ch33-p103 - 103. (a) From Eq. 33-1, 2 E 2 = Em sin( kx t )...

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22 2 . B B c tx ∂∂ = (b) From E E f kx t m () , ω = ∂± = 2 2 2 2 2 2 2 E t E fk x t t E df du mm uk x t and c E x cE x t t cEk du 2 2 2 2 2 2 2 2 = = Since = ck the right-hand sides of these two equations are equal. Therefore, 2 . EE c = Changing E to B and repeating the derivation above shows that BB m = ± satisfies 2 . B B c = 103. (a) From Eq. 33-1, = −= 2 2 2 2 2 E tt Ek x t x t sin( ) sin( ), ωω and c E x c x x t k ck x t E k x t 2 2 2 2 2 2 2 = = ) )
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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