Page 1. Question 19.1b Electric Charge II Question 19.2a Conductors I. ConcepTest Clicker Questions Chapter 19. Physics, 4 th Edition James S.

Transcription

1 ConTst Clikr ustions Chtr 19 Physis, 4 th Eition Jms S. Wlkr ustion 19.1 Two hrg blls r rlling h othr s thy hng from th iling. Wht n you sy bout thir hrgs? Eltri Chrg I on is ositiv, th othr is ngtiv both r ositiv ) both r ngtiv ) both r ositiv or both r ngtiv Coyright 2010 Prson Eution, In. ustion 19.1 Eltri Chrg I ustion 19.1b Eltri Chrg II Two hrg blls r rlling h othr s thy hng from th iling. Wht n you sy bout thir hrgs? on is ositiv, th othr is ngtiv both r ositiv ) both r ngtiv ) both r ositiv or both r ngtiv From th itur, wht n you onlu bout th hrgs? hv oosit hrgs hv th sm hrg ) ll hv th sm hrg ) on bll must b nutrl (no hrg) Th ft tht th blls rl h othr n tll you only tht thy hv th sm hrg, but you o not know th sign. So thy n b ithr both ositiv or both ngtiv. Follow-u: Wht os th itur look lik if th two blls r oositly hrg? Wht bout if both blls r nutrl? ustion 19.1b Eltri Chrg II ustion 19.2 Conutors I From th itur, wht n you onlu bout th hrgs? hv oosit hrgs hv th sm hrg ) ll hv th sm hrg ) on bll must b nutrl (no hrg) A mtl bll hngs from th iling by n insulting thr. Th bll is ttrt to ositiv-hrg ro hl nr th bll. Th hrg of th bll must b: ositiv ngtiv ) nutrl ) ositiv or nutrl ) ngtiv or nutrl Th GREEN n PINK blls must hv th sm hrg, sin thy rl h othr. Th YELLOW bll lso rls th GREEN, so it must lso hv th sm hrg s th GREEN (n th PINK). Pg 1 1

2 ustion 19.2 A mtl bll hngs from th iling by n insulting thr. Th bll is ttrt to ositiv-hrg ro hl nr th bll. Th hrg of th bll must b: Conutors I ositiv ngtiv ) nutrl ) ositiv or nutrl ) ngtiv or nutrl ustion 19.2b Conutors II Two nutrl onutors r onnt by wir n hrg ro is brought nr, but os not touh. Th wir is tkn wy, n thn th hrg ro is rmov. Wht r th hrgs on th onutors? ) ) ) 0 0 Clrly, th bll will b ttrt if its hrg is ngtiv. Howvr, vn if th bll is nutrl, th hrgs in th bll n b srt by inution (olriztion), ling to nt ttrtion. Rmmbr th bll is onutor! Follow-u: Wht hns if th mtl bll is rl by lsti bll? 0 0?? ustion 19.2b Conutors II Two nutrl onutors r onnt by wir n hrg ro is brought nr, but os not touh. Th wir is tkn wy, n thn th hrg ro is rmov. Wht r th hrgs on th onutors? ) ) ) 0 0 ustion 19.3 Wht is th mgnitu of th for F 2? F 1 = 3 N F 2 =? Coulomb s Lw I 1.0 N 1.5 N ) 2.0 N ) 3.0 N ) 6.0 N Whil th onutors r onnt, ositiv hrg will flow from th blu to th grn bll u to olriztion. On isonnt, th hrgs will rmin on th srt onutors vn whn th ro is rmov. 0 0 Follow-u: Wht will hn whn th onutors r ronnt with wir??? ustion 19.3 Coulomb s Lw I ustion 19.3b Coulomb s Lw II Wht is th mgnitu of th for F 2? F 1 = 3 N F 2 =? 1.0 N 1.5 N ) 2.0 N ) 3.0 N ) 6.0 N F 1 = 3 N F 2 =? If w inrs on hrg to 4, wht is th mgnitu of F 1? F =? 1 4 F =? 2 3/4 N 3.0 N ) 12 N ) 16 N ) 48 N Th for F 2 must hv th sm mgnitu s F 1. This is u to th ft tht th form of Coulomb s lw is totlly symmtri with rst to th two hrgs involv. Th for of on on th othr of ir is th sm s th rvrs. Not tht this souns susiiously lik Nwton s 3r lw!! Pg 2 2

