Tài Liệu Điện Quang .-Hoàng Trọng
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Transcript of Tài Liệu Điện Quang .-Hoàng Trọng
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I HC QUC GIA H NI
TRNG I HC KHOA HC T NHIN
VT L I CNG 2
IN QUANG
H Ni 31/12/2013
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I HC QUC GIA H NI
TRNG I HC KHOA HC T NHIN
VT L I CNG 2
IN QUANG
SINH VIN : HONG VN TRNG
LP : K54 a l
IN THOI : 0974 971 149
MAIL : [email protected]
H Ni 31/12/2013
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Li chia s
in Quang l mn hc thuc "Khi kin thc c bn chung ca nhm ngnh" v cc ngnh o to ca trng HKHTN, HQGHN u hc mn hc ny.
Ni dung ca in Quang gm 2 phn chnh c lin quan cht ch vi nhau:
+ in hc (bao gm in v T)
+ Quang hc (tnh cht sng v tnh cht ht ca nh sng)
Trong phn in hc l nhng ni dung c bn v trng in v trng t. Trc ht, cn nm c cc khi nim, hiu v nh cc nh lut gii quyt cc bi tp lin quan y l iu kin cn. tip thu kin thc ca mn hc bt kh khn th cn phi xem li nhng kin thc lin quan ti mn Ton nh: o hm vi phn, tch phn hm mt bin (mn Gii tch 1 gio trnh Ton hc cao cp tp 2), tch phn hm nhiu bin (mn Gii tch 2 gio trnh Ton hc cao cp tp 3), ton t rot, ton t div,
Cc nh lut pht biu cho in tch im, ht c bn, yu t dng,.nhng bi ton li cn tm gi tr tng hp. V th, trong qu trnh vn dng l thuyt vo bi tp th thng gp kh khn do lin quan n php tnh tch phn m c th l i tm biu thc di du tch phn.
Bn cht ca php ly tch phn ch l php cng: cng v s cc s hng trong
mi s hng c gi tr v cng nh:
n
1i
in
b
a
)f(xn
ablimdxf(x)
Ta s dng tch phn khi c mt hoc nhiu yu t bin i, v d in trng do cc in tch v tr khc nhau l khc nhau, cc in tch phn b lin tc. Mun tm biu thc di du tch phn th phi xc nh gi tr (v d in trng) do mt yu t vi phn v cng nh gy nn sau ly tch phn cng cc gi tr li vi nhau. Ngoi ra, biu thc di du tch phn c th xut hin vct tc l chiu ca cc vct thay i theo tng yu t vi phn. Nu c 1 yu t bin i th ta c tch phn ca hm 1 bin, nu c 2 yu t bin i th ta c tch phn mt, nu c 2 yu t bin i m bin s ny l hm ca bin s kia th c th a v tch phn ng,
Ngoi ra, mt s nh l v nh lut cn c th hin di dng vi phn v lin quan ti ton t rot, ton t div: nh l O G trong in trng, t trng; nh lut Faraday v hin tng cm ng in t, nh lut Ampere v dng ton phn, cc phng trnh ca Maxwell. V vy cn phi hiu c rot v div:
S
dlE
limErot L0S
(trong S l din tch gii hn bi ng cong kn L)
V
dSE
limEdiv S0V
(trong V l th tch gii hn bi mt kn S)
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Mt vn kh khn na l khi xc nh chiu theo quy tc bn tay phi th thng b gng tay. Cc bn c th thay bng quy tc vn inh c cho d tng tng, vi quy c: vn inh c xoay theo chiu kim ng h th inh c s chuyn ng tnh tin v pha trc v ngc li.
Phn Quang hc cng cha nhiu ni dung tng ng vi lch s pht trin ca n. Trong khun kh ca mn hc v cng phc v cho thi kt thc mn, chng ta nn tp trung vo bn cht sng ca nh sng (giao thoa, nhiu x, phn cc) v bn cht ht ca nh sng (hiu ng quang in, hiu ng Compton). Nhn chung, bi tp trong phn ny d hn phn in nhng l thuyt cn phi nh th kh nhiu.
File ny c cu trc nh sau: i vi mi phn th u tin l tm tt l thuyt vi cc cng thc hay s dng c bao quanh bi vin mu , sau l p dng l thuyt vo gii mt s bi ton lin quan. Cui file l mt s thi v kim tra. Cn hiu v ghi nh nhng ni dung l thuyt c bn, cch xy dng cng thc,v rt c th chng s xut hin trong bi thi kt thc mn hc.
Trn y l cht kin thc t i m mnh mun chia s cng cc bn. Do hn ch nhn thc v mn hc nn c th cn ni dung no vit cha ng hoc cha y , rt mong cc bn thng cm v gp mnh chnh sa thm.
Cc bn c iu g thc mc xin gi v a ch: [email protected]
Hoc ng kin ln page: THI HUS KHTN H NI ca trang web facebook.com cng trao i v tho lun.
Hong Vn Trng
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MC LC
PHN I: IN HC .................................................................................................................. 7 A. L THUYT ......................................................................................................................... 7
1. in trng ........................................................................................................................ 7 a. in tch ......................................................................................................................... 7 b. nh lut Coulomb ......................................................................................................... 7 c. in trng ..................................................................................................................... 8 d. ng sc in trng ................................................................................................... 8 e. Nguyn l chng cht in trng .................................................................................. 8
2. nh l Ostrogradski Gauss (O G) .............................................................................. 9 a. Thng lng in trng ................................................................................................ 9 b. nh l O G ................................................................................................................. 9
3. in th .............................................................................................................................. 9 a. Cng ca lc tnh in.................................................................................................... 9 b. Tnh cht th ca trng tnh in ............................................................................... 10 c. Th nng ca mt in tch trong in trng ............................................................. 10 d. in th - Hiu in th ................................................................................................ 10 e. Mt ng th ................................................................................................................. 11 g. Mi lin h gia in th v cng in trng ..................................................... 11
4. Nng lng in trng ................................................................................................... 12 a. Nng lng tnh in ca vt dn ................................................................................ 12 b. Nng lng ca t in ................................................................................................ 12 c. Nng lng v mt nng lng in trng ........................................................... 13
5. Dng in ......................................................................................................................... 13 a. Mt dng in .......................................................................................................... 13 b. Phng trnh lin tc .................................................................................................... 14 c. Lc l ............................................................................................................................ 14
6. T trng ......................................................................................................................... 14 a. nh lut Ampere v tng tc t gia hai yu t dng c bn .................................. 14 b. T trng ...................................................................................................................... 15 c. nh lut Biot Savart Laplace ................................................................................ 15 d. Lc tc dng ca t trng ln dng in ................................................................... 15
7. nh lut Ampere v dng ton phn v ng dng ......................................................... 16 a. nh lut Ampere v dng ton phn ........................................................................... 16 b. ng dng ca nh lut Ampere v dng ton phn .................................................... 16 c. nh l Ostrogradski Gauss trong t trng ............................................................ 17
8. Lc Lorentz Hiu ng Hall ........................................................................................... 17 a. Lc Lorentz ................................................................................................................... 17 b. Hiu ng Hall ............................................................................................................... 17
9. Cc nh lut v cm ng in t .................................................................................... 18 a. nh lut Faraday ........................................................................................................ 18 b. nh lut Lenz .............................................................................................................. 18
10. Hin tng h cm v hin tng t cm ...................................................................... 18 a. Hin tng h cm ....................................................................................................... 18 b. Hin tng t cm ........................................................................................................ 19
11. Nng lng t trng .................................................................................................... 19 a. Nng lng t trng ................................................................................................... 19 b. Mt nng lng t trng ....................................................................................... 19
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12. H phng trnh Maxwell v h qu ............................................................................. 19 B. BI TP .............................................................................................................................. 21
Bi 1: in trng ca in tch im ................................................................................. 21 Bi 2: in th ca in tch im ....................................................................................... 22 Bi 3: in trng ca lng cc in ................................................................................ 22 Bi 4: in th ca lng cc in ...................................................................................... 24 Bi 5: in trng v in th ca dy dn thng di tch in .......................................... 24
Trng hp 1: M nm trn ng trung trc ca si dy v cch trung im ca si dy
mt khong l z. Xt trng hp c bit khi a ....................................................... 24 Trng hp 2: M nm trn ng thng AB v cch trung im O mt khong l z...... 27 Trng hp 3: M nm trn ng thng vung goc vi mt u ca si dy v cch u si dy mt khong l z .................................................................................................... 29 Trng hp 4: M nm trn ng thng vung goc vi si dy ti im cch u si dy mt khong l h, M cch si dy mt khong l z. y l trng hp tng qut cho 3 trng hp trn. ............................................................................................................ 31
Bi 6: in trng v in th ca vng dy trn tch in ................................................ 33
Trng hp 1: Vng dy trn tm O bn knh R, tch in q, mt in di l . im M nm trn trc ca vng dy v cch O mt khong l z. Xt trng hp c bit khi z bng 0. .............................................................................................................................. 33 Trng hp 2: Mt ng hnh tr rt mng co chiu cao h, tch in dng vi mt
in mt l . im M nm trn trc ca hnh tr v cch tm ca mt y mt khong l z .................................................................................................................................... 35
Trng hp 3: Bn cu rng co bn knh R, tch in dng vi mt in mt . im M nm ti tm ca bn cu..................................................................................... 36
