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Physics 212 Lecture 3

Physics 212 Lecture 3. Today's Concepts: Electric Flux and Field Lines Gauss’s Law. Introduce a new constant:  0. k = 9 x 10 9 N m 2 / C 2 e 0 = 8.85 x 10 -12 C 2 / N m 2. 04. Plan for Today. A little more about electric field lines Electric field lines and flux An analogy

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Physics 212 Lecture 3

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  1. Physics 212 Lecture 3 Today's Concepts: Electric Flux and Field Lines Gauss’s Law

  2. Introduce a new constant: 0 k = 9 x 109 N m2 / C2 e0 = 8.85 x 10-12 C2 / N m2 04

  3. Plan for Today • A little more about electric field lines • Electric field lines and flux • An analogy • Introduction to Gauss’s Law • Gauss’ Law will make it easy to calculate electric fields for some geometries. 06

  4. Electric Field Lines Direction & Density of Lines representDirection & Magnitude ofE Point Charge: Direction is radial Density  1/R2 07

  5. Electric Field Lines Dipole Charge Distribution: Direction & Density simulation 08

  6. Checkpoint Preflight 3 Field lines are are denser near Q1 so  Q1  >  Q2  simulation 09

  7. 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

  8. Checkpoint Density of lines is greater at B than at A. Therefore, magnitude of field at B is greater than at A. Preflight 3 12

  9. Point Charges -2Q +Q +2Q -Q -Q -Q +Q +Q A B E C D -q +2q What charges are inside the red circle? 13

  10. 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 D simulation 15

  11. Electric Flux “Counts Field Lines” Flux through surface S Integral of on surface S 18

  12. Electric Field/Flux Analogy: Velocity Field/Flux Flux through surface S Integral of on surface S 20

  13. Checkpoint 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 cylinder in Case 1. F1=F2 F1=2F2 F1=1/2F2 none (D) (C) (A) (B) TAKE s TO BE RADIUS ! L/2 23

  14. Checkpoint 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 cylinder in Case 1. Definition of Flux: Case 1 Case 2 F1=F2 F1=2F2 F1=1/2F2 none (D) (C) (A) (B) TAKE s TO BE RADIUS ! L/2 E constant on barrel of cylinder E perpendicular to barrel surface (E parallel to dA) RESULT: GAUSS’ LAW F proportional to charge enclosed ! 26

  15. Direction Matters: E E E dA For a closed surface, A points outward dA E E dA dA E dA E E E 29

  16. Direction Matters: E E E dA For a closed surface, A points outward dA E E dA dA E dA E E E 30

  17. Trapezoid in Constant Field dA E A B C F2 < 0 F3 < 0 F4 < 0 F1 < 0 A B C A B C A B C F2 = 0 F3 = 0 F4 = 0 F1 = 0 F2 > 0 F3 > 0 F4 > 0 F1 > 0 y Label faces: 1: x = 0 2: z = +a 3: x = +a 4: slanted 4 3 1 2 x Define Fn = Flux through Face n z 31

  18. Trapezoid in Constant Field + Q 3 +Q 1 2 x Add a charge+Qat (-a,a/2,a/2) A B C A B C A B C F3 increases F increases F1 increases F3 decreases F decreases F1 decreases F1 remains same F3 remains same F remains same y Label faces: 1: x = 0 2: z = +a 3: x = +a x Define Fn = Flux through Face n F = Flux through Trapezoid z How does Flux change? 36

  19. Gauss Law E E E dA dA Q E E dA dA E dA E E E 41

  20. Checkpoint What happens to total flux through the sphere as we move Q ? (C) F stays same (A) F increases (B) F decreases The same amount of charge is still enclosed by the sphere, so flux will not change. 43 Physics 212 Lecture 3, Slide 20

  21. Checkpoint (C) dFA stays same dFB stays same (A) dFA increases dFB decreases (B) dFA decreases dFB increases 44

  22. Think of it this way: 2 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

  23. Things to notice about Gauss’s Law 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. 47

  24. Things to notice about Gauss’s Law 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… 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 48

  25. Prelecture 4 and Checkpoint 4 due Thursday • Homework 2 due next Monday 50

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