3 ustion 19.3b Coulomb s Lw II ustion 19.3 Coulomb s Lw III F 1 = 3 N F 2 =? If w inrs on hrg to 4, wht is th mgnitu of F 1? F =? 1 4 F =? 2 3/4 N 3.0 N ) 12 N ) 16 N ) 48 N Th for btwn two hrgs srt by istn is F. If th hrgs r ull rt to istn 3, wht is th for on h hrg? 9F 3F ) F ) 1/3F ) 1/9F Originlly w h: F 1 = k()()/r 2 = 3 N F F Now w hv: F 1 = k(4)()/r 2?? whih is 4 tims biggr thn bfor. Follow-u: Now wht is th mgnitu of F 2? 3 ustion 19.3 Coulomb s Lw III Th for btwn two hrgs 9F srt by istn is F. If 3F th hrgs r ull rt to ) F istn 3, wht is th for on ) 1/3F h hrg? ) 1/9F ustion 19.4 Two blls with hrgs n 4 r fix t srtion istn of. Is it ossibl to l nothr hrg bll 0 on th lin btwn th two hrgs suh tht th nt for on 0 will b zro? Eltri For I ys, but only if 0 is ositiv ys, but only if 0 is ngtiv ) ys, innnt of th sign (or vlu) of 0 ) no, th nt for n nvr b zro Originlly w h: F bfor = k()()/ 2 = F Now w hv: F ftr = k()()/(3) 2 = 1/9F? F F? 4 Follow-u: Wht is th for if th originl istn is hlv? 3 ustion 19.4 Two blls with hrgs n 4 r fix t srtion istn of. Is it ossibl to l nothr hrg bll 0 on th lin btwn th two hrgs suh tht th nt for on 0 will b zro? Eltri For I ys, but only if 0 is ositiv ys, but only if 0 is ngtiv ) ys, innnt of th sign (or vlu) of 0 ) no, th nt for n nvr b zro ustion 19.4b Eltri For II Two blls with hrgs n 4 r srt by. Whr shoul you l nothr hrg bll 0 on th lin btwn th two hrgs suh tht th nt for on 0 will b zro? A ositiv hrg woul b rll by both hrgs, so oint whr ths two rulsiv fors nl n b foun. A ngtiv hrg woul b ttrt by both, n th sm rgumnt hols. 4 b R 2R 4 Follow-u: Wht hns if both hrgs r? Whr woul th F = 0 oint b in this s? Pg 3 3

4 ustion 19.4b Two blls with hrgs n 4 r srt by. Whr shoul you l nothr hrg bll 0 on th lin btwn th two hrgs suh tht th nt for on 0 will b zro? b R Eltri For II 2R 4 ustion 19.4 Two blls with hrgs n 4 r fix t srtion istn of. Is it ossibl to l nothr hrg bll 0 nywhr on th lin suh tht th nt for on 0 will b zro? Eltri For III ys, but only if 0 is ositiv ys, but only if 0 is ngtiv ) ys, innnt of th sign (or vlu) of 0 ) no, th nt for n nvr b zro Th for on 0 u to is: F = k( 0 )()/R 2 Th for on 0 u to 4 is: F = k( 0 )(4)/(2R) 2 Sin 4 is 4 tims biggr thn, 0 ns to b frthr from 4. In ft, 0 must b twi s fr from 4, sin th istn is squr in Coulomb s lw. 4 ustion 19.4 Two blls with hrgs n 4 r fix t srtion istn of. Is it ossibl to l nothr hrg bll 0 nywhr on th lin suh tht th nt for on 0 will b zro? Eltri For III ys, but only if 0 is ositiv ys, but only if 0 is ngtiv ) ys, innnt of th sign (or vlu) of 0 ) no, th nt for n nvr b zro ustion 19.5 A roton n n ltron r hl rt istn of 1 m n thn rls. As thy roh h othr, wht hns to th for btwn thm? Proton n Eltron I it gts biggr it gts smllr ) it stys th sm A hrg (ositiv or ngtiv) n b l to th lft of th hrg, suh tht th rulsiv for from th hrg nls th ttrtiv 4 for from 4. Follow-u: Wht hns if on hrg is n th othr is? ustion 19.5 Proton n Eltron I ustion 19.5b Proton n Eltron II A roton n n ltron r hl rt istn of 1 m n thn rls. As thy roh h othr, wht hns to th for btwn thm? it gts biggr it gts smllr ) it stys th sm A roton n n ltron r hl rt istn of 1 m n thn rls. Whih rtil hs th lrgr lrtion t ny on momnt? roton ltron ) both th sm By Coulomb s lw, th for btwn th two hrgs is invrsly roortionl to th istn squr. So, th losr thy gt to h othr, th biggr th ltri for btwn thm gts! F = k r Follow-u: Whih rtil fls th lrgr for t ny on momnt? Pg 4 4

