De Cuong on Tap Vat Ly 11 HK1

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  • 8/22/2019 De Cuong on Tap Vat Ly 11 HK1

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    CNG N TP HC K I VT L 11Chng I. in tch in trng

    Bi 1: Hai qu cu ging nhau bng kim loi tch in tri du t cch nhau 20cm chng ht nhau bng 1 lc F1=4.10-3N. Cho 2 qu cu tip xc vi nhau sau li tch chng ra v tr c. Khi 2 qu cu y nhau bi 1 lc F 2=2,25.10-3N. Xc nh in tch ca mi qu cu trc khi cho chng tip xc nhau.

    Bi 2: Ba in tch im c cng ln, cng du q1 = q2 = q3 = 10-8 C c t ti ba nh ca tam

    gic u ABC AB = 10cm trong khng kh.a. Xc nh lc do in tch q1 tc dng ln q2.

    b. Xc nh lc do hai in tch q1 v q3 cng tc dng ln q2Bi 3: Ba im A,B,C to thnh mt tam gic vung (vung A); AC= 4 cm; AB=3cm nm trong mt in trng u c E song song vi cnh CA, chiu t C nA. im D l trung im ca AC.1) Bit UCD=100 V. Tnh E, UAB; UBC2) Tnh cng ca lc in khi mt e di chuyn :a) T C n Db) T C n B

    c) T B n AHD: Dng cc cng thc: AMN=q.UMN; E= UMN/ ''

    NM; UMN=VM-VN

    Bi 4: Hiu in th gia 2 bn ca mt t in phng l U=300 V. Mt htbi nm cn bng gia 2 bn ca t in v cch bn di ca t d 1=0,8 cm. Hisau bao lu ht bi s ri xung bn di ca t nu hiu in th gia 2 bngim i 60 V (0,09 s)Bi 5: Hai in tch q1=6,67.10-9C v q2=13,35.10-9C nm trong khng kh cchnhau 40 cm. Tnh cng cn thit a h in tch trn li gn nhau v cchnhau 25 cm.Bi 6: Mt t in khng kh c C=2000 pF c mc vo 2 cc ca ngun

    in c ht l U=5000 VTnh in tch ca t inS ( 10-5C)Bi 7: Cho hai ien tch q1 = 4. 10-10 C, q2 = -4. 10-10 C, at tai A va B trong khongkh biet AB = 2 cm. Xac nh vect cng o ien trng E tai:

    a. H, la trung iem cua AB.b. M, MA = 1 cm, MB = 3 cm.c. N, biet rang NAB la mot tam giac eu.

    s: 72. 103 V/m. 32. 103

    V/m. 9. 103 V/m.Bi 8: Hai ien tch q1 = 8. 10-8 C, q2 = -8. 10-8 C at tai A va B trong khong kh

    biet AB = 4 cm. Tm vect cng o ien trng tai C tren ng trung trccua AB va cach AB 2 cm, suy ra lc tac dung len ien tch q = 2. 10 -9 C at taiC. s: 12,7. 105 V/m. F = 25,4. 10-4 N.Bi 9: Mot ien tch iem dng Q trong chan khong gay ra mot ien trngco cng o E = 3. 104 V/m tai iem M cach ien tch mot khoang 30 cm. Tnho ln ien tch Q ? s: 3. 10-7 C.Bi 10: Mot ien tch iem q = 10-7 C at tai iem M trong ien trng cuamot ien tch iem Q, chu tac dung cua mot lc F = 3.10 -3 N. Cng o ientrng do ien tch iem Q gay ra tai M co o ln la bao nhieu s: 3. 104 V/m.

    Chng II. Dng in khng i

    Bi 1: Cng dng in khng i chy qua dy tc ca mt bng n lI=0,273A. Tnh in lng v s e dch qua tit din thng ca dy tc trongthi gian 1 pht. (1,02.1020 e)

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    E1

    r1

    E2

    r2

    R

    Hnh 2.14

    A BC

    R

    Hnh 2.15

    Bi 2. Hai ngun c sut in ng v in tr trong tng ng l E 1=3V; r1=0,6; E2=1,5V; r2=0,4 c mc vi in tr R = 2 thnh mch kn nh s hnh 2.14a. Tnh cng dng in trong mch.

    b. Tnh hiu in th gia hai cc ca mi ngunBi 3. Hai ngun c sut in ng e v in tr trong r c mc vi nhauthnh b v mc vi in tr R=11 to thnh mt mch in kn. Khi hai ngun mc ni tip th

    cng dng in chy qua in tr R l 0,4A cn khi hai ngun mc song song th cng dngin chy qua in tr R l 0,25A. Tnh sut in ng e v in tr trong r..Bi 4. Cho mch in nh hnh 2.15, cc ngun ging nhau,mi ngun c e=1,5V v r=1. Mch ngoi l in trR=6. Tnh:

    a. Sut in ng v in tr trong ca b ngun.b. Cng dng in qua mch ngoi.c. Hiu in th gia cc im B v A; A v C.

    Bi 5. Mt ngun in c sut in ng 3V, in tr trong 2.

    Mc song song hai bng n cng loi 3V-1,5W vo hai cc ca ngun in.

    a. Tnh cng sut tiu th ca mi n.

    b. Nu tho b mt n th n cn li sng mnh hay yu hn trc . V sao ?

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

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

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

    Bi 7. Hai bng n c in tr 5mc song song v ni vo mt ngun in c in tr trong 1th cng dng in trong mch l 12/7A. Khi tho mt n ra th cng dng in trong mchl bao nhiu.

    Bi 8. C mt s in tr ging nhau, mi in tr l R0=4. Tm s in tr t nht v cch mc c in tr tng ng l 6,4.

    Chng III.Dng in trong cc mi trng

    - Yu cu nm c bn cht ca dng in trong cc mi trng.

    Bi 1. Khi in phn dd ZnSO4 Vi Anot lm bng Zn. Trong thi gian 1 gi thu c Catot cabnh in phn 22,448g Zn. Hiu in th ca mng in mt chiu m bnh in phn mc vo lnhn hiu in th cn thit nh in phn lm vic bnh thng l 6V. Hi phi mc mt in tr R=

    bao nhiu,mc nh th no bnh hot ng bnh thng Zn = 65, n = 2.

    Bi 2. Mt bng n 220V-40 W c dy tc lm bng vnfram. in tr ca dytc n 200C l R0=121 . Tnh nhit t ca dy tc n khi sng bnh th-ng. Gi thit in tr ca dy tc n trong khong nhit ny tng bcnht theo nhit vi h s nhit in tr =4,5.10-3 K-1

    Bi 3. Mt b ngun in gm 30 pin mc thnh 3 nhm ni tip mi nhm c

    10 pin mc song song, mi pin c S e=0,9 V v TT 0,6 . Mt bnh inphn c in tr R=205 c mc vo 2 cc ca b ngun ni trn. Tnhkhi lng ng bm vo catt ca bnh trong 50 pht (0,013 g).

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    Bi 4. Mun m ng 1 tm st c din tch tng cng 200 cm 2 ngi ta dngtm st lm catt ca 1 bnh in phn ng dung dch CuSO4 v ant l 1thanh ng nguyn cht, cho dng in I=10 A chy qua trong thi gian2h40min50s. Tm chiu dy ca lp ng bm trn tm st. Bit nguyn t lngca ng l 64, KLR ca ng l 8,9 g/cm3.