-
21F
12F
q1.q2 >0
r
IN TCH IN TRNG Lp 11 nng cao - 2013 2014
(Thy NGUYN VN DN bin son) ------------
Ch 1: Lc tng tc tnh in
12 21;F F c:
- im t: trn 2 in tch. - Phng: ng ni 2 in tch. - Chiu: + Hng ra
xa nhau nu q1.q2 > 0 + Hng vo nhau nu q1.q2 < 0
- ln: 1 2
2
.
.
q qF k
r; k = 9.109
2
2
.N m
C
- Biu din
Bi 1: Hai in tch im bng nhau t trong chn khng, cch nhau mt on R
= 4cm. Lc y tnh in gia chng l F = 10-5N. Tm khong cch gia chng lc y
tnh in gia chng l F' = 2,5.10-6N. S: 8cm Bi 2: Xc nh lc tng tc gia
hai in tch im q1= 3.10-6C v q2= -3.10-6C cch nhau mt khong r = 3cm
trong hai trng hp
a. t trong chn khng b.t trong in mi c = 4 s: a. F = 90N; b. F=
22,5N Bi 3: Cho hai in tch im q1= 9.10-8C v q2= -4.10-8 C cch nhau
mt khong r = 6cm trong khng kh. a. Xc nh lc tng tc gia hai in tch
b. Khong cch gia hai in tch phi bng bao nhiu lc tng tc gia chng c
ln l 20,25.10-3N
s: a. F= 9.10-3N ; b. r = 4cm
Bi 4: Cho hai qu cu nh tch in q1= 9 C v q2= 4 C t cch nhau 10cm
trong khng kh. a. Tnh lc tng tc gia hai in tch
b. Khi t hai qu cu trong in mi c = 4 th khong cch gia hai qu cu
phi bng bao nhiu lc tng tc khng i s: a. F= 32,4N ; b. r = 5cm Bi 5:
Cho hai in tch q1 v q2 t cch nhau mt khong r = 30 cm trong khng kh,
lc tc dung gia chng l F0. Nu t chng trong du th lc ny yu i 2,25 ln.
Vy cn dch chuyn chng li mt khong bng bao nhiu lc tng tc gia chng vn
bng F S:: 10 cm
Bi 6: Cho hai qu cu nh tch in q1= 9 C v q2= 4 C t cch nhau 10cm
trong khng kh. a. Tnh lc tng tc gia hai in tch
b. Khi t hai qu cu trong in mi c = 4 th khong cch gia hai qu cu
phi bng bao nhiu lc tng tc khng i s: a. F= 32,4N ; b. r = 5cm Bi 7:
Hai qu cu nh ging nhau mang in tch c ln bng nhau, t cch nhau 3cm
trong khng kh th chng y nhau bng mt lc c ln 3,6.10-2N. Xc nh in tch
ca 2 qu cu ny. s: q1= q2= 6. 10-8C hay q1= q2= -6. 10-8C Bi 8: Hai
ht bui trong khng kh cch nhau mt on 3cm, mi ht mang in tch q =
-9,6.10-13C. a) Tnh lc tnh in gia hai ht. b) Tnh s electrn d trong
mi ht bui. S: a) 9,216.10-12N ; b) 6.106
-
Bi 9: Hai qu cu nh ging nhau, mi vt tha mt electrn. Tm khi lng
mi qu cu lc tnh in bng lc hp dn gia chng. Cho G = 6,67.10-11
m3/kg.s2
S: 1,86.10-9 kg Bi 10: Electrn quay u quanh ht nhn nguyn t Hir
theo qu o trn vi bn knh R = 5.10-11 m. a) Tnh ln lc hng tm t ln
electrn. b) Tnh vn tc v tn s chuyn ng ca electron. Coi electrn v ht
nhn trong nguyn t Hir tng tc theo nh lut tnh in. S: a) F 9.10-8N.b)
v=2,2.106 m/s; n 0,7. 1016 s-1
Bi 11: Hai qu cu bng kim loi ging ht nhau mang cc in tch q1 v
q2. Khi t chng cch nhau 10cm trong khng kh th chng ht nhau vi mt lc
F1 = 4,5N. Sau khi cho hai qu cu tip xc nhau v tch ra mt khong 20cm
th chng tc dung ln nhau nhng lc F2= 0,9N. Xc nh cc in tch q1 , q2
s: q1= 10-6C ; q2= 5.10-6C v ngc li Bi 12: Hai qu cu bng kim loi
ging ht nhau mang cc in tch q1 v q2. Khi t chng cch nhau 20cm trong
khng kh th chng ht nhau vi mt lc F1 = 5.10-7N.Ni hai qu cu bng mt
dy dn, xong b dy dn i th thy hai qu cu y nhau vi mt lc F2= 5.10-7N.
Xc nh cc in tch q1 , q2
s: q1= 810
3
C ; q2= 810
15
C v ngc li
Bi 13: C ba qu cu kim loi kch thc bng nhau. Qu cu A mang in tch
27C , qu cu B mang in tch -3C , qu cu C khng mang in. Cho hai qu cu
A v B chm nhau ri tch chng ra. Sau cho hai qu cu B v C chm nhau ri
tch ra. Tnh in tch trn mi qu cu. s : qA= 12C ; qB= qC= 6C Bi 14:
Hai qu cu nh mang in tch q1, q2 t cch nhau 3cm trong khng kh th y
nhau bng mt lc 6.10-3N. in tch tng cng ca 2 qu cu l -5.10-8N. Xc nh
in tch ca mi qu cu? Bit rng
1 2q q .
s: q1= -2.10-8C v q2= -3.10-8C Bi 15: Hai qu cu nh mang in tch
q1, q2 t cch nhau 3cm trong khng kh th ht nhau bng mt lc 2.10-2N.
in tch tng cng ca 2 qu cu l -1.10-8N. Xc nh in tch ca mi qu cu? Bit
rng
1 2q q
s: q1= -5.10-8C v q2= 4.10-8C
Bi 16: 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. s: 40,8 N. Bi 17:
Hai qu cu nh bng kim loi ging ht nhau, mang in tch nh nhau q t cch
nhau mt khong R, chng y nhau mt lc c ln 6,4 N. Sau khi cho chng tip
xc nhau ri tch ra mt khong 2R th chng y nhau mt lc bao nhiu ? s:
1,6 N. Bi 18: Hai hn bi bng kim loi ging nhau, hn bi ny c ln in tch
bng 5 ln hn bi kia. Cho x dch hai hn bi chm nhau ri t chng li v tr
cu. ln ca lc tng tc bin i th no nu in tch ca chng : a. cung du. b.
tri du. s: Tng 1,8 ln; Gim 0,8 ln. Bi 19: Hai hn bi bng kim loi
ging nhau c in tch cung du q v 4q cch nhau mt khong r. Sau khi cho
hai hn bi tip xc nhau, cho lc tng tc gia chng khng thay i, ta phi t
chng cch mt khong r. Tm r ?
