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BI DNG KIN THC VT L 11 - 1 - CHUYN 1: TNH IN HC
H v tn hc sinh :..Trng:THPT........................................
I. TM TT L THUYT:
I. B sung kin thc v vct lc
1. Lc
- c im ca vecto lc
+ im t ti vt
+ Phng ca lc tc dng
+ Chiu ca lc tc dng
+ ln t l vi ln ca lc tc dng
2. Cn bng lc: l cc lc cng tc dng vo mt vt v khng gy gia tc cho
vt
- Hai lc cn bng: l hai lc cng tc dng vo mt vt, cng gi cng
ln nhng ngc chiu
3. Tng hp lc:
- Quy tc tng hp lc: Quy tc hnh bnh hnh
Nu vt chu tc dng ca 2 lc 1 2,F F
th 1 2F F F= +
+ 1 2 1 2F F F F F = +
+ 1 2 1 2F F F F F =
+ 0 2 21 2 1 2( , ) 90F F F F F= = +
+ 2 21 2 1 2 1 2( , ) 2 osF F F F F F F c = = + +
Nhn xt: 1 2 1 2F F F F F +
Nu vt chu tc dng ca nhiu lc th tin hnh tng hp hai lc ri ly
hp lc ca 2 lc tng hp tip vi lc th 3
Lu : chng ta c th tm hp lc bng phng php chiu cc lc thnh phn
xung cc trc Ox, Oy trn h trc cc vung gc.
N TP, B TR KIN THC0
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BI DNG KIN THC VT L 11 - 2 - CHUYN 1: TNH IN HC
1 2F F F= +
lc ny, biu thc trn vn s dng trn cc trc ta chn Ox,
Oy:
1 2
1 2
Ox Ox Ox
Oy Oy Oy
F F F
F F F
= +
= +
ln: 2 21 2Ox OyF F F= +
4.Phn tch lc:
- Quy tc phn tch lc: Quy tc hnh bnh hnh
Ch : ch phn tch lc theo cc phng m lc c tc dng c th
5. iu kin cn bng ca cht im1
0n
ii
F=
=
II. H THC LNG TRONG TAM GIC.
+ nh l hm s cos.+ nh l hm s sin
+ H thc lng trong tam gic vung, h qu.
+ Cch tnh ng cho tam gic cn, u, hnh vung, hnh thoi, hnh ch
nht
III. CC DNG CHUYN NG.
+ Thng u.
+ Thng bin i u
+ Nm ngang, nm xin
IV. Ba nh lut newton.
V. Nng lng, cng, cng sut.
VI. Bng dn v o, cch qui i vi n v o.
II. Bi tp VN DNG
Bi 1: Tm hp lc ca cc lc trong cc trng hp sau (Cc lc c v theo
th t chiu quay ca kim ng h)
a. F1 = 10N, F2 = 10N, ( 1 2,F F
) =300
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BI DNG KIN THC VT L 11 - 3 - CHUYN 1: TNH IN HC
b. F1 = 20N, F2 = 10N, F3 = 10N,( 1 2,F
) =900, ( 2 3,F F
) =300, ( 1 3,F
) =2400
c. F1 = 20N, F2 = 10N, F3 = 10N, F4 = 10N, ( 1 2,F F
) =900, ( 2 3,F
) =900, ( 4 3,F F
)
=900, ( 4 1,F
) =900
d. F1 = 20N, F2 = 10N, F3 = 10N, F4 = 10N, ( 1 2,F
) =300, ( 2 3,F
) =600, ( 4 3,F
)
=900, ( 4 1,F
) =1800
p s: a. 19,3 N b. 28,7 N c. 10 N d. 24N
Bi 2: Mt cht im chu tc dng ng thi ca 2 lc c ln 20N v 30N,
xc nh gc hp bi phng ca 2 lc nu hp lc c gi tr:
a. 50N b. 10N c. 40N d. 20N
s: a. 00 b. 1800 c. 75,50 d. 138,50
Bi 3: Mt cht im chu tc dng ng thi ca 3 lc: F1 =
20N, F2 = 20N v F3. Bit gc gia cc lc l bng nhau v u bng 1200. Tm
F3 hp lc tc dng ln cht im bng 0?
p s: F3 = 20 N
Bi 4: Vt m = 5kg c t nm yn trn mt phng
nghing gc 300 so vi phng ngang nh hnh v. Xc nh
cc lc tc dng ln vt? Bit trng lc c xc nh bng
cng thc P = mg, vi g = 10m/s2.
p s: P = 50N; N = 25 3N; Fms = 25 N
Bi 5: Vt m = 3kg c gi nm yn trn mt phng nghing gc 450 so vi
phng ngang bng mt si dy mnh v nh, b qua ma st. Tm lc cng ca
si dy( lc m vt tc dng ln si dy lm cho si dy b cng ra) p s:
T = 15 2N
D l khi nh gi li lm ca ngi khc,
nhng kh l khi nhn ra sai lm ca chnh mnh.
m
m
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BI DNG KIN THC VT L 11 - 4 - CHUYN 1: TNH IN HC
H v tn:..Thpt.........
I. kin thc:
1. Vt nhim in_vt mang in, in tch_ l vt c kh nng ht c
cc vt nh.
C 3 hin tng nhim in l nhim in do c xt, nhim in do do tip
xc v nhim in do hng ng.
2. Mt vt tch in c kch thc rt nh so vi khong cch ti im ta
xt c gi l in tch im.
