MaiTrungHiuVtl11Bitp theo chuyn SutmvbinsonEmail:beaumup@gmail.comMclc1 intch-intrng 51.1 Lc tng tc tnh in . . . . . . . . . . . . . . . . . . . . . 61.1.1 Tng tc gia hai in tch im ng yn. . . . . . 61.1.2 ln in tch . . . . . . . . . . . . . . . . . . . . . 71.1.3 Tng tc ca nhiu in tch . . . . . . . . . . . . . . 81.1.4 Cn bng ca in tch . . . . . . . . . . . . . . . . . 101.2 in trng. . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.1 in trng do mt in tch im gy ra . . . . . . . 111.2.2 Cng in trng do nhiu in tch im gy ra 111.2.3 Cng in trng tng hp trit tiu . . . . . . . 131.2.4 Cn bng ca in tch trong in trng . . . . . . . 141.2.5 Cng in trng do vt c kch thc to nn . . 141.3 in th - Hiu in th - Bi ton vt di chuyn trong intrng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3.1 Cng ca lc in - Hiu in th . . . . . . . . . . . 151.3.2 Vector vn tc ca in tch cng phng vi ngsc in . . . . . . . . . . . . . . . . . . . . . . . . . . 171.3.3 Vector vn tc ca in tch vung gc vi ng scin . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3.4 Vectorvntccaintchxingcvingscin . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.4 T in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.4.1 Tnh ton cc i lng . . . . . . . . . . . . . . . . . 201.4.2 Ghp t cha tch in . . . . . . . . . . . . . . . . . 221.4.3 Ghp t tch in. . . . . . . . . . . . . . . . . . . 281.4.4 Hiu in th gii hn . . . . . . . . . . . . . . . . . . 281.4.5 T c cha ngun - T xoay . . . . . . . . . . . . . . . 291.4.6 Mch cu t . . . . . . . . . . . . . . . . . . . . . . . 321.4.7 Nng lng in trng . . . . . . . . . . . . . . . . . 341.5 p s chng 1 . . . . . . . . . . . . . . . . . . . . . . . . . 354 Mclc2 Dnginkhngi 412.1 Cng dng in - Hiu in th . . . . . . . . . . . . . . 412.1.1 Bi tp c bn . . . . . . . . . . . . . . . . . . . . . . 412.2 Bi tp lin quan n in tr . . . . . . . . . . . . . . . . . . 422.2.1 in tr ca dy dn - S ph thuc ca in tr vonhit . . . . . . . . . . . . . . . . . . . . . . . . . . 422.2.2 intrtngngcamchmcni tiphocsong song . . . . . . . . . . . . . . . . . . . . . . . . . 432.3 p s chng 2 . . . . . . . . . . . . . . . . . . . . . . . . . 43Chng 1intch-intrngNidung1.1 Lctngtctnhin. . . . . . . . . . . . . . 61.1.1 Tngtcgiahaiintchimngyn. . 61.1.2 lnintch . . . . . . . . . . . . . . . . . 71.1.3 Tngtccanhiuintch . . . . . . . . . . 81.1.4 Cnbngcaintch . . . . . . . . . . . . . 101.2 intrng. . . . . . . . . . . . . . . . . . . . . 111.2.1 intrngdomtintchimgyra . . . 111.2.2 Cngintrngdonhiuintchimgyra . . . . . . . . . . . . . . . . . . . . . . . 111.2.3 Cngintrngtnghptrittiu . . . 131.2.4 Cnbngcaintchtrongintrng . . . 141.2.5 Cngintrngdovtckchthctonn . . . . . . . . . . . . . . . . . . . . . . . . . 141.3 inth - Hiuinth - Bi tonvt dichuyntrongintrng . . . . . . . . . . . . . 151.3.1 Cngcalcin-Hiuinth . . . . . . . 151.3.2 Vectorvntccaintchcngphngvingscin . . . . . . . . . . . . . . . . . . 171.3.3 Vector vn tc ca in tch vung gc vi ngscin . . . . . . . . . . . . . . . . . . . . . . 191.3.4 Vector vn tc ca in tch xin gc vi ngscin . . . . . . . . . . . . . . . . . . . . . . 201.4 Tin . . . . . . . . . . . . . . . . . . . . . . . 201.4.1 Tnhtonccilng . . . . . . . . . . . . . 201.4.2 Ghptchatchin . . . . . . . . . . . . . 221.4.3 Ghpttchin. . . . . . . . . . . . . . . 286 Chng1 intch-intrng1.4.4 Hiuinthgiihn . . . . . . . . . . . . . . 281.4.5 Tcchangun-Txoay. . . . . . . . . . . 291.4.6 Mchcut . . . . . . . . . . . . . . . . . . . 321.4.7 Nnglngintrng . . . . . . . . . . . . . 341.5 pschng1 . . . . . . . . . . . . . . . . . . 351.1 Lctngtctnhin1.1.1 Tngtcgiahai intchimngynBi 1.1Hai in tchq1=2 108C vq2= 108C t cch nhau 20 cm trongkhng kh. Xc nh ln v v hnh lc tng tc gia chng?Bi 1.2Hai in tchq1=2 106C vq2= 2 106C t ti hai imA vBtrong khng kh. Lc tng tc gia chng l0,4 N. Xc nh khong cchAB, v hnh lc tng tc .Bi 1.3Haiintchtcchnhaumtkhongrtrongkhngkhthlctngtc gia chng l 2 103N. Nu vi khong cch m t trong in mith lc tng tc gia chng l103N.a. Xc nh hng s in mi ca in mi.b. lc tng tc gia hai in tch khi t trong in mi bng lctng tc khi t trong khng kh th phi t hai in tch cch nhaubao nhiu? Bit trong khng kh hai in tch cch nhau 20 cm.Bi 1.4Trong nguyn t hidro, e chuyn ng trn u quanh ht nhn theo quo trn c bn knh5 109cm.a. Xc nh lc ht tnh in giae v ht nhn.b. Xc nh tn s quay cae.Bi 1.5Mtqucuckhilngring=9,8 103kg/m3,bnknhR=1 cmtchinq= 106Cctreovoumtsidymnhcchiudil = 10 cm. Ti im treo c t mt in tch m q0= 106C. Tt c t1.1 Lctngtctnhin 7trong du c khi lng ring D= 0,8 103kg/m3, hng s in mi = 3.Tnh lc cng ca dy? Lyg= 10 m/s2.Bi 1.6Hai qucunh, gingnhau, bngkimloi. QucuAmangintch4,5 C; qucuBmangintch 2,4 C. Chochngtipxcnhauria chng ra cch nhau1,56 cm. Tnh lc tng tc in gia chng.1.1.2 lnintchBi 1.7Hai in tch im bng nhau, t trong chn khng, cch nhau 10 cm. Lcy gia chng l9 105N.a. Xc nh du v ln hai in tch .b. lctngccgiahaiintchtng3lnthphitnghaygimkhongcchgiahaiintchbaonhiuln?Vsao?Xcnh khong cch gia hai in tch lc .Bi 1.8Hai in tch c ln bng nhau, t cch nhau25 cm trong in mi chng s in mi bng 2 th lc tng tc gia chng l6,48 103N.a. Xc nh ln cc in tch.b. Nu a hai in tch ra khng kh v vn gi khong cch thlc tng tc gia chng thay i nh th no? V sao?c. lc tng tc ca hai in tch trong khng kh vn l 6,48 103Nth phi t chng cch nhau bao nhiu?Bi 1.9Hai vt nh tch in t cch nhau50 cm, ht nhau bng mt lc0,18 N.in tch tng cng ca hai vt l4 106C. Tnh in tch mi vt?Bi 1.10Hai in tch