s: r = 1,25 r. Bi 20: Hai qu cu kim loi ging nhau, c tch in
3.10-5 C v 2.10-5 C. Cho hai qu cu tip xc nhau ri t cch nhau mt
khong 1m. Lc in tc dung ln mi qu cu c ln l bao nhiu? s: 5,625 N. Bi
21: Hai in tch im t trong chn khng, cch nhau mt on R = 20cm. Lc tng
tc tnh in gia chng l F . Khi t trong du, cung khong cch, lc tng tc
tnh in gia chng gim 4 ln. Hi khi t trong du, khong cch gia cc in
tch phi l bao nhiu lc tng tc tnh in gia chng cung l F. S: 10cm
-
Bi 22: Hai vt nh mang in tch t trong khng kh cch nhau mt on R =
1m, y nhau bng F =1,8N. in tch tng cng ca hai vt l 3.10-5N. Tnh in
tch mi vt. S: q1 = 2.10-5C, q2 = 10-5N hoc ngc li Bi 23: Hai qu cu
kim loi nh nh nhau mang cc in tch q1, q2 t trong khng kh cch nhau R
= 2cm, y nhau bng lc F = 2,7.10-4N. Cho hai qu cu tip xc nhau ri a
v v tr cu, th chng y nhau bng lc F' = 3,6.10-4N. Tnh q1, q2. S: q1
= 6.10-9C, q2 = 2.10-9N hoc ngc li; q1 = -6.10-9C, q2 = -2.10-9N
hoc ngc li Bi 24: Hai qu cu kim loi nh nh nhau mang cc in tch q1,
q2 t trong khng kh cch nhau R=20cm, ht nhau bng lc F = 3,6.10-4N.
Cho hai qu cu tip xc nhau ri a v v tr cu, th chng y nhau bng lc F'
= 2,025.10-4N. Tnh q1, q2. S: q1 = 8.10-8C, q2 = -2.10-8N hoc ngc
li; q1 = -6.10-8C, q2 = 2.10-8N hoc ngc li Chiu t A ti BC, F =
8,4.10-4N Bi 25. 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 Hz
Bi 26. Mt qu cu c khi lng ring = 9,8.103 kg/m3, bn knh R = 1cm
tch in q = -10 -6 C c
treo vo u mt si dy mnh c chiu di l =10cm. Ti im treo c t mt in
tch m q0 = - 10 -6 C. Tt c t 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: 0,614N
Ch 2: Lc tng hp tc dng ln in tch q3
3 13 23 .....F F F
Xt trng hp mt in tch im chu hai lc tc dung
13 233F F F
Nu
13 23 3 13 23
13 23 3 13 23
2 213 23 3 13 23
2 213 23 3 13 23 13 23
13 23 3 13
+
+
+
+ ; 2 .cos
Khi F F 2. .cos2
F F F F F
F F F F F
F F F F F
F F F F F F F
F F
Bi 1: Ba in tch im q1 = -10-7C, q2 = 5.10-8C, q3 = 4.10-8C ln lt
t ti A, B, C trong khng kh. Bit AB = 5cm, AC = 4cm, BC = 1cm. Tm lc
tc dung ln in tch q3.
S: F hng t C A, ln F = 20,25.10-2N Bi 2: Ba in tch im q1 =
4.10-8C, q2 = -4.10-8C, q3 = 5.10-8C t trong khng kh ti ba nh ABC
ca mt tam gic u cnh a = 2cm. Xc nh vect lc tc dung ln q3.
S: F AB , ln F = 4,5.10-2N Bi 3: Ba in tch im q1 = q2 = q3 = q =
1,6.10-19C t trong chn khng ti ba nh ABC ca mt tam gic u cnh a =
16cm. Xc nh vect lc tc dung ln q3.
S: F c phng AB, ln F = 9 3 .10-27N Bi 4: Ba in tch im q1 =
27.10-8C, q2 = 64.10-8C, q3 = -10-7C t trong khng kh ti ba nh ABC
ca mt tam gic vung ti C. Bit AC = 30cm, BC = 40cm. Xc nh vect lc tc
dung ln q3.
S: F hng t CO trung im ca AB , ln F = 45.10-4N
-
Bi 5: Ba in tch im q1 = 6.10-9C, q2 = q3 = -8.10-9C t trong khng
kh ti ba nh ABC ca mt tam gic u cnh a = 6cm. Xc nh vect lc tc dung
ln q0 = 8.10-9C ti tm ca tam gic.
S: F c phng BC hng t ABC , ln F = 8,4.10- 4N Bi 6: Cho hai in
tch q1= q2 =16 C t ti hai im A, B trong khng kh cch nhau 1m trong
khng kh. Xc nh lc in tng hp tc dung ln q0 = 4 C t ti. a.im M : MA=
60cm ; MB= 40cm b.im N : NA= 60cm ; NB= 80cm
c.im P : PA= 60cm ; PB= 80cm d.im Q : QA=QB= 100cm s: a. F= 16N
; b.3,9N ; c. 10,4N, d. 0,98N Bi 7: Cho 3 in tch im q1 = 4.10-8C ;
q2 = -4.10-8C ; q3 = 5.10-8C t ti ba nh ca mt tam gic ABC u cnh a =
2cm trong khng kh. Xc nh vect lc tc dung ln q3. S: t ti C, Phng
song song vi AB, Chiu t A ti B, F = 45.10-3N Bi 8: Cho 3 in tch im
q1 = 6.10-9C ; q2 = -8.10-9C ; q3 = -8.10-9C t ti ba nh ca mt tam
gic ABC u cnh a = 6cm trong khng kh. Xc nh vect lc tc dung ln q0=
8.10-9C t ti tm tam gic S: t ti tm O, Phng vung gc vi BC, Bi 9: Bn
in tch ging nhau t 4 nh ca mt t din u cnh a. Tm ln lc in tc dung ln
mi in tch
S: F = 2
2
q6.k.
a
Bi 10: C hai in tch q1 = + 2.10-6 (C), q2 = - 2.10-6 (C), t ti
hai im A, B trong chn khng v cch nhau mt khong 6 (cm). Mt in tch q3
= + 2.10-6 (C), t trn ng trung trc ca AB, cch AB mt khong 4 (cm).
ln ca lc in do hai in tch q1 v q2 tc dung ln in tch q3 S: F = 17,28
(N).
Ch 3: Bi ton cn bng ca mt in tch, h in tch
Khi khao sat iu kin cn bng cua mt in tich ta thng gp hai trng
hp: . Xt trng hp mt in tch im chu hai lc tc dng V q3 cn bng nn
13 233 0F F F 2313F F
V tri t q3 l M. M s + trong on ni 2 in tch nu q1 v q2 cung du +
ngoi on ni 2 in tch nu q1 v q2 tri du Lu : q3 lun gn in tch yu v xa
in tch mnh . Trng hp co thm lc c hoc (trong lc, lc cng dy, ) - Xc
nh y phng, chiu, ln ca tt c cc lc tc dung ln vt mang in m ta xt. -
Tm hp lc ca cc lc c hoc v hp lc ca cc lc in.
