng rt ln hn mt electron. bn dn loi n, mt electron
rt ln hn mt l trng.- Lp tip xc n p c c im cho dng in i theo mt chiu
t p sang n. y gi l c tnh chnh lu. c tnh ny c dng ch to diot bn dn.-
Bn dn cn c dng ch to transistor c c tnh khuych i dng in.B . BI TON
V HIN TNG IN PHNPhng php: s dng cc nh lut Faray v hin tng in phn*
nh lut Faray I:m = kq = k.I.tTrong , k (Kg/C) l ng lng in ho ca cht
gii phng in cc.* nh lut Faray II:m = F1.nAItTrong : F = 96500 Kg/C.
m (g) khi lng gii phng in cc I (A) cng dng in qua bnh in phn t (s)
thi g ian dng in qua bnh in phnA: nguyn t lng ( khi lng mol)n: ha
tr ca cht thot ra in ccCh :1.Khi bi tonyu cu tm cng dng in qua bnh
in phn th lu :+ Nu bnh in phn c hin tng dng cc tan th xem nh in tr
thun.+ Nu bnh in phn khng c hin tng dng cc tan th xem nh l may thu
v p dng nh lut m trong trng hp c my thu.2. Trong trng hp cht gii
phng in cc l cht kh th ta vn p dng cng thc trn tm khi lng cakh thot
ra v t tm th tch ( iu kin chun 1mol kh chim th tch 22400cm3).C. BI
TP P DNGBi 1: Mt tm kim loi c em m niken bng phng php in phn. Bit
din tch b mt kim loi l 40cm2, cng dng in qua bnh l 2A, niken c khi
lng ring D = 8,9.103kg/m3, A =58, n=2. Tnh chiu dy ca lp niken trn
tm kinh loi sau khi in phn 30 pht. Coi niken bm u ln b mt tm kim
loi.Hng dn:S dng cng thc: m = F1.nAIt- Chiu dy ca lp m c tnh: d = .
.0,03 .. . . .V m AI tmmS SD F nSD Trng THPT Ng Quyn Ph Qu Bnh
ThunTi liu dy thm Vt L 11 Ban c bn GV: T Hng SnBi 2:in phn dung dch
H2SO4vi cc in cc platin, ta thu c khi hidro v xi in cc. Tnh th tch
kh thu c mi in cc ( iu kin tiu chun) nu dng in qua bnh in phn c cng
I = 5A v trong thi gian t = 32 pht 10 giy.Hng dn:- Khi lng Hir thu
c catot:m1 = F1.11AnIt = 0,1 g.- Th tch Hir thu c catot: V1 =
30,1.22400 12002m c . - Khi lng xi thu c l:m2 = F1.22AnIt = 0,8 g.-
Th tch xi thu c l:V2 = 30,8.22400 56032cm .Bi3: Chomachiennhhnhve.
Trongobonguonco10 nguongiongnhaumoi nguoncosuat ienong=4Vva ien tr
trong r = 0,2 mac thanh 2 day, moi day co 5 nguon. en co ghi (6V -
18W).Cac ien tr R1 = 5 ;R2 = 2,9 ; R3 = 3 ; RB = 5 va la bnh ien
phan ng dung dch Zn(NO3)2 co cc dng bang Zn. ien tr cua day noi
khong ang ke. Tnh :a) Cngodongienchaytrongmach chnh.b) Lng Zn
giaiphong ra cc am cua bnh am ienphantrongthi gian2gi8phut 40giay.
Biet Zn co hoa tr 2 va co nguyen t lng 65.c) Hieu ien the gia hai
iem A va M.Bi 4: Chomachiennhhnhve. Trongo en co ghi (6V - 6W) ; R1
= 3 ; R2 = R4 = 2 ; R3 = 6 ;RB = 4va la bnh ien phan ng dung dch
CuSO4coccdngbangong;bonguon gom 5 nguon giong nhau moi cai co suat
ien ong co ien tr trong r = 0,2 mac noi tiep. Biet en sang bnh
thng. Tnh :a) Suat ien ong cua moi nguon ien.b) Lng ong giai phong
cc am cua bnh ien phan sau thi gian 32 phut 10 giay. c) Biet ong co
hoa tr 2 va co nguyen t lng 64.d) Hieu ien the gia hai iem A va N.
