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

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

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! S E +q (One of Maxwell’s 4 equations!)

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

What is total charge inside?

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.

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!

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

Two Insulating Sheets

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

E does not pass through a conductor 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

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

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! E r E=k(15C)/r2 E=k(5C)/r2 E=0 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

Electric Fields With Insulating Sphere

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.