Bi 7: in trng v in th ca a trn tch in .......................................................... 39
Trng hp 1: a trn tm O bn knh R, tch in dng vi mt in mt . im M nm trn trc ca a v cch O mt khong l z. Xt trng hp c bit khi z >> R
v khi R . ................................................................................................................... 39 Trng hp 2: Hnh tr c bn knh R, tch in dng vi mt in khi l . im M nm trn trc ca khi tr v cch tm ca mt y mt khong l z. ....................... 41
Bi 8: Xc nh vc t cm ng t B do mt dy dn thng di c dng I chy qua ti im M cch dy mt khong l z. ................................................................................. 44
Bi 9: Xc nh vct cm ng t B do mt dng in trn tm O bn knh R, cng I gy ra ti im M nm trn trc ca vng dy trn v cch O mt khong l z. Tm cm ng t ti tm O ca vng dy trn. ................................................................................ 46
Bi 10: Na vng dy dn in bn knh R = 0,49m v khi lng m = 250g, c dng in
I = 25A chy qua (hnh v). Hi cn mt t trng B c hng v ln nh th no na vng dy trn l lng trong khng gian. ............................................................. 47
Bi 11: Mt dy cp ng trc c ng knh trong d1 = 2mm v ngoi bc ch ng knh
d2 = 8 cm, gia li v v bc l cht in mi c hng s in mi = 3. Trong li v v bc tch in tri du nhau vi mt in di = 3,14.10-4 C/m. Hy xc nh cng in trng ti cc im cch trc mt khong: ............................................. 48
(a) r1 = 3 cm. ..................................................................................................................... 48 (b) r2 = 10 cm. ................................................................................................................... 48
Bi 12: Cho qu cu khng dn in tm O, bn knh R = 15 cm c tch in u vi
mt in tch khi = 1,699.10-7 C/m3, c t trong chn khng. ........................ 50 (1) Xc nh cng in trng ti im M cch tm O mt on: (a) r1 = 10 cm; (b) r2 = 30 cm. ........................................................................................................................ 50 (2) Ly in th ti v cng bng 0. Xc nh in th ti P cch tm 20 cm. ................ 50
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Bi 13: Mt qu cu kim loi tm O, bn knh R = 15 cm. Ly in th ti v cng bng 0, tch in cho qu cu n hiu in th 1500V. Hy xc nh: ...................................... 53
(a) in tch v mt in tch trn mt qu cu. .......................................................... 53 (b) Cng in trng, hiu in th ti cc im M v N ln lt cch tm O mt khong l 5 cm v 45 cm. ................................................................................................. 53 (c) Mt in trng ti cc im M, N. ....................................................................... 53
Bi 14: Mt dng in thng di v hn c dng in khng i 1A chy qua. Mt khung dy hnh ch nht ABCD t trong mt phng i qua dng in. Cho cnh AB = 30cm, BC = 20cm. on AB song song vi dng in, cch dng in 10cm. Hy xc nh t thng i qua cun dy. Cho hng s t thm ca mi trng bng 1. ............................ 54
Bi 15: Mt dy tch in lin tc nm dc theo trc Ox t im x = x0 n +. Mt in tch di trn dy l 0. Tnh cng in trng v in th ti gc ta O. ... 56
Bi 16: Mt thanh dn hnh tr, khi lng 0.72 kg, bn knh tit din 6 cm, c dng in I = 48A chy qua theo chiu mi tn, nm trn hai thanh ray c di L = 45 cm t song song v cch nhau mt khong d = 12 cm. Ton b h c t trong mt t trng u c ln 0.24 T, hng vung gc vi mt phng cha thanh dn v thanh ray. Thanh dn ng yn mt u ca ray v bt u ln khng trt theo ray. Tnh tc ca thanh dn ti thi im ri khi u kia ca ray. .......................................... 57
Bi 17: Mt dy dn c un nh hnh v, c dng I = 5A chy qua. Bn knh cung trn l R = 3 cm. Xc nh ln v hng ca cm ng t ti tm ca cung trn. ............. 58
Bi 18: Mt solenoid vi n = 400 vng/m c dng in bin thin I = 30(1 e-1.6t) A chy qua. Mt cun dy c tng cng N = 250 vng, bn knh 6cm c t ng trc vo trong lng ca solenoid. Tm sut in ng cm ng xut hin trong cun dy. ......... 59
Bi 19: Mt thanh di 14 cm c tch in u, c din tch tng cng l 22 C. Xc nh cng in trng v in th ti im nm trn trc thanh, cch trung im ca thanh mt khong l 36 cm....................................................................................... 60
Bi 20: Mt thanh dn in c mt khi lng l 0.04 kg/m, c treo bng hai si dy dn mm cho dng in I chy qua, t trong t trng Bin = 3,6 T, hng vung gc vo trong mt phng. Dng in I phi c hng v ln nh th no khng c sc cng trn cc dy treo. ..................................................................................................... 62
Bi 21: Mt dy dn gm vng dy trn c bn knh R v hai on dy thng, di, nm trong cng mt mt phng. Dy dn c dng in I = 7A chy qua theo chiu mi tn (hnh v). Tm biu thc ca vct cm ng t ti tm ca vng dy. ........................... 63
Bi 22: Mt cun c 15 vng dy, bn knh R = 10 cm, c cun quanh mt solenoid c bn knh 2 cm v n = 1000 vng/m. Dng in chy trong solenoid theo chiu mi tn (hnh v) v bin thin theo quy lut I = 5sin(120t) A. Tm biu thc ca sut in ng cm ng trong cun c 15 vng dy. .............................................................................. 64
Bi 23: Mt qu cu c, bn knh 40 cm, tch in u trong ton b th tch vi in tch
tng cng l +26C. Tm ln v hng ca cng in trng ti nhng v tr cch tm qu cu mt khong: ........................................................................................ 65
(a) 0 cm. ............................................................................................................................ 65 (b) 10 cm. .......................................................................................................................... 65
(c) 40 cm. .......................................................................................................................... 65 (d) 60 cm. .......................................................................................................................... 65
Bi 24: Mt thanh c di L nm dc theo trc x (hnh v). u bn tri ca thanh c
t ti gc ta . Thanh c tch in khng u vi mt in tch di l = .x ( l mt hng s dng). ............................................................................................... 67
(a) n v ca l g........................................................................................................ 67 (b) Tm in th ti im M cch gc ta mt khong d............................................. 67
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Bi 25: Bn dy dn thng song song di v hn c cng dng in I = 5A (hnh v). Cc dng in A v B hng vung gc vo trong mt phng hnh v. Cc dng C v D hng vung gc ra bn ngoi mt phng hnh v. Tm ln v hng ca cm ng t
B ti im M nm tm hnh vung c cnh 0,2m. ..................................................... 68 Bi 26: Mt qu cu khng dn in ng knh 8 cm, tch in u trong ton b th tch
vi in tch tng cng l +5,7C. Tnh in tch cha trong cc mt cu ng tm vi qu cu c bn knh. ........................................................................................................ 69
(a) r1 = 2 cm. ..................................................................................................................... 69
(b) r2 = 6 cm. ..................................................................................................................... 69 Tm ln v hng ca cng in trng ti cc mt cu ng tm . ............... 69
Bi 27: Tnh cng in trng v in th ti im P nm trn trc ca bn vnh khn
tch in u vi mt in tch mt (hnh v). ........................................................ 71 Bi 28: Mt vt dn hnh tr di v hn, bn knh R, c dng in I chy qua (hnh v) vi
mt dng J khng u trn tit din vt dn, J = br (vi b l hng s v r l khong
cch tnh t trc ca hnh tr). Tm ln cm ng t B ti nhng im nm cch trc hnh tr mt khong: ....................................................................................................... 73
(a) r1 < R. ...................................................................................................................... 73
(b) r2 > R. ..................................................................................................................... 73 Bi 29: Thanh dn c th trt khng ma st trn hai thanh ray song song, t cch nhau
mt khong l. Ton b h c t trong t trng u B hng vung gc vo trong
mt phng hnh v. Mt lc khng i c ln 1NFapp lm thanh dn trt u
sang phi vi tc 2m/s. B qua lc ma st. ............................................................... 74
(a) Tnh cng dng in chy trong in tr R = 8. ............................................... 74 (b) Tnh cng sut ta ra trn in tr R. ......................................................................... 74
Bi 30: Mt electrn chuyn ng trn qu o trn (hnh v) c ng nng E = 22,5 eV (1eV = 1,6.10-19 J), cm ng t B = 4,55.10-4 T. ............................................................ 75
(a) Tnh bn knh qu o in t, bit khi lng electrn m = 9,1.10-31 kg v in tch q = 1,6.10-19 C. .................................................................................................................. 75 (b) Chu k chuyn ng ca electrn. .............................................................................. 75
Bi 31: Mt si dy thng t nm ngang c dng I = 28A. Hi chiu v ln ca t trng bng bao nhiu n gy ra mt lc cn bng vi trng lng ca si dy. Cho bit khi lng trn mt n v chiu di ca si dy l: m/L = 46,6 g/m. .................... 77
Bi 32: Mt dy dn thng c tch thnh hai na vng trn nh nhau, c dng I chy qua. Xc nh cng t trng ti tm O ca vng trn. .......................................... 77
Bi 33: Tnh vct cm ng t ti tm C ca hnh c dng di y (hnh v) khi c dng I chy qua. ......................................................................................................................... 78
Bi 34: Hai vng dy dn mt ln mt nh t song song vi nhau (hnh v). Trong vng ln c dng I ang tng. Hi: .......................................................................................... 79
(a) Chiu ca dng in cm ng trong cun nh. ........................................................... 79 (b) Chiu ca lc tc dng ln cun nh. ......................................................................... 79
PHN II: QUANG HC .......................................................................................................... 81 A. L THUYT ....................................................................................................................... 81
1. C s quang hnh hc ...................................................................................................... 81 2. Giao thoa nh sng ........................................................................................................... 82
a. nh sng l mt song ................................................................................................... 82 b. Cng sng .............................................................................................................. 82 c. Giao thoa vn khng nh x ........................................................................................ 83 d. Giao thoa vn nh x .................................................................................................. 85 e. Cc giao thoa k ........................................................................................................... 88
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3. Nhiu x nh sng ............................................................................................................ 90 a. Nguyn l Hugen Fresnel .......................................................................................... 90 b. Nhiu x Fresnel ........................................................................................................... 90 c. Nhiu x Fraunhofer ..................................................................................................... 92