- Dung iu kin cn bng: 0R F R F (hay ln R = F).
Bi 1: Hai in tch im q1 = 10-8 C, q2 = 4. 10-8 C t ti A v B cch
nhau 9 cm trong chn khng. Phi t in tch q3 = 2. 10-6 C ti u in tch
q3 nm cn bng (khng di chuyn) ? s: Ti C cch A 3 cm; cch B 6 cm. Bi
2: Hai in tch im q1 = q2 = -4. 10-6C, t ti A v B cch nhau 10 cm
trong khng kh. Phi t in tch q3 = 4. 10-8C ti u q3 nm cn bng? s: CA
= CB = 5 cm. Bi 3: Hai in tch q1 = 2. 10-8 C t ti A v q2 = -8. 10-8
C t ti B, chng cch nhau mt on AB = 15 cm trong khng kh. Phi t mt in
tch q3 ti M cch A bao nhiu n cn bng? s: AM = 10 cm. Bi 4: Cho hai
in tch q1= -2.10-8 C v q2=1,8.10-7 C t ti hai im A, B cch nhau 8cm
trong khng kh.
a. t in tch q0 ti u q0 cn bng b. Du v ln ca q0 q1 v q2 cung cn
bng s: a. AC = 4cm ; BC= 12cm ; b. q0= 4,5.10-8C Bi 5: Cho hai in
tch q1= 2.10-8 C v q2=8.10-8 C t ti hai im A, B cch nhau 9cm trong
khng kh.
-
a. t in tch q0 ti u q0 cn bng b.Du v ln ca q0 q1 v q2 cung cn
bng s: a. AC = 3cm ; BC= 6cm ; b. q0= 8/9.10-8C Bi 6: Ti ba nh ca
mt tam gic u cnh a ngi ta t ba in tch ging nhau q1 = q2 =q3 =
6.10-7C, Phi t in tch q0 ti u v c in tch bng bao nhiu h cn bng s:
Ti tm ; q0= - 3,46.10-7C Bi 7: Mt qu cu nh c m= 1,6g, q1= 2.10-7C c
treo bng mt si dy mnh. pha di cch q1 30cm cn phi t mt in tch q2 bng
bao nhiu lc cng dy gim i mt na s: q2= 4.10-7C Bi 8: Treo hai qu cu
nh c khi lng bng nhau m= 0,6g bng nhng dy c cng chiu di l= 50cm.
Khi hai qu cu tch in bng nhau v cng du th chng y nhau v cch nhau
6cm.
a. Tnh in tch ca cc qu cu
b. Nhng c h vo ru c hng s in mi l = 27. Tnh khong cch gia hai qu
cu khi cn bng. B qua lc y Acsimet. Ly g= 10m/s2a
s: a. 912.10q C ; b. 2cm
Bi 9: Hai in tch im q1 = 2.10-8C, q2 = -8.10-8C t ti A, B trong
khng kh, AB = 8cm. Mt in tch q3 t ti C. Hi: a) C u q3 cn bng? b) Du
v ln ca q3 q1, q2 cung cn bng? S: CA = 8cm, CB = 16cm; q3 = -8.
10-8 C
Bi 10: trong tm ca mt tam gic u ngi ta t mt in tch q1 = C610.3 .
Xc nh in tch q cn
t mi nh ca tam gic cho c h trng th cn bng? s: -3. 10-6 C. Bi 11:
Mt h in tch c cu to gm mt ion dng +e v hai ion m nm cn bng. Bit
khong cch gia hai ion m l a. B qua trong lng cc ion. a) Hy cho bit
cu trc ca h v khong cch gia ion dng v ion m (theo a) b) Tnh in tch
ca mt ion m (theo e) S: a) Ba ion nm trn cung mt ng thng, ion dng
nm chnh gia b) q =- 4e Bi 12: Ti ba nh ca mt tam gic u, ngi ta t 3
in tch dng ging nhau c ln l q1 = q2 = q3 = q. Phi t in tch q0 u, du
v ln (theo q) nh th no h cn bng?
S: q0 trong tm ca tam gic q0 = - 3
q3
Bi 13: Cho hai in tch q1 = 6q, q2 = 3q/2ln lt t ti A v B cch
nhau mt mt khong a (cm). Phi t mt in tch q0 u v c tr s th no n cn
bng? s: Nm trn AB, cch B: a/3 cm. Bi 14: Hai qu cu nh kim loi ging
nhau c treo bi hai dy cung chiu di vo cung mt im, c tch in nh nhau
v cch nhau mt on a = 5cm (a ). Chm nh tay vo mt qu cu. Tnh khong
cch gia chng sau .
S: 3
a
4
Bi 15: Hai qu cu nh ging nhau bng nhm khng nhim in, mi qu cu c
khi lng 0,1 kg v c treo vo hai u mt si ch t di 1m ri mc vo cung mt
im c nh sao cho hai qu cu va chm vo nhau. Sau khi chm mt vt nhim in
vo mt trong hai qu cu th thy chng y nhau v tch ra xa nhau mt khong
r = 6 cm. Xc nh in tch ca mi qu cu? s: 0,035. 10-9 C. Bi 16: Hai qu
cu nh khi lng ging nhau treo vo 1 im bi 2 dy di = 20cm. Truyn cho 2
qu
cu in tch tng cng q= 8.10-7C, chng y nhau, cc dy treo hp thnh gc
= 90. Cho g = 10m/s2 a/ Tm khi lng mi qu cu. b/ Truyn thm cho 1 qu
cu in tch q , 2 qu cu vn y nhau nhng gc gia 2 dy treo gim cn
60. Tnh q. S: a/ 1,8g; b/ - 2,85.10-7C
Ch 4: Xc nh cng in trng to bi in tch im.
-
M
+ Vct cng in trng E 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
> 0 Hng vo Q nu Q < 0
- ln: 2.