im c ln bng nhau t trong chn khng, cch nhaumt khong5 cm, gia chng xut hin lc yF= 1,6 104N.a. Hy xc nh ln ca hai in tch im trn?b. lc tng tc gia chng l 2,5 104N th khong cch gia chngl bao nhiu?8 Chng1 intch-intrngBi 1.11Hai vt nh t trong khng kh cch nhau mt on1 m, y nhau mtlcF= 1,8 N. in tch tng cng ca hai vt l3 105C. Tm in tchca mi vt.Bi 1.12Hai qu cu kim loi nh nh nhau mang cc in tchq1vq2t trongkhngkh cchnhau2 cm, ynhaubngmtlc2,7 104N. Chohaiqu cu tip xc nhau ri li a v v tr c, ch y nhau bng mt lc3,6 104N. Tnhq1vq2?Bi 1.13Hai qu cu nh ging nhau bng kim loi c khi lng 50 g c treo vocng mt im bng hai si ch nh khng gin di 10 cm. Hai qu cu tipxc nhau tch in cho mt qu cu th thy hai qu cu y nhau cho nkhi hai dy treo hp vi nhau mt gc60.Tnh in tch m ta truyncho cc qu cu qu cu. Chog= 10 m/s2.Bi 1.14Mt qu cu nh cm=60 g, in tchq=2 107C c treo bng sit mnh. pha di n 10 cm cn t mt in tch q2 nh th no sccng ca si dy tng gp i?Bi 1.15Hai qu cu nh tch in q1= 1,3 109C, q1= 6,5 109C t cch nhaumt khongrtrong chn khng th y nhau vi mt nhng lc bngF.Cho 2 qu cu y tip xc nhau ri t cch nhau cng mt khong r trongmt cht in mi th lc y gia chng vn lF.a. Xc nh hng s in mi ca cht in mi .b. BitF= 4,5 106N,tmr.1.1.3 TngtccanhiuintchBi 1.16Cho hai in tch imq1=2 107C, q2= 3 107C t ti hai imA vBtrong chn khng cch nhau5 cm. Xc nh lc in tng hp tcdng lnq0= 2 107C trong hai trng hp:a. q0t tiC, viCA = 2 cm;CB= 3 cm.b. q0t tiD, viDA = 2 cm;DB= 7 cm.1.1 Lctngtctnhin 9Bi 1.17Haiintchimq1=3 108C, q2=2 108CttihaiimAvBtrongchnkhng, AB=5 cm. intchq0= 2 108Ctti M,MA = 4 cm,MB= 3 cm. Xc nh lc in tng hp tc dng lnq0.Bi 1.18Trong chn khng, cho hai in tchq1=q2=107C t ti hai imAvBcch nhau10 cm. Ti imCnm trn ng trung trc caABvcch AB5 cm ngi ta t in tchq0=107C. Xc nh lc in tnghp tc dng lnq0.Bi 1.19C 3 din tch imq1=q2=q3=q=1,6 106C t trong chn khngti 3 nh ca mt tam gic uABCcnha=16 cm. Xc nh lc intng hp tc dng ln mi in tch.Bi 1.20Ba qu cu nh mang in tch q1= 6 107C, q1= 2 107C, q1= 106Ctheo th t trn mt ng thng nhng trong nc nguyn cht c = 81.Khongcchgiachnglr12=40 cm, r23=60 cm. Xcnhlcintng hp tc dng ln mi qu cu.Bi 1.21Baintchimq1=4 108C, q2= 4 108C, q3=5 108Cttrongkhngkh ti banhcamttamgicucnh2 cm. Xcnhvect lc tc dng lnq3?Bi 1.22Hai in tchq1= 8 108C,q2= 8 108C t ti A v B trong khngkh (AB= 10 cm). Xc nh lc tc dng lnq3= 8 108C, nu:a. CA = 4 cm,CB= 6 cm.b. CA = 14 cm,CB= 4 cm.c. CA = CB= 10 cm.d. CA = 8 cm,CB= 6 cm.Bi 1.23Ngi ta t 3 in tchq1= 8 109C,q2= q3= 8 109C ti ba nhcamttamgicucnh6 cmtrongkhngkh.Xcnhlctcdngln in tchq0= 6 109C t tmO ca tam gic.10 Chng1 intch-intrng1.1.4 CnbngcaintchBi 1.24Hai in tchq1= 2 108C,q2= 8 108C t tiA vBtrong khngkh,AB= 8 cm. Mt in tchq0t tiC. Hi:a. C u q0cn bng?b. Du v ln caq0q1,q2cng cn bng?Bi 1.25Hai intchq1= 2 108C, q2= 1,8 107Ctti AvBtrongkhng kh,AB= 8 cm. Mt in tchq0t tiC. Hi:a. C u q0cn bng?b. Du v ln caq0q1,q2cng cn bng?Bi 1.26Hai qu cu nh ging nhau, mi qu c in tch q v khi lng m = 10 gc treo bi hai si dy cng chiu di l=30 cm vo cng mt imO.Gi qu cu 1 c nh theo phng thng ng, dy treo qu cu 2 s blch gc = 60so vi phng thng ng. Chog= 10 m/s2. Tmq?Bi 1.27Hai in tch imq1=108C,q2=4 108C t tiA vBcch nhau9 cm trong chn khng.a. Xc nh lc tng tc gia hai in tch?b. Xc nh vecto lc tc dng ln in tch q0= 3 106C t ti trungimAB.c. Phi tintchq3=2 106Cti uintchq3nmcnbng?Bi 1.28Hai in tch imq1= q2= 4 106C, t tiA vBcch nhau10 cmtrong khng kh. Phi t in tchq3=4 108C ti u q3nm cnbng?Bi 1.29Hai in tch q1= 2 108C, q2= 8 108C t ti A v B trong khngkh,AB= 8 cm. Mt in tchq3t tiC. Hi:a. C u q3cn bng?b. Du v ln caq3q1vq2cng cn bng?1.2 intrng 11Bi 1.30Ba qu cu nh khi lng bng nhau v bng m, c treo vo ba si dynh, khngdn, cngchiudi l vcbucvocngmtim. Khic tch mt in tch q nh nhau, chng y nhau v xp thnh mt tamgic u c cnh a. Tnh in tch q ca mi qu cu?Bi 1.31Cho ba qu cu ging ht nhau, cng khi lngm v in tchq. trngthi cnbngv tr baqucuvimtreochungOtothnhtdinu. Xc nh in tch mi qu cu?1.2 intrng1.2.1 intrngdomtintchimgyraBi 1.32Mt in tch imq= 106C t trong khng kha. Xc nh cng in trng ti im cch in tch 30 cm, v vectcng in trng ti im ny.b. tintchtrongchtlngchngsinmi =16.imccng in trng nh cu a cch in tch bao nhiu.Bi 1.33Chohai imAvBcngnmtrnmtngsccaintrngdomt in tch imq> 0 gy ra. Bit ln ca cng in trng tiA l36 V/m, tiBl9 V/m.a. Xc nh cng in trng ti trung imMcaAB.b. Nu t tiMmt in tch imq0= 102C th ln lc intc dng lnq0l bao nhiu? Xc nh phng chiu ca lc.Bi 1.34Qu cu kim loi bn knh R = 5 cm c tch in q, phn b u. t =qSl mt in mt,Sl din tch hnh cu. Cho= 8,84 105C/m2.Tnh ln cng in trng ti im cch mt cu5 cm?(Ch cng thc tnh din tch xung quanh ca hnh cu:S= 4R2)1.2.2 Cngintrngdonhiuintchimgyra12 Chng1 intch-intrngBi 1.35Chohai intchq1=4 1010C, q2= 4 1010CtA, Btrongkhng kh,AB= a = 2 cm. Xc nh vc t cng in trng ti:a. Hl trung im caAB.b. McchA1 cm, cchB3 cm.c. Nhp viA,Bthnh tam gic u.Bi 1.36Haiintchq1=8 108C, q2= 8 108Ctti A, Btrongkhngkh,AB= 4 cm. Tm vct cng in trng tiCvi:a. CA = CB= 2 cm.b. CA = 8 cm;CB= 4 cm.c. CtrntrungtrcAB, cchAB2 cm, suyralctcdnglnq=2 109C t tiC.Bi 1.37Hai in tch +q v q(q> 0) t ti hai im A v B vi AB= 2a. Mlmt im nm trn ng trung trc caABcchABmt onx.a. Xc nh vect cng in trng tiMb. Xc nhx cng in trng tiMcc i, tnh gi tr .Bi 1.38Haiintchq1=q2=q>0tti AvBtrongkhngkh.ChobitAB= 2aa. Xc nh cng in trng ti im Mtrn ng trung trc caAB cchABmt onh.b. nhh EMcc i. Tnh gi tr cc i ny.Bi 1.39Ti ba nhA, B, Cca t din uS.ABCcnha trong chn khng cbaintchimqgingnhau(q0tA, C;haiintchq3=q4= qtB

