# Lect29 - Physics 212 Lecture 29 Course Review • The Topics You Want to See – – – – – – – – Electric Fields/Gauss’

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Unformatted text preview: Physics 212 Lecture 29 Course Review • The Topics You Want to See – – – – – – – – Electric Fields/Gauss’ Law/Potential (34%) Faraday’s Law (12%) RC/RL Circuits (14%) AC Circuits (11%) Geometric Optics (13%) Magnetic Fields & Forces (7%) Electromagnetic Waves/Polarization (8%) DC Circuits (3%) Physics 212 Lecture 29, Slide 1 Music Who is the Artist? A) B) C) D) E) Marc Ribot Marc Ribot Bill Frisell Bill Frisell Jake Hertzog Pat Metheny Pat Metheny John Scofield John Scofield Why?? I think he is my favorite jazz guitarist !! We bookend the course with incredible guitar players Jeff Beck & Bill Frisell Jeff Frisell Physics 212 Lecture 29, Slide 2 BB E from top arc points down Horizontal components cancel E from bottom arc points up Top arc produces smaller horizontal components Etotal points down Calculation: Etop = ∫ λ= dq 4πε0 r Q 2rθtop Etop = 2 θ cos θ Etop = Q sinθ θ 1 4πε0 Q r 2 2rθtop sin θtop 4πε0 r 2 θtop θ top 2 ∫ rdθ cos θ 0 θ Physics 212 Lecture 29, Slide 3 BB Potential Energy is a measure of work done by E field Spherical symmetry & Gauss’ law law E = 0 inside shell inside E = 0 inside shell inside No work done to move q No No change in potential energy ! Physics 212 Lecture 29, Slide 4 BB ALWAYS START ALWAYS FROM DEFINITION OF POTENTIAL OF ar r ∆V = − ∫ E ⋅ dr (A) 1 2Q 3Q − 4πε 0 a b (B) 1 3Q 2Q − 4πε 0 b a (C) 0 (D) (E) 1 2Q 2Q − 4πε 0 a b 1 2Q 2Q − 4πε 0 b a 2Q ∆V = − 4πε0 b Spherical symmetry & Spherical Gauss’ law determines E Gauss 1 2Q a < r < b: E = 4πε0 r 2 a dr ∫2 br ∆V = 2Q 1 1 − 4πε0 a b Physics 212 Lecture 29, Slide 5 BB (A) 1 2Q 4πε 0 a (B) − (C) 0 1 2Q 4πε 0 a (E) Spherical symmetry & Spherical Gauss’ law determines E Gauss r < a: (D) E =0 1 3Q 4πε 0 b − 1 3Q 4πε 0 b ar r ∆V = − ∫ E ⋅ dr = 0 0 Physics 212 Lecture 29, Slide 6 Charge must be Charge induced to insure E = 0 within conducting shell conducting BB Spherical symmetry & Spherical Gauss’ law determines E Gauss r r Qenclosed ∫ E ⋅ dA = ε0 (A) (B) − Q1 4πa 2 + Q1 4πa 2 (C) (D) Q2 − Q1 4π (b − a ) 2 2 (E) − Q1 4πb 2 E ⋅ 4πr 2 = 0 + Q1 4πb 2 Qenclosed = Q1 + (−Q1 ) σ= − Q1 4πb 2 Physics 212 Lecture 29, Slide 7 r < a: E = 0 Eliminate (a) and (c) BB b < r < c: E = 0 Eliminate (e) a < r < b: E = kQ1/r2 r > c: E = k(Q1+Q2)/r2 Q1 = -3µC Q1 + Q2 = +2µC For r > c, For E must be less than the continuation of E from a to b continuation Physics 212 Lecture 29, Slide 8 BB ALWAYS START ALWAYS FROM DEFINITION OF POTENTIAL OF br r ∆V = − ∫ E ⋅ dr 0 Break integral into two pieces ar r br r ∆V = − ∫ E ⋅ dr − ∫ E ⋅ dr 0 conductor: = 0 insulator: ∫ 0 insulator: a same for same conductor & insulator insulator Physics 212 Lecture 29, Slide 9 1 L dI1 + IR − V = 0 dt I – I1 I I1 2 L dI1 − ( I − I1 ) R = 0 dt IR = V − L 1 30% 2 L dI1 dI − V + L 1 + I1R = 0 dt dt 2L Strategy: Back to First Principles • The time constant is determined from The a differential equation for the current through the inductor. through • Equation for current through inductor Equation obtained from Kirchhoff’s Rules obtained dI1 dt dI1 + I1R = V dt τ= " L" 2 L = " R" R Physics 212 Lecture 29, Slide 10 10 BB 76% Current induced because flux is changing Flux is changing beause B is changing Flux beause At t = 5 seconds, B is positive, but decreasing Lenz’ law: emf induced to oppose change that brought it into being law: emf Induced current must produce positive B field Positive B field produced by counterclockwise current Physics 212 Lecture 29, Slide 11 11 BB 62% rr Flux definition: Φ = ∫ B ⋅ dA = BA = Bwh Faraday’s law: ε = − dB = −2T / s dt dΦ dB = − wh dt dt I= ε R = 9/5 = .012 A 150 Physics 212 Lecture 29, Slide 12 12 BB 65% Current is determined by time rate of change of the flux dΦ/dt is determined by dB/dt dB/dt (6s) = dB/dt (5 s) = -2 T/s (5 The induced currents at t = 5s and t = 6s are equal Physics 212 Lecture 29, Slide 13 13 Phasor diagram at t = 0 What is VC at t = π/(2ω) at (A) α + VC max sin α (C) + VC max cos α (B) − VC max sin α (D) − VC max cos α BB Phasor diagram at t = π/(2ω) diagram VR α VC VL Voltage is equal to Voltage projection of phasor phasor along vertical axis Physics 212 Lecture 29, Slide 14 14 ...
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## This note was uploaded on 02/09/2012 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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