Chapter 22FluxElectric Flux, FSimple CasesFrom S toGauss’s LawMathematicallyFluxes, Fluxes, Fluxes3 ShapesSphereSphere ExampleCylinderCylinder ExamplePillboxCharge Isolated Conductor in Electrostatic EquilibriumElectric field on an infinitely large sheet of chargeElectric field on a conducting sheetA differential view of Gauss’s LawDivergence Theorem (aka Gauss’s Thm or Green’s Thm)Div(E)1Chapter 222FluxNumber of objects passing through a surface3Electric Flux, is proportional to the number of electric field lines passing through a surfaceAssumes that the surface is perpendicular to the linesIf not, then we use a cosine of the angle between them to get the components that are parallelMathematically: AEAEcos4Simple CasesE=EAEAA=0=EAcosEAE cos5From to A represents a sum over a large a collection of objectsIntegration is also a sum over a collection of infinitesimally small objects, in our case, small areas, dASoAdEytechnicallthendxdyrepresentsAdS inceAdE,,6Gauss’s LawThe field lines emitted by a charge are proportional to the size of the charge.Therefore, the electric field must be proportional to the size of the chargeIn order to count the field lines, we must enclose the charges in some geometrical surface (one that we choose)7Mathematicallyq0q0enclosedqAdE Charge enclosed within bounding limits of this closed surface integral8Fluxes, Fluxes, Fluxes93 ShapesSphereCylinderPillbox10SphereWhen to use: around spherical objects (duh!) and point chargesHey! What if an object is not one of these objects?Closed surface integral yields:r is the radius of the geometrical object that you are creating)4(2rEAdE11Sphere Example305022532334)4()4(34)34)((:rAErArErEAdErArArqVqspheretheInsid eenclosedenclosedWhat if you had a sphere of radius, b, which contained a material whose charge density depend on the radius, for example, =Ar2 where A is a constant with appropriate units?2505022532334)4()4(34)34)((:rbAEbArErEAdEbAbAbqVqspheretheOutsideenclosedtotalAt r=b, both of these expressions should be equal12CylinderWhen to use: around cylindrical objects and line chargesClosed surface integral yields:r is the radius of the geometrical object that you are creating and L is the length of the cylinder)2( rLEAdEL13Cylinder ExampleWhat if you had an infinitely long line of charge with a linear charge density, ?rrEo rrELrLELqrLEAdEenclosedˆ22)2()2(00014PillboxWhen to use: around flat surfaces and sheets of chargeClosed surface integral yields:A is the area of the pillboxEAAdE 15Charge Isolated Conductor in Electrostatic EquilibriumIf excess charge is placed on an isolated conductor, the charge resides on the surface. Why?If there is an E-field inside the conductor then it would exert forces on the free electrons which would then be in motion. This is NOT electrostatic.Therefore, if there is no E-field inside, then, by Gauss’s Law, the charge enclosed inside must be zeroIf the charges are not on the outside, you are only left with the surfaceA caveat to this is that E-field lines must be perpendicular to the surface else free charges would move.16Electric field on an infinitely large sheet of chargeAqLet +++++++++++++++++++++++++++++++++++++++++++++AEEnEorEAEASoAqEAAEEAAdEenclosedˆ2222))((00017Electric field on a conducting sheetAqLet +++++++++++++++++++++++++++++++++++++++++++++AEnEorEAEASoAqEAEAAdEenclosedˆ0000So a conductor has 2x the electric field strength as the infinite sheet of charge18A differential view of Gauss’s LawRecall the Divergence of a field of vectorsvv)(DivDiv=0Div=+largeHow much the vector diverges around a given point19Divergence Theorem (aka Gauss’s Thm or Green’s Thm) AdvdvSuspiciously like LHS of Gauss’s LawSum of the faucets in a volume = Sum of the water going thru the surfaceA place of high divergence is like a faucetBounded surface of some region20Div(E) 000EsodqandqdEqAdEenclosedenclosedenclosedSo how the E-field spreads out from a point depends on the amount of charge density at that
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