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# Lect04 - Physics 212 Lecture 4 Today's Concepts Conductors...

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Unformatted text preview: Physics 212 Lecture 4 Today's Concepts: Conductors + Using Gauss’ Law Physics 212 Lecture 4, Slide 1 Physics Music Who is the Artist? BB A) B) C) D) E) Professor Longhair Dr. John Allen Touissant David Egan Henry Butler Theme of week: New Orleans piano players If you have a chance to ever see him, DO I never miss him at Jazzfest… Physics 212 Lecture 4, Slide 2 Physics Your Comments “I thought I understood the prelecture and then I came to these questions and I feel very uestions confused and not very confident. “i don't understand how to pick a gaussian surface or even when to pick it really :( ” “Conductors in general, I also don't really get why the electric field is 0 inside a conductor!! “ “I miss when Stelzer would put or comments up on the big screen at the beginning of class. i the miss the funny ones..... “ My Response 40 Most students are having difficulties with this topic: The performance on the preflights show this clearly. I believe the problem is mostly that this whole way of thinking (Gauss’ Law) is very unfamiliar to you. The solution? Hang in there… I’m confident that the more you work with these concepts, the more familiar you will become with them and you will master them. TODAY’S PLAN: 04 30 20 10 0 Confused Confident • Do Preflights! Try to understand the reasoning • Do a calculation using Gauss’ Law Physics 212 Lecture 4, Slide 3 Physics 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). If E <>0, then charge feels force and moves! Claim: Excess charge on conductor only on surface at equilibrium Why? • Apply Gauss’ Law • Take Gaussian surface to be just inside conductor surface • E = 0 everywhere inside conductor • Gauss’ Law: ∫ E idA = Qenc ε0 Qenc = 0 SIMULATION 2 07 E=0 ∫ E idA = 0 Physics 212 Lecture 4, Slide 4 Physics Gauss’ Law + Conductors + Induced Charges ∫ E idA = Qenc ε0 ALWAYS TRUE! If choose a Gaussian surface that is entirely in metal, then E=0 so Qenclosed must also be zero! Qenc E= Aε 0 How Does This Work?? Charges in conductor move to surfaces to make Qenclosed = 0. We say charge is induced on the surfaces of conductors 9 Physics 212 Lecture 4, Slide 5 Physics 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 inner conducting inner Qouter and outer surfaces of the sphere? and A) inner = –Q, outer = +Q A) Q, Qinner B) inner = –Q/2 , outer = +Q/2 Q/2 C) inner = 0, C) outer = 0 Q D) inner = +Q/2, outer = -Q/2 D) E) inner = +Q, E) outer = -Q Since E=0 in conductor • Gauss’ Law: 11 ∫ E idA = Qenc ε0 Qenc = 0 Physics 212 Lecture 4, Slide 6 Physics 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 (λi) and on the outer surface (λo)? λi: +2.5 C/m λo: -6.5 C/m A) 15 -4 C/m -2.5 C/m -2.5 C/m 0 +2.5 C/m -1.5 C/m C) D) λι -4 C/m B) λο 0 E) Physics 212 Lecture 4, Slide 7 Physics Gauss’ Law ∫ E idA = Qenc ε0 ALWAYS TRUE! In cases with symmetry can pull E outside and get E = Qenc Aε 0 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 8 Physics Gauss’ Law Symmetries ∫ E idA = Qenc ε0 ALWAYS TRUE! In cases with symmetry can pull E outside and get Spherical Cylindrical A = 4π r Qenc E= 4π r 2ε 0 2 22 Qenc E= Aε 0 Planar A = 2π rL λ E= 2π rε 0 A = 2π r 2 σ E= 2ε 0 Physics 212 Lecture 4, Slide 9 Physics BB Preflight 2 70 60 50 40 30 20 10 0 “A sphere is needed so that the sphere field lines are perpendicular to the surface “ (D) The field cannot be calculated using Gauss’ Law (E) None of the above THE CUBE HAS NO GLOBAL SYMMETRY ! THE FIELD AT THE FACE OF THE CUBE THE IS NOT PERPENDICULAR OR PARALLEL PERPENDICULAR 23 3D 2D 1D POINT LINE PLANE Ø Ø Ø SPHERICAL CYLINDRICAL PLANAR “The cube would be best because The all of its surfaces are perpendicular to the field lines coming from the cube .” cube “Since the field lines from the cube are not symmetric and not always perpendicular to the surface, the field from am charged cube cannot be calculated using Gauss' law. “ Physics 212 Lecture 4, Slide 10 Physics BB Preflight 6 60 50 40 30 20 10 0 out in zero “E-field lines point from a positive