4. Phn cc nh sng ............................................................................................................ 95 5. Lng t nh sng ........................................................................................................... 97
a. Bc x nhit .................................................................................................................. 97 b. Cc nh lut bc x ca vt en tuyt i .................................................................. 97 c. Thuyt lng t nh sng ca Einstein ........................................................................ 98 d. Hin tng quang in ................................................................................................. 99 e. Hiu ng Compton ........................................................................................................ 99
B. BI TP ............................................................................................................................ 100 Bi 1: Chiu chm nh sng trng xung bn mng c chit sut n = 1,33 trong khng kh
vi gc ti 60o, nh sng c bc sng 550nm phn x cho cng cc i vi bc giao thoa bng 2. Hy xc nh b dy ca bn mng. Ngoi nh sng trn cn nh sng n sc no khi phn x cng cho cng cc i. ................................................... 100
Bi 2: Trn b mt ca mt quang c lm bng thy tinh c chit sut n = 1,7 ngi ta ph
mt lp trong sut c chit sut n1 = 1,7 . Hy xc nh b dy ti thiu ca lp trong
sut nh sng c bc sng 550nm khng b phn x. Coi nh sng chiu vung gc.101 Bi 3: Mt cu ca thu knh phng li tip xc vi bn thy tinh. Bn knh cong ca thu
knh l R = 100cm. Chiu chm sng n sc c bc sng = 0,5m ti vung gc vi bn thy tnh sao cho vn Newton xut hin mt trn mt cong ca thu knh. Cho bit chit sut ca vt liu lm thu knh l n1 = 1,5 v chit sut ca thy tinh l n3 = 1,7. ................................................................................................................................. 102
a) Hy xc nh bn knh ca vn ti th 5. ................................................................... 102 b) Khng gian gia thu knh v bn cha y sulphua cacbon c chit sut n2 = 1,63. Hy xc nh bn knh ca vn ti th 5. .............................................................. 102
Bi 4: Hai tm thy tinh di 120 mm c mt u chm nhau cn u kia cch nhau 48 m
to thnh nm khng kh. Chiu chm nh sng n sc c bc sng = 0,48 m xung vung gc vi mt di ca nm. Hy xc nh: .............................................. 104
a) Khong vn. ................................................................................................................ 104 b) S vn giao thoa quan st c. ................................................................................. 104
Bi 6: Mt ngun sng im pht nh sng n sc bc sng = 0,5m c t trn trc vung gc i qua tm ca l trn truyn sng, bn knh r = 1 mm, cch l trn mt khong a = 1m. Hy xc nh khong cch b t mn n im quan st i vi im l trn cha ng 3 i Fresnel. .............................................................................. 105
Bi 7: Chiu chm nh sng n sc bc sng = 0,44 m ti vung gc vi khe hp b rng a. Trn mn quan st t cch khe hp 1m ngi ta o c khong cch t cc tiu nhiu x th 2 n cc i chnh gia l 50cm. Hy xc nh: ............................. 106
a) Gc nhiu x ng vi cc tiu th 2. ......................................................................... 106 b) B rng a ca khe hp. ............................................................................................... 106
Bi 8: Trong mt th nghim nhiu x ca sng phng qua mt khe hp di v hn, b rng khe l a = 1200nm, khong cch t mn n khe hp l 1m, nh sng c bc sng 600nm. Ly chnh gia mn lm gc, hy xc nh v tr gc v v tr trn mn ca cc i ph v cc tiu th nht (v pha gc m) trong cc trng hp khi chm sng ti:107
a) Vung gc vi khe hp. .............................................................................................. 108 b) To vi php tuyn ca khe hp mt gc 30o. ........................................................... 108
Bi 9: Chiu chm sng n sc song song bc sng = 0,55 m ti hai khe hp ging nhau c b rng a = 0,25mm; khong cch hai khe l d = 1,55mm. Mn quan st cch mt phng cha hai khe on D = 1m........................................................................... 109
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a) Xc nh khong cch gia cc cc i giao thoa. ..................................................... 109 b) C bao nhiu vn sng quan st c trong cc i trung tm ca bao hnh nhiu x. ........................................................................................................................................ 109
Bi 10: Chiu mt chm sng n sc bc sng 600nm ti vung gc vi mt cch t c hng s (chu k) l d = 1900nm v s khe l N = 104. Sau cch t t mt thu knh hi t, mn quan st t mt phng tiu din ca thu knh. Hy xc nh: .................... 111
a) V tr v b rng gc ca vch quang ph bc 2. ....................................................... 111 b) Trn mn quan st c bao nhiu vch quang ph. ................................................. 111
Bi 11: Chiu chm nh sng pht ra t ngun Natri ti vung gc vi cch t c cc thng s nh sau: hng s d = 1900 nm v s khe N = 104. Natri c hai nh sng n sc bc sng 589 nm v 589,59 nm. Hy xc nh: .................................................... 113
a) Khong cch gc gia hai vch quang ph bc 2 ca hai nh sng trn. ................... 113 b) Cch t c phn bit c hai vch quang ph bc 1 ca hai nh sng trn khng. Ti sao? ................................................................................................................................. 113
Bi 12: Chiu chm nh sng t nhin c cng Io ti h gm knh phn tch A v knh phn cc P. Hy xc nh gc gia hai quang trc ca hai knh P v A nh sng i qua h I = Io/8. B qua hin tng hp th nh sng khi qua hai knh. ........................ 114
Bi 13: Mt ngi thng thng nhy cm nht i vi nh sng c bc sng = 550 nm. Hy xc nh nhit ca mt hc en tuyt i mt ngi nhn r nht nh sng do n pht ra. ........................................................................................................ 114
Bi 14: Ph bc x ca mt tri cc i bc sng max = 480 nm. Coi b mt ca mt tri nh vt en tuyt i. Hy xc nh nhit b mt v nng sut bc x ton phn
ca mt tri. Cho h s Stefan Boltzmann = 5,67.10-8 W/m2K4. H s dch chuyn Wien b = 2898 m.K. ................................................................................................... 115
Bi 15: Mt ngun sng im cng sut 3W pht nh sng n sc bc sng 589 nm. Hy xc nh s photon i qua tit din 1cm2 theo phng vung gc vi phng truyn, cch ngun 1,75 m. ....................................................................................................... 115
Bi 16: Mt photon nng lng 150 keV tn x n hi trn electron t do ng yn di
gc tn x 90o. Cho bc sng Compton = 2,42.10-12 m, h = 6,625.10-34 J.s, c = 3.108 m/s. Hy xc nh:......................................................................................................... 116
a) Nng lng photon tn x. ......................................................................................... 116 b) ng nng v vn tc ca electron Compton (sau tn x). ........................................ 116
Bi 17: Mt photon nng lng 58 keV tn x n hi trn electron t do ng yn, sau tn x bc sng photon tng ln 25%. Hy xc nh: ....................................................... 117
a) Gc tn x. .................................................................................................................. 117 b) Bc sng v nng lng photon tn x. ................................................................... 117 Cho h = 6,625.10-34 J.s, c = 3.108 m/s, k = 2,43.10-12 m v ly 1eV = 1,6.10-19J. .......... 117
MT S THI V KIM TRA .................................................................................. 119 1. thi cui k h nm 2013 ........................................................................................... 119 2. thi cui k II nm hc 2012 2013 ......................................................................... 120 3. thi cui k h nm 2012 ........................................................................................... 122
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PHN I: IN HC
A. L THUYT
1. in trng
a. in tch
- L mt trong nhng thuc tnh vt l c bn ca vt cht.
- nh lut bo ton in tch: Trong mt h c lp, in tch c bo ton.
- in tch nguyn t: L in tch nh nht c bit trong t nhin. ln:
e = 1,6. 10-19 C (in tch ca electrn)
- Phn loi vt cht theo tnh cht dn in:
+ Vt dn in: l vt cht m cc in tch c th chuyn ng t do trong ton b th tch ca vt.
+ Vt cch in: l vt cht trong cc in tch khng th chuyn ng t do.
+ Vt bn dn: l vt cht ch dn in c mt iu kin xc nh.
b. nh lut Coulomb
- Th hin mi quan h tng tc gia hai in tch im ng yn.
- Ni dung:
Lc tng tc gia hai in tch im c phng nm trn ng thng ni hai in tch, c chiu l lc ht nu hai in tch tri du v l lc y nu hai in tch cng du, c ln t l thun vi tch ln ca hai in tch v t l nghch vi bnh phng khong cch gia chng.
- Biu thc:
Nr
r.
r
qqkF
2
21 (I.1)
Trong :
F l lc tng tc gia hai in tch im, c n v N.
q1, q2 c n v C.
r l khong cch gia hai in tch im, c n v m. Vct r c hng t in tch gy nn tc dng n in tch b tc dng.
k l hng s tnh in,0 l hng s in:
2
29
0
2
212
0C
Nm8,99.10
4
1k;
Nm
C8,85.10
- Lc Coulomb tun theo nguyn l chng cht:
n
1i
iFF
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c. in trng
- Khi nim: in trng l mi trng vt cht tn ti xung quanh in tch v tc dng lc in ln in tch khc t trong n.
- Vct cng in trng: Cng in trng ti mt im l mt i lng vct, c o bng lc tc dng ca in trng ln mt n v in tch dng ti im .
C
N
r
r.
r4
q
q
FE
2
00 (I.2)
+ Hng ca E trng vi hng ca lc tc dng F ln q0
E hng ra ngoi in tch im nu q > 0.
E hng vo trong in tch im nu q < 0.
+ ln:
200 r4
q
q
FE
Vc t r hng t in tch q n im cn kho st in trng.
d. ng sc in trng
- ng sc in trng l ng m tip tuyn ti mi im ca n trng vi phng ca vct cng in trng ti im . Chiu ca ng sc l chiu ca
vct cng in trng E .
- Quy c: S ng sc i qua mt n v din tch t vung gc vi ng
sc th bng ln ca vct cng in trng E .
- Tp hp cc ng sc in trng gi l in ph.
Tnh cht:
+ S ng sc i qua mt n v din tch vung gc vi chng th t l vi ln ca vect cng in trng ti .
+ Hai ng sc bt k khng th ct nhau.
+ ng sc b gin on ti ni c in tch. Cc ng sc i ra t in tch dng v i vo in tch m.
+ Cc ng sc n vung gc vi b mt ca vt dn trong iu kin tnh in.
e. Nguyn l chng cht in trng
- Cng in trng do h cc in tch im gy ra ti mt im bng tng vct cng in trng do tng in tch im gy ra ti im .
C
NEE
n
1i
i (I.3)
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2. nh l Ostrogradski Gauss (O G)
a. Thng lng in trng
- Thng lng in trng gi qua yu t din tch dS l tch v hng gia E
v dS.
E.dS.cosdS.Ed
Trong o:
+ l gc gia E v dS.
+ dSl vct php tuyn ca yu t din tch dS.
+ ln ca vct dSbng ln din tch ca yu t dS.
- Thng lng in trng gi qua mt kn S l:
S
E dS.E (I.4)
Trong trng hp ny, dSl hng ca php tuyn ngoi ca mt S ti yu t dS.
b. nh l O G
Thit lp mi quan h gia in tch v in trng.
- Ni dung: Thng lng in trng gi qua mt kn S bng tng i s cc in
tch cha trong mt kn chia cho hng s in 0.
- Biu thc:
n
1i
i
0S
E q1
dS.E
(I.5)
- Dng vi phn ca nh l (nu in tch phn b lin tc trong th tch ca vt):
0
),,(),,(Ediv
zyxzyx
( l mt in tch khi ti im c ta (x, y, z))
3. in th
a. Cng ca lc tnh in
- Cng cn thit a in tch th q0 t M n N l:
NM0
0
N
M
0MNr
1
r
1
4
qqdlEqA
(I.6)
rM v rN ln lt l khong cch t in tch q ti M v N.
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- Cng ny khng ph thuc vo dng ng cong dch chuyn m ch ph thuc vo v tr im u v im cui.
- Nu c n in tch th cng dch chuyn s l:
n
1i NiMi0
0iMN
r
1
r
1
4
qqA
(I.7)
b. Tnh cht th ca trng tnh in
- Nu mt mi trng c cng dch chuyn in tch trn mt ng cong khp kn bng 0 th mi trng c tnh cht th.
