Lect07 - Physics 212 Le cture7 Today's C pt: once C...

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Unformatted text preview: Physics 212 Le cture7 Today's C pt: once C onductors and C apacitance Exam1 in 8 days !! Go to Grade book and chooseC onflict Exam(5:15) if de d sire 50 40 30 20 10 0 Confused Avg = 3.0 Confident Physics 212 Le cture7, S 1 lide Music Who is the Artist? A) B) C) D) E) Miles Davis Wynton Marsalis Chris Botti Nina Simone Chet Baker Why? I played this last night working on the lecture DVD Must-See: Twilight of his career/life Twilight Van Morrison does “Send in the Clowns” Van Physics 212 Lecture 7, Slide 2 Physics 212 Le cture7 Today's C pt: once C onductors and C apacitance Exam1 in 8 days !! Go to Grade book and chooseC onflict Exam(5:15) if de d sire 50 40 30 20 10 0 Confused Avg = 3.0 Confident Physics 212 Le cture7, S 3 lide S m of your que um ary stions: 1) Adding conductor be e theplate of a capacitor twe n s This was tough.. Wewill do som work on it today e 2) Capacitors and Calculations What is a Capacitor? Calculation of C apacitance? What Wewill calculatethecapacitanceof a cylindrical capacitor We Q C≡ V Physics 212 Le cture7, S 4 lide Conductors You did we on theque ll stions on chargedistributions on conductors TheMain Points • • • • C harge fre to m se ove E = 0 in a conductor S urface= Equipote ntial E at surfacepe ndicular to surface rpe 5 Physics 212 Le cture7, S 5 lide Pre flight 2 80 60 40 20 “We know that the potential is inversely We proportional to radius. since conductor A's radius is bigger than that of conductor B's, the potential of A will be smaller than the potential of B. “ potential 0 6 Physics 212 Le cture7, S 6 lide Pre flight 4 100 80 60 40 20 0 “The charge will flow and over time the The two spheres will will act as 1 conductor and have the same potential. “ and 7 Physics 212 Le cture7, S 7 lide Pre flight 6 80 60 “Charge flows from higher potential to lower Charge potential so charge will flow towards A. “ potential 40 20 0 8 Physics 212 Le cture7, S 8 lide Pre flight 8 AAARRGGHHH Poorly state !! d n ove ar re s ove Whe • Whe +q m d ne to conducting sphe , charge will m How? • How? gativecharge in conductor will m toward +q s ove Ne • Ne What te ine • What de rm s thefinal distribution? ust qual to ze in conductor ro t otal • total E m bee rs…. and • Cand D arenon-starte tte al and • A and B be r, but can’t bere ost gativecharge( 1023) still insidesphe re • m ne Physics 212 Le cture7, S 9 lide xce ust • e ss chargem beon surface Pre flight 10 BB BB THI SONE WASTOUGH THE PLAN: e • I ’ll show you som e xplanations ’ll • We vote ’ll e We • We do som work ’ll We • We voteagain “Since the distance between plates decreases and the potential remains the same, then the Since charge increases, so Q1>Q0. Therefore, since the potential remains the same but the charge increases, the capacitance also increases, so C1>C0. “ increases, “Capacitance stays the same because it just depends on the area of the plates and the Capacitance distance in between them. “ distance “The insertion of the conducting plate cancels the electric field in the region of the plate and The the e-field travels a lesser distant; therefore, the capacitance of c1 is lesser than that of c0.” the Physics 212 Le cture7, S 10 lide 10 Capacitance C apacitanceis de d for any pair of spatially se fine parate conductors d