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Ch 24: Gauss’s Law 24.1 Electric Flux

Ch 24: Gauss’s Law 24.1 Electric Flux.

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Ch 24: Gauss’s Law 24.1 Electric Flux

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  1. Ch 24: Gauss’s Law 24.1 Electric Flux Think of flowing water as an analogy where the flow lines would be mass flow (mass/time) per area. Here E is field lines per area. We define a quantity called electric flux, E which is EA in this simple case. Electric flux is total number of lines through the area. In the water analogy we would have mass flow.

  2. 24.1 Electric Flux The total electric flux E which goes through the vertical plane A also goes through the diagonal plane. Noting that EA = EAcos, we see that E = E•A, where A is a vector normal to the area of value equal to the area, is a consistent definition for E. For small area patches dE = E•dA

  3. E = E•A = EAcosi

  4. E is positive. Outflow is positive.

  5. E is negative. Inflow is negative.

  6. E is zero. Inflow equals outflow.

  7. Ch 24: Gauss’s Law 24.2 Gauss’s Law

  8. CT1: S1 equals CT2: S2 equals CT3: S3 equals CT4: S4 equals -3Q • -3Q/0 • -2Q /0 • -Q /0 • 0 • Q /0 • 2Q /0 • 3Q /0

  9. CT5: A icosahedron has 20 equal triangular faces as pictured above. Assume a charge q is placed at the center of the icosahedron (an equal distance from each face). Using the symmetry of the situation, determine how much electric flux goes through each face. • q/0 • q/40 • q/60 • q/200 • none of the above

  10. Gauss's law: using superposition qin is the sum of the charges enclosed by the Gaussian surface. E = = qin / 0 A. Symmetries: 1) spherical 2) cylindrical (linear) 3) planar B. Method: 1) note symmetry 2) draw appropriate Gaussian surface 3) calculate electric flux E 4) set E = qin / 0 5) solve for E

  11. 24.4 Perfect Conductors inElectrostatic Equilibrium • E = 0 inside perfect conductors • The charge must reside on the surface of a perfect conductor • E =  / 0n where n is a unit vector normal to the surface and pointing outward. •  is greatest at points of least radii of curvature (i.e. pointy)

  12. Before Class Assignment 2/6/08 13 correct 1 I don’t know 1 no explanation

  13. P24.44 (p. 689)

  14. 24.3 Application of Gauss’s Law to Various Charge Distributions • CT6: At a distance r >c from the common center, the electric field is • 2Q/40r2 • -3Q/40r2 • -Q/40r2 • -Q/40r • constant 2Q -3Q

  15. P24.39 (p. 689)

  16. CT7: The electric field in the conducting cylinder is • 0 • /20 • /40 • /20r • /40r • r/40 • r/40

  17. CT8: The electric field between the wire and the cylinder is • 0 • /20 • /40 • /20r • /40r • r/40 • r/40

  18. P24.35 (p. 688) Left 2 2 1 3 3 E 1 Gaussian Surface 1: A squat cylinder Gaussian Surface 2: A squat cylinder Right

  19. A. B. C. D. E. conducting plates so  on each side

  20. Table 24-1, p.754

  21. Summary • Spherical Symmetry: use concentric, spherical Gaussian surface • Cylindrical Symmetry: use concentric, cylindrical Gaussian surface, assume far from ends of long thin cylinder • Planar Symmetry: use a disk shaped Gaussian surface, assume far from edges of planar surface Gauss’s Law: E = = qin/0 E will be constant over the surface and either parallel or normal to the area so will either be EA or 0. You may have to integrate to get qin.

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