0dl.Eqdl.FA 0
0dl.E (I.8)
- Gi tr dl.E gi l lu s ca vct cng in trng. Lu s ca vct cng in trng dc theo ng cong kn th bng 0.
0Erot
c. Th nng ca mt in tch trong in trng
- Th nng ca mt in tch q0 ti mt im trong in trng l mt gi tr bng cng ca lc tnh in dch chuyn in tch t im ang xt ra xa v cng.
- Th nng ca mt in tch im q0 trong in trng do q gy ra:
r4
qqW
0
0
(I.9)
- Nu h c n in tch th:
n
1i i0
0in
1i
ir4
qqWW
(I.10)
d. in th - Hiu in th
- in th ti mt im c tnh bng cng ca lc tnh in dch chuyn mt n v in tch dng t im ra xa v cng.
- in th ti mt im cch in tch q mt khong r, in tch th q0 l:
r4
q
q
Wdl.EV
00M
(I.11)
- Nu c nhiu in tch qi th:
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n
1i i0
in
1i
ir4
qVV
(I.12)
- Do cng dch chuyn in tch q0 t M n N:
)V(VqA NM0MN
- Hiu in th:
Hiu in th gia hai im M v N trong in trng l mt i lng v tr s bng cng ca lc tnh in dch chuyn mt n v in tch dng t M n N.
NM0
MNMN VV
q
AU (I.13)
0V
AV MM
e. Mt ng th
- Mt ng th l tp hp nhng im c cng mt in th.
- Mt ng th ca in trng do in tch im gy ra l cc mt cu c tm nm ti ni c in tch.
- Tnh cht:
+ Cc mt ng th khng ct nhau.
+ Cng ca lc tnh in dch chuyn mt in tch trn mt ng th bng 0.
+ Vct cng in trng ti mi im trn mt ng th th vung gc vi mt ng th ti im .
g. Mi lin h gia in th v cng in trng
- Tnh in th t cng in trng:
M
M dl.EV (I.14)
Trong dl l vi phn ca yu t chiu di theo ng dch chuyn.
- Tnh cng in trng t in th:
Vc t cng in trng ti mt im bt k trong in trng bng v ngc du vi gradient ca in th ti im .
VgradE (I.15)
Ex = Vx ; Ey = Vy ; Ez = Vz
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4. Nng lng in trng
a. Nng lng tnh in ca vt dn
- Tnh cht ca vt dn:
+ Vct cng in trng iE ti mi im bn trong vt dn bng 0.
+ Thnh phn tip tuyn tE ca vct cng in trng ti mi im trn b mt vt dn bng 0.
+ in trng E ch cn li thnh phn vung gc: nEE
+ Mt vt dn l mt mt ng th. Vt dn cn bng tnh in l mt khi ng th.
+ Mt vt dn khc nm trong mt vt dn rng s khng b nh hng ca in trng ngoi.
+ S phn b in tch trn b mt vt dn ch ph thuc vo hnh dng ca vt dn . in tch q ch phn b trn b mt ca vt dn.
- Nng lng ca vt dn tch in;
2
2
CV2
1
C
q
2
1qV
2
1W (I.16)
Trong C l in dung ca vt dn. in dung ca vt dn bng s in tch cn truyn cho vt dn in th ca n tng ln 1V.
+ Nu vt dn l mt cu bn knh R th: R4C 0
+ T in phng: d
SC 0 (S din tch mt mt t, d khong cch)
+ T in tr:
1
2
0
R
Rln
2C
l (l chiu di t tr; R1, R2 bn knh
trong v ngoi)
+ T in cu: d
SC 0 (S din tch mt mt cu, d khong cch gia
hai bn t)
b. Nng lng ca t in
- T in l mt h gm hai vt dn c t rt gn nhau. Cc vt dn to nn t c gi l cc bn t.
- in dung ca t:
U
qC (I.17)
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Trong , q l in tch trn mt bn t v U l hiu in th gia hai bn t.
- Ghp t:
+ Ghp ni tip:
n
1i iC
1
C
1
+ Ghp song song:
n
1i
iCC
- Nng lng ca t tch in:
2
2
CU2
1
C
q
2
1qU
2
1W (I.18)
c. Nng lng v mt nng lng in trng
- Nng lng in trng:
.VE2
1W 20 (I.19)
Trong , V l th tch phn khng gian c in trng.
- Mt nng lng in trng:
2
0E2
1
V
Ww (I.20)
5. Dng in
a. Mt dng in
- Dng in: l dng chuyn di c hng ca cc ht mang in tch.
- Cng dng in l lng in tch chuyn qua trong mt n v thi gian.
dt
dqI
- Chiu ca dng in c quy c l chiu chuyn ng ca cc ht mang in tch dng.
- Vct mt dng in:
Vct mt dng in ti mi im c ln bng cng dng dI chuyn qua yu t mt dSn t vung gc vi hng chuyn ng ca cc ht mang in ti chia cho dSn:
ndS
dIj (I.21)
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+ Hng ca vct j trng vi hng chuyn ng ca cc ht mang in tch
dng.
- Do , cng dng qua mt mt c tit din S l:
s
dS.jI (I.22)
b. Phng trnh lin tc
Mt kn S nm trong mi trng c dng in chy qua th:
dVt
dVjdiv
dt
dqdS.j
V VS
t
jdiv
(I.23)
- i vi dng dng th:
0jdiv
c. Lc l
Lc l c c trng bng cng m n thc hin c a mt n v in tch dng i c theo mch in.
Vq
A (I.24)
6. T trng
a. nh lut Ampere v tng tc t gia hai yu t dng c bn
- Yu t dng c bn: l mt on v cng ngn ca dy dn c dng chy qua
(k hiu l dlI ) trong dlc chiu trng vi chiu dng in.
- nh lut Ampere: lc t do yu t dng 11dlI tc dng ln yu t
dng 22 dlI cng t trong chn khng l mt i lng vct 12dF c:
+ Phng vung gc vi mt phng cha yu t 22 dlI v php tuyn n ( n c
chiu sao cho 1dl , r , n to thnh mt tam din thun).
+ C chiu sao cho 2dl , n , 12dF to thnh mt tam din thun.
+ ln:
2111222
12r
sindl.IsindlIkdF
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Trong :
n;dlI
r;dlI
222
111
; k l h s t l: 4
k 0
Tng qut: 311220
12r
]rdl[IdlI.
4
dF
(I.25)
b. T trng
- nh ngha: T trng l mi trng vt cht tn ti xung quanh in tch chuyn ng v tc dng lc t ln in tch khc chuyn ng trong n.
- Cm ng t do mt in tch q chuyn ng vi vn tc v l:
30
r
]rv[q.
4
B
(I.26)
Trong , r l khong cch t q n im cn xc nh cm ng t.
c. nh lut Biot Savart Laplace
Vct cm ng t dB do yu t dng dlI gy ra ti im P, cch yu t dng mt khong r l mt i lng vct c:
+ Gc ti P.
+ Phng vung gc vi mt phng cha phn t dng in dlI v P.
+ Chiu sao cho dl , r , dB to thnh mt tam din thun.
+ ln:
20
r
sinIdl.
4
dB vi r,dlI
Tng qut:
30
r
]rdl[I.
4
dB
(I.27)
d. Lc tc dng ca t trng ln dng in
- Lc Ampere: lc t tc dng ln dng in I bng lc tc dng ln cc ht e chuyn ng c hng trong dy dn.
- nh lut Ampere v lc tc dng ln yu t dng dlI :
]Bdl[IdF (I.28)
Vy lc tc dng ln c dng in l: dng
dFF
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7. nh lut Ampere v dng ton phn v ng dng
a. nh lut Ampere v dng ton phn
- Lu s ca vc t cm ng t theo ng cong kn L:
L
dlBL (I.29)
- Ni dung nh lut: lu s ca vct cm ng t theo ng cong kn bng tch
0 vi tng i s cc dng in xuyn qua mt S gii hn bi ng cong kn .
n
1k
k0
L
IdlB (I.30)
+ Nu trong din tch gii hn bi ng cong kn L c mt dng l j th:
S
dSjI
jBrot 0 (I.31)
b. ng dng ca nh lut Ampere v dng ton phn
- T trng do mt dy in thng co kch thc ngang ng k, bn knh tit din l R.
+ Ti im cch trc ng dy mt khong r < R:
B c phng l tip tuyn ca ng trn bn knh r.
B c chiu c xc nh theo quy tc vn inh c.
ln: 20
R2
I.rB (I.32)
+ Ti im cch trc ng dy mt khong r > R:
Chiu c xc nh theo quy tc vn inh c.
ln: R2
IB 0 (I.33)
- T trng ca mt ng dy in thng (Solenoid)
B c phng song song vi trc ca ng dy.
B c chiu c xc nh theo quy tc vn inh c.
ln: n.IB 0 (I.34)
(vi n s vng dy trn mt n v di)
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- T trng trong cun dy hnh xuyn (Toroid)
B c phng tip tuyn vi trc ca li.
B c chiu c xc nh theo quy tc vn inh c.
ln: n.IB 0 (I.35)
(vi n s vng dy trn mt n v di)
c. nh l Ostrogradski Gauss trong t trng
Thng lng t trng qua mt kn bt k th bng 0.
0dSBS
B (I.36)
0Bdiv (I.37)
8. Lc Lorentz Hiu ng Hall
a. Lc Lorentz
- Lc t tc dng ln in tch q chuyn ng vi vn tc v:
]Bv[qF (I.38)
ln:
sinqvBF
+ Lc Lorentz khng sinh cng do vF
+ Khng th thay i nng lng ca cc ht in tch bng cch tc dng ln n mt t trng c nh.
+ Nu c ng thi E v B tc ng th:
]Bv[qEqF (I.39)
b. Hiu ng Hall
- Mt bn kim loi c dng mt chiu chy qua c t trong t trng B vung gc vi mt bn ca bn. Quan st thy c hiu in th gia mt trn v di ca bn.
bjBRU HH (I.40)
Trong o:
+ RH l h s Hall: ne
1RH
+ j l ln mt dng in
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+ B l ln ca cm ng t.
+ b l rng ca bn.
9. Cc nh lut v cm ng in t
a. nh lut Faraday
Sut in ng cm ng trong mt vng dy dn bng v tri du vi tc bin thin theo thi gian ca t thng gi qua vng dy .
(V)dt
d B
(I.41)
Nu c N vng dy th:
(V)dt
dN B
(I.42)
b. nh lut Lenz
- Ni dung: Sut in ng cm ng lun to ra dng cm ng sao cho t trng m n sinh ra chng li s bin thin ca thng lng t trng sinh ra n.
