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Physics 2112 Unit 3: Electric Flux and Field Lines

Physics 2112 Unit 3: Electric Flux and Field Lines. Today’s Concepts: A) Electric Flux B) Field Lines. Gauss’ Law. Electric Field Lines. First….a comment. What the point of all this flux stuff?. To create Guass’ Law so we can find the E field without doing an integral. Guass’ Law. + Q.

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Physics 2112 Unit 3: Electric Flux and Field Lines

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  1. Physics 2112Unit 3: Electric Flux and Field Lines Today’s Concepts: A) Electric Flux B) Field Lines Gauss’ Law

  2. Electric Field Lines First….a comment What the point of all this flux stuff? To create Guass’ Law so we can find the E field without doing an integral Guass’ Law +Q Always true, sometimes not useful

  3. Electric Field Lines Electric Field Lines Electric Field is a “vector field” “Density” represents magnitude

  4. Electric Field Lines Electric Field Lines in 3D • Point Charge: • Direction is radial • 3D Density  area of sphere(1/R2)

  5. Electric Field Lines Electric Field Lines for multiple charges - + Just add E-Field vectors at every point

  6. CheckPoint: Field Lines - Two Point Charges 1 Compare the magnitudes of the two charges • |Q1| > |Q2| • |Q1| = |Q2| • |Q1| < |Q2| • There is not enough information to determine the relative magnitudes of the charges.

  7. CheckPoint: Field Lines - Two Point Charges 1 Compare the magnitudes of the two charges |Q1| > |Q2| |Q1| = |Q2| |Q1| < |Q2| There is not enough information to determine the relative magnitudes of the charges.

  8. CheckPoint: Field Lines - Two Point Charges 3 Compare the magnitudes of the electric fields at points A and B. |EA| > |EB| |EA| = |EB| |EA| < |EB| There is not enough information to determine the relative magnitudes of the fields at A and B.

  9. Flux E Q A Same number of field lines but different areas. Flux  Amount of electric field passing through an surface.  Count field lines. F = |E|*|A|*cosQ.

  10. Electric Flux “Counts Field Lines” SAMPLE DATA: PLEASE REPLACE WITH OWN “I’m very confused by the general concepts of flux through surface areas. please help” Flux through surface S Integral of on surface S

  11. Example 3.1: Flux through a plate |E|=12N/C Q A 30o A 3cm X 3cm copper plate is placed in a constant horizontal electric field with a strength of 12N/C at an angle of 30o to the horizontal. What is the electric flux through the plate?

  12. CheckPoint: Flux from Uniformly Charged Rod An infinitely long charged rod has uniform charge density of l, and passes through a cylinder (gray). The cylinder in case 2 has twice the radius and half the length compared to the cylinder in case 1. • Compare the magnitude of the flux through the surface of the cylinder in both cases. • Φ1 = 2 Φ2 • Φ1 = Φ2 • Φ1 = 1/2 Φ2 • None of these TAKE s TO BE RADIUS !

  13. Direction Matters: For a closed surface, points outward

  14. Direction Matters: For a closed surface, points outward

  15. Gauss’ Law for simple point charge Gauss Law

  16. CheckPoint Results: Flux Point Charge (Sphere) 1 A positive charge (blue) is contained inside a spherical shell (black). • How does the electric flux through the two surface elements, dΦA and dΦB change when the charge is moved from position 1 to position 2? • dΦA increases and dΦB decreases • dΦA decreases and dΦB increases • Both dΦA and dΦB do not change

  17. CheckPoint: Flux from Point Charge (Sphere) 2 A positive charge (blue) is contained inside a spherical shell (black). • How does the flux ΦE through the entire surface change when the charge is moved from position 1 to position 2? • ΦE increases • ΦE decreases • ΦE does not change

  18. Think of it this way: +Q +Q 2 1 The total flux is the same in both cases (just the total number of lines) The flux through the right hemisphere is smaller for case 2.

  19. Things to notice about Gauss 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.

  20. Things to notice about Gauss Law “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 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.

  21. A solid conducting sphere with net charge –Q and a radius r1. What is the electric field at within the sphere some distance r2 from the center? (r2<r1) Question -Q r1 r2 • -kQ/r12 • -kQ/r22 • Zero • None of the above

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