Gauss's Law and Electric Flux

1 P05 -Class 05: OutlineHour 1:Gauss’ LawHour 2:Gauss’ Law2 P05 -Six PRS Questions On Pace and Preparation3 P05 -Last Time:Potential and E Field4 P05 -E Field and Potential: CreatingA point charge qcreates a field and potential around it:2ˆ;eeqqkVkrr==Er􀁇Use superposition for systems of charges;BBAAVVVVd=−∇∆≡−=−⋅∫EEs􀁇􀁇􀁇They are related:5 P05 -E Field and Potential: Effectsq=FE􀁇􀁇If you put a charged particle, q, in a field:To move a charged particle, q, in a field:WUqV=∆=∆6 P05 -Two PRS Questions:Potential & E Field7 P05 -Gauss’s LawThe first Maxwell Equation A very useful computational technique This is important!8 P05 -Gauss’s Law –The IdeaThe total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside9 P05 -Gauss’s Law –The Equation0SsurfaceclosedεinEqd=⋅=Φ∫∫AE􀁇􀁇Electric flux ΦE(the surface integral of E over closed surface S)is proportional to charge inside the volume enclosed by S10 P05 -Now the Details11 P05 -Electric Flux ΦECase I: E is constant vector field perpendicular to planar surface S of area A∫∫⋅=ΦAE􀁇􀁇dEEEAΦ=+Our Goal: Always reduce problem to this12 P05 -Electric Flux ΦECase II: E is constant vector field directed at angle θto planar surface S of area AcosEEAθΦ=∫∫⋅=ΦAE􀁇􀁇dE13 P05 -PRS Question:Flux Thru Sheet14 P05 -Gauss’s Law0SsurfaceclosedεinEqd=⋅=Φ∫∫AE􀁇􀁇Note: Integral must be over closed surface15 P05 -Open and Closed SurfacesA rectangle is an open surface —it does NOT contain a volume A sphere is a closed surface —it DOES contain a volume 16 P05 -Area Element dA: Closed SurfaceFor closed surface, dAis normal to surface and points outward( from inside to outside)ΦE > 0 if E points outΦE < 0 if E points in17 P05 -Electric Flux ΦECase III: E not constant, surface curved EddΦ=⋅EA􀁇􀁇SAdE∫∫Φ=ΦEEd18 P05 -Example: Point ChargeOpen Surface19 P05 -Example: Point ChargeClosed Surface20 P05 -PRS Question:Flux Thru Sphere21 P05 -Electric Flux: Sphere Point charge Qat center of sphere, radius rE field at surface:20ˆ4Qrπε=Er􀁇Electric flux through sphere:20Sˆˆ4QdArπε=⋅∫∫rr􀁷SEdΦ=⋅∫∫EA􀁇􀁇􀁷rAˆdAd=􀁇0Qε=22044Qrrππε=20S4QdArπε=∫∫􀁷22 P05 -Arbitrary Gaussian Surfaces 0SsurfaceclosedεQdE=⋅=Φ∫∫AE􀁇􀁇For all surfaces such as S1, S2or S323 P05 -Applying Gauss’s Law1.Identify regions in which to calculate E field.2.Choose Gaussian surfaces S: Symmetry3.Calculate 4.Calculate qin, charge enclosed by surface S5.Apply Gauss’s Law to calculate E:0SsurfaceclosedεinEqd=⋅=Φ∫∫AE􀁇􀁇∫∫⋅=ΦSAE􀁇􀁇dE24 P05 -Choosing Gaussian SurfaceChoose surfaces where Eis perpendicular & constant. Then flux is EA or -EA.Choose surfaces where Eis parallel.Then flux is zeroOREAEA−EExample: Uniform FieldFlux is EA on topFlux is –EA on bottomFlux is zero on sides25 P05 -Symmetry & Gaussian SurfacesSymmetryGaussian SurfaceSphericalConcentric