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Physics 212 Lecture 3

Physics 212 Lecture 3. Today's Concepts: Electric Flux and Field Lines Gauss’s Law. Music. Who is the Artist? Professor Longhair Henry Butler David Egan Dr. John Allen Toussaint. WHY? Mardi Gras is 4 weeks from today. Our Comments. Lecture. Worked Example. Prelecture. E. E. E.

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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. Music • Who is the Artist? • Professor Longhair • Henry Butler • David Egan • Dr. John • Allen Toussaint WHY? Mardi Gras is 4 weeks from today

  3. Our Comments Lecture Worked Example Prelecture E E E Finite line of charge (non-constant l) Infinite line of charge Finite line of charge (constant l) Homework E WORKS FOR ALL !! Arc of charge (constant l) • SmartPhysics scores will not be visible in the main gradebook for a few weeks • You will need to understand integrals in this course!!! • Forces and Fields are Vectors • Always Draw a Picture First… What do the Forces/Fields Look Like? The “1/4 shell problem” is not special !! It’s the same as all the others!

  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 & Densitymuch more interesting… simulation 08

  6. Checkpoint 3.1 The following three questions pertain to the electric field lines due to the two charges shown. Compare the magnitude of the two charges A. |Q1| > |Q2| B. |Q1| = |Q2| C. |Q1| < |Q2|D. There isn’t enough information to determine the relative magnitude of the charges Preflight 3 “more field lines emanating from the charge 1, so must be greater in magnitude.” simulation 09

  7. Checkpoint 3.3 What do we know about the signs of the charges from looking at the picture? A. Q1 and Q2 have the same sign B. Q1 and Q2 have the opposite sign C. There isn’t enough information to determine the relative signs of the charges “The field lines leave one and go to the other..” simulation 10

  8. Checkpoint 3.5 Compare the magnitude of the electric field at point A and B. A. |EA| > |EB| B. |EA| = |EB| C. |EA| < |EB|D. There isn’t enough information to determine the relative magnitude of the electric field at points A and B Preflight 3 “Since the electric field lines are denser near point B the magnitude of the electric field is greater at point B than at point A.” 12

  9. Point Charges -2Q +Q +2Q -Q -Q -Q +Q +Q A B E C D “Telling the difference between positive and negative charges while looking at field lines. Does field line density from a certain charge give information about the sign of the charge?” -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” “I still feel a little unclear about flux integrals. Can you go over some more example problems?” Flux through surface S Integral of on surface S 18

  12. Checkpoint 1 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 ! “The flux is directly proportional to the charge enclosed by the Gaussian surface, therefore Case 1 will have 2 times the flux of of Case 2 because it has two charges enclosed, versus 1 in Case 2. “ “The flux is the measure of field lines passing through the surface. Once calculated, it can be shown that the surface area of both cylinders are the same.” L/2 23

  13. Checkpoint 1 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. WHAT IS WRONG WITH THIS REASONING?? “The flux is the measure of field lines passing through the surface. Once calculated, it can be shown that the surface area of both cylinders are the same.” Definition of Flux: Case 1 Case 2 F1=F2 F1=2F2 F1=1/2F2 none (D) (C) (A) (B) TAKE s TO BE RADIUS ! THE E FIELD AT THE BARREL SURFACE IS NOT THE SAME IN THE TWO CASES !! 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

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

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

  16. 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

  17. 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

  18. Gauss Law E E E dA dA Q E E dA dA E dA E E E 41

  19. Checkpoint 2.3 “The flux throughout the entire surface does not change as the charge moves, for it is still in the sphere. However, if it was moved outside of the sphere, the fluxes would be different because one would be zero!“ (C) F stays same (A) F increases (B) F decreases 43 Physics 212 Lecture 3, Slide 19

  20. Checkpoint 2.1 “charge is closer, field is larger, more lines through a so increase in flux.“ “as long as the charge is within the shell, the flux will remain the same” (C) dFA stays same dFB stays same (A) dFA increases dFB decreases (B) dFA decreases dFB increases 44

  21. 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

  22. 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

  23. 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… “Gausses law and what concepts are most important that we know.” 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

  24. Final Thoughts…. • Keep plugging away • Prelecture 4 and Checkpoint 4 due Thursday • Homework 2 due next Tuesday … 7 questions mostly Gauss’ law – try them today! 50

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