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ROCHESTER PHY 217 - PHY 217 Lecture 7B

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18 September 2002 Physics 217, Fall 2002 1Today in Physics 217 E as a vector field Gauss’ Law (Maxwell equation #1)zrφαdqdEσ18 September 2002 Physics 217, Fall 2002 2Divergence of E, and Gauss’ LawE for an arbitrary, static 3-D charge distribution:Take the Cartesian components of to be X,Y,Z, as those of r are x,y,z. Then()()() ()2222ˆˆ, soˆˆ.VVVdq dddρτρτ ρτ′′==′′ ′′⋅=⋅ = ⋅∫∫∫∫Er rEr rrrrrrrrr—— —′=−rrrXxxXXYyyYYZZzzZ ∂∂∂ ∂ ∂∂∂∂ ∂∂∂ ∂ == = = ∂∂∂∂ ∂ ∂∂∂ ∂∂∂∂ r——18 September 2002 Physics 217, Fall 2002 3Divergence of E, and Gauss’ Law (continued)Recall from Friday thatand we getIntegration of this over volume, and use of the divergence theorem, yields a more familiar result:()()332ˆ44 ,πδ πδ′⋅= = −rrrrrr—()()()344.Vdπδ ρ τ πρ′′′⋅= − =∫Errr r—Gauss’ Law, differential form()enclosed44VVSdddQτπρ τπ⋅=⋅=∫∫∫ErEav—Gauss’ Law, integral form18 September 2002 Physics 217, Fall 2002 4The curl of EAs before,Call the spherical components of () ()()222ˆˆˆ.VVVdddρτ ρτρτ′′ ′′×=× = ×′′=×∫∫∫Er rrrrrrrrr—— —— , , and ; then,θφrr2222222ˆˆˆ ˆ ˆ ˆ ˆ1sinsin sinˆˆˆ0, so0.φθ φθθθθ φ θφθ∂∂ ∂∂  ×= − + −  ∂∂ ∂∂  ∂∂+−∂∂=×=θφErr r r r rrrrrrrrrrrrrrrrrr——=0=0=0 =0=0=018 September 2002 Physics 217, Fall 2002 5The curl of E (continued)For any finite distribution of charges, Edecreases with as rapproaches infinity. This, taken together with means that Helmholtz’s theorem for irrotational fields applies (see Monday’s lecture notes): is the gradient of a scalar potential: .0 (just shown).is independent of path, given and .0.Vdd=−∇× =⋅⋅=∫∫baEEEEl a bElv—Electric potential18 September 2002 Physics 217, Fall 2002 6Summary of electrostatics, so far…expressed in the language of field theory, with all the empirical facts (like Coulomb’s Law) built in.()()00enclosed enclosed4 in MKS4 in MKS00SdQ QdVπρ ρ επε⋅= =⋅= =×=⋅==−∫∫EEaEElEvv———18 September 2002 Physics 217, Fall 2002 7Use of Gauss’ Law in integral formBy choosing a closed surface, over which to integrate, that fitsthe symmetry of the charge distribution, one can use Gauss’ Law to compute fields much more easily than with Coulomb’s Law. (As I’m sure you’re aware…)ExampleCalculate the electric field from an infinite plane, parallel tox-y,with constant charge per unit area σ; first, with Coulomb’s Law, and second, with Gauss’ Law. The answer, as you probably remember, is (+ above the plane, -below.)enclosed4SdQπ⋅=∫Eavˆ2.πσ=±Ez18 September 2002 Physics 217, Fall 2002 8zrφαUsing Coulomb’s Law:Break the plane into annuli withradius r and width dr, and break theannuli into segments of width The charge of each segment isHorizontal componentsof field from segmentsat φand φ+πcancel, and their vertical components add, so above the plane, we have:dqdE.rdφdqrdrdσφ=22222ˆˆ2cos 2dqrzddrdzrzrσαφ==++Ez zr()22123222 3200ˆˆˆˆ2212zzuE z d r r z dr z u du zπσ φ πσ πσ πσ∞∞∞−−−=+===−∫∫ ∫zzzz18 September 2002 Physics 217, Fall 2002 9Using Gauss’ LawFirst note that E must point perpendicularto, and away from, the plane, sincethe plane is infinite and there’s no difference between the view to the right and theview to the left. Thendraw a cylinder, bisected by the plane. By symmetry, E is perpendicular to the area element vectors on the cylinder walls, parallel to those on the circular faces, and constant on those faces, soHarder setup (finding and exploiting symmetry), easier math.zzEr222enclosed24 4, orˆ2.dEr Q rππ πσπσ⋅= =


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