vD

.Tnh ln cng in trng ti tmO ca hnh lp phng.1.2 intrng 131.2.3 CngintrngtnghptrittiuBi 1.41Cho hai in tch im cng du c ln q1= 4q2 t ti A, B cch nhau12 cm. im c vect cng in trng doq1 vq2 gy ra bng nhau v tr.Bi 1.42Cho hai in tch tri du, c ln in tch bng nhau, t ti A, B cchnhau12 cm. im c vect cng in trng doq1vq2gy ra bngnhau v tr. ( s: r = r = 6cm)Bi 1.43Cho hai in tchq1= 9 108C,q1= 16 108C t tiA,Bcch nhau5 cm. im c vec t cng in trng vung gc vi nhau vE1= E2( s: r = 3cm, r = 4cm)Bi 1.44Ti banhA, B, CcahnhvungABCDcnha=6 cmtrongchnkhng, t ba in tch imq1= q3= 2 107C vq2= 4 107C. Xcnhintchq4tti Dcngintrngtnghpgybihin tch ti tmO ca hnh vung bng 0.Bi 1.45Cho hnh vung ABCD, ti A v C t cc in tch q1= q3= q. Hi phit Bin tch bao nhiu cng in trng D bng khng.Bi 1.46TihainhA, BcatamgicuABCcnhathaiintchimq1=q2=4 109Ctrongkhngkh.Hiphitintchq3cgitrbaonhiuti Ccngintrnggybih3intchtitrngtmG ca tam gic bng 0.Bi 1.47BnimA, B, C, Dtrongkhngkh tothnhhnhchnhtABCDcnhAD = a = 3 cm,AB= b = 4 cm. Cc in tchq1,q2,q3 c t lnltti A, B, C. Bitq2= 12,5 108Cvcngintrngtnghp ti D bng 0. Tnhq1,q2.14 Chng1 intch-intrng1.2.4 CnbngcaintchtrongintrngBi 1.48Mt qu cu nh khi lng m = 0,1 g mang in tch q= 108C c treobng si dy nh, khng gin v t vo in trng u

Ec ng scnmngang. Khi qucucnbng, dytreohpvi phngthngngmt gc = 45. Lyg= 10 m/s2. Tnh:a. ln ca cng in trng.b. Tnh lc cng dy.Bi 1.49in trng gia hai bn ca mt t in phng t nm ngang c cngE= 4900 V/m. Xc nh khi lng ca ht bi t trong in trngny nu n mang in tchq= 4 1010C v trng thi cn bng.Bi 1.50Mthnbi nhbngkimloi cttrongdu. Bi cthtchV =10 mm3, khi lng m = 9 105kg. Du c khi lng ring D = 800 kg/m3.Tt c c t trong mt in trng u,

Ehng thng ng t trnxung, E= 4,1 105V/m. Tm in tch ca bi n cn bng l lng trongdu. Chog= 10 m/s2.Bi 1.51Hai qu cu nh A v B mang nhng in tch ln lt l q1= 2 109Cvq2= 2 109C c treo u hai si dy t cch in di bng nhau.Hai im treoMvNcch nhau2 cm. Khi cn bng, cc dy treo hp viphng thng ng gc . Hi a cc dy treo tr v v tr thng ngngi ta phi dng mt in trng u c hng no v ln bao nhiu?1.2.5 CngintrngdovtckchthctonnBi 1.52Mt bn mt phng rt ln t thng ng, tch in u vi mt inmt.a. Xc nh vector cng in trng

