hw3sol.pdf - HW III Solution to 3.12 In 2.52 we saw that...

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HW III Solution to 3.12: In 2.52 we saw that two wires of linear charges + λ, - λ would generate symmetric equipotential surfaces of + V 0 , - V 0 . The equations were ( y 0 = d in this problem) d = acoth (2 π 0 V 0 ) , R = acsch (2 π 0 V 0 ) . The unknowns are the vector a = ( a, 0 , 0) which represents the displacement of the image wires, and the charge density λ . The system above can be solved by dividing the two equations to obtain λ = 2 π 0 V 0 arcosh ( d/R ) , a = d 2 - R 2 . The potential is then V ( r ) = λ 2 π 0 ln | r + a | | r - a | . Solution to 3.19: cos 3 θ = 4 cos 3 θ - 3 cos θ . Since the expression contains only odd functions, only P 1 and P 3 are needed to express this potential. A system is built aP 3 = 4 , aP 3 + bP 1 = - 3 , with solutions a = 8 / 5 , b = - 3 / 5 . There is no need to do the integrals, the result is inside the sphere ( y = r/R ) V ( r, θ ) = k ( - 3 5 yP 1 + 8 5 y 3 P 3 ) . Applying boundary conditions, there will be two terms also outside the sphere V ( r, θ ) = k ( - 3 5 y - 1 P 1 + 8 5 y - 3 P 3 ) . The charge density can be found by taking derivatives σ ( θ ) = 0 ( ∂V ∂r out - ∂V ∂r in ) .
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  • Fall '15
  • Giovani Bonvicini
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