Lect04 - Physics 212 Le cture4 Today's C pts once C onductors Using Gauss’ Law 50 40 30 20 10 0 Confused Avg = 2.9 Confident Physics 212 Le

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Unformatted text preview: Physics 212 Le cture4 Today's C pts: once C onductors + Using Gauss’ Law 50 40 30 20 10 0 Confused Avg = 2.9 Confident Physics 212 Le cture4, S 1 lide Music Who is the Artist? BB A) B) C) D) E) Diana Krall Nora Jones kd lang Madeline Peyroux Edith Piaf Great version of River Great (joni mitchell) with kd lang with WHY?? She’s singing at jazz Festival in Grant Park Friday night See you there????? Other great albums Physics 212 Lecture 4, Slide 2 Physics 212 Le cture4 Today's C pts: once C onductors + Using Gauss’ Law 50 40 30 20 10 0 Confused Avg = 2.9 Confident Physics 212 Le cture4, S 3 lide Ove w rvie • Pre cture4 le te ine lds m trical chargedistributions – UseGauss’ Law to de rm fie for sym e e phe ym e ly d re • 3-Dim nsions: S rical S m try: Uniform charge solid sphe e ym e • 2-Dim nsions: Cylindrical S m try: I nfinitesolid cylindrical conductor e • 1-Dim nsion: Planar sym e m try: I nfiniteshe t of charge e upe e – S rposition: Two infiniteshe ts of charge • Le cturePlan vie s – Re w difficultie ctric fie and chargedistributions in conductors lds • Ele vie flights – Re w pre – Do Gauss’ Law calculation ld • E fie hargedistribution •C Physics 212 Le cture4, S 4 lide C onductors (What m s?) ove Do positivecharge m in a re m tal conductor ? No, but… s ove al e The net result of this can be viewed in two equivalent ways: The equivalent 05 Physics 212 Le cture4, S 5 lide C onductors (What m s?) ove Do positivecharge m in a re m tal conductor ? No, but… s ove al e The net result of this can be viewed in two equivalent ways: The equivalent And… 05 Physics 212 Le cture4, S 6 lide C onductors (What m s?) ove Do positivecharge m in a re m tal conductor ? No, but… s ove al e The net result of this can be viewed in two equivalent ways: The equivalent And… 05 Physics 212 Le cture4, S 7 lide Conductors = charges free to move Claim: E = 0 inside any conductor at equilibrium Charges in conductor move to make E field zero inside. (Induced charge distribution) E <>0, then charge feels force and moves! Claim: Excess charge on conductor only on surface at equilibrium Why? • Apply Gauss’ Law • TakeGaussian surfaceto bejust insideconductor surface ve re • E = 0 e rywhe insideconductor • Gauss’ Law: E=0 r r Qenc dA ∫ Eg = ε0 Qenc = 0 S MULATI ON 2 I rr dA ∫ Eg = 0 07 Physics 212 Le cture4, S 8 lide Gauss’ Law + Conductors + I nduce Charge d s r r Qenc dA ∫ Eg = ε0 ALWAYSTRUE! Q I f choosea Gaussian surfacethat is e ly in m tal, the E=0 so E = enc ntire e n Aε 0 Qe nclose m also beze d ust ro! 9 Physics 212 Le cture4, S 9 lide BB Charge in Cavity of Conductor A particle with charge +Q is placed in the center of an uncharged conducting hollow sphere. How much charge will be induced on the Q inner and outer surfaces of the sphere? A) inner = –Q, outer = +Q B) inner = –Q/2 , outer = +Q/2 C) inner = 0, outer = 0 D) inner = +Q/2, outer = ­Q/2 E) inner = +Q, outer = ­Q outer Qinner Q • Gauss’ Law: r r Qenc dA ∫ Eg = ε0 Since E=0 in conductor Qenc = 0 11 Physics 212 Lecture 4, Slide 10 10 BB Infinite Cylinders A long thin wire has a uniform positive charge density of 2.5 C/m. Concentric with the wire is a long thick conducting cylinder, with inner radius 3 cm, and outer radius 5 cm. The conducting cylinder has a net linear charge density of -4 C/m. What is the linear charge density of the induced charge on the inner surface of the conducting cylinder? A) -6.5 C/m B) -4 C/m C) -2.5 C/m D) -1.5 C/m E) 0 What is the linear charge density of the induced charge on the outer surface of the conducting cylinder? A) -4 C/m B) -2.5 C/m C) -1.5 C/m D) 0 C/m E) +2.5 C/m 15 Physics 212 Lecture 4, Slide 11 11 Gauss’ Law r r Qenc dA ∫ Eg = ε0 ALWAYS TRUE! E= Qenc Aε 0 In cases with symmetry can pull E outside and get In General, integral to calculate flux is difficult…. and not useful! To use Gauss Law to calculate E, need to choose surface carefully! 