Q C≡ V How do weunde rstand this de finition ??? r xce • Conside two conductors, onewith e ss charge= +Q and theothe with e ss charge= -Q r xce and +Q d -Q se s ate le ld twe n m The • The charge cre an e ctric fie in thespacebe e the gratethee ctric fie be e the to find thepote le ld twe n m ntial diffe ncebe e the re twe n • Wecan inte conductor conductor ntial diffe nceshould beproportional to Q re This • This pote ntial diffe nceis thecapacitanceand only de nds on the re pe • Theratio of Q to thepote ge e of theconductors om ge try 9 E V Physics 212 Le cture7, S 11 lide 11 Exam (donein Pre cture7) ple le Exam First de rm E fie produce by charge conductors: te ine ld d d y +Q d x E -Q σ E= ε0 What is σ ?? What ?? Q σ= A A = are of plate a S cond, inte e grateE to find thepote ntial diffe nceV re d d d Q V = − ∫ E ⋅ dy V = − ∫ (− Edy ) =E ∫ dy = d ε0 A 0 0 0 As prom d, V is proportional to Q !! ise 12 Q Q C≡ = V Qd /(ε 0 A) ε0 A C= d Cde rm d by ge e !! te ine om try Physics 212 Le cture7, S 12 lide 12 Que stion Re d to Pre late flight +Q I nitial chargeon capacitor = Q0 d BB 0 0 -Q +Q I nse uncharge conductor rt d C hargeon capacitor now = Q1 d 1 1 t How is Q1 re d to Q0 ?? late How A. B. C. 14 -Q Plate not conne d to anything s cte Q1 < Q0 Q1 = Q0 Q1 > Q0 C HARGE C ANNOT CHANGE !! Physics 212 Le cture7, S 13 lide 13 Whe to S re tart?? +Q d 0 0 t BB -Q What is thetotal chargeinduce on thebottomsurfaceof theconductor? d A. B. C. D. E. +Q0 -Q0 0 Positivebut them agnitudeunknown Ne gativebut them agnitudeunknown 17 Physics 212 Le cture7, S 14 lide 14 WHY ?? WHY +Q 0 0 E -Q +Q +Q 0 E=0 E 0 WHAT DO WE KNOW ??? -QE must be= 0 in conductor !! C harge insideconductor m to cance E fie fromtop & bottomplate s ove l ld s 19 Physics 212 Le cture7, S 15 lide 15 C alculateV y Now calculateV as a function of distancefromthebottomV ( y ) = − E ⋅ dy ∫ conductor. conductor 0 d BB +Q +Q 0 y t E -E0 t y d E=0 V 0 21 What is ∆ V = V(d)? -Q A) ∆ V = E0d B) ∆ V = E0(d – t) C ∆ V = E0(d + t) ) y The integral = area under the curve Physics 212 Le cture7, S 16 lide 16 Back to Pre flight 10 BB What havewele d? arne C = Q0/V0 = ε 0A/d 0 If charge arethesam s e Q1 = Q0 V0 = E0d V1 = E0(d – t) 40 What do the re se sults te us about how ll What C1 is re d to C ?? 0 is late 30 Preflight Results 20 10 0 Physics 212 Le cture7, S 17 lide 17 Back to Pre flight 10 BB Same V: What is Q1 interms of Q0? Leave as exercise! Wecan de rm Cfrome r case te ine ithe sam V (pre e flight) sam Q (le e cture ) Cde nds only on ge e !! pe om try S eQ: am V0 = E0d V1 = E0(d – t) C0 = Q0/E0d C1 = Q0/(E0(d – t)) E0 = Q0/ ε 0A C0 = ε 0A/d C1 = ε 0A/(d – t) Physics 212 Le cture7, S 18 lide 18 Ene in Capacitors rgy BANG 31 Physics 212 Le cture7, S 19 lide 19 cross-section cross-section a a a a C alculation A capacitor is constructe from two conducting cylindrical d she of radii a , a , a , and a and le lls ngth L (L >> a). m tal e m tal e once • C ptual Analysis: Q C≡ V But what is Q and what is V? They are not given?? portant Point: Cis a prope of theobje (conce rty ct!! ntric cylinde he ) rs re • Im ee ., r) • Assum som Q (i.e +Q on oneconductor and –Q on theothe se s ate ld gion be e conductors twe n • The charge cre E f ie in re ld te ine ntial diffe nceV be e theconductors re twe n • This E f ie de rm s a pote . • V should beproportional to Q; theratio Q/V is thecapacitance 33 33 Physics 