QE k
r k = 9.109
2
2
.N m
C
- Biu din:
Bi 1: Xc nh vect cng in trng ti im M trong khng kh cch in tch im
q = 2.10-8 C mt khong 3 cm. s: 2.105 V/m. Bi 2: Mt in tch im dng Q
trong chn khng gy ra mt in trng c cng E = 3. 104 V/m ti im M cch in
tch mt khong 30 cm. Tnh ln in tch Q ? s: 3. 10-7 C. Bi 3: Mt in tch
im q = 10-7 C t ti im M trong in trng ca mt in tch im Q, chu tc
dung ca mt lc F = 3.10-3 N. Cng in trng do in tch im Q gy ra ti M c
ln l bao nhiu?
s: 3. 104 V/m. Bi 4: Qu cu nh c khi lng m=0,25g mang in tch q =
2,5.10-9C c treo bi mt si dy v c
t vo trong mt in trng u E c phng nm ngang v E =106 V/m. Tnh gc
lch ca dy treo so vi phng thng ng. Cho g=10m/s2
s: = 450 Bi 5: Mt qu cu c khi lng m=1g treo trn mt si dy mnh,
cch in. Qu cu c t vo trong
mt in trng u E c phng nm ngang v E = 2.103 V/m. Khi dy treo hp
vi phng thng ng gc 600. Tnh lc cng dy v in tch ca qu cu. Cho g
=10m/s2
s: q= 8,76 C; T = 0,02N Bi 6: 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 ca AB? Cho bit
A, B cng nm trn mt ng sc s: 16 V/m Bi 7: Mt qu cu nh khi lng m=
0,25 g mang in tch q= 2,5. 10-9C c treo bi mt dy v t
trong mt in trng u E
. E
c phng nm ngang v c ln E= 106 V/m. Tnh gc lch ca dy treo so vi
phng thng ng. Ly g= 10 m/s2.
s: = 450. Bi 8: in tch im q = 10-5 C t tai im O trong khng kh a)
Tnh vect cng in trng ti im M cch in tch im mt on 10cm b) Xc nh lc
in trng do in tch im q tc dung ln in tch im q' = -10-7C t ti M. Suy
ra lc in do in tch im q' tc dung ln in tch im q
S: a) E OM ; E = 9.106 V/m
b) F E ; F = 0,9N; /
F F
Bi 9: in tch im q = -10-5 C t tai im O trong khng kh a) Tnh vect
cng in trng ti im M cch in tch im mt on 10cm b) Xc nh lc in trng do
in tch im q tc dung ln in tch im q' = -10-7C t ti M.
S: a) E OM ; E = 9.106 V/m
b) F E ; F = 0,9N
ME q > 0
-
Ch 5: Xc nh cng in trng tng hp
+ Nguyn l chng cht in trng:
1 2 ..... nE E E E
Xt trng hp ti im ang xt ch c 2 cng in trng thnh phn:
1 2E E E
Nu
1 2 1 2
1 2 1 2
2 2
1 2 1 2
2 2
1 2 1 2 1 2
1 2 1
+ E .
+ .
+
+ ; 2 .cos
2. .cos2
E E E E
E E E E E
E E E E E
E E E E E E E
E E E E
Bi 1: Cho hai in tch q1 = 4. 10-10 C, q2 = -4. 10-10 C, t ti A v
B trong khng kh bit AB = 2 cm. Xc
nh vect cng in trng E
ti: a. H, l trung im ca AB. b. M, MA = 1 cm, MB = 3 cm.
c. N, bit rng NAB l mt tam gic u. s: 72. 103 V/m. 32. 103 V/m.
9. 103 V/m. Bi 2: Gii li bi ton s 1 vi q1 = q2 = 4. 10-10 C. s: 0
V/m. 40. 103 V/m. 15,6. 103 V/m. Bi 3: Cho in tch q1 = 4 C t ti A
trong khng kh
1. Tnh cng in trng ti im B cch in tch mt khong 5 cm 2. t ti B
thm mt in tch q2 = 1 C tnh lc in tc dung ln q2 3. Tnh cng in trng
tng hp ti M a. MA = 2cm, MB= 3cm b. MA = 7cm, MB= 2cm c: MA = 3cm,
MB= 4cm d. MA = MB = 5cm
4. Tm v tr im M m ti cng in trng bng 0 s : 1. E= 144.105 V/m ;
2. F= 14,4N ; 3.a. EM= 8.107 V/m ; b. EM = 29,8.106V/m; c. EM=
40,4.106
V/m; d. 16,5.106V/m; 4. MA= 10/3 cm, MB= 5/3 cm
Bi 4: Cho in tch q1 = -9 C t ti A trong khng kh 1. Tnh cng in
trng ti im B cch in tch mt khong 10 cm 2. t ti B thm mt in tch q2 =
4 C tnh lc in tc dung ln q2 3. Tnh cng in trng tng hp ti M nm trn
trung trc ca AB
a. nm ti trung im AB b. M cch AB 5cm c: M A = MB=10cm
4. Tm v tr im M m ti cng in trng bng 0 s : 1. E= 81.105 V/m 2.
F= 32,4N ; 3.a. EM= 46,8.106 V/m ; b. EM = 17,7.106V/m; c. EM=
7,03.106V/m; 4. MA= 30cm, MB= 20cm
Bi 5: Cho hai in tch bng nhau q = 10 C t ti A, B cch nhau 20cm.
Tnh cng in trng ti M a, Tam gic MAB vung cn ti A b, Tam gic MAB
vung ti A c gc M bng 600 c, Tnh v v lc in tc dung ln q1= 5 C ti M
trong 2 trng hp s: a. EM= 3,14.106V/m; b. EM= 7,73.106V/m; c. Fa=
15,7N; Fb= 38,65N
Bi 6: Hai in tch im q1 = q2 = q = 10-9 C t c nh ti A, B; vi AB =
2cm. Xc nh vc t cng in trng ti a) M trung im ca on AB b) N cch A
1cm cch B 3cm.
c) C hp vi A, B thnh tam gic u S: a) EM =0; b) EN = 105V/m hng t
N ra xa A
-
c) EC = 42
q3k 3,9.10
aV/m, E AB
Bi 7: Hai in tch im q1 = 10-10 C, q2 = -9.10-10 C t c nh ti A,
B; vi AB = a = 2cm. Xc nh vc t cng in trng ti a) M trung im ca on
AB b) N cch A 1cm cch B 3 cm.
S:a) EM =72.103 V/m hng t M n A; b) EN = 0 V/m
Bi 8: Ti 3 nh ca mt tam gic u, cnh a = 10cm c 3 in tch im ging
nhau q = 10nC. Hy xc nh cng in trng ti a) trung im mi cnh tam gic.
b) tm ca tam gic
S: EM = 2
q4k
3a=12000 V/m; EG = 0
Bi 9: Ti ba nh ca tam gic vung ABC, AB = 30 cm, AC = 40cm t q1 =
-2,7.10-9C, ti B t q2. Bit
E tng hp ti C c phng song song vi AB, xc nh q2 v E ti C s: q2=
5,27.10-9C; E = 12,66 V/m Bi 10: Hai in tch q1 = 8. 10-8 C, q2 =
-8. 10-8 C t ti A v B trong khng kh bit AB = 4 cm. Tm vect cng in
trng ti C trn ng trung trc ca AB v cch AB 2 cm, suy ra lc tc dung
ln in tch q = 2. 10-9 C t ti C. s: 12,7. 105 V/m. F = 25,4. 10-4 N.