Trng THPT Ng Quyn Ph Qu Bnh ThunTi liu dy thm Vt L 11 Ban c bn GV:
T Hng SnBi 5:Chomachiennhhnhve. Trongobonguonco10 nguongiongnhau,
moi nguoncosuat ienong=3,6V, ien tr trong r = 0,8 mac thanh 2 day,
moi day co 5 nguon. en co ghi (6V - 3W).Cac ien tr R1 = 4 ;R2 = 3 ;
R3 = 8 ; RB = 2 va la bnh ien phan ng dung dch CuSO4 co cc dng bang
Cu. ien tr cua day noi va ampe ke khong ang ke, cua von ke rat
ln.a) Xac nh so ch cua ampe ke va von ke.b) Tnh lng Cu giai phong
ra cc am cua bnhamienphantrongthi gian32phut 10 giay. Biet Cu co
hoa tr 2 va co nguyen t lng 64.c) Cho biet en co sang bnh thng
khong ? Tai sao ?Bi 6:Cho mach ien nh hnh ve. Ngun c co suat ien
ong = 24V, r = 1, in dung t C = 4F . en co ghi (6V - 6W). Cac ien
tr R1 = 6 ; R2 = 4 ;Rp = 2 va la bnh ien phan ng dung dch CuSO4 co
cc dng bang Cu. a. Tnh in tr tng ng ca on mch.b. Tnh lng Cu giai
phong ra cc am cua bnh am ien phan trong thi gian 16 phut 5 giay.
Biet Cu co hoa tr 2 va co nguyen t lng 64.c. Tnh in tch trn t
C.Trng THPT Ng Quyn Ph Qu Bnh ThunE,rBAR1R2CRpMNMaiccn Faray (1791
1867)Chng no loi ngi cn cn s dng in, th chng mi ngi cn ghi nh cng
lao ca Mai cn FarayTi liu dy thm Vt L 11 Ban c bn GV: T Hng SnBI TP
TNG HP CHNG I+ II + III.Bi 1:Chomachiennhhnhve. Trongobo
nguonco8nguongiongnhau, moi nguonco suat ien ong e = 1,5V, ien tr
trong r = 0,5, mac thanh 2nhanh, moi nhanh co4nguon mac noi tiep.en
co ghi (3V 3W) ; R1= R2= 3 ; R3 = 2 ; R4 = 1 . Tnh :a) Cng o dong
ien chay trong mach chnh va qua tng ien tr.b) Hieu ien the gia hai
iem M va N.c) Hay cho biet en co sang bnh thng hay khong?Tai sao?Bi
2:Cho mch in nh hnh v . E1 = E2 = 12 V ; r1 = r2 = 4 ; R1 =12 ;R2 =
24; R3 =8 a)Tnh Eb v rb ca b ngun. b)Tnh cng dng in qua R1. c)Tnh
cng sut tiu th ca in tr R3 .Hng dna )E b = E1 = E2 = 12 Vr b= 2242
22 1r r b).( R1 // R2 ) nt R3 ==> R t =32 12 1RR RR R++ = 8 + 8
= 16 AR rEI It bb3216 2123 12++ Trng THPT Ng Quyn Ph Qu Bnh
ThunR3R2E2R1E1Ti liu dy thm Vt L 11 Ban c bn GV: T Hng Sn 12 12 12R
I U316832 VR1 // R2==> U1 = U2 = U1244 , 0123 / 1611 RUI Ac )c).
P = R3I32 = 8(2/3)2 = 3,56 WBi 3:Ti A trong khng kh t in tch Q =
3.10-4Ca. Ti B cch A 1cm trong khng kh t in tch q =-5.10-6C. Xc nh
vect lc tc dng ln in tch q.b. Xc nh vect E ti B. c. Xc nh vc t cng
in trng ti C cch u A, B khong 1cm.B i 4: Cho mch in nh hnh vBit
E1=2V; E2=8V; r1= r2= 0,5; R1= 10; R2= 9 a. Tnh Eb v rb, xc nh dng
in trong mch v dng in qua R1;R2.b. Tnh nhit lng ta ra in tr R1;R2 v
ca mch ngoitrong 3s.c. Xc nh hiu in th hai u mch ngoi v mi cc ca
ngun in.d. Xc nh cng sut v hiu sut ca b ngun in.Bi 5: Cho mch in nh
hnh: E= 13,5V, r = 1 ; R1 = 3 ; R3 = R4 = 4. Bnh in phn ng dung dch
CuSO4, ant bng ng, c in tr R2 = 4. Hy tnh : a) in tr tng ng RMNca
mch ngoi, cng dng in qua ngun, qua bnh in phn.b) Khi lng ng thot ra
catt sau thi gian t = 3 pht 13 giy. Cho Cu = 64, n =2.c) Cngsut
cangunvcngsut tiuthmch ngoi.S : a) RMN = 2 ; I = 4,5A ; Ib = 1,5A ;
b) m = 0,096g ; c) PE = 60,75W ; PN = 40,5W.B i 6: Cho haiin tch im
Q1 =-21Q2 =- 3.10-8C,t ti haiim A, B trong khng kh cach nhau mt
khong AB = 6 (cm). Xac nh cng in trng tng hp do haiin tch o gay ra
tai trung iem M cua oan thang AB va lc tac dung len ien tch iem Q3=
4.10-6C at tai M.Trng THPT Ng Quyn Ph Qu Bnh ThunR2R1E, rMR3R4N
R2E2E1r2r1E2E1r1R1R2E2E1r2r1E2E1r1E2E1r2r1E2E1r1R1