5 ustion 19.5b Proton n Eltron II A roton n n ltron r roton hl rt istn of 1 m ltron n thn rls. Whih ) both th sm rtil hs th lrgr lrtion t ny on momnt? ustion 19.5 A roton n n ltron r hl rt istn of 1 m n thn lt go. Whr woul thy mt? Proton n Eltron III in th mil losr to th ltron s si ) losr to th roton s si Th two rtils fl th sm for. Sin F = m, th rtil with th smllr mss will hv th lrgr lrtion. This is th ltron. F = k r ustion 19.5 Proton n Eltron III ustion 19.6 Fors in 2D b A roton n n ltron r hl rt istn of 1 m n thn lt go. Whr woul thy mt? in th mil losr to th ltron s si ) losr to th roton s si Whih of th rrows bst rrsnts th irtion of th nt for on hrg u to th othr two hrgs? 2 By Nwton s 3r lw, th ltron n roton fl th sm for. But, sin F = m, n sin th roton s mss is muh grtr, th roton s lrtion will b muh smllr! 4 Thus, thy will mt losr to th roton s originl osition. Follow-u: Whih rtil will b moving fstr whn thy mt? ustion 19.6 Fors in 2D Whih of th rrows bst rrsnts th irtion of th nt for on hrg 2 u to th othr two hrgs? b ustion 19.7 You r sitting rtin istn from oint hrg, n you msur n ltri fil of E 0. If th hrg is oubl n your istn from th hrg is lso oubl, wht is th ltri fil strngth now? Eltri Fil 4E 0 2E 0 ) E 0 ) 1/2E 0 ) 1/4E 0 Th hrg 2 rls towr th right. Th hrg 4 rls uwr, but with strongr for. Thrfor, th nt for is u n to th right, but mostly u. Follow-u: Wht woul hn if th yllow hrg wr 3? Pg 5 5

6 ustion 19.7 You r sitting rtin istn from oint hrg, n you msur n ltri fil of E 0. If th hrg is oubl n your istn from th hrg is lso oubl, wht is th ltri fil strngth now? Eltri Fil 4E 0 2E 0 ) E 0 ) 1/2E 0 ) 1/4E 0 ustion 19.8 Btwn th r n th blu hrg, whih of thm xrins th grtr ltri fil u to th grn hrg? Fil n For I 1 2 ) th sm for both Rmmbr tht th ltri fil is: E = k/r 2. Doubling th hrg uts ftor of 2 in th numrtor, but oubling th istn uts ftor of 4 in th nomintor, bus it is istn squr!! Ovrll, tht givs us ftor of 1/ Follow-u: If your istn is oubl, wht must you o to th hrg to mintin th sm E fil t your nw osition? ustion 19.8 Fil n For I ustion 19.8b Fil n For II Btwn th r n th blu hrg, whih of thm xrins th grtr ltri fil u to th grn hrg? 1 2 ) th sm for both Btwn th r n th blu hrg, whih of thm xrins th grtr ltri for u to th grn hrg? 1 2 ) th sm for both Both hrgs fl th sm ltri fil u to th grn hrg bus thy r t th sm oint in s! E = k 2 r ustion 19.8b Fil n For II ustion 19.9 Surosition I Btwn th r n th blu hrg, whih of thm xrins th grtr ltri for u to th grn hrg? 1 2 ) th sm for both Wht is th ltri fil t th ntr of th squr? ) E = 0 b Th ltri fil is th sm for both hrgs, but th for on givn hrg lso ns on th mgnitu of tht sifi hrg. F = qe Pg 6 6