Bi 11: Hai in tch q1 = -10-8 C, q2 = 10-8 C t ti A v B trong khng
kh, AB = 6 cm. Xc nh vect cng in trng ti M nm trn ng trung trc ca
AB cch AB 4 cm. s: 0,432. 105 V/m. Bi 12: Ti ba nh ca mt tam gic
vung ti A cnh a= 50 cm, b= 40 cm, c= 30 cm.Ta t ln lt cc in tch q1
= q2 = q3 = 10-9 C. Xc nh vect cng in trng ti H, H l chn ng cao ke
t A. s: 246 V/m. Bi 13: Trong chn khng, mt in tch im q = 2. 10-8C t
ti mt im M trong in trng ca mt in tch im Q = 2. 10-6C chu tc dung
ca mt lc in F = 9.10-3N. Tnh cng in trng ti M v khong cch gia hai
in tch? s: 45.104V/m, R = 0,2 m. Bi 14: Trong chn khng c hai in tch
im q1= 3. 10-8C v q2= 4.10-8C t theo th t ti hai nh B v C ca tam
gic ABC vung cn ti A vi AB = AC = 0,1 m. Tnh cng in trng ti A. s:
45. 103 V/m. Bi 15: Trong chn khng c hai in tch im q1 = 2. 10-8C v
q2= -32.10-8C t ti hai im A v B cch nhau mt khong 30 cm. Xc nh v tr
im M ti cng in trng bng khng. s: MA = 10 cm, MB = 40 cm. Bi 16: Ti
cc nh A,C ca hnh vung ABCD ta t cc in tch dng q1=q2= q . Hi phi t
ti B mt in tch q3 nh th no cng in trng ti D bng 0.
s: q3= - 2 2 Bi 17: Cho hai qu cu nh mang in tch q1, q2 t ti a v
B trong khng kh cch nhau 2cm. Ti im C cch q1 6cm v cch q2 8cm c cng
in trng tng hp bng 0. Tm q1, q2 bit in tch tng cng ca chng l
7.10-8C. s: q1= -9.10-8C, q2= 16.10-8C Bi 18: Bn im A, B, C, D
trong khng kh to thnh mt hnh ch nht ABCD cnh AD = a= 3 cm, AB= b= 1
cm. Cc in tch q1, q2, q3 c t ln lt ti A, B, C. Bit q2 = - 12,5.
10-8C v cng in
trng tng hp D 0DE . Tnh q1 v q3?
s: q1 = 10-7 C, q2 = 3,6.10-8 C. Bi 19: Cho hai in tch im q1 v
q2 t A v B trong khng kh, AB = 100 cm. Tm im C m ti cng in trng bng
khng vi: a. q1= 36. 10
-6C, q2= 4. 10-6C.
b. q1= - 36. 10-6C, q2= 4. 10
-6C.
s: a. CA= 75cm, CB= 25cm. b. CA= 150 cm, CB= 50 cm.
-
Bi 20: Cho hai in tch im q1, q2 t ti A v B, AB= 2 cm. Bit q1 +
q2 = 7. 10-8C v im C cch q1 l 6 cm, cch q2 l 8 cm c cng in trng bng
E = 0. Tm q1 v q2 ? s: q1= -9.10-8C, q2= 16.10-8C. Bi 21: Cho hnh
vung ABCD, ti A v C t cc in tch q1 = q3 = q. Hi phi t B mt in tch
bao nhiu cng in trng D bng khng?
s: q2 = - q.22
Bi 22: Mt in tch im q = 10-9 C t c nh ti A trong in trng u c E
hng t trn xung theo phng thng ng Oy, ln E = 4.104V/m. Xc nh ln cng
in trng tng hp ti M
cch A mt on r = 1,73 cm = 3 cm. (Bit OA nm trn phng ngang
Ox)
S: EM = 5.104 V/m Bi 23: Cho bn in tch cung ln q t ti bn nh hnh
vung cnh a. Tm cng in trng ti tm O hnh vung trong trng hp bn in tch
ln lt c du: a) +, +, +, + b) +, -, +, - c) +, -, -, +
Bi 24: Mt in tch im q= 2,5 C c t ti im M. in trng ti M c 2 thnh
phn Ex = 6000
V/m, Ey = - 63 .103 V/m. Hi : a/ Gc hp bi vecto lc tc dung ln in
tch q v truc Oy? b/ ln ca lc tc dung ln in tch q?
S: 150; 0,03N Bi 25: Mt ht bui tch in m c khi lng m = 10-8 g nm
cn bng trong in trng u c hng thng ng xung di v c cng E = 1000
V/m.
a/ Tnh in tch ht bui. b/ Ht bui mt ht mt s in tch bng in tch ca
5.105 electron, mun ht bui vn cn bng th
cng in trng phi bng bao nhiu? Cho me = 9,1.10-31kg. S: a/ -10-13
;b/ 5000 V/m
Bi 26: Mt giot cht lng tch in c khi lng 2.10-9 gam nm cn bng
trong in trng u c phng thng ng, c E = 1,25.105V/m. Tnh in tch ca
giot cht lng v s electron tha hoc thiu trn giot cht lng. S:
1,6.10-16 C ; 1000 Bi 27: Mt hn bi nh bng kim loi c t trong du. Bi
c th tch V= 10 mm3, khi lng m = 9.10-5
kg. Du c khi lng ring D = 800 kg/m3. Tt c c t trong in trng
u,
E hng t trn xung, E = 4,1.105 V/m. Tm in tch ca bi n cn bng l
lng trong du.
S: - 2.10-9 C Bi 28: Mt qu cu kim loi nh, bn knh R= 1cm tch in
dng qo nm l lng trong du, trong c in trng u hng thng ng t di ln trn
v c cng E = 2.104V/m. Bit trong lng ring ca kim loi v ca du ln lt l
87840 N/m3 v 7840 N/m3. Tm qo. S: 1,67.10-8 C Bi 29: Hai in tch im
q1=-9.10-5C v q2=4.10-5C nm c nh ti hai im AB cch nhau 20 cm trong
chn khng.
1) Tnh cng in trng tai im M nm trn ng trung trc ca AB cch A 20
cm 2) Tm v tr ti cng in trng bng khng. Hi phi t mt in tch q0 u n nm
cn bng? S: Cch q2 40 cm
==============================
T IN NNG LNG T IN Khi 11 nng cao 2013 2014
Thy Nguyn Vn Dn bin son ==============
Ch 1: Cu to t in
in dung
-
+ C = Q
U (n v l F, vi mF = 10-3 F; F = 10-6 F ; nF = 10-9 F )
Cng thc tnh in dung ca t in phng:
9
.
9.10 .4 .
SC
d. Vi S l phn din tich i din gia 2 ban.
Ghi ch : Vi mi mt tu in c 1 hiu in th gii hn nht nh, nu khi s
dung m t vo 2 bn tu ht ln hn hiu in th gii hn th in mi gia 2 bn b
nh thng.
Bi 1. Mt tu in c in dung 500 pF c mc vo 2 cc ca 1 my pht in c
hiu in th 220V. Tnh in tch ca tu in. S: 11.10-8 (C) Bi 2. Hai bn ca
tu in phng c dng hnh trn bn knh R=60 cm,khong cch gia 2 bn l 2 mm.