charge (in this case the solid sphere) to sphere) a negative charge (in this case the shell). “ negative 30 “The field would point inwards because the negative charges and positive ositive charges attract each other. “ Careful: what does inside mean? Careful: inside mean? “E iinside conductor is zero “ nside This is always true for a solid conductor (within the material of the conductor) (within Physics P” Here we have a charge “insidehysics 212 Lecture 4, Slide 11 Here BB Preflight 8 60 50 40 30 20 10 0 out in zero What is direction of field OUTSIDE the red sphere? “To be honest, I have yet to see E field lines point inward in any of the of prelectures...everytime they are pointed outward “ “Field lines point towards a negative charge” “By placing a Gaussian surface enclosing both t conductors, the net et enclosed charge will be zero. Therefore, the field will be zero. “ enclosed 30 Physics 212 Lecture 4, Slide 12 Physics BB Preflight 4 50 40 30 20 10 0 “when r<a, E=q/3Eor which simplifies to r/Eo” “Gauss' law states that integral (E dot dA) = q enclosed / epsilonnot. Since our nnot. gaussian surface in this case is a sphere with radius r < a, effectively our gaussian gaussian ectively surface is enclosing no charge! Therefore the magnitude of the E field is just zero. “ surface field “E is equal to the charge enclosed over e. The charge enclosed is jjust the charge ust density times the volume. “ 33 Physics 212 Lecture 4, Slide 13 Physics BB Preflight 10 In which case is E at point P the biggest? A) A B) B C) the same “There is an overall right E field in Case A, in Case B the negatiive ve and positive planes cancel out. “ 50 40 “The sum of the fields is greater in case B. The negative is closer er than the rightmost positive so it has a greater effect. “ than 30 20 “Charged infinite planes create constant electric fields. In Case A, A, there is one positive field, and in Case B there are two positive there and one negative. In Case B, one negative and one positive cancel out, leaving only one positive electric field. “ 27 10 0 Physics 212 Lecture 4, Slide 14 Physics Superposition: NET + Case A 28 + - + Case B Physics 212 Lecture 4, Slide 15 Physics BB Calculation y r2 neutral conductor +3Q r1 Point charge +3Q at center of neutral conducting shell of inner radius r1 and outer radius r2. a) What is E everywhere? x First question: Do we have enough symmetry to use Gauss’ Law to determine E? Yes.. Spherical Symmetry (what does this mean???) A Magnitude of E is fcn of r A Direction of E is along B Magnitude of E is fcn of (r-r1) B Direction of E is along ˆ x ˆ y C Magnitude of E is fcn of (r-r2) C Direction of E is along ˆ r D None of the above D None of the above Physics 212 Lecture 4, Slide 16 Physics y BB Calculation Point charge +3Q at center of neutral conducting shell of inner radius r1 and outer radius r2. a) What is E everywhere? We know: x magnitude of E is fcn of r direction of E is along r ˆ neutral conductor r2 +3Q r1 We can use Gauss’ Law to determine E Use Gaussian surface = sphere centered on origin ∫ E idA = r < r1 E idA = E 4π r 2 ∫ r1 < r < r 2 E= 1 3Q 4πε 0 r 2 A E= B 1 3Q E= 4πε 0 r 2 E= 1 3Q 4πε 0 r12 B E= C E=0 C E=0 ε0 r > r2 A Qenc = +3Q Qenc 1 3Q 4πε 0 r 2 1 3Q 4πε 0 ( r − r2 ) 2 Physics 212 Lecture 4, Slide 17 Physics y BB r2 Calculation neutral conductor +3Q r1 x Point charge +3Q at center of neutral conducting shell of inner radius r1 and outer radius r2. a) What is E everywhere? We know: r < r1 E= r > r2 ∫ E idA = Qenc E=0 r1 < r < r 2 ε0 1 3Q 4πε 0 r 2 b) What is charge distribution at r1? A σ <0 B Gauss’ Law: σ =0 C σ >0 E=0 r2 +3Q r1 Qenc = 0 −3Q σ1 = 4π r12 Similarly: σ2 = +3Q 4π r22 Physics 212 Lecture 4, Slide 18 Physics y BB r2 Calculation -Q Suppose give conductor a charge of -Q a) What is E everywhere? b) What are charge distributions at r1 and r2? conductor +3Q r1 x ∫ E idA = + + r2 Qenc ε0 1 3Q E= 4πε 0 r 2 A 1 3Q E= 4πε 0 r 2 B E= 1 2Q 4πε 0 r 2 B E= C E= C E= 1 Q 4πε 0 r 2 + +3Q + A + + + r > r2 + + + r < r1 + r1 + -3Q + ++ + +2Q + + r1 < r < r 2 E =0 1 2Q 4πε 0 r 2 1 Q 4πε 0 r 2 Physics 212 Lecture 4, Slide 19 Physics ...
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