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Physics 2113 Lecture 10

Physics 2113 Jonathan Dowling. Flux Capacitor (Operational). Physics 2113 Lecture 10. Gauss ’ Law III. Carl Friedrich Gauss 1777 – 1855. Constant Area:. A. dA. Changing Area:. E. What? The Flux!. S. +q. E. Gauss ’ Law: General Case.

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Physics 2113 Lecture 10

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  1. Physics 2113 Jonathan Dowling Flux Capacitor (Operational) Physics 2113Lecture 10 Gauss’ Law III Carl Friedrich Gauss 1777 – 1855

  2. Constant Area: A dA Changing Area: E What? The Flux!

  3. S +q E Gauss’ Law: General Case • Consider any ARBITRARY CLOSED surface S -- NOTE: this “Gaussian Surface” does NOT have to be a “real” physical object! • The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED! • The results of a complicated integral is a very simple formula: it avoids long calculations! (One of Maxwell’s 4 equations!)

  4. Examples What is Flux Through Surfaces: S1 = S2 = S3 = S4 = +q/ε0 –q/ε0 0 0

  5. What is total charge inside?

  6. Gauss’ Law: Insulating Plate • Infinite INSULATING plane with uniform charge density s • E is NORMAL (perpendicular) to plane • Construct Gaussian box as shown Surface Charge; σ = q/A Units: [C/m2] For an insulator, E=σ/2ε0, and for a conductor, E=σ/ε0.

  7. Recall Disk of Charged Sheet From Last Week! • Add the Vectors! • Horrible Integral! • Trig Substitution! • So Hard We Didn’t Do It! If the Disk Has Large Radius (R>> z) … Blah, Blah, Blah… With Gauss’s Law We Got Same Answer With Two Lines of Algebra!

  8. Q Q/2 Insulating and Conducting Planes Insulating Plate: Charge Distributed Homogeneously. Conducting Plate: Charge Distributed on the Outer Surfaces. Electric Field Inside a Conductor is ZERO!

  9. Two Insulating Sheets

  10. The field from the plates cancels out so ignore them and use Coulombs inverse square law for the central charge only.

  11. Two Conducting Sheets E does not pass through a conductor Formula for E different by Factor of 2 7.68 4.86 4.86 7.68 12.54

  12. Gauss’ Law: Spherical Symmetry • Consider a POINT charge q & pretend that you don’t know Coulomb’s Law • Use Gauss’ Law to compute the electric field at a distance r from the charge • Use symmetry: • place spherical surface of radius R centered around the charge q • E has same magnitude anywhere on surface • E normal to surface r q

  13. +10 C E -15C E=k(15C)/r2 E=k(5C)/r2 E=0 r Electric Fields With Spherical Symmetry: Shell Theorem A spherical shell has a charge of +10C and a point charge of –15C at the center. What is the electric field produced OUTSIDE the shell? If the shell is conducting? Field Inside a Conductor is ZERO! And if the shell is insulating? Charged Shells Behave Like a Point Charge of Total Charge “Q” at the Center Once Outside the Last Shell! Conducting

  14. Electric Fields With Insulating Sphere

  15. Summary • Electric Flux: a Surface Integral (Vector Calculus!); Useful Visualization: Electric Flux Lines Like Wind Through a Window. • Gauss’ Law Provides a Very Direct Way to Compute the Electric Flux.

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