Edo mt ny gy ra ti mtim cch mt mt onh. Nu c im ca in trng ny.b. Mt qu cu nh khi lngm tch inq cng du vi in tch camt phng, c treo vo mt im c nh gn mt phng bng dynh, khng dn, chiu dil. Gi sqkhng nh hng n phn bintchtrnmtphng.Khicnbng,dytreonghinggcso1.3 inth-Hiuinth-Bitonvtdichuyntrongintrng 15vi phng thng ng. Tnhq.Bi 1.53Tnh cng in trng gy ra bi hai mt phng rng v hn:a. t song song, mt in mt> 0 v .b. Hp vi nhau gc v c cng mt in mt> 0.Bi 1.54Mt bn phng rng v hn c tch in v t vo mt in trng u.Bitcngintrnghaibnmtbnl

E1,

E2cnghngvunggc vi mt, lnE1,E2. Hy tnh mt in mt ca bn v lc intc dng ln mt n v din tch ca bn.Bi 1.55Tnh cng in trng gy ra bi mt dy thng di v hn tch inu vi mt in di.Bi 1.56Haidydnthngdivhntsongsongtrongkhngkh,cchnhaumt ona, tch in cng du vi mt in di.a. Xc nh

E ti mt im trong mt phng i xng gia hai dy, cchmt phng ch hai dy mt khongh.b. Xc nh khongh ln

Eln nht v tnh gi tr ln nht ny.1.3 inth-Hiuinth-Bi tonvtdichuyntrongintrng1.3.1 Cngcalcin-HiuinthBi 1.57Ba imA,B,Cto thnh mt tam gic vung tiC.AC= 4 cm,BC=3 cm v nm trong mt in trng u. Vector cng in trng songsong viAC, hng tA nCv c lnE= 5000 V/m. Tnh:a. UAC,UCB,UAB.b. Cng ca in trng khi mt electron (e) di chuyn tA nB?16 Chng1 intch-intrngBi 1.58Tam gic ABC vung ti A c t trong in trng u

E, = ABC=60,

ABngc hng vi

E. BitBC= 6 cm,UBC= 120 V.a. TmUAC,UBAv cng in trngE?b. tthmCintchimq =9 1010C. Tmcngintrng tng hp tiA.Bi 1.59Mt in tch imq= 4 108C di chuyn dc theo chu vi ca mt tamgic MNPvung ti P, trong in trng u, c cng 200 V/m. CnhMN=10 cm,

MNcng hng

E, NP=8 cm. Mi trng l khng kh.Tnh cng ca lc in trong cc dch chuyn sau caq:a. tMnN.b. tNnP.c. tPnM.d. Theo ng kn (chu tuyn)MNPM.Bi 1.60Mt in trng u c cng E= 2500 V/m. Hai im A, B cch nhau10 cm khi tnh dc theo ng sc. Tnh cng ca lc in trng thc hinmt in tchq khi n di chuyn tA nB, ngc chiu ng sc. Giibi ton khi:a. q= 106C.b. q= 106CBi 1.61Cho3bnkimloi phngA, B,C c tch in v t song songnh hnh. Cho d1= 5 cm, d2=8 cm. Coi in trng gia cc bn lu v c chiu nh hnh v. CngintrngtngnglE1=4 104V/m , E2= 5 104V/m. Tnhin th ca bn B v bn C nu lygc in th l in th bnA.A B C

