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Gauss’s Law

Gauss’s Law. PH 203 Professor Lee Carkner Lecture 5. Gauss. F = q/ e 0 = ∫EdA Note that: Flux only depends on net q internal to surface For a uniform surface and uniform q, E is the same everywhere on surface so ∫EdA = EA. Cylinder. E field is always radially outward

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Gauss’s Law

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  1. Gauss’s Law PH 203 Professor Lee Carkner Lecture 5

  2. Gauss F = q/e0 = ∫EdA • Note that: • Flux only depends on net q internal to surface • For a uniform surface and uniform q, E is the same everywhere on surface • so ∫EdA = EA

  3. Cylinder • E field is always radially outward • We want to find E a distance r away • To solve Gauss’s Law: • F = q/e0 = ∫EdA = EA • q is lh • Solve for E

  4. Plane • We can again capture the flux with a cylindrical Gaussian surface • Useful for large sheet or point close to sheet

  5. Spherical Shell • Consider a spherical shell of charge of radius r and total charge q

  6. Surface within Sphere • What if we have total charge q, uniformly distributed with a radius R? • What if surface is inside R? • If we apply r3/R3 to the point charge formula we get, E = (q/4pe0R3)r

  7. Conductors and Charge • The charges in the conductor are free to move and so will react to each other • Like charges will want to get as far away from each other as possible • No charge inside conductor

  8. Charge Distribution • How does charge distribute itself over a surface? • e.g., a sphere • No component parallel to surface, or else the charges would move • Excess charge there may spark into the air

  9. Conductors and External • The positive charges will go to the surface “upfield” and the negative will go to the surface “downfield” • The field inside the conductor is zero • The charges in the conductor cancel out the external field • A conductor shields the region inside of it

  10. Conducting Ring No E Field Inside Field Lines Perpendicular to Surface Charges Pushed To Surface

  11. Faraday Cage • If we make the conductor hollow we can sit inside it an be unaffected by external fields • Your car is a Faraday cage and is thus a good place to be in a thunderstorm

  12. Next Time • Read 24.1-24.6 • Problems: Ch 23, P: 24, 36, 45, Ch 24, P: 2, 4

  13. A uniform electric field of magnitude 1 N/C is pointing in the positive y direction. If the cube has sides of 1 meter, what is the flux through sides A, B, C? • 1, 0, 1 • 0, 0, 1 • 1, 0, 0 • 0, 0, 0 • 1, 1, 1 B A C

  14. Consider three Gaussian surfaces. Surface 1 encloses a charge of +q, surface 2 encloses a charge of –q and surface 3 encloses both charges. Rank the 3 surfaces according to the flux, greatest first. 1, 2, 3 1, 3, 2 2, 1, 3 2, 3, 1 3, 2, 1 +q -q 1 2 3

  15. Rank the following Gaussian surfaces by the amount of flux that passes through them, greatest first (q is at the center of each). 1, 2, 3 1, 3, 2 2, 1, 3 3, 2, 1 All tie q q q 1 2 3

  16. Rank the following Gaussian surfaces by the strength of the field at the surface at the point direction below q (where the numbers are). 1, 2, 3 1, 3, 2 2, 1, 3 3, 2, 1 All tie q q q 1 2 3

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