212 Lecture 7, Slide 20 20 cross-section a a a a C alculation A capacitor is constructe from two conducting cylindrical d she of radii a , a , a , and a and le lls ngth L (L >> a). m tal e C≡ m tal e Q V trate • S gic Analysis: – – – – Put +Q on oute she and –Q on inne she r ll r ll C ylindrical sym e UseGauss’ Law to calculateE e rywhe m try: ve re Inte grateE t o ge V t Takeratio Q/V: should ge e ssion only using ge e param te (ai, L) t xpre om tric e rs Physics 212 Le cture7, S 21 lide 21 cross-section + + + + m tal e C alculation A capacitor is constructe from two conducting cylindrical BB d she of radii a , a , a , and a and le lls ngth L (L >> a). + + + + + + -Q + + + m tal e Whe is +Q on oute conductor locate re r d? (A) at r=a4 (B) at r=a3 (C both surface (D) t hroughout she ) s ll (A) Why? Gauss’ law: Q dA ∫ Eg = ε Weknow that E = 0 in conductor (be e a3 and a4) twe n + + + a a + a a +Q + + C≡ Q V enclosed 0 Q enclosed =0 Q enclosed =0 +Q m beon insidesurface(a ), ust Physics 212 Le cture7, S 22 lide 22 cross-section + + + m tal e C alculation A capacitor is constructe from two conducting cylindrical BB d she of radii a , a , a , and a and le lls ngth L (L >> a). + + + -Q − − + + − − + − − metal − − + − + + +++ Whe is -Q on inne conductor locate re r d? (A) at r=a2 (B) at r=a1 (C both surface (D) t hroughout she ) s ll (A) Why? Gauss’ law: Q dA ∫ Eg = ε Weknow that E = 0 in conductor (be e a1 and a2) twe n − − − − − + + a a −−− − + a a +Q + C≡ Q V enclosed 0 Q enclosed =0 Q enclosed =0 -Q m beon outsidesurface(a ), ust Physics 212 Le cture7, S 23 lide 23 cross-section + + + m tal e C alculation A capacitor is constructe from two conducting cylindrical she BB d lls of radii a , a , a , and a and le ngth L (L >> a). + + -Q − − + + − − + − − metal − − + − + + +++ a2 < r < a3: What is E(r)? (A) 0 (A) Why? Gauss’ law: (B) (B) 1Q (C ) 4πε r 2 0 Q dA ∫ Eg = ε + − − − − − + + a a −−− − + a a +Q + C≡ Q V 1Q (D) Lr 2πε 0 1 2Q (E) 2πε Lr 0 1Q 4πε r 2 0 enclosed 0 Q dA ∫ Eg = ε Q E 2πrL = ε0 enclosed 0 E= 1Q 2πε Lr 0 Dire ction: Radially I n Physics 212 Le cture7, S 24 lide 24 cross-section + + + m tal e C alculation A capacitor is constructe from two conducting cylindrical she d lls of radii a , a , a , and a and le ngth L (L >> a). BB + + + -Q − − + + − − + − − metal − − + − + + +++ r < a2: E(r) = 0 sinceQenclosed = 0 • What is V? What ntial diffe nce re • Thepote be e theconductors twe be n What is thesign of V = V − − − − − + + a a −−− − + a a +Q + C≡ Q V a2 < r < a3: E = 1Q 2πε Lr 0 Q (2πε 0a2L) -V ? Physics 212 Le cture7, S 25 lide 25 cross-section + + + m tal e 1 4 C alculation A capacitor is constructe from two conducting cylindrical she d lls of radii a , a , a , and a and le ngth L (L >> a). BB + + + -Q − − + + − − + − − metal − − + − + + +++ What is V ≡ V V= Q a ln 2πε L a 0 -V V= Q (2πε 0a2L) − − − − − + + a a −−− − + a a +Q + C≡ Q V a2 < r < a3: E = 1Q 2πε r 0 ? Q a ln 2πε L a 0 4 V= 1 Q a ln 2πε L a 0 3 V= 2 Q a ln 2πε L a 0 2 3 −Q dr V = −∫ 2πε L r a3 a2 0 Q dr V= ∫ 2πε L r a3 0 a2 V= Q a ln 2πε L a 0 3 2 V proportional to Q, as prom d ise C≡ Q 2πε L = V ln(a / a ) 0 3 2 Physics 212 Le cture7, S 26 lide 26 ...
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This note was uploaded on 03/05/2011 for the course PHYS 212 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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