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Download Topic 50 Notes 50 Applications and and interpretation of Stokes theorem
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Topic 50 Notes Jeremy Orloff 50 Applications and and interpretation of Stokes theorem Read section V15 in the supplementary notes. Divergence and Stokes’ Z Z Theorems Zin Z Zdel notationZ Z Z Divergence Theorem: F · n dS = divF dA = ∇ · F dA. S D D I ZZ ZZ Stokes’ Theorem: F · dr = curlF · n dS = ∇ × F · n dS. C Maxwell’s Equations S = closed surface C1 = closed curve E = electric field Q = charge inside S ρ = charge density inside S S S D S1 B = = = interior of S surface capping C1 magnetic field J = current density k and k1 are constants depending on units. Integral Z Z form of Maxwell’s equations 1’) E · n dS = 4πkQ Z ZS 2’) B · n dS = 0 I S ZZ d 3’) E · dr = − B · n dS dt IC1 Z Z S1 ZZ k1 d E · n dS 4’) B · dr = k1 J · n dS + k dt S1 C1 S1 Gauss-Coulomb Law. Gauss’ law of magnetism. Faraday’s law. 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. ∂B 3) ∇ × E = − Faraday’s law. ∂t k1 ∂E 4) ∇ × B = k1 J + Ampere’s law. k ∂t 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.) 1 50 APPLICATIONS AND AND INTERPRETATION OF STOKES THEOREM 2 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. We discussed Gauss’ law in topic 46 with respect to gravitation. Here’s a quick recap for electricity. hx, y, zi For a charge q at the origin the electric field is Eq = k (here ρ is distance ρ3 from the origin, not charge density). By direct computation the flux of Eq through a sphere centered on the origin is 4πkq. Everywhere but the origin ∇ · Eq = 0, so the extended divergence theorem gives flux ( 4πkq if S goes around q of Eq through S = 0 otherwise. In general, if q is placed anywhere else the same formula holds. Now suppose q1 , q2 , . . . , qn are charges with fields E1 , E2 , . . . , En . Let E = E1 + . . . + En be the net electric field. The net flux through S is the sum of the flux contributed by each field 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. RRR Theorem: Assume h(x, y, z) is a continuous function and h dV = 0 for every D volume D then h(x, y, z) = 0. Proof: We argue by contradiction. Suppose for some point (x0 , y0 , z0 ) we have h(x0 , y0 , z0 ) > 0. Since h is continuous we can put aRRR small ball D around (x0 , y0 , z0 ) such that h(x, y, z) > 0 throughout D. This implies h dV > 0, which contradicts D our assumption. Therefore it can’t be positive. Likewise it can’t be negative. QED RRR RRR Corollary: Assume f (x, y, z) and g(x, y, z) are continuous and f dV = g dV D D for 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. ZZ ZZZ The divergence theorm implies E · n dS = ∇ · E dV . S D ZZZ The definition of density says: 4πkQ = 4πk ρ dV . D ZZ ZZZ ZZZ Thus, E · n dS = 4πkQ ⇔ ∇ · E dV = 4πk ρ dV . S D D Since this is true for every D the above corollary says it is equivalent to ∇ · E = 4πkρ. Showing (2), (3) and (4) are equivalent to (2’), (3’) and (4’) respectively is similar. QED 50 APPLICATIONS AND AND INTERPRETATION OF STOKES THEOREM 3