E1

E2d1d2Bi 1.62Mt electron di chuyn c mt on1 cm, dc theo mt ng sc in,di tcdngcamtlcintrongmtintrnguccng1000 V/m. Hy xc nh cng ca lc in?1.3 inth-Hiuinth-Bitonvtdichuyntrongintrng 17Bi 1.63Khi bay t im M n im N trong in trng, electron tng tc, ngnng tng thm 250eV (1 eV = 1,6 1019J). TmUMN?1.3.2 Vectorvntccaintchcngphngvi ngscinBi 1.64Gia hai bn ca t in t nm ngang cch nhaud = 40 cm c mt intrnguE=60 V/m.Mthtbickhilngm=3 gvintchq =8 105Cbtuchuynngttrngthi ngh tbntchindng v pha tm tch in m. B qua nh hng ca trng trng. Xcnh vn tc ca ht ti im chnh gia ca t in.Bi 1.65Mtelectron bay vo trongmtintrng theohngngcvihngng sc vi vn tc2000 km/s. Vn tc ca electron cui on ngs l bao nhiu nu hiu in th u v cui on ng l15 V.Bi 1.66Mtelectronbtuchuynngdctheochiungscintrngcamttinphng,haibncchnhaumtkhongd=2 cmvgiachng c mt hiu in thU= 120 V. Electron s c vn tc l bao nhiusau khi dch chuyn c mt qung ng1 cm.Bi 1.67Mt electronbayvointrngcamt tinphngtheophngsong song cng hng vi cc ng sc in trng vi vn tc ban ul8 106m/s. Hiuinthtphi cgitr nhnhtlbaonhiuelectron khng ti c bn i din?Bi 1.68Hi bi c khi lng m = 1012g nm cn bng gia in trng u giahai bn t. Bit hiu in th v khong cch gia hai bn l U= 125 V vd = 5 cm.a. Tnh in tch ht bi?b. Nu ht bi mt i5e th mun ht bi cn bng, hiu in th giahai bn lc ny phi c gi tr bao nhiu?18 Chng1 intch-intrngBi 1.69Mtelectroncvntcbanuv0=3 106m/schuynngdctheochiu ng sc ca mt in trng c cng E=1250 V/m. B quatc dng ca trng trng, e chuyn ng nh th no?Bi 1.70Mt electron chuyn ng vi vn tc ban u104m/s dc theo ng scca mt in trng u c mt qung ng10 cm th dng li.a. Xc nh cng in trng.b. Tnh gia tc ca electron.Bi 1.71Mt electron chuyn ng dc theo ng sc ca mt in trng u ccng 364 V/m. Electron xut pht t im Mvi vn tc 3,2 106m/s.Hi:a. Electron i c qung ng di bao nhiu th vn tc ca n bng0?b. Sau bao lu k t lc xut pht electron tr v imM?Bi 1.72Mt electron bay t bn m sang bn dng ca mt t in phng. intrng trong khong hai bn t c cng E= 6 104V/m. Khong cchgia hai bn t ld = 5 cm.a. Tnh gia tc ca electron.b. Tnh thi gian bay ca electron bit vn tc ban u bng 0.c. Tnh vn tc tc thi ca electron khi chm bn dng.Bi 1.73Gia hai bn kim loi t song song nm ngang tch in tri du c mthiu in thU1=1000 V khong cch gia hai bn ld=1 cm. nggi hai bn c mt git thy ngn nh tch in dng nm l lng. tnhin hiu in th gim xung ch cnU2=995 V. Hi sau bao lu gitthy ngn ri xung bn dng?1.3 inth-Hiuinth-Bitonvtdichuyntrongintrng 191.3.3 Vectorvntccaintchvunggcvi ngscinBi 1.74Mtelectroncbnvi vntcu2 106m/svomtintrngutheophngvunggcvi ngscin. Cngintrngl100 V/m. Tnhvntccaelectronkhi nchuynngc 107strong in trng. Electron c in tch l 1,6 1019C v khi lng l9,1 1031kg.Bi 1.75Mt electron c bn vi vn tc u 4 107m/s vo mt in trng utheophngvunggcvi ccngscin. Cngintrngl103V/m. Tnh:a. Gia tc ca electron.b. Vntccaelectronkhi nchuynngc2 107strongintrng.Bi 1.76Cho hai bn kim loi phng c di l=5 cm t nm ngang song songvinhau,cchnhaumtkhongd=2 cm.Hiuinthgiahaibnl910 V.Mtelectronbaytheophngngangvogiahaibnvivntcbanuv0=5 107m/s.Bitelectronrakhicintrng.Bquatc dng ca trng trng.a. Vit phng trnh qu o ca e trong in trng.b. Tnh thi gian electron i trong in trng? Vn tc ca n ti imbt u ra khi in trng?c. Tnhlchcaelectronkhi phngbanukhi rakhi intrng?Bi 1.77Mtelectronbaytrongintrnggiahaibncamttintchin v t cch nhau2 cm vi vn tc3 107m/s theo phng song songviccbncatin.Hiuinthgiahaibnphilbaonhiuelectron lch i2,5 mm khi i c on ng5 cm trong in trng.Bi 1.78SaukhictngtcbihiuinthU=200 V,mtintbayvochnh gia hai bn t theo phng song song hai bn. Hai bn c chiu dil =10 cmvcchnhaud=1 cm.Tmhiuinthgiagiahaibn20 Chng1 intch-intrngin t khng ra khi uc t?Bi 1.79Mt electron c ng nng 11,375 eV bt u vo in trng u nm giahai bn theo phng vung gc vi ng sc v cch u hai bn.a. Tnh vn tcv0ca electron lc bt u vo in trng?b. Thi gian i ht qung ngl = 5 cm ca bn.c. dchtheophngthngngkhi electronrakhi intrng,bit hiu in th v khong cch gia hai bn ln lt lU= 50 V,d = 10 cm.d. ng nng v vn tc electron ti cui bn.Bi 1.80in t mang nng lng1500 eV bay vo t phng theo hng song songhai bn. Hai bn dil = 5 cm, cch nhaud = 1 cm.Tnh hiu in th giahai bn in t bay ra khi t theo phng hp cc bn gc11.1.3.4 Vectorvntccaintchxingcvi ngscinBi 1.81Hai bn kim loi ni vi ngun in khng i c hiu in thU= 228 V.Ht electron c vn tc ban u v= 4 107m/s, bay vo khong khng giangiahaibnqualnhObndng,theophnghpvibndnggc = 60.a. Tm qu o ca electron sau .b. Tnhkhongcchgnbnmnhtmelectrontti,bquatc dng ca trng lc.Bi 1.82Hai bn kim loi tch in tri du t cch nhaud = 3 cm, chiu di mibn l = 5 cm. Mt in t lt vo gia hai bn hp bn dng gc 30. Xcnh Usao cho khi chui ra khi bn in t chuyn ng theo phng songsong vi hai bn?1.4 Tin1.4.1 Tnhtoncci lng1.4 Tin 21Bi 1.83T in phng gm hai bn t c din tch0,05 m2t cch nhau0,5 mm,in dung ca t l3 nF. Tnh hng s in mi ca lp in mi gia haibn t.Bi 1.84Mt t in khng kh nu c tch in lng 5,2 109C th in trnggia hai bn t l20 000 V/m. Tnh din tch mi bn t.Bi 1.85Mttinphngindung12 pF,inmilkhngkh.Khongcchgia hai bn t 0,5 cm. Tch in cho t in di hiu in th 20 V. Tnh:a. in tch ca t in.b. cng in trng trong t.Bi 1.86Mt t in phng khng kh, in dung 40 pF, tch in cho t in hiuin th120 V.a. Tnh in tch ca t.b. Sau tho b ngun in ri tng khong cch gia hai bn