- Cc cch co th to ra sut in ng cm ng nh sau, da vo (I.43):
+ Thay i cm ng t B
+ Thay i din tch A ca vng dy
+ Thay i gc gia vng dy v t trng.
(V)cosBdSdt
d
S
(I.43)
10. Hin tng h cm v hin tng t cm
a. Hin tng h cm
Khi t hai ng dy gn nhau th dng in trong cun th nht bin thin s to nn sut in ng cm ng cun th 2. Sut in ng ny t l vi tc bin thin ca t thng qua cun 2 do dng trong cun 1 gy ra.
21221 NM gi l h s h cm, c n v Henri
- Sut in ng cm ng trong cun 2 l:
dt
dN 2122
dt
dIM 1212 (I.44)
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b. Hin tng t cm
- L hin tng cm ng xy ra trong chnh cun dy c lp c dng xoay chiu.
- Dng in qua cun dy bin thin, t thng do dng sinh ra cng bin thin v sinh ra sut in ng cm ng c chiu chng li nguyn nhn sinh ra n.
L.IN
L: h s t cm c n v Henri, ph thuc vo thng s hnh hc ca si dy v li vt liu t trong lng cun dy.
dt
dIL (I.45)
11. Nng lng t trng
a. Nng lng t trng
2
B LI2
1W (I.46)
b. Mt nng lng t trng
0
2
BB
B
2
1
V
Ww (I.47)
Vi V l th tch trong lng cun cm.
12. H phng trnh Maxwell v h qu
a. S to thnh in trng do t trng bin thin. Phng trnh Maxwell Faraday
- Bt k mt t trng no bin i theo thi gian cng sinh ra mt in trng
xoy BE . Sut in ng bng lu s ca vct theo vng dy dn kn:
dlE B
Theo nh lut cm ng in t ca Faraday:
S
B dSBdt
ddlE
Lu s ca vct cng in trng xoy dc theo vng dy kn bt k bng v tri du vi tc bin thin theo thi gian ca t thng gi qua din tch gii hn bi vng dy kn o.
- Biu thc in trng trong trng hp tng qut:
t
BErot
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b. S to thnh t trng do in trng bin thin
- Dng in dch l dng in ng vi in trng bin i theo thi gian v phng din sinh ra t trng.
- nh l dng ton phn trong trng hp tng qut:
D0 IIdlB
Trong , I v ID ln lt l cng dng in dn, cng dng in dch.
Dng vi phn ca nh l:
t
EjBrot 00
Trong , j v t
Ej 0D
ln lt l mt dng in dn v mt dng in dch.
c. Cc phng trnh Maxwell
- in trng v t trng ng thi tn ti trong khng gian to thnh mt trng thng nht gi l in t trng. H qu chnh ca l thuyt Maxwell l kt lun v bn cht sng in t ca nh sng.
- Cc phng trnh Maxwell:
+ T trng bin thin theo thi gian sinh ra in trng xoy:
t
BErot
(I.48) (phng trnh Maxwell Faraday)
+ T thng gi qua mt kn bt k bng 0:
0Bdiv (I.49) (nh l O G trong t trng)
+ in trng xoy sinh ra dng in dch v t sinh ra t trng bin i:
t
EjBrot 00 (I.50) (phng trnh Maxwell Ampere)
+ in tch ngoi l ngun gc ca in trng:
0
Ediv (I.51) (nh l O G trong in trng)
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d. Sng in t
- T trng bin thin sinh ra in trng bin thin v in trng bin thin ny li sinh ra t trng bin thin. Kt qu l c mt h trng in t bin i lan truyn trong khng gian. l sng in t.
- Vn tc truyn song in t trong mi trng:
cv (I.52)
Trong : v ln lt l hng s in mi v hng s t thm ca mi trng,
(m/s)458792299
1c
00
l vn tc nh sng trong chn khng.
- Nng lng ca song in t. Vct Pointing (khng cn hc)
+ Mt nng lng ca sng in t:
0
22
0
0
22
0
BE
B
2
1E
2
1
+ Vct Pointing biu th nng lng sng in t truyn qua mt n v din tch trong mt n v thi gian:
BE
1S
0
(I.53)
ln trung bnh ca S :
0
00
0
2
02
00
BE
2
1
Bc
2
1Ec
2
1S (I.54)
Trong , E0 v B0 ln lt l gi tr cc i ca in trng v t trng.
B. BI TP
Bi 1: in trng cua in tich im
- Khi c 1 in tch im q: th cng in trng E ti im P cch in tch q mt khong l r c xc nh nh sau:
+ E c im t ti im P.
+ E c phng nm trn ng thng ni in tch q vi im P.
+ E c chiu t in tch q ti P nu q > 0 v c chiu ngc li nu q < 0.
+ ln ca E c xc nh theo cng thc (vi l hng s in mi ca mi trng):
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 22
20r4
qE
C
N
Vy:
r
r
r4
qE
2
0 (I.50)
- Khi c n in tch im:
in trng tng cng v tr ang xt c xc nh theo nguyn l chng cht.
n21 E...EEE
Bi 2: in th cua in tich im
- Khi c 1 in tch im q: th in th V ti im P cch in tch q mt khong l r c xc nh nh sau:
- in th ti mt im c tnh bng cng ca lc tnh in dch chuyn mt n v in tch dng t im ra xa v cng. Th ti mt im l i lng v hng.
P
P dsEV vi ds l vi phn ca qung ng dch chuyn.
ds.s4
qV
r
2
0
P
vi s l khong cch t yu t ds ti in tch p.
1
r
1
4
q
s
1
4
qV
0r0
P
r.4
qV
0
P
Coulomb
JuneVol (I.51)
- Th ti im P quanh mt in tch dng th c gi tr dng, quanh mt in tch m th c gi tr m.
- Khi c n in tch im:
Ta c th tnh th tng cng mt im gy ra bi mt nhm in tch im nh nguyn l chng cht. Vi n in tch, th tng cng l:
n
1i i0
in
1i
ir4
qVV
Vol (I.52)
Bi 3: in trng cua lng cc in
- Lng cc in l mt h gm 2 in tch im c ln bng nhau v ngc nhau v du, t cch nhau mt khong l d.
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 23
- Tnh in trng ca lng cc in ti im M nm trn trc ca lng cc (ng thng ni hai in tch im) v cch trung im ca n mt khong l z:
in trng gy ra ti M do in tch dng v in tch m ca lng cc gy ra. Ap dng nguyn l chng cht in trng ta c:
)()( EEE
+ E c phng nm trn ng thng ni hai in tch.
+ E c chiu t in tch m n in tch dng.
+ V ln (vi trng hp trn hnh v):
22
2
0
2
0
2
0
)()(
2z
d1
2z
d1
z4
q
2
dz
q.
4
1
2
dz
q.
4
1
EEE
Khi im M cch rt xa trung im ca lng cc (z >> d) th 12z
d . Khai
trin biu thc trn theo cng thc Taylor Maclaurin ta c:
...z
d1
2z
d1
2
; ...z
d1
2z
d1
2
Do o:
z
2d.
z4
q
z
d1
z
d1
z4
qE
2
0
2
0
30 z
qd.
2
1E
C
N (I.53)
- Tch qd c gi l mmen ca lng cc in. Mmen ca lng cc in l mt i lng vect, c ln bng tch qd, c chiu hng dc theo trc ca lng cc t in tch m n in tch dng.
+ - q q
M d z
)(E )(E
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 24
- in trng gy ra bi lng cc in t l nghch vi lp phng khong cch t im cn xc nh n trung im ca lng cc.
Bi 4: in th cua lng cc in
Xt in th ti im M cch trung im ca lng cc mt on l z v cch in tch dng v m mt khong l z+ v z-
Theo nguyn l chng cht, in th ti M l tng in th do in tch dng v in tch m gy ra:
)()(0
)()(
2
1i
iz
q
z
q
4
1VVVV
)()(
)()(
0 .zz
zz
4
1V
Nu z >> d (d l khong cch gia hai in tch) th ta c mt cch gn ng:
2
)()(
)()(
z.zz
d.coszz
Do :
qdz4
cosV
2
0 Vol (I.54)
- Tch qd gi l m men ca lng cc in.
- in th M s khng thay i nu quay M quanh trc ca lng cc sao cho z
v khng thay i.
- in th bng 0 ti mi im nm trn mt phng vung gc vi trc ca lng cc ti trung im ca n.
Bi 5: in trng v in th cua dy dn thng di tich in
Si dy AB c di 2a, tch in dng vi mt din di .Tnh in trng ti im M.
Trng hp 1: M nm trn ng trung trc ca si dy v cch trung im ca
si dy mt khong l z. Xt trng hp c bit khi a
q q
M
d
z (+)
z()
+
z
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 25
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im c trong si dy.
- Chia si dy tch in thnh nhng yu t c di dx v cng nh (nh ti mc coi nh l mt in tch im) v cch trung im O ca si dy mt on l x. tnh in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx
gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
Mt s bn thc mc: sao khng t vi phn l dx v khong cch t dx ti tm l d hoc t vi phn l dl v khong cch t dl ti tm l x. L do l trong khi nim ch c in trng do cc in tch im gy ra, v th ta phi chia si dy thnh v s cc in tch im v dx c chiu di rt nh, nh ti mc coi in tch trn n ging nh in tch im. on vi phn c chiu di rt nh th c th t l dx, dl, dk, nhng bin khi ly tch phn phi t tng ng vi n: t on vi phn l dx th khong cch t dx ti O l x, t on vi phn l dl th khong cch t n ti O l l.
Ch : nu y l mt hm ca x th dx v dy l hon ton khc nhau v: dy = ydx.
- Tnh in trng do yu t dx gy ra ti M:
+ C gc t ti M.
+ in tch ca yu t dx l: dq = .dx
+ in trng Ed c chiu nh hnh v, c phng hp vi ng trung trc ca
AB mt gc .
