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week5

# week5 - Physics 1C Winter Quarter 2009 W Gekelman Notes-5...

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1 Physics 1C Winter Quarter 2009 W. Gekelman – Notes-5© Let us write down Maxwell’s equations in both integral and differential form. Differential Form Integral Form ! ! E = " # 0 ! E i ˆ ndA = q ! 0 " ! ! B = 0 ! B i ˆ ndA = 0 ! ! " ! E = # \$ ! B \$ t ! E i d ! l = ! d dt " " ! B i ˆ ndA " ! " ! B = μ 0 ! j + μ 0 # 0 \$ ! E \$ t ! B i d ! l = μ 0 " ! i + μ 0 " 0 d dt ! E i ˆ ndA ! What are the curl the gradient and the divergence? In rectangular coordinates: ! " ! E = ˆ i ˆ j ˆ k # # x # # y # # z E x E y E z \$ % & & & & & ( ) ) ) ) ) = # E z # y * # E y # z + , - . / 0 ˆ i + # E x # z * # E z # x + , - . / 0 ˆ j + # E y # x * # E x # y + , - . / 0 ˆ k CURL ! i ! E = " E x " x + " E y " y + " E z " z DIVERGENCE ! 2 ! E = " 2 E x " x 2 + " 2 E y " y 2 + " 2 E z " z 2 These equations are all familiar except for the term in red. This is Maxwell’s great contribution. How do you go from integral to differential form? Take Faraday’s law: ! E i d ! l = ! d dt " " ! B i ˆ ndA " This is integral form. Then use the mathematical theorem: ! A i d ! l = ! " ! A ( ) # " # i ˆ ndA This is true for any vector A ! " ! E ( ) i ˆ ndA = # d dt " \$ ! B i ˆ ndA \$ and equating the integrands: ! " ! E = # \$ ! B \$ t For Coulombs law:

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2 ! E i ˆ ndA = q ! 0 " Now use the vector divergence theorem: ! i ! E ( ) " dV = ! E i ˆ ndA " Since the charge is the integral over the volume charge density: q = ! dV " we arrive at: ! i ! E ( ) " dV = ! E i ˆ ndA " = 1 # 0 \$ dV " . Again equating the integrands: ! ! E = " # 0 Consider a capacitor when it is charging Inside the capacitor no charge flows and Ampere’s law with Maxwell’s term becomes: (1) ! B i d ! l = μ 0 " ! " 0 # # t ! E i ˆ ndA + μ 0 ! i = " 0 μ 0 # # t ! E i ˆ ndA ! We can integrate both sides: 2 ! rB " = μ 0 # 0 dE dt ! r 2 B " = μ 0 # 0 dE dt r 2 for r<a There is a magnetic field inside of the capacitor. Part of right hand side of the integral in equation (1) must have the dimension of current; I = ! 0
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