t lngp i. Tnh hiu in th mi gia hai bn t. Bit rng in dungca t in phng t l nghch vi khong cch gia hai bn ca n.Bi 1.87T in phng khng kh c in dungC= 500 pF c tch in n hiuin th300 V.a. Tnh in tchQ ca t in.b. Ngt t in khi ngun ri nhng t in vo cht in mi lng chng s in mi = 2. Tnh in dung C1, in tch Q1 v hiu inthU1ca t in lc .c. Vnni tinvi ngunnhngnhngtinvochtinmilng c hng s in mi =2. Tnh in dungC2, in tchQ2vhiu in thU2ca t in lc .Bi 1.88Tinphngkhngkh indung2 pFctchinhiuinth600 V.a. Tnh in tchQ ca t.b. Ngttkhingun,ahaiutraxakhongcchtnggpi. Tnh in dungC1, in tchQ1 v hiu in thU1 ca t in22 Chng1 intch-intrnglc .c. Vn ni t vi ngun, a hai bn t ra xa khong cch tng gpi. Tnh in dungC2, in tchQ2 v hiu in thU2 ca t inlc .Bi 1.89Tinphngcccbnthnhtrnbnknh10 cm. Khongcchvhiu in th gia hai bn ld=1 cm,U=108 V. Gia hai bn l khngkh. Tm in tch ca t in?Bi 1.90T in phng gm hai bn t hnh vung ccha=20 cm t cch nhau1 cm. Cht in mi gia hai bn l thy tinh c hng s in mi =6.Hiu in th gia hai bnU= 50 V.a. Tnh in dung ca t in.b. Tnh in tch ca t in.c. Tnh nng lng ca t in, t in c dng lm ngun in ckhng?Bi 1.91T in cu to bi qu cu bn knh R1 v v cu bn knh 4R2 (R1< R2).Tnh in dung ca qu cu ny?1.4.2 GhptchatchinBi 1.92Mt t in phng in dungC=0,12 F c lp in mi dy0,2 mm chng s in mi = 5. T c t di mt hiu in thU= 100 V.a. Tnh din tch cc bn ca t in, in tch v nng lng ca t.b. Sau khi c tch in, ngt t khi ngun ri mc vo hai bn catinC1=0,15 Fchactchin.Tnhintchcabtin, hiu in th v nng lng ca b t.Bi 1.93Mt t in6 F c tch in di mt hiu in th12 V.a. Tnh in tch ca mi bn t.b. Hi t in tch ly mt nng lng cc i l bao nhiu?c. Tnh cng trung bnh m ngun in thc hin a 1 electron tbn mang in tch dng sang bn mang in tch m?1.4 Tin 23Bi 1.94Tnh in dung tng ng, in tch, hiu in th trong mi t in cc trng hp sau (hnh v):BAC1C2C3Hnh 1AC1C2C3BHnh 2C2C3C1A BHnh 3AC1C2C3BHnh 4Hnh 1:C1= 2 F,C2= 4 F,C3= 2 F vUAB= 100 V.Hnh 2:C1= 1 F,C2= 1,5 F,C3= 3 F vUAB= 120 V.Hnh 3:C1= 0,25 F,C2= 1 F,C3= 3 F vUAB= 12 V.Hnh 4:C1= C2= 2 F,C3= 1 F vUAB= 10 V.Bi 1.95C3tinC1=10 F, C2=5 F, C3= 4 F c mc vo ngunin c hiu in thU= 38 V.a. TnhindungCcabtin, in tch v hiu in thtrn cc bn ca t in.b. TC3b "nhthng". Tmin tch v hiu in th trntC1.UC2C1C3Bi 1.9624 Chng1 intch-intrngCho b t mc nh hnh v:C1=1 F, C2=3 F, C3=6 F, C4=4 F,UAB=20 V. Tnhindungca b t, in tch v hiu in thmi t khi:a. khaKm.b. khaKng.A BC1C2C3C4KBi 1.97Trong hnh bn: C1=3 F, C2=6 F, C3=C4=4 F, C5=8 F,U= 900 V. Tnh hiu in th giaA vB.+ C5C1AC2C3BC4UBi 1.98Cho mch in nh hnh v:C1=C2=C3=C4=C5=1 F,U=15 V.Tnh in dung ca b t, in tch v hiu in th hai u mi t khi:a. Kh.b. Kng.KC1C2C5 +UC3C41.4 Tin 25Bi 1.99Cho mch in nh hnh v:C2= 2C1,UAB= 16 V. TnhUMB.BC1MAC2C2C1C1Bi 1.100Cho 4 t in ging nhau c cng in dungC.a. KhaK v tr no th in dung ca b t in ln hn?b. Nu in dung ca cc t l khc nhau, in dung ca b t tronghai v tr ca khaKl nh nhau th cn c h thc lin h g?KAC4+C1C2C3B UBi 1.101Tnh in dung ca b t in, in tch v hiu in th hai u mi ttrong cc trng hp sau:a. C1= 2 F,C2= 4 F,C3= 6 F,U= 100 V.C3+C1C2Ub. C1= 1 F,C2= 1,5 F,C3= 3 F,U= 120 V.+C1C2C3c. C1= 0,25 F,C2= 1 F,C3= 3 F,U= 12 V.26 Chng1 intch-intrng+C2C3C1d. C1= C2= 2 F,C3= 1 F,U= 10 V.+C1C2C3Bi 1.102Hai t in khng kh phng c in dung lC1=0,2 F vC2=0,2 Fmc song song. B t c tch in n hiu in thU= 450 V ri ngtkhi ngun. Sau lp y khong gia hai bn t inC2 bng in mic hng s in mi l = 2. Tnh in th ca b t v in tch ca mit?Bi 1.103Hai t in phng c C1= 2C2, mc ni tip vo ngun U khng i. Cng in trng trong C1 thay i bao nhiu ln nu nhng C2 vo cht inmi c hng s in mi = 2.Bi 1.104Batmkimloi phnggingnhautsongsongvi nhaunhhnhv.Din tch ca mi bn lS= 100 cm2. Khong cch gia hai bn lin tipld = 0,5 cm. NiA vBvi ngunU= 100 V.a. Tnhindungcabtvin tch ca mi bn.b. Ngt Av Bra khi ngunin.DchchuynbnBtheophng vung gc vi cc bntinmt onl x. Tnhhiu in th gia A v B theox. p dng khix =d2.A B1.4 Tin 27Bi 1.105Bn tmkimloi phng gingnhau nh hnh v. Khong cchBD=2AB=2DE. BvDcni vi nguninU =12 V, sau ngt ngun i. Tm hiu in thgiaBvD nu sau :a. NiA viB.b. Khng ni A vi Bnhng lpy khong giaBvD bnginmi chngsinmi = 3.ABDE+Bi 1.106Tinphngkhngkh C=2 pF.Nhngchmmtnavotronginmi lng = 3. Tm in dung ca t in nu khi nhng, cc bn t:a. Thng ng.b. Nm ngang.Bi 1.107Cho b t in mc nh hnh v bn.Chng minh rng nu c:C1C2=C3C4Th khi K ng hay K m, in dungca b t u khng thay i.AC1MC2BC3NC4KBi 1.108Mt t in phng c in dung C0. Tm in dung ca t in khi a vobn trong t mt tm in mi c hng s in mi , c din tch i dinbng mt na din tch mt tm, c chiu dy bng mt phn ba khongcch hai tm t, c b rng bng b rng tm t, trong hai trng hp sau:28 Chng1 intch-intrngA B(a)A B(b)1.4.3 GhpttchinBi 1.109em tch in cho t inC1= 3 F n hiu in thU1= 300 V, cho tinC2= 2 F n hiu in thU2= 220 V ri:a. Ni cc tm tch in cng du vi nhau.b. Ni cc tm tch in khc du vi nhau.c. Mc ni tip hai t in (hai bn m c ni vi nhau) ri mc vohiu in thU= 400 V.Tm in tch v hiu in th ca mi t trong trong trng hp trn.Bi 1.110em tch in cho t inC1=1 F n hiu in thU1=20 V, cho tinC2=2 F n hiu in thU2=9 V. Sau ni hai bn m hai tvi nhau, 2 bn dng ni vi hai bn ca tC3= 3 F cha tch in.a. Tnh in tch v hiu in th mi bn sau khi ni?b. Xc nh chiu v s electron di chuyn qua dy ni hai bn m haitC1vC2?1.4.4 Hiuinthgii hnBi 1.111Hai