7 ustion 19.9 Surosition I ustion 19.9b Surosition II Wht is th ltri fil t th ntr of th squr? b Wht is th ltri fil t th ntr of th squr? b ) E = 0 ) E = 0 For th ur hrg, th E fil vtor t th ntr of th squr oints towr tht hrg. For th lowr hrg, th sm thing is tru. Thn th vtor sum of ths two E fil vtors oints to th lft. Follow-u: Wht if th lowr hrg wr 2 C? Wht if both hrgs wr 2 C? ustion 19.9b Wht is th ltri fil t Surosition II b th ntr of th squr? ) E = 0 ustion 19.9 Wht is th irtion of th ltri fil t th osition of th X? Surosition III - b Th four E fil vtors ll oint outwr from th ntr of th squr towr thir rstiv hrgs. Bus thy r ll qul, th nt E fil is zro t th ntr!! Follow-u: Wht if th ur two hrgs wr 2 C? Wht if th right-hn hrgs wr 2 C? ustion 19.9 Surosition III - Wht is th irtion of th ltri fil t th osition of th X? b Th two hrgs giv rsultnt E fil tht is own n to th right. Th hrg hs n E fil u n to th lft, but smllr in mgnitu. Thrfor, th totl ltri fil is own n to th right. Follow-u: Wht if ll thr hrgs rvrs thir signs? ustion Two hrgs r fix long th x xis. Thy rou n ltri fil E irt long th ngtiv y xis t th init oint. Whih of th following is tru? Fin th Chrgs hrgs r qul n ositiv hrgs r qul n ngtiv ) hrgs r qul n oosit ) hrgs r qul, but sign is untrmin ) hrgs nnot b qul 1 y E 2 x Pg 7 7

8 ustion Fin th Chrgs ustion Uniform Eltri Fil Two hrgs r fix long th x xis. Thy rou n ltri fil E irt long th ngtiv y xis t th init oint. Whih of th following is tru? hrgs r qul n ositiv hrgs r qul n ngtiv ) hrgs r qul n oosit ) hrgs r qul, but sign is untrmin ) hrgs nnot b qul In uniform ltri fil in mty s, 4 C hrg is l n it fls n ltri for of 12 N. If this hrg is rmov n 6 C hrg is l t tht oint inst, wht for will it fl? 12 N 8 N ) 24 N ) no for ) 18 N Th wy to gt th rsultnt PINK vtor is to us th GREEN n BLUE vtors. Ths E vtors orrson to qul hrgs (bus th lngths r qul) tht r both ngtiv (bus thir y E irtions r towr th hrgs). 1 2 x Follow-u: How woul you gt th E fil to oint towr th right? ustion Uniform Eltri Fil ustion Eltri Fil Lins I In uniform ltri fil in mty 12 N s, 4 C hrg is l n it fls n ltri for of 12 N. If this hrg is rmov n 6 C hrg is l t tht oint inst, wht for will it fl? 8 N ) 24 N ) no for ) 18 N Wht r th signs of th hrgs whos ltri fils r shown t right? ) ) ) no wy to tll Sin th 4 C hrg fls for, thr must b n ltri fil rsnt, with mgnitu: E = F / q = 12 N / 4 C = 3 N/C On th 4 C hrg is rl with 6 C hrg, this nw hrg will fl for of: F = q E = (6 C)(3 N/C) = 18 N Follow-u: Wht if th hrg is l t iffrnt osition in th fil? ustion Eltri Fil Lins I ustion 19.12b Eltri Fil Lins II Wht r th signs of th hrgs whos ltri fils r shown t right? ) Whih of th hrgs hs th grtr mgnitu? ) ) both th sm ) no wy to tll Eltri fil lins origint on ositiv hrgs n trmint on ngtiv hrgs. Pg 8 8

9 ustion 19.12b Eltri Fil Lins II Whih of th hrgs hs th grtr mgnitu? ) both th sm Th fil lins r nsr roun th r hrg, so th r on hs th grtr mgnitu. Follow-u: Wht is th r/grn rtio of mgnitus for th two hrgs? Pg 9 9