18.02A Topic 50: Applications and interpretations of Stokes’ theorem.Author: Jeremy OrloffRead: SN: V15Read section V15 in the supplementary notes.Divergence and Stokes’ Theorems in del notationDivergence Theorem:ZZSF · n dS =ZZZDdivF dA =ZZZD∇ · F dA.Stokes’ Theorem:ICF · dr =ZZScurlF · n dS =ZZS∇ × F · n dS.Maxwell’s EquationsS = closed surface D = interior of SC1= closed curve S1= surface capping C1E = electric field B = magnetic fieldQ = charge inside Sρ = charge density inside S J = current densityk and k1are constants depending on units.Integral form of Maxwell’s equations1’)ZZSE · n dS = 4πkQ Gauss-Coulomb Law.2’)ZZSB · n dS = 0 Gauss’ law of magnetism.3’)IC1E · dr = −ddtZZS1B · n dS Faraday’s law.4’)IC1B · dr = k1ZZS1J · n dS +k1kddtZZS1E · n dS Ampere’s law.Differential form of Maxwell’s equations (equivalent to the integral form).1) ∇ · E = 4πkρ Gauss-Coulomb Law.2) ∇ · B = 0 Gauss’ law of magnetism.3) ∇ × E = −∂B∂tFaraday’s law.4) ∇ × B = k1J +k1k∂E∂tAmpere’s law.In words these statements say the following:1,1’) Flux of E through S = 4πk×charge enclosed.2,2’) There is no magnetic monopole. (No net magnetic charge enclosed.)3,3’) A changing magnetic field induces an electric field.4,4’) Magnetic fields are induced by either a current or a changing electric field.118.02A topic 50 2We discussed Gauss’ law in topic 46 with respect to gravitation. Here’s a quick recap forelectricity.For a charge q at the origin the electric field is Eq= khx, y, ziρ3(here ρ is distance from theorigin, not charge density).By direct computation the flux of Eqthrough a sphere centered on the origin is 4πkq.Everywhere but the origin ∇ · Eq= 0, so the extended divergence theorem gives flux of Eqthrough S =4πkq if S goes around q0 otherwise.In general, if q is placed anywhere else the same formula holds.Now suppose q1, q2, . . . , qnare charges with fields E1, E2, . . . , En. Let E = E1+ . . . + Enbe the net electric field. The net flux through S is the sum of the flux contributed by eachfield Ej. That is, the flux of E through S = 4πk · ×(total charge inside S).For a continuous charge distribution we replace sums by integrals and find the same thing.This proves (1’).Before showing (1) and (1’) are equivalent we state a very reasonable theorem.Theorem: Assume h(x, y, z) is a continuous function andRRRDh dV = 0 for every volumeD then h(x, y, z) = 0.Proof: We argue by contradiction. Suppose for some point (x0, y0, z0) we haveh(x0, y0, z0) > 0. Since h is continuous we can put a small ball D around (x0, y0, z0) suchthat h(x, y, z) > 0 throughout D. This impliesRRRDh dV > 0, which contradicts ourassumption. Therefore it can’t be positive. Likewise it can’t be negative. QEDCorollary: Assume f(x, y, z) and g(x, y, z) are continuous andRRRDf dV =RRRDg dVfor every volulme D then f(x, y, z) = g(x, y, z).Proof: Let h = f − g and apply the theorem.Now we show (1) and (1’) are equivalent.The divergence theorm impliesZZSE · n dS =ZZZD∇ · E dV .The definition of density says: 4πkQ = 4πkZZZDρ dV .Thus,ZZSE · n dS = 4πkQ ⇔ZZZD∇ · E dV = 4πkZZZDρ dV .Since this is true for every D the above corollary says it is equivalent to ∇ · E = 4πkρ. QEDShowing (2), (3) and (4) are equivalent to (2’), (3’) and (4’) respectively is similar.End of topic 50
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