+ ln ca Ed l:
220
2
0
zx
dx
4
1
r
dq
4
1dE
- in trng do c on dy gy ra ti M:
a
a
y
a
a
x
a
a
EdEdEdE
x
-a
a
M
dx
xEd
yEdEd
z O
r
A
B
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 26
Do tnh cht i xng ca si dy nn 0Eda
a
y
. Suy ra:
a
a
xEdE
+ E c phng v chiu trng vi xEd
+ V ln:
a
a
xdEE
3/2220
2222
022
x
xz
dx
4
z
xz
z
xz
dx
4
1
xz
zdEdEcosdE
a
a
3/2220
a
a
x
xz
dx
4
zdEE
Ta c: dcos
zdxz.tanx
2
Do o:
3/2
2
3
2
0
3/2232
0
3/22222
0
a
a
3/2220
cos
1z
d
cos
z
4
z
tan1z
d
cos
z
4
z
.tanzz
d
cos
z
4
z
xz
dx
4
zE
.sinz
2
1sin
z2
dcos
z4
2
dcosz4
z
.dcos
cos
z
4
z
00
000
-0
3
3
2
0
220 za
a.
z2
E
C
N (I.55)
- Khi a th a >> z, sin = 1 hoc
1
za
alim
22av in trng ti
M l:
z2
E
0
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 27
in th:
in th ti M do yu t vi phn dx gy ra c tnh bng cng m lc in trng ca dx dch chuyn mt n v in tch dng t M ra xa v cng.
in th ti M do dx gy ra l:
22
00 zx4
dx
r.4
dqdV
in th ti M do c on dy gy ra l: p dng nguyn l chng cht (i vi i lng v hng) ta c:
a
a22
0
n
1i
in zx4
dxdVVlimV
a
0
22
0022
0
zxxln2
zx
dx
4
2V
a
zlnzaaln2
V 22
0
z
zaaln.
2
V
22
0
Vol (I.56)
- Khi a th a >> z v
z
zaa 22
. Do , V
Trng hp 2: M nm trn ng thng AB v cch trung im O mt khong l z
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn si dy.
- Chia si dy tch in thnh nhng yu t c di dx v cng nh (nh ti mc coi nh l mt in tch im) v cch u A ca si dy mt on l x. tnh
z
A
B
-a
a
dx
x
M
Ed
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 28
in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx gy ra
ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do yu t dx gy ra ti M:
+ C gc t ti M.
+ in tch ca yu t dx l: dq = .dx
+ in trng gy ra ti M l Ed c phng trng vi ng thng AB v c chiu nh hnh v.
+ ln ca Ed l:
20
2
0 x
dx
4
x
dq
4
1dE
- in trng do c on dy gy ra ti M:
B
A
EdE
+ ln:
az
az0
az
az
2
0
B
Ax
1.
4
x
dx
4
dEE
az
1
az
1.
4
E
0
C
N (I.57)
in th:
in th ti M do yu t vi phn dx gy ra c tnh bng cng m lc in trng ca dx dch chuyn mt n v in tch dng t M ra xa v cng.
in th ti M do dx gy ra l:
x
dx
4
x.4
dx
x.4
dqdV
000
in th ti M do c on dy gy ra l: p dng nguyn l chng cht (i vi i lng v hng) ta c:
az
az00
xln4
x
dx
4
dVV
az
az
az
azln
4
V
0
Vol (I.58)
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 29
Trng hp 3: M nm trn ng thng vung gc vi mt u ca si dy v cch
u si dy mt khong l z
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn si dy.
- Chia si dy thnh nhng yu t vi phn dx v cng nh, cch A mt on l x. tnh in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx
gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do yu t dx gy ra ti M:
+ C gc t ti M.
+ Ed c chiu nh hnh v, c phng hp vi ng thng AM mt gc .
+ ln ca Ed l:
22020 zxdx
4
1
r
dq
4
1dE
(v dq = .dx)
- in trng do c on dy gy ra ti M:
2a
0
y
2a
0
x
2a
0
EdEdEdE
+ V ln:
2a
0
2
0
22
0
2a
0
xx cos.xz
dx
4
1dE.cosdEE
a
M: dcos
zdxz.tanx
2
Do o:
A
dx
z
r
M
Ed
B
0
x
xEd
yEd
2a
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 30
220000
000
2222
0
x
z4az
2a
4
sin
z4
.sin
z4
dcosz4
tanzzcos
dz.cos
4
E
220000
000
2222
0
2a
0
22
0
2a
0
2a
0
yy
z4az
z
4
cos
z4
.cos
z4
dsinz4
tanzzcos
dz.sin
4
.sinzx
dx
4
1dE.sindEE
Du " " trong Ey th hin rng vct yE c chiu hng xung di so vi trc
tung Oy.
2
y
2
x EEE
)z(4az
z
)z(4az
4a.
4
E
222
2
222
2
0
z4
E
0
C
N (I.59)
+ Chiu ca E hp vi phng ngang mt gc sao cho: x
y
E
Etan
in th:
in th ti M do yu t vi phn dx gy ra c tnh bng cng m lc in trng ca dx dch chuyn mt n v in tch dng t M ra xa v cng.
in th ti M do dx gy ra l:
22
000 zx
dx
4
r.4
dx
r.4
dqdV
in th ti M do c on dy gy ra l: p dng nguyn l chng cht (i vi i lng v hng) ta c:
2a
0
22
0
2
022
0
zxxln4
zx
dx
4
dVV
a
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 31
z
z4a2aln
4
V
22
0
Vol (I.60)
Trng hp 4: M nm trn ng thng vung gc vi si dy ti im cch u si dy mt khong l h, M cch si dy mt khong l z. y l trng hp tng qut cho 3 trng hp trn.
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn si dy.
Chia si dy thnh nhng yu t vi phn dx v cng nh, cch H mt on l x. tnh in trng ti M do si dy gy ra, ta phi tnh in trng do tng yu t dx
gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do yu t dx gy ra ti M:
+ C im t ti M.
+ Ed c chiu nh hnh v, c phng hp vi ng thng HM mt gc .
+ ln ca Ed l:
22020 zxdx
4
1
r
dq
4
1dE
(do dq = dx)
- in trng do c on dy gy ra ti M:
B
A
y
B
A
x
B
A
EdEdEdE
+ V ln:
a2h 2
22
0
B
A
xx cos.xz
dx
4
1dE.cosdEE
h
ah
h
z
dx
M
Ed
A
B
0
a
x
xEd
yEd
h
H
r
2
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 32
M:
dcos
zdxz.tanx
2
Do :
220000
000
2222
0
x
zh)(2a
h2a
z4
sin
z4
.sin
z4
dcosz4
cos
tanzzcos
zd
4
E
220000
000
2222
0
a2h
22
0
a2h
HA
yy
zh)(2a
z
z4
cos
z4
.cos
z4
dsinz4
sin
tanzzcos
zd
4
.sinzx
dx
4
1dE.sindEE
h
B
2
y
2
x EEE
cossinz4
E 22
0
z4
E
0
C
N (I.61)
+ Chiu ca E hp vi phng ngang mt gc sao cho: x
y
E
Etan
in th:
in th ti M do yu t vi phn dx gy ra c tnh bng cng m lc in trng ca dx dch chuyn mt n v in tch dng t M ra xa v cng.
in th ti M do dx gy ra l:
22
000 zx
dx
4
r.4
dx
r.4
dqdV
in th ti M do c on dy gy ra l: p dng nguyn l chng cht (i vi i lng v hng) ta c:
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 33
2ah
h
22
0
2ah
h22
0
zxxln4
zx
dx
4
dVV
22
22
0 zhh
zh)(2ah)(2aln
4
V
Vol (I.62)
Bi 6: in trng v in th cua vng dy trn tich in
Tnh in trng v in th ti im M.
Trng hp 1: Vng dy trn tm O bn knh R, tch in q, mt in di l . im M nm trn trc ca vng dy v cch O mt khong l z. Xt trng hp c
bit khi z bng 0.
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn vng dy trn.
Chia vng dy thnh nhng yu t vi phn dx v cng nh. tnh in trng ti M do vng dy gy ra, ta phi tnh in trng do tng yu t dx gy ra ti M l
cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do yu t dx gy ra ti M:
+ C im t ti im M.
+ in trng gy ra ti M l Ed c phng v chiu nh hnh v.
+ ln ca Ed :
in tch ca yu t dx l: dq = .dx = .Rd
22020 zRRd
4
1
r
dq
4
1dE
- in trng do c vng dy gy ra ti M:
2
0
y
2
0
x
2
0
EdEdEdE
M
dx
z
R
xEd
EdyEd
O d
q
r
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 34
Do tnh cht i xng ca vng dy nn 0Ed2
0
x
. Suy ra: 2
0
yEdE
+ E c phng v chiu trng vi yEd
+ V ln:
2
0
2
0
22
0
2
0
y .coszR
Rd
4
1dE.cosdEE
+ Ta c:
22 zR
zcos
2
02222
0 zR
z
zR
Rd
4
1E
2/3220 zRRz
2
1E
C
N (I.63)
Khi z = 0 th ta c in trng ti tm ca vng dy trn tch in l: E = 0.
in th:
in th ti M do yu t vi phn dx gy ra c tnh bng cng m lc in trng ca dx dch chuyn mt n v in tch dng t M ra xa v cng.
in th ti M do dx gy ra l:
22
000 zR
d
4
R
r.4
Rd
r.4
dqdV
in th ti M do c on dy gy ra l: p dng nguyn l chng cht (i vi i lng v hng) ta c:
22
0
2
022
0 zR4
R.2
zR
d
4
RdVV
22
0 zR2
RV
Vol (I.64)
Khi z = 0 th in th ti tm ca vng dy trn tch in l: 02
V
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 35
Trng hp 2: Mt ng hnh tr rt mng c chiu cao h, tch in dng vi mt
in mt l . im M nm trn trc ca hnh tr v cch tm ca mt y mt khong l z
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn mt xung quanh ca hnh tr.
Chia ng tr thnh cc nhn trn c rng dx v cng nh. tnh in trng ti M do ng tr gy ra, ta phi tnh in trng do tng nhn trn c rng dx gy
ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do nhn trn co b rng dx gy ra ti M, tm vng trn cch M mt khong x l:
+ C im t ti im M.
+ Ed c phng v chiu nh hnh v.
+ ln ca Ed :
in tch ca nhn trn b rng dx l: dq = 2R.dx (R l bn knh ng tr)
Do o theo trng hp 1 ca bi ny ta c:
dx
xR
R.x
2
1dx
xR
2R.x
4
1dE
3/2220
3/2220
- in trng do c ng tr gy ra ti M:
z
h-z
EdEdE
O
y x
h
z
M
z
Ed
Hnh tr rng
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 36
+ E c phng v chiu trng vi Ed
+ V ln:
z
h-z
3/222
22
0
z
h-z
3/2220
z
h-z xR
)Rd(x
4
Rdx
xR
x
2
RdEE
z
hz
22
0 xR
1
2
RE
22220 zR
1
h)-z(R
1
2
RE
C
N (I.65)
in th:
in th ti M do v s cc nhn trn c b rng v cng nh dx gy ra.
in th ti M do mt nhn trn b rng dx gy ra l: theo trng hp 1 bi ny th:
dxxR2
Rdx
xR4
R2.dV
22
0
22
0
in th ti M do c ng tr gy ra l:
z
hz
22
0
z
h-z22
0
xRxln2
R
xR
dx
2
RdVV
22
22
0 h)-(zRh)(z
zRzln
2
RV
Vol (I.66)
Trng hp 3: Bn cu rng c bn knh R, tch in dng vi mt in mt . im M nm ti tm ca bn cu.