3. Cc in tch cng du th y nhau, tri (ngc) du th ht nhau.
4. nh lut Cu_Lng (Coulomb): Lc ht hay y gia hai in tch im
t trong chn khng c phng trng vi ng thng ni hai in tch im
, c ln t l thun vi tch ln ca hai in tch v t l nghch vi bnh
phng khong cch gia chng
Cng thc:2
21.
r
qqkF= Vi k = 9
0
10.9.4
1=
(
2
2.
C
mN )
q1, q2 : hai in tch im (C )
r : Khong cch gia hai in tch (m)
5.Lc tng tc ca cc in tch trong in mi (mi trng ng tnh)
in mi l mi trng cch in.
Cc th nghim chng t rng, lc tng tc gia cc in tch im t
trong mt in mi ng cht, chim y khng gian xung quanh in tch, gim
i ln khi chng c t trong chn khng:
HIN TNG NHIM IN - I CNG V TNG TC IN-S 1
1
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BI DNG KIN THC VT L 11 - 5 - CHUYN 1: TNH IN HC
2
21
.
.
r
qqkF
= : hng s in mi ca mi trng. (chn khng
th = 1)
6. Thuyt electron (e) da vo s c tr v di chuyn ca cc e gii thch
cc hin tng in v cc tnh cht in ca cc vt. Trong vic vn dng thuyte gii thch cc hin tng nhim in (do c xt, tip xc, hng ng), ta
tha nhn ch c e c th di chuyn t vt ny sang vt kia hoc t im ny n
im kia trn vt.
7.cht dn in l cht c nhiu in tch t do,cht cch in(in mi)
8. nh lut bo ton in tch: Trong mt h vt c lp v in, tng i s
ca cc in tch l khng i.DNG 1: I CNG V TNG TC IN - ( TNH TON CC I LNGTHNG GP TRONG CNG THC)
PP chung:
TH ch c hai (2) in tch im q1 v q2.
- p dng cng thc ca nh lut Cu_Lng :2
21
.
.
r
qqkF
= (Lu n v ca cc
i lng)- Trong chn khng hay trong khng kh = 1. Trong cc mi trng khc
> 1.
DNG 2: TNH LNG IN TCH TRONG KHI CHT.
- PP:
Tnh s ht nguyn t, phn t trong khi cht theo cng thc N = m.NA/M
Tnh s ht prtn, s e trong nguyn t, phn t => in tch.DNG 3: S TRUYN IN TCH, TNG TC V TNG TC
GIA2 QU CU GING NHAU
i vi dng bi tp ny, Hs cn vn dng :
nh lut bo ton in tch: Trong mt h c lp v in, tng i s cc in
tch lun lun l mt hng s.
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BI DNG KIN THC VT L 11 - 6 - CHUYN 1: TNH IN HC
II. Bi tp t lun:
1. Hai in tch im dng q1 v q2 c cng ln in tch l 8.10-7 C c t
trong khng kh cch nhau 10 cm.
a. Hy xc nh lc tng tc gia hai in tch .
b. t hai in tch vo trong mi trng c hng s in mi l =2 th
lc tng tc gia chng s thay i th no ? lc tng tc gia chng l
khng i (bng lc tng tc khi t trong khng kh) th khong cch gia
chng khi t trong mi trng c hng s in mi =2 l bao nhiu ?
s: 0,576 N, 0,288 N, 7 cm.
2. Hai in tch im nh nhau t trong chn khng cch nhau mt on 4 cm,
lc y tnh in gia chng l 10-5 N.
a. Tm ln mi in tch.
b. Tm khong cch gia chng lc y tnh in gia chng l 2,5. 10-6 N.
s: 1,3. 10-9 C, 8 cm.
3. Mi prtn c khi lng m= 1,67.10
-27
kg, in tch q= 1,6.10
-19
C. Hi lcy gia hai prtn ln hn lc hp dn gia chng bao nhiu ln ?
s: 1,35. 1036
4. Hai vt nh ging nhau, mi vt tha 1 electron. Tm khi lng mi vt
lc tnh in bng lc hp dn.
s: 1,86. 10-9 kg.
5. Hai vt nh t trong khng kh cch nhau mt on 1m, y nhau mt lc F=1,8 N. in tch tng cng ca hai vt l 3.10-5 C. Tm in tch ca mi vt.
s: q1= 2. 10-5 C, q2 = 10
-5 C (hoc ngc li)
6. Hai in tch im q1 = q2 = 5.10-10 C t trong khng kh cch nhau mt on
10 cm.
a. Xc nh lc tng tc gia hai in tch?
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BI DNG KIN THC VT L 11 - 7 - CHUYN 1: TNH IN HC
b. em h hai in tch ny t vo mi trng nc ( = 81), hi lc
tng tc gia hai in tch s thay i th no ? lc tng tc gia hai in
tch khng thay i (nh t trong khng kh) th khong cch gia hai in tch
l bao nhiu?
7. Cho hai in tch q1 v q2 t cch nhau mt khong r = 30 cm trong khng
kh, lc tc dng 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.
8. Hai qu cu kim loi nh nh nhau mang cc in tch q1 v q2 t trong
khng kh cch nhau 2 cm, y nhau bng mt lc 2,7.10-4 N. Cho hai qu cu
tip xc nhau ri li a v v tr c, ch y nhau bng mt lc 3,6.10 -4 N. Tnh
q1, q2 ? s: 6.10-9 C , 2. 10-9 C, -6. 10-9 C, -2. 10-9 C.
9. 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.
10. 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.
11. 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 c. ln ca lc tng tc bin i th no nu in tch ca chng :
a. cng du.
b. tri du. s: Tng 1,8 ln, gim 0,8 ln.
12. Hai hn bi bng
