(Ch. 22.5) 2. What is the magnitude (in pc) of a point charge whose electric field 50 cm away has a magnitude of 2V/m?


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1 Em I Solutions PHY049 Summe 0 (Ch..5). Two smll, positively chged sphees hve combined chge of 50 μc. If ech sphee is epelled fom the othe by n electosttic foce of N when the sphees e.0 m pt, wht is the chge (in μc ) on the sphee with the smllest chge? (A).6 Suppose tht the chges on the two sphees e nd. Then nd F / mens tht At this stge, note tht if we define nd s espectively nd y μc, then we hve +y 50 nd y 4000/ Thee e sevel wys to solve these eutions, most diect being the substitution. The nswes fo (,y) e.6 μc nd 8.4 μc. (Ch..5). Wht is the mgnitude (in pc) of point chge whose electic field 50 cm wy hs mgnitude of V/m? (A) 55.6 ` E / which mens tht E / C (Ch..7). A poton is distnce d/ diectly bove the cente of sue of side d. Wht is the mgnitude of the electic flu (in nn.m /C) though the sue? (A) It would be sith of the totl flus though cube which encloses the poton. The nswe is /6ε o.009 Nm /C. (Ch. 4.4) 4. wht is the net electic potentil (in mv) t the oigin due to the cicul c of chge Q +7. pc nd the two pticles of chges Q 4.00Q nd Q .00Q? The c's cente of cuvtue is t the oigin nd its dius is R.00 m; the ngle indicted is θ 0.0. The potentil is ten to be zeo t infinity.
2 (A).5 The potentil is (Ch. 5.68) Q 4 V ( ) R + Q R 5. The cpcitnces of the fou cpcitos shown in Fig. 55 e given in tems of cetin untity C. In tio to C, wht is the euivlent cpcitnce between points A nd B? (Hint: Fist imgine tht bttey is connected between those two points; then educe the cicuit to n euivlent cpcitnce.). (A) 0.8 Hee 4C nd 6C e in seies ( C/5), the combintion is pllel to C ( C/5) nd tht combintion is in seies with C ( C/7). The euivlent cpcitnce tio / CQ 6. In the bove poblem, now conside bttey connected between points A nd D. Wht fction of the chge is stoed on the 4C cpcito? Epess you nswe s tio to the chge stoed on the C cpcito. (A) 0.7 Now C nd 6C e in seies, the combo is pllel to 4C etc. The euivlent cpcitnce between A nd D is C/. Tht mens tht the chge in C, Q CV/. Whee V volt bttey is ttched between A nd D. The potentil dop coss cpcito C is then V/ nd coss 4C then, it is V V/ V/. The chge on 4C is 8CV/. In tio to the chge on Q, tht would be 8/ 0.7. (e,sum0) 7. Two chges,  C nd  4 C, e plced long the is distnce pt with chge t the oigin nd t. A thid chge, +4/9 C, is lso plced long the is such tht thee is no net Coulomb foce on ny of the chges. Wht is the position of this chge long the is in units of, i.e., wht is /? (A) / The foce on chge consists of epulsion due to chge nd the foce due to the unnown chge in the middle. Chge must be of opposite sign to cncel the epulsive foce due to.
3 F One cn lso get simil conditions by putting the foces on chge nd to be zeo. They e F F 0 : 0 : + Notice tht the thee nswes fo / e the sme. This is gunteed by the selection of the thee chges. The chges hee e ll mgnitudes since thei signs e ledy used in detemining the cnceling diections of the foces. (s) 8. The figue shows unifomly chged, nonconducting spheicl shell of inne dius nd oute dius. If the electic field t the oute dius is E, wht is the electic field t point P with dius.5? (A) 0.6E It is unifomly chged nonconducting sphee. Hee the electic field cn be finite inside (in conducting sphee, the electic field vnishes nd the chges eside t the sufce). Suppose the totl chge is Q. The the field just outside the shell is E Q/(). The mount of chge enclosed by Gussin sufce of dius (between < < ) Q (  )/[() ]. The electic field t P is: E P EP E 4 7 Q (s0) 9. A nonunifom electic field given by ~E (5.5ˆi.ˆj + (4.6z )ˆ) N/C pieces cube 7
4 with sides m, s shown in the figue. The cube hs its e cone t the oigin. Wht is the totl chge inside the cube? (A) +. nc A cube hs sides with nomls long the Ctesin es. The totl flu being the sme on the sides with nomls long nd y (ecept fo thei signs) will vnish. Only the sides up nd down will contibute nd they give, Φ /ε o [4.6 ] >. nc. (e. f09) 0. In the cicuit shown ll cpcitos e 6.0μF nd the powe supply is V. The chge (in μc) on the cpcito lbeled is: (A) 9 Conside only the lowe bnch which is diectly connected to the bttey. It is pllel to the uppe bnch nd hs the potentil diffeence of V. In this bnch, cpcito is pllel to the two in seies nd the pllel combintion is in seies with nothe one. This lst one being the one net to the bttey will hve ll the chge of the euivlent cpcito of he lowe bnch, which is 8/5 μf. It s chge is theefoe 8/5 4. μc. The potentil diffeence coss its two pltes will be 4./6 7. V. The potentil diffeence coss the cpcito is then ( 7.) 4.8 V. Tht goes to tell you tht the chge in must be μc. (e.7, f0). A coppe wie nd nichome wie of the sme length nd cosssection e connected in seies coss lge bttey. If the esistivity of the coppe wie is Ω m nd the esistivity of the nichome wie is. 0 6 Ω m, wht is the powe dissipted in the coppe wie divided by the powe dissipted in the nichome wie? (A) 0.05 The wies e connected in seies nd hve the sme cuent. Powe being I R, nd R being popotionl to esistivity ρ, it follows tht if the e nd length e the sme, P coppe /P nichome
5 ρ coppe /ρ nichome (e.0, f09). In the multiloop cicuit shown the cuent though the.0ω esisto is (in ma), (A). Kichhoff's ules e designed to eep tc of ntul ccounting nd in fct led to n nswe tht is diffeent fom one you might guess. et's guess fist: Two btteies e the sme nd connected to 8 Ohms in seies (??). Hence the cuent in ech cicuit might be /4A nd the totl cuent is theefoe.5a. Not so s you see clely but why? Now the ight solution: Suppose tht the cuents in the left nd ight bnches e I nd I, both in ma nd going up. Though the middle bnch, it must be (I +I ) in ma nd going down. The loop on the left gives 6 6I (I +I ) 0. The loop on the ight gives 66I (I +I ) 0. Add the two eutions nd you get, 0(I +I ) 0 o (I +I ).A.
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