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im trn b mt ca bn cu.
R
dx d
z M
Ed
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 37
Chia bn cu thnh cc vng nhn trn c rng dx v cng nh. tnh in trng ti M do bn cu rng gy ra, ta phi tnh in trng do tng vng nhn trn
c rng dx gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do vng nhn trn co b rng dx gy ra ti M, tm vng nhn trn cch M mt khong z v bn knh vng nhn l r:
+ C im t ti im M.
+ Ed c phng v chiu nh hnh v.
+ ln ca Ed :
in tch ca vng nhn trn b rng dx l: dq = 2r.dx
Do o theo trng hp 1 ca bi trc v in trng do vng dy trn tch in gy ra, ta c:
dx
zr
rz
2
1dx
zr
2r.z
4
1dE
3/2220
3/2220
- in trng do c bn cu gy ra ti M:
EdE
+ E c phng v chiu trng vi Ed
+ V ln:
dEE
Ta c:
Rcosz
Rsinr
Rddx
(vi d l gc tm m cung dx chn)
Do :
2
00
2
0
3/2222
3
0
2
0
3/222220
.d.cossin2
d
)cos(sinR
.cossinR
2
dcosRsinR
.R.RcosRsin
2
dEE
2
0
2
0
2
002
sin
2
)(sin.dsin
2
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 38
04
E
C
N (I.67)
Cch khc:
Ta c th khng cn chia thnh cc vng nhn trn c b rng dx. u tin tnh in trng do yu t vi phn din tch dS gy ra ti M, sau ly tch phn theo mt ca bn cu.
- in trng do yu t vi phn din tch dS gy ra c ln:
20 R
dq
4
1dE
M: .dx.rddq (vi l gc m dx chn tm ng trn bn knh r)
dd.sin.Rd.Rsin.Rddq 2
Do : dd..sin4
R
dq
4
1dE
0
2
0
Khi ly tch phn trn ton mt bn tr th s c c d v d cng bin thin. tnh tch phn 2 lp ny th phi chuyn v tch phn 1 lp bng cch p dng h qu ca nh l Fubini.
in trng do c bn cu gy ra ti M l:
0
2
00
2
0
2
00
S0S
4
.d.cossin4
2d.d.cossin
4
dd.cos.sin4
dE.cosE
in th:
in th ti M do v s cc vng nhn trn c b rng v cng nh dx gy ra.
in th ti M do mt vng nhn trn b rng dx gy ra l: theo trng hp 1 v in th ca vng dy trn tch in th:
dxxr2
rdx
xr4
r2.dV
22
0
22
0
M:
Rcosz
Rsinr
Rddx
nn ta c:
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Hong Vn Trng 0974.971.149 39
dsin
2
RRd
cosRsinR2
RsindV
02222
0
in th ti M do c bn tr rng gy ra l:
200
2
00
cos2
Rdsin
2
RdVV
02
RV Vol (I.68)
Bi 7: in trng v in th cua a trn tich in
Tnh in trng v in th ti im M.
Trng hp 1: a trn tm O bn knh R, tch in dng vi mt in mt . im M nm trn trc ca a v cch O mt khong l z. Xt trng hp c bit
khi z >> R v khi R .
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im nm trn din tch b mt ca a trn.
Chia a trn thnh cc yu t xuyn c rng dr v cng nh, bn knh r. tnh in trng ti M do a gy ra, ta phi tnh in trng do tng yu t xuyn c
b rng dr gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do yu t xuyn co b rng dr gy ra ti M, tm yu t xuyn cch M mt khong z l:
+ C im t ti im M.
+ Ed c phng v chiu nh hnh v.
+ ln ca Ed :
in tch ca yu t xuyn b rng dr l: dq = 2r.dr (r l bn knh yu t xuyn tng ng)
O d
M
z
xEd
EdyEd
R
q
dr r
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Hong Vn Trng 0974.971.149 40
Do o theo trng hp 1 ca bi trc v vng dy trn tch in ta c:
dr
zr
r.z
2
1dr
zr
2r.z
4
1dEdE
3/2220
3/2220
y
- in trng do c a trn gy ra ti M:
yx dEdEdEE
Do tnh cht i xng ca a trn m: 0dEx nn: ydEE
+ E c phng v chiu trng vi ydE
+ V ln:
R
0
3/222
22
00
3/2220
y
zr
)zd(r
4
zdr
zr
r
2
zdEE
R
R
0
220
R
0
220 zr
1
2
z
zr2
1
1
4
zE
220 zR
1
z
1
2
zE
220 zR
z1
2
E
C
N (I.69)
Khi z >> R th 1zR
z22
v in trng ti M l: 0E . Khi a
trn tch in coi nh mt in tch im gy ra mt in trng cch xa v cng.
Khi R th 0zR
z22
v in trng ti M l:
02
E
. Khi
a trn tch in coi nh mt mt phng rng v hn.
in th:
in th ti M do v s cc vng xuyn c b rng v cng nh dr gy ra.
in th ti M do mt vng xuyn b rng dr gy ra l: theo trng hp 1 ca bi trc v in th do vng dy trn tch in gy ra th:
drzr2
.rdr
zr4
r2.dV
22
0
22
0
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In xong 31/12/2013
Hong Vn Trng 0974.971.149 41
in th ti M do c a trn gy ra l:
R
022
22
0
R
022
0 zr
)zd(r
4
zr
drr
2
dVV
R0
22
0
R
0
22
0
zr2
2
1
zr
4
zzR2
V 22
0
Vol (I.70)
Khi z >> R th zzR22 v in th ti M l: 0V . Khi a trn
tch in coi nh mt in tch im c th v cng bng 0.
Khi R th 22 zR v in th ti M l: V . Khi a trn
tch in coi nh mt mt phng rng v hn gy ra mt in th im rt gn b mt ca n.
Trng hp 2: Hnh tr c bn knh R, tch in dng vi mt in khi l . im M nm trn trc ca khi tr v cch tm ca mt y mt khong l z.
in trng:
- in trng gy ra ti M l in trng tng cng gy ra bi tt c cc in tch im ca khi tr.
z
z
Hnh tr c
O
y x
h
M
Ed
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Hong Vn Trng 0974.971.149 42
Chia khi tr thnh cc a trn rt mng, c rng dx v cng nh, khong cch t M ti a trn l x. tnh in trng ti M do khi tr gy ra, ta phi tnh
in trng do tng a trn c b rng dx gy ra ti M l cc Ed sau cng cc Ed li vi nhau theo nguyn l chng cht.
- in trng do a trn co b rng dx gy ra ti M:
+ C im t ti im M.
+ Ed c phng v chiu nh hnh v.
+ ln ca Ed :
in tch ca a trn b rng dx l: dq = .R2.dx
Do o theo trng hp 1 ca bi ny ta c:
220
y
xR
x1
2
.dxdEdE
- in trng do c khi tr gy ra ti M:
yx dEdEdEE
Do tnh cht i xng trc ca khi tr m: 0dEx nn: ydEE
+ E c phng v chiu trng vi dE
+ V ln:
z
hz22
22z
hz0
z
hz22
z
hz0
z
hz22
0
xR
)xd(R
2
1dx
2
dx
xR
xdx
2
dxxR
x1
2
dEE
22220
z
hz
22z
hz0
h)-(zRzRhzz2
xRx2
22220
h)-(zRzRh2
E
C
N (I.71)
in th:
in th ti M do v s cc a trn c b rng v cng nh dx gy ra.
in th ti M do mt a trn b rng dx gy ra l: theo trng hp 1 ca bi ny th ta c:
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In xong 31/12/2013
Hong Vn Trng 0974.971.149 43
xxR2
.dxdV 22
0
in th ti M do c khi tr gy ra l:
dxxxR2
dVV
z
h-z
22
0
z
hz
z
h-z
22
0
z
h-z
z
h-z
22
0
x2
1dxxR
2
dxxdxxR
2
h2
1dxxR
2
z
h-z
22
0
Tnh: dxxRIz
h-z
22
t:
xvdxdv
xR
xduxRu
22
22
Suy ra:
zhz
222z
hz
22
z
h-z22
2z
hz
22
z
h-z22
2z
h-z
22z
hz
22
z
h-z22
222z
hz
22
z
h-z22
2z
hz
22
xRxlnRxRx2I
dxxR
RxRx2I
dxxR
RdxxRxRx
dxxR
RRxxRx
dxxR
xxRxI
22
2222222
h)-(zRh)(z
zRzlnRh)-(zRh)(zzRz
2
1I
h
h)-(zRh)(z
zRzlnRh)-(zRh)(zzRz
4
V
22
2222222
0
-
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Hong Vn Trng 0974.971.149 44
Vol (I.72)
(kt qu lng nhng qu, khng bit tnh ton c ng khng)
Bi 8: Xc nh vc t cm ng t B do mt dy dn thng di c dng I chy qua ti im M cch dy mt khong l z.
Cm ng t ti im M gy ra bi dng in thng AB c xc nh bng cm
ng t tng hp gy ra bi cc yu t dng in dlI trn dng in thng .
- Cm ng t ti M do yu t dng in dlI gy ra l mt i lng vct dB c xc nh theo nh lut Biot Savart Laplace:
+ C phng vung gc vi mt phng cha dlI v EM
+ C chiu sao cho dlI , EM v dB to thnh mt tam din thun.
Mun xc nh chiu ca dB ta s dng quy tc vn inh c (d s dng hn quy tc bn tay phi). t u nhn ca inh c ti im M sao cho inh c vung gc
vi mt phng cha dlI v EM (hy tng tng theo hnh hc khng gian). Xoay
inh c sao cho chiu quay ca inh c t vct dlI n vct EM th chiu chuyn
ng tnh tin ca inh c s l chiu ca vc t dB .
Khi ta xoay inh c m mun bit inh c s chuyn ng tnh tin theo hng no th da vo quy c: nu inh c quay cng chiu kim ng h (nhn t u to n u nhn ca inh) th inh c s chuyn ng tnh tin v pha trc v ngc li. Nh vy, trng hp trn ta xoay inh c cng chiu kim ng h nn inh c s chuyn ng tnh tin theo hng t ngoi xuyn vo trong trang giy.
d hnh dung chiu chuyn ng ca inh c cc bn lm th nghim sau: dng t vt xoy inh c vo mt cnh ca bng g. Khi ta xoay t vt theo chiu kim ng h th inh c s i su vo cnh ca. Mun ly inh c ra ta phi xoay t vt ngc chiu kim ng h.
