Lectures_Lect03

Lectures_Lect03 - Physics 212 Lecture 3 Today's Concepts...

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Unformatted text preview: Physics 212 Lecture 3 Today's Concepts: Electric Flux and Field Lines Gauss's Law Physics 212 Lecture 3, Slide 1 Introduce a new constant: 0 r q E =k 2r ^ r k = 9 x 109 N m2 / C2 k 1 4 0 0 = 8.85 x 10-12 C2 / N m2 r E= 1 q r 2 ^ 4 0 r Physics 212 Lecture 3, Slide 2 04 Plan for Today A little more about electric field lines Electric field lines and flux Introduction to Gauss's Law An analogy These questions will turn out to be among your favorites because Gauss' Law will make it easy to calculate electric fields 06 Physics 212 Lecture 3, Slide 3 Electric Field Lines Direction & Density of Lines represent Direction & Magnitude of E Point Charge: Direction is radial Density 1/R2 07 Physics 212 Lecture 3, Slide 4 Electric Field Lines Dipole Charge Distribution: Direction & Density 08 simulation Physics 212 Lecture 3, Slide 5 Checkpoint Preflight 3 Field lines are are denser near Q1 so Q1 > Q2 simulation 09 Physics 212 Lecture 3, Slide 6 Checkpoint The electric field lines connect the charges. A test charge will move towards one charge and away from the other. So charges 1 and 2 have opposite signs. 10 Physics 212 Lecture 3, Slide 7 Checkpoint Preflight 3 Density of lines is greater at B than at A. Therefore, magnitude of field at B is greater than at A. 12 Physics 212 Lecture 3, Slide 8 Point Charges -q +2q What charges are inside the red circle? Q + Q 13 + Q Q +2 Q 2 Q + Q Q A B C D E Physics 212 Lecture 3, Slide 9 Which of the following field line pictures best represents the electric field from two charges that have the same sign but different magnitudes? A B C 15 D Physics 212 Lecture 3, Slide 10 simulation Electric Flux "Counts Field Lines" r r S = E dA S Flux through surface S r r Integral of E dA on surface S 18 Physics 212 Lecture 3, Slide 11 Electric Field/Flux Analogy: Velocity Field/Flux r r S = v dA = flowrate S Flux through surface S r r Integral of v dA on surface S 20 Physics 212 Lecture 3, Slide 12 Checkpoint An infinitely long charged rod has uniform charge density and passes through a cylinder (gray). The cylinder in Case 2 has twice the radius and half the length compared TAKE s TO BE RADIUS ! with the cylinder in Case 1. L/2 1=22 (A) 1=2 (B) 1=1/22 (C) (D) none 23 Physics 212 Lecture 3, Slide 13 Checkpoint Definition of Flux: An infinitely long charged rod has uniform charge density l and passes through a cylinder (gray). The cylinder in Case 2 has twice the radius and half the length compared with the TAKE s TO BE RADIUS ! cylinder in Case 1. surface r r E dA L/2 E constant on barrel of cylinder E perpendicular to barrel surface (E parallel to dA) r = E dA = EAbarrel barrel 1=22 E1 = 2 0 s 26 Case 1 (A) 1=2 1=1/22 RESULT: GAUSS' LAW none (D) proportional to charge enclosed ! (B) (C) Case 2 A1 = ( 2s ) L 1 = L 0 E2 = 2 0 ( 2s ) A2 = (2 (2 s )) L / 2 = 2sL 2 = ( L / 2) 0 Physics 212 Lecture 3, Slide 14 Direction Matters: E E dA dA E dA E E dA dA E E E E For a closed surface, A points outward r r S = E dA > 0 S 29 Physics 212 Lecture 3, Slide 15 Direction Matters: E E dA dA E dA E E dA dA E E E E For a closed surface, A points outward r r S = E dA < 0 S 30 Physics 212 Lecture 3, Slide 16 Trapezoid in Constant Field y La b e l fa c e s : 1 : x = 0 2 : z = +a 3 : x = +a 4 : s la nte d De fine n = Flux through Face n z 4 r ^ E = E0 x 3 1 2 x dA r r E dA < 0 E A B C 1 < 0 1 = 0 1 > 0 A B C 2 < 0 2 = 0 2 > 0 A B C 3 < 0 3 = 0 3 > 0 A B C 4 < 0 4 = 0 4 > 0 31 Physics 212 Lecture 3, Slide 17 Trapezoid in Constant Field + Q r y La b e l fa c e s : 1 : x = 0 2 : z = +a 3 : x = +a De fine n = Flux through Face n = Flux through Trapezoid z ^ E = E0 x 3 r E0 x +Q 1 2 Add a charge +Q at (a,a/2,a/2) How does Flux change? A B C 1 increases 1 decreases 1 remains same A B C 3 increases 3 decreases 3 remains same A B C increases decreases remains same 36 Physics 212 Lecture 3, Slide 18 Gauss Law E E dA Q dA E E dA dA E E dA E E E r r Qenclosed S E dA = o closed surface 41 Physics 212 Lecture 3, Slide 19 Checkpoint What happens to total flux through the sphere as we move Q ? (A) increases (B) decreases (C) stays same The same amount of charge is still enclosed by the sphere, so flux will not change. 43 Physics 212 Lecture 3, Slide 20 Checkpoint (A) dA increases dB decreases (B) dA decreases dB increases (C) dA stays same dB stays same 44 Physics 212 Lecture 3, Slide 21 Think of it this way: 1 The total flux is the same in both cases (just the total number of lines) The flux through the right (left) hemisphere is smaller (bigger) for case 2. 45 2 Physics 212 Lecture 3, Slide 22 Things to notice about Gauss's Law surface If Qenclosed is the same, the flux has to be the same, which means that the integral must yield the same result for any surface. r r Qenclosed S E dA = o closed 47 Physics 212 Lecture 3, Slide 23 Things to notice about Gauss's Law r r Qenclosed E dA = o closed surface In cases of high symmetry it may be possible to bring E outside the integral. In these cases we can solve Gauss Law for E r r Qenclosed E dA = EA = o closed surface Qenclosed E= A 0 So - if we can figure out Qenclosed and the area of the surface A, then we know E ! This is the topic of the next lecture... 48 Physics 212 Lecture 3, Slide 24 Prelecture 4 and Checkpoint 4 due Thursday Homework 2 due next Monday 50 Physics 212 Lecture 3, Slide 25 ...
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This document was uploaded on 03/08/2012.

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