Gia 2 bn l khng kh.
a) Tnh in dung ca tu in b) C th tch cho tu in mt in tch ln nht l
bao nhiu tu in khng b nh thng. Bit
ct ln nht m khng kh chu c l 3.106 V/m . Hiu in th ln nht gia 2
bn tu l bao nhiu? S: a. 5.10-9F; b. 6.103V; 3.10-5C
Bi 3. Tu phng c cc bn hnh trn bn knh 48 cm, khong cch v hiu in
th 2 bn l 4 cm v 100 V. Gia 2 bn l khng kh. Tnh in tch tu in. S: Q3
= 16.10-9 (C). Bi 4. Cho 1 tu in phng m 2 bn c dng hnh trn bn knh
2cm v t trong khng kh. Hai bn cch nhau 2mm.
a/ Tnh in dung ca tu in. b/ C th t 1 hiu in th ln nht l bao nhiu
vo 2 bn tu in ? Cho bit in trng nh thng i vi khng kh l 3.106 V/m.
S: a. 5,56 pF; b. 6.104 (V)
Bi 5. Mt tu in phng c mc vo 2 cc ca 1 ngun in c hiu in th 50 V.
Ngt tu in ra khi ngun ri ko cho khong cch ca 2 bn tu in tng gp 2
ln. Tnh hiu in th ca tu in . S: 100 (V) Bi 6. Mt tu in c in dung 24
nF c tch in n hiu in th 450 V th c bao nhiu electron di chuyn n bn
tch in m ca tu in ? S: 6,75.1013 ht
Bi 7. Mt tu in phng c in dung 7 nF cha y in mi. Din tch mi bn
bng 15 cm2 v khong cch gia 2 bn bng 1.10-5 m. Hi hng s in mi ca cht
in mi trong tu in ? S: 5,28 Bi 8. Mt tu in phng c cc bn tu hnh trn
bn knh r= 10cm khong cch gia hai bn tu l d =1cm . Hiu in th gia hai
bn tu l U = 108V. Gia hai bn l khng kh. Tm in tch ca tu in. Nu lp y
hai bn tu bng in mi c hng s in mi l 7 th in tch ca tu thay i nh th
no? /S: 3.10-9 C
Bi 9. Hai tu in khng kh phng c in dung l C1= 0,2 F v C2= 0,4 F
mc song song. B c tch in n hiu in th U=450V ri ngt khi ngun. Sau lp
y hai bn tu in C2 bng in mi c hng s in mi l 2. Tnh in th ca b tu v
in tch ca mi tu
/S: 270V; 5,4.10-5C v 2,16.10-5C
Bi 10. Tu in phng khng kh C = 2pF. Nhng chm mt na vo trong in mi
lng 3. Tm in dung ca tu in nu khi nhng, cc bn t : a) Thng ng b) Nm
ngang
/S: a) 4pF; b)3pF
-
Bi 12. Hiu in th gia 2 bn ca mt tu in phng l U=300 V. Mt ht bui
nm cn bng gia 2 bn ca tu in v cch bn di ca tu d1=0,8 cm. Hi sau bao
lu ht bui s ri xung bn di ca tu nu hiu in th gia 2 bn gim i 60 V S:
0,09 s Bi 13. Mt e chuyn ng doc theo ng sc ca mt in trng u c cng
364 V/m. e xut pht t im M vi vn tc 3,2. 106 m/s,Hi: a. e i c qung
ng di bao nhiu th vn tc ca n bng 0 ? b. Sau bao lu k t lc xut pht e
tr v im M ?
s: 0,08 m, 0,1 s.
Bi 14. Mt e c bn vi vn tc u 2. 10-6 m/s vo mt in trng u theo
phng vung gc vi ng sc in. Cng in trng l 100 V/m. Tnh vn tc ca e khi
n chuyn ng c 10-7 s trong in trng. in tch ca e l 1,6. 10-19C, khi
lng ca e l 9,1. 10-31 kg.
s: F = 1,6. 10-17 N. a = 1,76. 1013 m/s2 vy = 1, 76. 106 m/s, v
= 2,66. 106 m/s.
Ch 2: Mch t in
GHP NI TIP GHP SONG SONG
Cch mc : Bn th hai ca tu 1 ni vi bn th nht ca tu 2, c th tip
tuc
Bn th nht ca tu 1 ni vi bn th nht ca tu 2, 3, 4
in tch QB = Q1 = Q2 = = Qn QB = Q1 + Q2 + + Qn Hiu in th UB = U1
+ U2 + + Un UB = U1 = U2 = = Un
in dung
B 1 2 n
1 1 1 1...
C C C C
CB = C1 + C2 + + Cn
Ghi ch CB < C1, C2 Cn CB > C1, C2, C3
Bi 1. Hai tu in c in dung C1 = 0,4 F, C2 = 0,6 F ghp song song
vi nhau. Mc b tu in vo ngun in c hiu in th U < 60 V th mt trong
2 tu in c in tch bng 3.10-5 C. Tnh :
a/ Hiu in th U; b/ in tch ca tu in kia. S: a. 50 (V); b. 2.10-5
C
Bi 2. in dung ca 3 tu in ghp ni tip nhau l C1= 20pF, C2= 10pF,
C3= 30pF. Tnh in dung ca b tu in . S: 60/11 pF
Bi 3. Cho mt s tu in ging nhau c in dung l C0= 3 F .Nu cch mc
dung t nht cc tu in trn mc thnh b tu c in dung l C= 5 F . V s cch
mc ny?
Bi 4. Cho b tu in nh hnh v sau y: C2= 2C1; UAB= 16V. Tnh UMB
S: 4 (V) Bi 5. Mt b gm 3 tu in ghp song song C1= C2= C3. Khi c
tch in bng ngun c hiu in th 45V th in tch ca b tu bng 18.10-4 C.
Tnh in dung ca cc tu in. S: C1 = C2 = 100 F; C3 = 20 F
Bi 6. Hai tu in c in dung C1= 2 F, C2= 3 F c mc ni tip. a/ Tnh
in dung ca b tu in. b/ Tch in cho b tu in bng ngun in c hiu in th
50 V. Tnh in tch v hiu in th ca cc tu in trong b.
S: a. C1 = 1,2 F; Q1 = Q2 = 6.10-5 (C); U1 = 30 (V); U1 = 20
(V).
