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WKU PHYS 260 - Chapter 22 FLUX

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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 222FluxNumber of objects passing through a surface3Electric Flux, is proportional to the number of electric field lines passing through a surfaceAssumes that the surface is perpendicular to the linesIf not, then we use a cosine of the angle between them to get the components that are parallelMathematically: AEAEcos4Simple CasesE=EAEAA=0=EAcosEAE cos5From  to A represents a sum over a large a collection of objectsIntegration is also a sum over a collection of infinitesimally small objects, in our case, small areas, dASoAdEytechnicallthendxdyrepresentsAdS inceAdE,,6Gauss’s LawThe 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 chargeIn order to count the field lines, we must enclose the charges in some geometrical surface (one that we choose)7Mathematicallyq0q0enclosedqAdE Charge enclosed within bounding limits of this closed surface integral8Fluxes, Fluxes, Fluxes93 ShapesSphereCylinderPillbox10SphereWhen to use: around spherical objects (duh!) and point chargesHey! 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(2rEAdE11Sphere Example305022532334)4()4(34)34)((:rAErArErEAdErArArqVqspheretheInsid eenclosedenclosedWhat 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)((:rbAEbArErEAdEbAbAbqVqspheretheOutsideenclosedtotalAt r=b, both of these expressions should be equal12CylinderWhen to use: around cylindrical objects and line chargesClosed surface integral yields:r is the radius of the geometrical object that you are creating and L is the length of the cylinder)2( rLEAdEL13Cylinder ExampleWhat if you had an infinitely long line of charge with a linear charge density, ?rrEo rrELrLELqrLEAdEenclosedˆ22)2()2(00014PillboxWhen to use: around flat surfaces and sheets of chargeClosed surface integral yields:A is the area of the pillboxEAAdE 15Charge Isolated Conductor in Electrostatic EquilibriumIf 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 zeroIf the charges are not on the outside, you are only left with the surfaceA 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))((00017Electric field on a conducting sheetAqLet +++++++++++++++++++++++++++++++++++++++++++++AEnEorEAEASoAqEAEAAdEenclosedˆ0000So a conductor has 2x the electric field strength as the infinite sheet of charge18A differential view of Gauss’s LawRecall 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)  AdvdvSuspiciously 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) 000EsodqandqdEqAdEenclosedenclosedenclosedSo how the E-field spreads out from a point depends on the amount of charge density at that


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