1) Want E to be constant and equal to value at location of interest OR 2) Want E dot A = 0 so doesn’t add to integral 19 Physics 212 Lecture 4, Slide 12 12 Gauss’ Law Symmetries r r Qenc dA ∫ Eg = ε0 ALWAYS TRUE! Qenc E= Aε 0 In cases with symmetry can pull E outside and get Spherical Cylindrical Planar A = 4π r Qenc E= 4π r 2ε 0 2 22 A = 2π rL λ E= 2π rε 0 A = 2π r 2 σ E= 2ε 0 Physics 212 Lecture 4, Slide 13 13 Preflight 2 80 60 BB 40 20 0 (D) The field cannot be calculated using Gauss’ Law (E) None of the above “The Cube inside the Sphere will The yield an electric field perpindicular to the sphere at all points on it.” to “The cube is best suited for a cube The because they have the same surfaces always perpendicular to the field.” the “The cube does not have symmetry in a way that makes calculating the flux integral easy “ Physics 212 Lecture 4, Slide 14 14 THE CUBE HAS NO GLOBAL SYMMETRY ! THE FIELD AT THE FACE OF THE CUBE THE IS NOT PERPENDICULAR OR PARALLEL PERPENDICULAR 3D 2D 1D POINT LINE PLANE v v v SPHERICAL CYLINDRICAL PLANAR 23 Preflight 6 BB 60 50 40 30 20 10 0 out in zero “The field points from the positively charged inner sphere to the The negatively charged outer sphere. Therefore the field points radially outward. “ outward. “The negative and positive spheres are attracted to each other so the The electric field points inward.” electric “The positive and negative charges will create a field of zero “ 30 Physics 212 Lecture 4, Slide 15 15 Preflight 4 BB 50 40 30 20 10 0 “The electric field is described by the charge enclosed divided by epsilon.” “The charge enclosed in the sphere is zero, and since EdA is proportional to The q, E must also be zero. “ q, “Since this is an insulator, B is not a possible answer. Therefore, C would Since make the most sense because it incorporates all the parameters presented in this problem. C also fits the model for electric field outside an insulator, which Physics 212 Lecture 4, Slide 16 16 is very similar to Coulomb's law. “ i33 s Preflight 10 BB In which case is E at point B the biggest? A) A B) B C) the same 60 50 40 30 20 10 0 “Case A because the negatively charged plane in the middle of Case case B cancels out the electric field produced by the left one. “ case “Case B has 2 panels pulling it to the right even though the 3rd Case panel is pushing it to the left and in Case A, there is only 1 panel pushing to the right. “ pushing “The positive electric fields in the Case B cancel out each other. The Thus, there is only eletric field going to the right at point P. There is eletric field going to the right at point P in Case A, as well. “ i27 s Physics 212 Lecture 4, Slide 17 17 Superposition: NET + Case A + - + Case B 28 Physics 212 Lecture 4, Slide 18 18 BB y Calculation Point charge+3Q at ce r of ne nte utral conducting she of inne ll r radius r1 and oute radius r2. r a) What is E e rywhe ? ve re x r2 neutral conductor +3Q r1 First que stion: Do wehavee nough sym e to useGauss’ Law m try t o de rm E? te ine Ye S rical S m try (what doe this m an???) s.. phe ym e s e A B C D Magnitudeof E is fcn of r Magnitude of E is fcn of (r-r ) Magnitude of E is fcn of (r-r ) None of the above A B C D Dire ction of E is along Direction of E is along Direction of E is along None of the above ˆ x ˆ y ˆ r Physics 212 Lecture 4, Slide 19 19 BB y Calculation Point charge+3Q at ce r of ne nte utral conducting she of inne ll r radius r1 and oute radius r2. r a) What is E e rywhe ? ve re Weknow: x m agnitudeof E is fcn of r dire ction of E is along ˆ r Wecan useGauss’ Law to de rm E te ine UseGaussian surface= sphe ce re on origin re nte d r2 neutral conductor +3Q r1 r r Qenc dA ∫ Eg = ε0 rr E g = E 4π r 2 ∫ dA Qenc = +3Q r < r1 r1 < r < r2 r > r2 A E= E= 1 3Q 4πε 0 r 2 1 3Q 4πε 0 r12 A E= E= 1 3Q 4πε 0 r 2 1 3Q 4πε 0 (r − r2 ) 2 Physics 212 Lecture 4, Slide 20 20 B B 1 3Q E= 4πε 0 r 2 C E=0 C E=0 BB y Calculation Point charge+3Q at ce r of ne nte utral conducting she of inne ll r radius r1 and oute radius r2. r a) What is E e rywhe ? ve re Weknow: r < r1 r > r2 r2 neutral conductor +3Q r1 x E= 1 3Q 4πε 0 r 2 r r Qenc dA ∫ Eg = ε0 b) What is chargedistribution at r1? A B C r1 < r < r2 E=0 σ <0 σ =0 Gauss’ Law: r2 E=0 +3Q r1 Similarly: Qenc = 0 σ1 = −3Q 4π r12 σ >0 σ2 = +3Q 4π r22 Physics 212 Lecture 4, Slide 21 21 BB y Calculation S upposegiveconductor a chargeof -Q a) What is E e rywhe ? ve re b) What arechargedistributions at r1 and r2? x + + r2 -Q conductor +3Q r1 r r Qenc dA ∫ Eg = ε0 + + r2 + + + + + + + + +3Q -3Q ++ r1 + + + + +2Q r < r1 r > r2 A 1 3Q E= 4πε 0 r 2 E= E= 1 2Q 4πε 0 r 2 1Q 4πε 0 r 2 A 1 3Q E= 4πε 0 r 2 E= E= 1 2Q 4πε 0 r 2 1Q 4πε 0 r 2 r1 < r < r2 E =0 B B C C Physics 212 Lecture 4, Slide 22 22 ...
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This note was uploaded on 04/04/2011 for the course PHYSICS 212 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.

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