18 02A Topic 50 Applications and interpretations of Stokes theorem Author Jeremy Orloff Read SN V15 Read section V15 in the supplementary notes Divergence and Stokes Theorems in del notation ZZ ZZZ ZZZ Divergence Theorem F n dS divF dA F dA S D D ZZ ZZ I F n dS curlF n dS F dr Stokes Theorem C Maxwell s Equations S closed surface C1 closed curve E electric field Q charge inside S 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 form of Maxwell s equations ZZ 1 E n dS 4 kQ Z ZS B n dS 0 2 I S ZZ d 3 E dr B n dS dt IC1 Z Z S1 ZZ k1 d 4 B dr k1 J n dS E 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 4 k 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 4 k 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 1 18 02A topic 50 2 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 is distance from the 3 origin not charge density By direct computation the flux of Eq through a sphere centered on the origin is 4 kq Everywhere but the origin Eq 0 so the extended divergence theorem gives flux of Eq 4 kq if S goes around q through S otherwise 0 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 4 k 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 D h dV 0 for every 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 putRRR a small ball D around x0 y0 z0 such that h x y z 0 throughout D This implies D h dV 0 which contradicts 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 D f dV D g dV 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 ZZZ D The definition of density says 4 kQ 4 k dV D ZZ ZZZ ZZZ Thus E n dS 4 kQ E dV 4 k dV S D D Since this is true for every D the above corollary says it is equivalent to E 4 k Showing 2 3 and 4 are equivalent to 2 3 and 4 respectively is similar End of topic 50 notes QED
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