SphereCylindricalCoaxial CylinderPlanarGaussian “Pillbox”Use Gauss’s Law to calculate E field from highly symmetric sources26 P05 -PRS Question:Should we use Gauss’ Law?27 P05 -Gauss: Spherical Symmetry+Quniformly distributed throughout non-conducting solid sphere of radius a. Find Eeverywhere 28 P05 -Gauss: Spherical SymmetrySymmetry is Spherical Use Gaussian SpheresrEˆE=􀁇29 P05 -Gauss: Spherical SymmetryRegion 1: r> aDraw Gaussian Sphere in Region 1 (r> a)Note: ris arbitrary butis the radius for which you will calculate the E field!30 P05 -Gauss: Spherical SymmetryRegion 1: r> aTotal charge enclosed qin= +QSEdAEA==∫∫􀁷SEdΦ=⋅∫∫EA􀁇􀁇􀁷()24Erπ=2004inEqQrEπεεΦ===204QErπε=20ˆ4Qrπε⇒=Er􀁇31 P05 -Gauss: Spherical SymmetryGauss’s law:QarQarqin⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛=33333434ππRegion 2: r< aTotal charge enclosed:()24EErπΦ=304QrEaπε=30ˆ4Qraπε⇒=Er􀁇ORinqVρ=3300inqrQaεε⎛⎞==⎜⎟⎝⎠32 P05 -PRS Question:Field Inside Spherical Shell33 P05 -Gauss: Cylindrical SymmetryInfinitely long rod with uniform charge density λFind Eoutside the rod.34 P05 -Gauss: Cylindrical SymmetrySymmetry is Cylindrical Use Gaussian CylinderrEˆE=􀁇Note: ris arbitrary butis the radius for which you will calculate the E field!􀁁is arbitrary and should divide out35 P05 -Gauss: Cylindrical Symmetry􀁁λ=inqTotal charge enclosed:()002inqErλπεε===􀁁􀁁SSEdEdAEAΦ=⋅==∫∫∫∫EA􀁇􀁇􀁷􀁷02Erλπε=0ˆ2rλπε⇒=Er􀁇36 P05 -Gauss: Planar SymmetryInfinite slab with uniform charge density σFind Eoutside the plane37 P05 -Gauss: Planar SymmetrySymmetry is Planar Use Gaussian PillboxxEˆE±=􀁇xˆGaussianPillboxNote: Ais arbitrary (its size and shape) and should divide out38 P05 -Gauss: Planar SymmetryAqinσ=Total charge enclosed:NOTE: No flux through side of cylinder, only endcaps()002inqAEAσεε===SSEEndcapsdEdAEAΦ=⋅==∫∫∫∫EA􀁇􀁇􀁷􀁷++++++++++++σE􀁇E􀁇xA{}0ˆto rightˆ-to left2σε⇒=xEx􀁇02Eσε=39 P05 -PRS Question:Slab of Charge40 P05 -Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). Find Eeverywhere.xˆ41 P05 -PRS Question:Slab of Charge42 P05 -Potential from E43 P05 -Potential for Uniformly Charged Non-Conducting Solid SphereFrom Gauss’s Law2030ˆ,4ˆ,4QrRrQrrRRπεπε⎧>⎪⎪=⎨⎪<⎪⎩rEr􀁇BBAAVVd−=−⋅∫Es􀁇􀁇Use()0BVV=−∞􀀈􀀉􀀊Region 1: r> a204rQdrrπε∞=−∫014Qrπε=Point Charge!44 P05 -Potential for Uniformly Charged Non-Conducting Solid Sphere()()RrRdrErRdrErR∞=−>−<∫∫Region 2: r< a()0DVV=−∞􀀈􀀉􀀊220138QrRRπε⎛⎞=−⎜⎟⎝⎠230044RrRQQrdrdrrRπεπε∞=−−∫∫()22300111442QQrRRRπεπε=−−45 P05 -Potential for Uniformly Charged Non-Conducting Solid Sphere46 P05 -Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). If V=0 at x=0 (definition) then what is V(x) for x>0?xˆ47 P05 -Group Problem: Spherical ShellsThese two spherical shells have equal but opposite charge.Find E everywhereFind V everywhere (assume V(∞) = 0)