C1
A
B C1 C1
C2 C2
M
-
Bi 7. Tnh in dung ca b tu in, in tch v hiu in th ca mi tu trong
cc trng hp sau y:
a) C1=2 F; C2 = 4 F; C3 = 6 F; U = 100V
b) C1 = 1 F; C2= 1,5 F C3 = 3 F; U = 120V
c) C1=0,25 F; C2=1 F C3 = 3 F; U = 12V
S: C=12 F; U1 = U2= U3 = 100V; Q1 = 2.10
-4 C; Q2= 4.10-4C
Q3 = 6.10-4C
/S: C = 0,5 F; U1 = 60V; U2 = 40V; U3 = 20V; Q1 = Q2 = Q3 =
6.10-5 C
/S: C=1 F; U1 =12V; U2 = 9 V U3 = 3V; Q1=3.10
-6 C; Q2 = Q3=
910-6 C
Bi 8. Bn tu in c mc thnh b theo s nh hnh bn. C1= 1 F ; C2= C3= 3
F. Khi ni hai im M, N vi ngun in th tu in C1 c in tch Q1= 6 C v c b
tu in c in tch Q= 15,6 C. Hi :
a/ Hiu in th t vo b tu in ? b/ in dung ca tu in C4 ? S: a. U = 8
(V); b. C4 = 2 F Bi 9. C 3 tu in C1= 3 nF, C2= 2 nF, C3= 20 nF c mc
nh hnh bn. Ni b tu in vi 2 cc mt ngun in c hiu in th 30 V.
a/ Tnh in dung ca c b, in tch v hiu in th trn cc tu in. b/ Tu in
C1 b nh thng. Tm in tch v hiu in th trn 2 tu in
cn li.
S: a. C = 4 F; Q3 = 12.10-8 (C); Q2 = 4,8.10-8 (C); Q1 =
7,2.10-8 (C); U3 = 6 (V); U1 = U2 = 24 (V). b. U2 = 0 (V); U3 = 30
(V); Q2 = 0; Q3 = 6.10
-7 (C).
Bi 10. Tu in phng gm 2 bn tu hnh vung cnh a = 20cm t cch nhau d
= 1 cm. Cht in mi gia 2
bn l thy tinh c = 6. Hiu in th gia 2 bn U = 50 V. a/ Tnh in dung
ca tu in. b/ Tnh in tch ca tu in. c/ Tnh nng lng ca tu in. Tu in ny
c dung l ngun in c khng ?
S: a. C = 100/471 nF; b. Q3 = 10,6.10-9 (C); c. W = 0,265 (J) Bi
11. Tu in phng khng kh c in dung C= 2pF c tch in hiu in th U=
600V
a) Tnh in tch ca tu in b) Ngt tu ra khi ngun, a hai bn tu ra xa
khong cch gia hai bn tu tng gp i. Tnh in
dung ca tu, in tch ca tu v hiu in th gia hai bn tu c) Vn ni tu
vi ngun, a hai bn tu ra xa khong cch gia hai bn tu tng gp i. Tnh
in
dung ca tu, in tch ca tu v hiu in th gia hai bn tu /S: a)
1,2.10-9C; b) 1pF; 1,2.10-9C; 1200V; c) 1pF; 0,6.10-9C; 600V
Bi 12. Tu phng khng kh c in dung C= 500pF c tch in n hiu in th
U= 300V. a/ Tnh in tch Q ca tu.
b/ Ngt tu in khi ngun. Nhng tu vo cht in mi lng c = 2. Tnh in
dung C1, in tch Q1, hiu in th U1 ca tu lc .
c/ Vn ni tu in vi ngun. Nhng tu in vo cht in mi lng = 2. Tnh C2,
Q2, U2 ca tu khi . S: a. Q = 150.10-9 (C); b. C1 = 1 nF; Q1 =
150.10-9 (C); U1 = 150 (V); c. C2 = 1 nF; Q2 = 300.10
-9 (C); U2 = 300 (V).
Bi 13. Tu phng khng kh in dung C= 2 pF tch in hiu in th U= 600
V. a/ Tnh in tch Q ca tu. b/ Ngt tu ra khi ngun, a 2 bn tu ra xa
khong cch tng 2 ln. Tnh C1, Q1, U1 ca tu.
M N
C1 C2
C3 C4
C1
C2
C3
C1 C2 C3 C1 C2 C3
C1
C2 C3
-
c/ Vn ni tu vi ngun, a 2 bn tu ra xa khong cch tng 2 ln. Tnh C2,
Q2, U2 ca tu. S: a. Q = 1,2.10-9 (C); b. C1 = 1 pF; Q1 = 1,2.10-9
(C); U1 = 1200 (V); c. C2 = 1 pF; Q2 = 0,6.10
-9 (C); U2 = 600 (V).
Bi 14. Tu phng c in tch mi bn l S= 100cm2, khong cch gia 2 bn d=
1mm, gia 2 bn l khng kh. Tm hiu in th ti a c th t vo 2 bn tu v in
tch cc i m tu c th tch c. Bit in trng gii hn i vi khng kh l 3.106
V/m. S: Umax = 3000 (V); Qmax = 26,55.10-8 (C)
Bi 15. Hai tu in c in dung v hiu in th gii hn C1= 5 F, U1gh =
500 V, C2 = 10 F, U2gh = 1000 V. Ghp 2 tu in thnh b. Tm hiu in th
gii hn ca b tu in nu hai tu :
a/ Ghp song song. b/ Ghp mi tip. S: a. Umax = 500 (V); b. Umax =
750 (V)
Bi 16. Ba tu in c in dung ln lt l: C1=1 F; C2=2 F; C3 =3 F c th
chu c cc hiu in th ln nht tng ng l: 1000V;200V; 500V. em cc tu in
ny mc thnh b 1) Vi cch mc no th b tu in c th chu c hiu in th ln nht
2) Tnh in dung v hiu in th ca b tu in
S: C1nt(C2//C3); 1200 V; 5/6 F
Bi 17. Cho mch in nh hnh v: C1 = 6 F, C2 = 3 F, C3 = 18 F, UAB =
18 V.
Tnh Cb v in tch mi tu. S: C = 20 F; b. Q1 = 3,6.10-6 (C); Q2 =
3,6.10-6 (C); Q3 = 32,4.10-6 (C).
Bi 18. Cho mch in : C1 = 1 F, C2 = 2 F, C3 = 3 F, C4 = 4 F, UAB
= 100V.
Tnh Cb v in tch mi tu. S: C = 2,4 F; b. Q1 = 3,6.10-6 (C); Q2 =
3,6.10-6 (C); Q3 = 32,4.10
-6 (C); Q4 = 2,4.10-4 (C).
Bi 19. Cho mch in:C1 = 2 F, C2 = 3 F, C3 = 4 F,C4 = 6 F, UAB =
1400 V. Tnh :
a/ Cb. b/ in th v hiu in th mi tu. S: C = 15/14 F; b. Q1 = Q2 =
15.10-4 (C); Q3 = 6.10-4 (C); Q4 = 9.10
-4 (C); U1 = 750 (V); U2 = 500 (V); U3 = U4 = 150 (V).
Bi 20. Cho mch in : C1 = C3 = C5 = 1F, C2 = 4F,C4 = 1,2F, UAB =
30 V.
Tnh: a/ in dung ca b tu. b/ in tch v hiu in th mi tu. S: C =
1,72 F; b. Q1 = Q2 = 9,6.10-5 (C); Q3 = 1,2.10-5 (C); Q4 =
2,16.10-5 (C); Q5 = 3.10-5 (C); U1 = 9,6 (V); U2 = 2,4 (V); U3 = 12
(V); U4 = 18 (V); U5 = 30 (V)
Bi 21. Cho b tu in: C1 = 3 F, C2 = 6 F, C3 = C4 = 4F, C5 = 8 F,
U = 900 V. Tnh Hiu in th gia 2 im AB.