2 B
I
z
dlI
1 A
E
M
dB
r
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In xong 31/12/2013
Hong Vn Trng 0974.971.149 45
Lu : Trong thc t c mt vi inh c khi xoay ngc chiu kim ng h th n s chuyn ng tnh tin v pha trc v ngc li: van bnh gas, c gi cnh qut khi quay,. Nhng quy tc vn inh c m chng ta s dng ch tun theo quy c trn.
+ ln:
20
EM
sinIdl.
4
dB vi EM,dlI
di ca yu t vi phn dl tnh theo gc l:
Do di dl v cng nh nn c th coi gc PKQ . Khi dl l cnh huyn
ca tam gic PKQ nn: sin
PQdl
Do gc d v cng nh nn c th coi PQ l cung trn ca ng trn tm M bn
knh r. Khi : PQ = r.d
Suy ra: sin
zd
sin
rddl
2 . M
sin
zEM
Vy: dsinz4
I
sin
zsin
sinIzd
EM
sinIdl
4
dB 0
2
22
2
0
- Cm ng t ti M do dng in thng gy ra l B :
dBB
+ B c im t ti M.
+ B c phng v chiu trng vi phng v chiu ca dB
+ ln:
2
1
2
1
0
0 cosz4
Idsin
z4
IdBB
210 coscosz4
IB (Tesla) (I.73)
K
dlI
d
M
Q
P
r
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 46
Khi dy dn di v hn th 1 00 v 2 1800. Khi , cm ng t do dy dn di v hn gy ra ti M cch dy mt khong z l:
z2
IB 0 (Tesla) (I.74)
Bi 9: Xc nh vct cm ng t B do mt dng in trn tm O bn knh R, cng I gy ra ti im M nm trn trc cua vng dy trn v cch O mt khong l z. Tm cm ng t ti tm O cua vng dy trn.
Cm ng t ti im M gy ra bi dng in trn c xc nh bng cm ng t
tng hp gy ra bi cc yu t dng in dlI trn dng in trn .
- Cm ng t ti M do yu t dng in dlI gy ra l mt i lng vct dB c xc nh theo nh lut Biot Savart Laplace:
+ C phng vung gc vi mt phng cha dlI v AM
+ C chiu sao cho dlI , AM v dB to thnh mt tam din thun.
+ ln:
20
AM
sinIdl.
4
dB vi 090AM,dlI
220
zR
Idl.
4
dB
- Cm ng t ti M do dng in trn gy ra l B :
yx dBdBdBB
Do tnh cht i xng ca dng in trn m 0dBx nn: ydBB
+ B c im t ti M.
+ B c phng vung gc vi vng in trn ti tm O.
dlI
R
O
z
M
dB
xdB
ydB
A
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In xong 31/12/2013
Hong Vn Trng 0974.971.149 47
+ B c chiu trng vi chiu ca ydB t M ti O.
+ ln:
R2
0
R2
02
322
0
2222
0
R2
0
22
0y
dlzR4
RI
zR
R
zR
Idl
4
sinzR
Idl
4
dB.sindBB
R.2
zR4
RI
23
22
0
2
322
2
0
zR2
IRB
(Tesla) (I.75)
Khi z 0 th 323
22 RzR v cm ng t ti tm O ca dng in trn l:
2R
IB 0 (Tesla) (I.76)
Bi 10: Na vng dy dn in bn kinh R = 0,49m v khi lng m = 250g, c
dng in I = 25A chy qua (hnh v). Hi cn mt t trng B c hng v ln nh th no na vng dy trn l lng trong khng gian.
- Na vng dy l lng trong khng gian th n phi cn bng v lc, hay tng cc vct lc tc dng vo n bng 0.
Na vng dy c khi lng m nn n chu tc dng ca trng lc:
gmP (vi g l gia tc trng trng)
ln: P = mg = 10m (ly g 10 m/s2)
- Na vng dy l lng khi t trng tc dng lc t ln na vng dy v lc ny c ln bng trng lc P, c phng trng vi phng ca trng lc v c chiu
hng ln trn. Tc l F c phng v chiu nh hnh v:
I
P
F
dlI
I
O
M B
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 48
Gi s rng t trng ngoi l t trng u. Ti im M, cho dlI , B , F to
thnh mt tam din thun th B phi c phng vung gc vi mt phng giy v chiu i t ngoi xuyn vo trong t giy.
- Khi c t trng B th t trng s tc dng lc t ln na vng dy, lc t c ln l:
R
0
B.sindlIdFF vi 090B,dl
RIB.
- Vy khi na vng dy l lng th:
P-F
+ ln:
RIB.mg
PF
RI.
mgB (I.77)
(T)0,06549,0.14159,3.25
10.25,0B
Bi 11: Mt dy cp ng trc c ng kinh trong d1 = 2mm v ngoi bc ch
ng kinh d2 = 8 cm, gia li v v bc l cht in mi c hng s in mi
= 3. Trong li v v bc tich in tri du nhau vi mt in di = 3,14.10-4 C/m. Hy xc nh cng in trng ti cc im cch trc mt khong:
(a) r1 = 3 cm.
(b) r2 = 10 cm.
d2 d1 S1
S2
O
r2
r1 = 3
-
In xong 31/12/2013
Hong Vn Trng 0974.971.149 49
Gi s dy cp ng trc rt di, li gia tch in dng v v bc tch in m.
Phn (a):
Xc nh cng in trng ti im M1 cch trc ca dy cp mt khong l r1 = 3cm.
Do c li gia v v bc u l cc vt dn nn cng in trng trong lng vt dn rng bng 0. V vy, im M1 nm gia hai vt dn nhng ch chu tc dng ca in trng do li gy ra.
Do tnh cht i xng trc ca dy cp nn cng in trng ti cc im nm trn mt tr c trc l trc ca dy cp th c ln bng nhau:
- Cng in trng 1E ti im M1:
+ C phng nm trn ng ni M1 vi tm O.
+ C chiu t O ti M1 (do li tch in dng).
+ ln:
Chn mt Gauss l mt cu S1, c trc l trc ca dy cp v cha im M1. p dng nh l O G trong in trng ta c: Thng lng in trng gi qua mt kn S1 bng tng i s cc in tch nm trong mt kn S1 chia cho hng s in mi.
Biu thc:
0
1
S
11
QdSE
1
(Q1 l tng i s in tch trong mt kn S1)
0
1
Squanh xung
1
QdSE
1
(v khng c in thng qua hai mt y)
0
1
Squanh xung
1
QdSE
1
(v E1 bng nhau ti mi im trn mt
xung quanh)
0
11
.hhr2E
(vi h l chiu di cp in)
01
1r2
E
(I.78)
7
12-
-4
1 10.28,68,85.103.3.2.3,14.0,0
3,14.10E
C
N
Phn (b):
Xc nh cng in trng ti im M2 cch trc ca dy cp mt khong l r1 = 10 cm.
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In xong 31/12/2013
Hong Vn Trng 0974.971.149 50
Do c li gia v v bc u l cc vt dn nn cng in trng trong lng vt dn rng bng 0. V vy, im M2 nm ngoi v bc nn khng chu tc dng ca in trng do li gia gy ra. (ng sc b gin on ti ni c in tch)
Do tnh cht i xng trc ca dy cp nn cng in trng ti cc im nm trn cng mt tr c trc l trc ca dy cp th c ln bng nhau:
- Cng in trng 2E ti im M2:
+ C phng nm trn ng ni M2 vi tm O.
+ C chiu t M2 ti O (do v tch in m).
+ ln:
Chn mt Gauss l mt cu S2, c trc l trc ca dy cp v cha im M2. p dng nh l O G trong in trng ta c: Thng lng in trng gi qua mt kn S2 bng tng i s cc in tch nm trong mt kn S2 chia cho hng s in mi.
Biu thc:
0
2
S
22
QdSE
2
(Q2 l tng i s in tch trong mt
kn S2)
0
2
Squanh xung
2
QdSE
2
(v khng c in thng qua hai mt y)
0
2
Squanh xung
2
QdSE
2
(v E2 bng nhau ti mi im trn mt
xung quanh)
0
22
.hhr2E
(vi h l chiu di cp in)
02
2r2
E
(I.79)
7
12-
-4
2 10.88,18,85.103..2.3,14.0,1
3,14.10E
C
N
Bi 12: Cho qu cu khng dn in tm O, bn kinh R = 15 cm c tich in
u vi mt in tich khi = 1,699.10-7 C/m3, c t trong chn khng.
(1) Xc nh cng in trng ti im M cch tm O mt on: (a) r1 = 10 cm; (b) r2 = 30 cm.
(2) Ly in th ti v cng bng 0. Xc nh in th ti P cch tm 20 cm.
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Hong Vn Trng 0974.971.149 51
Phn (1a): Xc nh cng in trng 1E ti im M1 cch O mt khong 10 cm:
Do tnh cht i xng ca qu cu tch in nn cng in trng ti cc im nm trn cng mt mt cu tm O bn knh bt k th c ln bng nhau.
+ 1E c phng nm trn ng thng M1O.
+ 1E c chiu t O ti M1.
+ ln:
- Chn mt Gauss l mt cu S1 tm O bn knh r1. Ap dng nh l O G (i vi in trng) ta c: T thng gi qua mt cu kn S1 bng tng i s in tch nm
trong mt kn chia cho hng s in mi 0.
- Biu thc:
0
1
S
11
QdSE
1
(vi Q1 l tng in tch nm trong khi cu bn knh r1)
0
1
S
1
QdSE
1
0
1
S
1
QdSE
1
(do E1 bng nhau ti mi im trn mt cu S1)
0
3
12
11
r3
4.
r4E
0
11
3
r.E (I.80)
38,20103.8,85.10
0,1.3,14.1,699.10E
12-
-7
1
C
N
Phn (1b): Xc nh cng in trng 2E ti im M2 cch O mt khong 30 cm:
2E
1E
S2
S1
M2
O R r1
M1
r2
-
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Hong Vn Trng 0974.971.149 52
Do tnh cht i xng ca qu cu tch in nn cng in trng ti cc im nm trn cng mt mt cu tm O bn knh bt k th c ln bng nhau.
+ 2E c phng nm trn ng thng M2O.
+ 2E c chiu t O ti M2.
+ ln:
- Chn mt Gauss l mt cu S2 tm O bn knh r2. Ap dng nh l O G (i vi in trng) ta c: T thng gi qua mt cu kn S2 bng tng i s in tch nm
trong mt kn chia cho hng s in mi 0.
- Biu thc:
0
2
S
21
QdSE
2
(vi Q2 l tng in tch nm trong khi cu bn knh r2)
0
2
S
2
QdSE
2
0
2
S
2
QdSE
2
(do E2 bng nhau ti mi im trn mt cu S2)
0
3
2
22
R3
4.
r4E
220
3
2r3
R.E (I.81)
89,7