homework_4_solution

homework_4_solution - y x 1 N+ mv+ 1 = 2 v+ c2 Ima 1 ( q 1...

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ˆ y ˆ x N + mv + 1 ° 1 v 2 + c 2 N mv 1 ° 1 v 2 c 2 = Ima q ( 1 ° 1 v 2 + c 2 1 ° 1 v 2 c 2 ) N + mv + N mv =0 Ima q ( 1 ° 1 v 2 + c 2 1 ° 1 v 2 c 2 ) = Ia q 1 c 2 E = Ia q 1 c 2 · ( E x Lq ) = 1 c 2 · E x · ( IaL )= 1 c 2 · E x · 2 μ
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E E x ˆ x μ Φ ( x, y, z )= 1 4 π° 0 λ ˆ L/ 2 0 dl 1 ° x 2 + y 2 +( l z ) 2 1 4 π° 0 λ ˆ 0 L/ 2 dl 1 ° x 2 + y 2 +( l z ) 2 = 1 4 π° 0 λ · [ ˆ L/ 2 z z dl ° 1 ° x 2 + y 2 + l 2 ˆ z L/ 2 z dl ° 1 ° x 2 + y 2 + l 2 ] = 1 4 π° 0 λ · [ln L 2 z + ± x 2 + y 2 +( L 2 z ) 2 z + ° x 2 + y 2 + z 2 ln z + ° x 2 + y 2 + z 2 L 2 z + ± x 2 + y 2 +( L 2 + z ) 2 ] ± p = ± e z ˆ L/ 2 L/ 2 ρ ( l ) · ldl = ± e z λ ( ˆ L/ 2 0 l · dl ˆ 0 L/ 2 l · dl )= λ · L 2 4 ± e z Q ij = ˆ (3 x ° i x ° j r ° 2 δ ij ) ρ ( ± x ° ) d 3 x ° ρ ( ± x ° )=0 x ° 1 ° =0 x ° 2 ° =0 Φ ( ± x )= 1 4 π° 0 ( q r + ± p · ± x r 3 + 1 2 ² i,j Q ij x i x j r 5 + o (( L/r ) 3 ) 1 4 π° ± p · ± x r 3 = 1 4 π° λ L 2 4 z r 3
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This note was uploaded on 12/29/2011 for the course PHYSICS 606 taught by Professor Bedaque during the Spring '11 term at Maryland.

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homework_4_solution - y x 1 N+ mv+ 1 = 2 v+ c2 Ima 1 ( q 1...

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