PHY2049: Chapter 231Chapter 23: Gauss’ LawPHY2049: Chapter 232Coulomb’s LawGauss’ LawDerivation given in Sec. 23-5 (Read yourself)Not derived in this book (Requires vector calculus) Equivalent!We will focus on what is Gauss’ law and how we use it.Two Equivalent Laws of Static ElectricityPHY2049: Chapter 233Electric FluxÎSimple definition of electric flux (E constant, flat surface) E at an angle θ to planar surface, area A Units = N m2/ C (SI units)ÎSimple example Let E = 104N/C pass through 2m x 5m rectangle, 30° to normal φE= 104* 10 * cos(30°) = 100,000 * 0.866 = 86,600 Nm2/CÎMore general ΦEdefinition (E variable, curved surface)cosEEAθΦ≡⋅ =EAESdΦ≡ ⋅∫EANormalEPHY2049: Chapter 234Example of Constant FieldN/C ˆ4i=E2m )ˆ3ˆ2( ji +=AC/Nm 8 )ˆ3ˆ2(ˆ4Φ2=+⋅=⋅≡ jiiEAEPHY2049: Chapter 235Flux Through Closed SurfaceÎSurface elements dA always point outward!ÎAs a result, sign of ΦE isE outward (+)E inward (−)ΦE< 0ΦE= 0ΦE> 0PHY2049: Chapter 236Example: Flux Through Cubic SurfaceÎE field is constant:Flux through front face?Flux through back face?Flux through top face?Flux through whole cube?ˆEEz=rPHY2049: Chapter 237Example: Flux Through Cylindrical SurfaceÎAssume E is constant, to the rightFlux through left face?Flux through right face?Flux through curved sideTotal flux through cylinder?PHY2049: Chapter 238Example: Flux Through Spherical SurfaceÎPoint charge +Q at centerÎE points radially outward (normal to surface!)()22044ESdkQrrQkQππεΦ= ⋅===∫EAForeshadowing of Gauss’ Law!Does not depend on the radius of the sphere!PHY2049: Chapter 239Gauss’ LawÎGeneral statement of Gauss’ lawÎCan be used to calculate E fields. But remember Outward E field, flux > 0 Inward E field, flux < 0ÎCan be useful in finding charge distribution (as we shall see)ÎConsequences of Gauss’ law (as we shall see) Excess charge on conductor is always on surface E is always normal to surface on conductor (Excess charge distributes on surface in such a way)ConductorIntegration over closedsurfaceqencis charge insidethe surfaceCharges outside surface have no effect0encεqdS=⋅∫AE(This does not mean they do not contribute to E.)PHY2049: Chapter 2310Reading QuizÎWhat is the electric flux through a sphere of radius R surrounding a charge +Q at the center?1) 02) +Q/ε03) -Q/ε04) +Q5) -QPHY2049: Chapter 2311QuestionHow does the flux ФEthrough the entire surface change when the charge +Q is moved from position 1 to position 2?a) ФE increasesb) ФE decreasesc) ФE doesn’t changedSdS12+Q+QJust depends on charge,not positionPHY2049: Chapter 2312Power of Gauss’ Law: Calculating E FieldsÎValuable for cases with high symmetry E = constant, ⊥ surface E || surfaceÎSpherical symmetry E field vs r for point charge E field vs r inside uniformly charged sphere Charges on concentric spherical conducting shellsÎCylindrical symmetry E field vs r for line charge E field vs r inside uniformly charged cylinderÎRectangular symmetry E field for charged plane E field between conductors, e.g. capacitorsEAdS±=⋅∫AE0=⋅∫SdAEPHY2049: Chapter 2313ExampleÎ4 Gaussian surfaces: 2 cubes and 2 spheresÎRank magnitudes of E field on surfacesÎWhich ones have variable E fields?ÎWhat are the fluxes through each of the Gaussian surfaces(a) E falls as radius increases(b) E non-constant on cube (r changes)(c) Fluxes are same, +Q/ε0PHY2049: Chapter 2314Derive Coulomb’s Law From Gauss’ LawÎCharge +Q at a point By symmetry, E must be radiallysymmetricÎDraw Gaussian surface around point Sphere of radius r E field has constant magnitude, ⊥ to Gaussian surface2204QkQErrπε==Gauss’ LawSolve for E02)4(εQrπEdS==⋅∫AEPHY2049: Chapter 2315ExampleÎCharges on a ball and spherical shells, eachuniformly charged:+Q (ball at center)+3Q (middle shell)-2Q (outer shell)ÎFind fluxes through the three Gaussian surfaces(a) Inner +Q/ε0(b) Middle +4Q/ε0(c) Outer
View Full Document
Unlocking...