bncamttinphngcdnghnhtrnbnknhR=30 cm,khong cch gia hai bn ld = 5 mm, gia hai bn l khng kh.a. Tnh in dung ca t.b. Bit rng khng kh ch cch in khi cng in trng ti a lEgh= 3 105V/m. Hi:1.4 Tin 29Hiu in th gii hn ca t in.C th tch cho t in mt in tch ln nht l bao nhiu t khng b nh thng?Bi 1.112Hai tincindunglnltC1=5 1010FvC2=15 1010F,c mc ni tip vi nhau. Khong cch gia hai bn ca mi t in ld=2 mm. intrnggii hncami tEgh=1800 V/m. Tnhhiuin th gii hn ca b t.Bi 1.113BatincindungC1=2 nF; C2=4 nF; C3=6 nFcmcnitip thnh b. Hiu in th nh thng ca mi t in l4000 V. Hi btintrncthchuchiuinthU=11 000 Vkhng?Khihiu in th t trn mi t l bao nhiu?Bi 1.114Mt b t gm 5 t in ging ht nhau ni tip mi t c C= 10 F cni vo hiu in th100 V.a. Hi nng lng ca b thay i ra sao nu 1 t b nh thng.b. Khittrnbnhthngthnnglngcabtbtiuhaodophng in. Tm nng lng tiu hao .Bi 1.115Hai t cC1=5 F, C2=10 F; Ugh1=500 V, Ugh2=1000 V. Ghp hait in thnh b. Tm hiu in th gii hn ca b t in nu hai t:a. Ghp song song.b. Ghp ni tip.1.4.5 Tcchangun-TxoayBi 1.116Chomchnhhnhv. Bit C1=2 F, C2=10 F, C3=5 F; U1=18 V, U2=10 V. Tnhintchvhiu in th trn mi t?+C1C2+ U1U2NC3M30 Chng1 intch-intrngBi 1.117Chomachnhhnhv. Bit U1=12 V, U2=24 V; C1=1 F, C2=3 F. Lc u khoKm.a. Tnh in tch v hiu in thtrn mi t?b. Kho Kng li. Tnhinlng qua khoK.+C1C2 +U1U2NMKBi 1.118Cho mach nh hnh v. Bit UAB= 24 V, C1= 1 F, C2= 3 F, C3= 4 F,C4= 2 F.a. Tnh in tch cc t khiKm.b. KhaKng li, tnh in lng qua khaK.AC1C2BC3C4KBi 1.119Cho mt s t in ging nhau c in dung lC0=3 F. Nu cch mcdng t nht cc t in trn mc thnh b t c in dung l C= 5 F.V s cch mc ny?Bi 1.120Cho b t nh hnh v. Tnh in dung ca b t, hiu in th gia hai bnt in v in tch ca cc t. Cho bit: C1= C3= C5= 1 F; C2= 4 F;C4= 1,2 F vU= 30 V.1.4 Tin 31+C4C2C1C3C5Bi 1.121Cho b t in nh hnh v sau y:C2= 2C1;UAB= 24 V. TnhUMB.AC2C2MBC1C1C1Bi 1.122Cho mch t nh hnh, bit:C1= 6 F,C2= 4 F,C3= 8 F,C4= 5 F,C5= 2 F. Hy tnh in dung ca b t.AC1C2BC3C4C5Bi 1.123T xoay gmn tm hnh bn nguyt ng knhD= 12 cm, khong cchgia hai tm lin tipD= 0,5 mm. Phn i din gia hai bn c nh vbn di chuyn c dng hnh qut vi gc tm l0< < 180.a. Bit in dung cc i ca t l1500 nF. Gi tr can l bao nhiu?32 Chng1 intch-intrngb. TnivihiuinthU=500 Vvvtrgc=120.Tnhin tch ca t?c. Sau ngt t v iu chnh. Xc nh c s phng in giahai bn. BitEgh= 3 106V/m.Bi 1.124T xoay c Cmax= 490 pF v in dung cc tiu Cmin= 10 pF ng vi gc20 c to bin = 10 l kim loi hnh bn nguyt gn vo trc chung iqua tm ng trn v lt vo gia 11 l c nh c cng kch thc.a. in mi l khng kh, khong cchd gia mt bn c nh v bngn n nht l0,5 mm. Hy tnh bn knhR ca mi bn?b. Tnh in dung ca t xoay khi cho cc l chuyn ng quay mt gc k t v tr ng gi tr cc iCmax?c. t C v tr ng gi tr cc i Cmax v t hiu in th U= 60 Vvo hai cc b t. Sau b ngun i v xoay cc l chuyn ng mtgc . Xc nh hiu in th ca t theo , xt trng hp = 60?1.4.6 MchcutBi 1.125Cho b t in mc nh hnh v bn.Chng minh rng nu c:C1C2=C3C4Th khi K ng hay K m, in dungca b t u khng thay i.AC1MC2BC3NC4KBi 1.126Cho mch t nh hnh, bit:C1= 6 F,C2= 4 F,C3= 8 F,C4= 6 F,C5= 2 F. Hy tnh in dung ca b t.1.4 Tin 33AC1C2BC3C4C5Bi 1.127Cho mch t nh hnh, bit:C1= 6 F,C2= 4 F,C3= 8 F,C4= 5 F,C5=2 F. Hy tnh in dung ca b t v hiu in th hai u mi tnu t vo hai uA, Bmt hiu in thUAB= 12 VAC1C2BC3C4C5Bi 1.128Cho mch t nh hnh, bit:C1= 6 F,C2= 6 F,C3= 2 F,C4= 4 F,C5=4 F. Hy tnh in dung ca b t v hiu in th hai u mi tnu t vo hai uA, Bmt hiu in thUAB= 20 VAC1C2BC3C4C534 Chng1 intch-intrng1.4.7 NnglngintrngBi 1.129T phng khng kh tch in ri ngt khi ngun. Hi nng lng ca tthay i th no khi nhng t vo in mi c = 2.Bi 1.130Mt t in c in dung C1= 2 F , khong cch gia hai bn l d1= 5 cmc np in n hiu in thU= 100 V.a. Tnh nng lng ca t in.b. Ngt t ra khi ngun in. Tnh bin thin nng lng ca t khidch hai bn gn li cn cch nhaud2= 1 cm.Bi 1.131T phng cS= 200 cm2, in mi l bn thy tinh dyd = 1 mm, = 5,tchindihiuinthU=300 V.Rtbnthytinhkhit.Tnh bin thin nng lng ca t v cng cn thc hin. Cng ny dng lm g? Xt trong cc trng hp:a. T c ngt khi ngun.b. T vn ni vi ngun.Bi 1.132Hai t in phng khng kh ging nhau c in dungCmc song song vc tch n hiu in thUri ngt khi ngun. Hai bn ca mt t cnh, cn hai bn ca t kia c th chuyn ng t do. Tm vn tc ca ccbn t do ti thi im m khong cch gia chng gim i mt na. Bitkhi lng ca mi bn t lM, b qua tc dng ca trng lc.Bi 1.133Tphngkhngkh C=1010F,ctchinnhiuinthU=100 Vringtkhingun.Tnhcngcnthchintngkhongcchhai bn t ln gp i?Bi 1.134T phng khng khC= 6 106F c tch nU= 600 V ri ngt khingun.a. Nhngtvochtinmi c=4ngp2/3dintchmi bn.Tnh hiu in th ca t?b. Tnh cng cn thit nhc t in ra khi in mi. B qua trnglng t?1.5 pschng1 35Bi 1.135Mt t in phng m in mi c = 2 mc vo ngun in c hiu inthU= 100 V; khong cch gia hai bn ld = 0,5 cm; din tch mt bnl25 cm2.a. Tnh mt nng lng in trng trong t (0,707J/m3)b. Saukhi ngttrakhi ngun,intchcatinphngqualpin mi gia hai bn t n lc in tch ca t bng khng. Tnhnhit lng to ra in mi (4,42.10-8J)Bi 1.136T phngkhng khc dintchi dingiahai bnlS,khongcchhai bn lx, ni vi ngun c hiu in thUkhng i.a. Nng lng t thay i th no khix tng.b. Bit vn tc cc bn tch xa nhau lv. Tnh cng sut cn tchcc bn theox.c. Cng cn thit v bin thin nng lng ca t bin thnh dngnng lng no?1.5 pschng11.14,5 105N1.2 30 cm1.3 = 2;14,14 cm1.49 108N;0,7 1016Hz1.50,614 N1.61.7q1=q2=108Choc q1=q2= 108C; Gim 3ln, r

