# emsol05 - Electromagnetic Theory I Solution Set 5 Due 26...

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Unformatted text preview: Electromagnetic Theory I Solution Set 5 Due: 26 September 2011 17. Following the method discussed in class for the Dirichlet Green function, obtain the mixed Dirichlet/Neumann Green function for the region between two concentric spheres of radii a < b with a) Neumann conditions ( ∂G/∂r = 0) on the inner shell and Dirichlet conditions G = 0 on the outer shell, and b) vice versa. Solution : a) For this case the combination ( l + 1) r l + la 2 l +1 /r l +1 has vanishing radial derivative at r = a . (For l = 0, note that this expression is just a constant!) Thus we start with the empty space expansion and make the substitution: r l < r l +1 > → C parenleftbigg r l < + la 2 l +1 ( l + 1) r l +1 < parenrightbiggparenleftbigg 1 r l +1 > − r l > b 2 l +1 parenrightbigg = C parenleftbigg r l < r l +1 > − ( rr ′ ) l b 2 l +1 + la 2 l +1 ( l + 1)( rr ′ ) l +1 − la 2 l +1 r l > ( l + 1) b 2 l +1 r l +1 < parenrightbigg = C parenleftbigg r l < r l +1 > − ( rr ′ ) l b 2 l +1 + la 2 l +1 ( l + 1)( rr ′ ) l +1 − la 2 l +1 r l ( l + 1) b 2 l +1 r ′ ( l +1) − la 2 l +1 r ′ l ( l + 1) b 2 l +1 r l +1 + la 2 l +1 r l < ( l + 1) b 2 l +1 r l +1 > parenrightbigg To produce the correct delta function on the right side of the Green function equation we simply match coefficients of r l < /r l +1 > : C = (1 + la 2 l +1 / ( l + 1) b 2 l +1 ) − 1 . Thus the appropriate Green function is: G ND ( r , r ′ ) = 1 4 π ∞ summationdisplay l =0 ( r l < + la 2 l +1 / ( l + 1) r l +1 < )( 1 /r l +1 > − r l > /b 2 l +1 ) 1 + la 2 l +1 / ( l + 1) b 2 l +1 P l (cos γ ) Where γ is the angle between r and r ′ , r · r ′ = rr ′ cos γ . The 1 / 4 π arranges the normalization in our lecture notes. b) For this case we need the combination and l/r l +1 +( l +1) r l /b 2 l +1 whose normal derivative 1 vanishes at r = b , and make the substitution: r l < r l +1 > → C parenleftbigg r l < − a 2 l +1 r l +1 < parenrightbiggparenleftbigg 1 r l +1 > + ( l + 1) r l > lb 2 l +1 parenrightbigg = C parenleftbigg r l < r l +1 > + ( l + 1)( rr ′ ) l lb 2 l +1 − a 2 l +1 ( rr ′ ) l +1 − ( l + 1) a 2 l +1 r l > lb 2 l +1 r l +1 < parenrightbigg = C parenleftbigg r l < r l +1 > + ( l + 1)( rr ′ ) l lb 2 l +1 − a 2 l +1 ( rr ′ ) l +1...
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## This note was uploaded on 01/08/2012 for the course PHY 6346 taught by Professor Staff during the Spring '08 term at University of Florida.

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emsol05 - Electromagnetic Theory I Solution Set 5 Due 26...

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