S: UAB = - 100 (V) Bi 22: Cho b tu mc nh hnh v vi C1 = 1 F; C2 =
3 F; C3 = 6 F; C4 = 4 F; UAB = 20 (V). Tnh in dung b tu; in tch v
hiu in th mi tu nu a. K m; b. K ng
S: a. C =3,15 F; Q1 = Q2 = 15 (C); Q3 = Q4 = 48 (C); U1 = 15
(V); U2 = 5 (V); U3 = 8 (V); U4 = 12 (V).
C3 C1 C2
C1
C2
C3
C4
C1 C2
C3
C4
C5
C3 B C4
C1 A C2
+
- C5 C3 C
1
C4 C2
-
b. C =3,5 F; Q1 = 10 (C); Q2 = 30 (C); Q = 60 (C); Q4 = 40 (C);
U1 = U3 = U2 = U4 = 10 (V).
Bi 23: Cho bo tu nh hnh ve: C1 = 2 F; C2 = 3 F; C3 = 6 F; C4 =
12 F; UAB = 80 V. Tnh:
a. nang lng cua bo tu CAB ? b. hieu ien the gia M va N ?
s: 5,33 V.
Bi 24: Cho C1 = 1 F; C2 = 3 F; C3 = 4 F; C4 = 2 F; UAB = 24 V.
a. Tnh ien tch cua cac tu khi K m.
b. Tm ien lng qua khoa K khi K ong va cho biet electron i t
M
en N hay ngc lai.
s: 18 C; 32 C; 10 C ; t M en N.
Bi 25: Hai tu in c in dung ln lt C1 = 5.10-10F v C2 = 15.10-10F,
c mc ni tip vi nhau. Khong cch gia hai bn ca mi tu in l d = 2mm. in
trng gii hn ca mi tu Egh = 1800V. Tnh hiu in th gii hn ca b tu.
Ugh=4,8V
Bi 26: Ba tu in c in dung C1=0,002 F; C2=0,004 F; C3=0,006 F c
mc ni tip thnh b. Hiu in th nh thng ca mi tu in l 4000 V.Hi b tu in
trn c th chu c hiu in th U=11000 V khng? Khi hiu in th t trn mi tu
l bao nhiu? S: Khng. B s b nh thng; U1=6000 V; U2=3000 V; U3=2000
V
Ch 3: Nng lng t in
- Nng lng ca t in: 2 2. .
2 2 2
QU C U QW
C
- Nng lng in trng: Nng lng ca tu in chnh l nng lng ca in trng
trong tu in.
- T in phng 2
9
. .
9.10 .8.
E VW
- Mt nng lng in trng: 2
8
W Ew
V k
Bi 1. Tu in khng kh c d = 5 mm; S = 100 cm2. Nhit lng do tu ta
ra khi phng in l 4,19.10-3 (J). Tm hiu in th khi np in? S: 21,7
(kV) Bi 2. B tu in trong 1 chic n chup nh c in dung 750 F c tch in
n hiu in th 330 V. a/ Xc nh nng lng m n tiu thu trong mi ln n le
sng. b/ Mi ln n le sng tu in phng in trong thi gian 5 ms. Tnh cng
sut phng in trung bnh ca tu in. S: a. 40,8 (J); b. 8,16 (mW)
Bi 3. Mt tu in c in dung C= 2 F c tch in, in tch ca tu l 103 C.
Ni tu in vo b c
qui c sut in ng E =50V. Bn tch in dng ni vi cc dng. Hi khi nng
lng ca b c qui tng ln hay gim i? Tng hay gim bao nhiu? S: gim
0,2475 (J) Bi 4. Mt b tu gm 5 tu in ging ht nhau ni tip mi tu c C =
0,2 F. B c tch in vi nng lng ca b l 2.10-4 (J). Tnh hiu in th mi
tu? S: 20 (V) Bi 5. Tu in C1= 0,5 F c tch in n hiu in th U1= 90V ri
ngt tu ra khi ngun. Sau tu C1 c ni song song vi tu C2 = 0,4 F cha
tch in. Tnh nng lng ca tia la in pht ra khi ni hai tu vi nhau.
C1 C2
C3 C4
B A M
N
C1 C2
C3 C4
B A
M
N
-
/S: 900 J
Bi 6: Tu phng khng kh tch in ri ngt khi ngun. Hi nng lng ca tu
thay i th no khi nhng tu vo in mi c =2.
S: gim mt na
Bi 7: Mt tu in c in dung C1 = 0,2 F , khong cch gia hai bn l d1
= 5cm c np in n hiu
in th U = 100V.
a. Tnh nng lng ca tu in
b. Ngt tu ra khi ngun in. Tnh bin thin nng lng ca tu khi dch hai
bn gn li cn cch nhau d2 = 1cm.
. S: a. W=10-3J; b. W=0,8.10-3J.
Bi 8: Tu phng c S = 200cm2, in mi l bn thy tinh dy d = 1nn, = 5,
tch in di hiu in th U = 300V. Rt bn thy tinh khi tu. Tnh bin thin
nng lng ca tu v cng cn thc hin. Cng ny dung lm g? Xt trong cc trng
hp: a. Tu c ngt khi ngun. b. Tu vn ni vi ngun. S: a. 1,593.10-4J;
b. 3,18.10-5J
Bi 9: Hai tu in phng khng kh ging nhau c in dung C mc song song
v c tch n hiu in th U ri ngt khi ngun. Hai bn ca mt tu c nh, cn hai
bn ca tu kia c th chuyn ng t do. Tm vn tc ca cc bn t do ti thi im m
khong cch gia chng gim i mt na. Bit khi lng ca mi bn tu l M, b qua
tc dung ca trong lc.
S:3
Cv U
M
Bi 10: Tu phng khng kh C=10-10 F, c tch in n hiu in th U=100V ri
ngt khi ngun. Tnh cng cn thc hin tng khong cch hai bn tu ln gp i?
S: 5.10-7J Bi 11: Tu phng khng kh C=6.10-6 F c tch n U=600V ri ngt
khi ngun. a.Nhng tu vo cht in mi c = 4 ngp 2/3 din tch mi bn. Tnh
hiu in th ca tu? b.Tnh cng cn thit nhc tu in ra khi in mi. B qua
trong lng tu? S;A.U=200V; b.0,72J
Bi 12: Mt tu in phng m in mi c =2 mc vo ngun in c ht U=100 V;
khong cch gia 2 bn l d=0,5 cm; din tch mt bn l 25 cm2 1) Tnh mt nng
lng in trng trong tu 2) Sau khi ngt tu ra khi ngun,in tch ca tu in
phng qua lp in mi gia 2 bn tu n lc in tch ca tu bng khng. Tnh nhit
lng to ra in mi S: a. 0,707J/m3; b. 4,42.