5,77 cm1.8 |q1| = |q2| =3 107C;tng2ln;rkk= rm. 35,36 cm1.9q1= 106C, q2=5 106Choc ngc li.1.10667 nC;0,0399 m1.11q1=2 105C, q2=105Choc ngc li.1.12q1= 2 109C, q2=6 109C hoc ngc li.q1= 2 109C, q2= 6 109Choc ngc li.1.13q= 3,33 C1.14q2= 3,33 C36 Chng1 intch-intrng1.15 = 1,8;r = 1,3 cm1.16F0= 1,5 N;F0= 0,79 N1.17F0 5,23 103N1.18F0 0,051 N1.191.201.211.221.23F= 7,2 105N1.24CA = 8 cm, CB = 16 cm;q0= 8 108C1.25CA = 4 cm, CB = 12 cm;q0= 4,5 108C1.26q= l

mgk= 106C1.271.281.291.30q=

ma3gk

3(3l2a2)1.31q= l

ma3gk61.321.33EM= 16 V/m;F0= 0,16 N1.34E= 2,5 106V/m1.35EH= 72 103V/m; EM=32 103V/m;EN= 9000 V/m1.361.37EM=2kqa(a2+x2)3/2; Emax=2kqa21.38EM=2kqh(h2+x2)3/2, Emax=4kq33a21.39ES=kq6a21.40EO=16kq633a21.41r1= 24 cm,r2= 12 cm1.42r1= r2= 6 cm1.43r1= 3 cm,r2= 4 cm1.44q4= 4 107C1.45q2= 22q1.46q1= 4 109C1.47q1= 2,7 108C, q2=6,4 108C1.48E= 105V/m,1.5 pschng1 371.49m = 0,2 mg1.50q= 2 109C1.51 Hng t Asang B, E =4,5 104V/m1.52 in trng u,

E vung gcvi mt phng, E =20; q =20mg tan 1.53 Giahaimt: E=0,ngoihai mt E=0; Tronggc: E=0sin 2, ngoi gc:E=0cos 21.54= 0(E1+E2); F=02(E21+E22)1.55E=20r1.56

Ec phng vung gc vimt phng cha hai dy v c lnE=0.hh2+a24; Emax=0a khih =a21.57UAC=200 V, UCB=0 V,UAB= 200 V;A = 3,2 1017J1.58UAC=0 V, UBA=120 V,E= 4000 V/m;E= 5000 V/m1.59AMN= 8 107J; ANP=5,12 107J APM= 2,88 107J;AMNPM= 0 J1.60AAB= 25 105J; AAB=25 105J1.61VB= 2000 V, VC=2000 V1.62A = 1,6 1018J1.63UMN= 250 V1.64v= 0,8 m/s1.65v= 3,04 106m/s1.661.67U 1821.681.69a = 2,2 1014m/s2, s =2 cm1.70E = 284 105V/m, a =5 107m/s21.71d = 0,08 m,t = 0,1 s1.72a=1,05 1016m/s2; t=3 ns;v= 3,2 107m/s1.731.74F = 1,6 1017N, a =1,76 1013m/s2, vy=1,76 106m/s,v= 2,66 106m/s1.75a = 3,52 1014m/s2, v =8,1 107m/s38 Chng1 intch-intrng1.76y=0.64x2; t =107s; =0,4 cm1.771.78U 2 V1.791.80U= 120 V1.811.82U= 47,9 V1.83 = 3.41.84S= 0,03 m21.85Q = 24 1011C; E =4000 V/m1.86Q = 48 1010C;U

= 240 V1.87Q=150 nC; C1=1000 pF,Q1=150 nC, U1=150 V; C2=1000 pF,Q2= 300 nC,U1= 300 V1.88Q = 1,2 109nC;C1= 1 pF,Q1=1,2 109nC, U1=1200 V;C2= 1 pF,Q2= 0,6 109nC,U1=600 V1.89Q = 3 109C1.90C=212,4 pF; Q=10,6 nC;W= 266 nJ1.91C=R1R2R1 +R21.92S= 0,54 m2, Q = 12 C, W=0,6 mJ; Qb=12 C, Ub=44,4 V,Wb= 0,6 mJ1.93Q = 7,2 105C; W =4,32 104J;W= 9,6 1019J1.941.95Cb= 3,16 F, Q1=8 105C, Q2=4 105C, Q3=1,2 105C, U1=U2=8 V, U3=30 V;Q1= 3,8 104C,U1= 38 V1.961.971.981.994 V1.100CA=43CB;C4=C1C2C1 +C21.101C=12 F, U1=U2=U3=100 V,Q1=2 104C, Q2=4 104C,Q3= 6 104CC=0,5 F, U1=60 V, U2=40 V,U3= 20 V, Q1= Q2= Q3=6 105CC=0,5 F, U1=12 V, U2=9 V,U3=3 V, Q1=3 106C, Q2=Q3= 9 106CC = 1,2 F, U1= 6 V, U2=U3=4 V, Q1=1,2 105C, Q2=8 106C,Q2= 4 106C1.102270 V, Q1= 5,4 105C,Q2= 2,16 105C1.103 Tng 1,5 ln1.5 pschng1 391.104C = 3,54 1011F, Q1=1,77 109C, Q2= 3,54 109C;U

= U.d2x2d2,75 V1.1058 V,6 V1.1064 pF;3 pF1.1071.1081.1091.1101.1111.112Ugh= 4,8 V1.113 Khng. B s b nh thng;U1=6000 V, U2=3000 V, U3=2000 V1.1141.1151.1161.1171.1181.1191.1201.1211.1221.123n=16; Q=5 107C;