Jackson 2.25 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

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1 Jackson 2.25 Hoework Proble Solution Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: Two conducting planes at zero potential eet along the z axis, aking an angle β between the, as in Fig A unit line charge parallel to the z axis is located between the planes at position (ρ', φ'). (a) Show that (4πε 0 ) ties the potential in the space between the planes, that is, the Dirichlet Green function G(ρ, φ; ρ', φ'), is given by the infinite series G(ρ,ϕ,ρ ',ϕ ')=4 = ρ π/β < ρ π/β > sin ( π ϕ/β)sin ( π ϕ' /β) (b) By eans of coplex-variable techniques or other eans, show that the series can be sued to give a closed for, G(ρ,ϕ,ρ ',ϕ ')=ln[ (ρ)2π /β +(ρ') 2π/β 2(ρρ') π/β cos[π(ϕ+ϕ ')/β] (ρ) 2π /β +(ρ') 2π/β 2(ρρ') π/β cos[π(ϕ ϕ ')/β]] (c) Verify that you obtain the failiar results when β = π and β = π/2. SOLUTION: (a) Split the space between the planes into two regions, region I where ρ < ρ' and region II where ρ > ρ'. Each region is charge free, so we can solve the Laplace equation in two-diensions in polar coordinates. The line charge will coe into play when we apply boundary conditions linking the two regions. ρ ρ( ρ Φ ρ ) + 2 Φ ρ 2 ϕ =0 2 Try a solution of the for, =R to find R R = 2 2 Both sides are now independent and can be set to a constant. R = 2 R and 2 = 2 2

2 The general solution is: Φ(ρ,ϕ)=(a 0 lnρ)(a 0 + B 0 ϕ)+ (a ν ρ ν +b ν ρ ν )( A ν e i ν ϕ +B ν e i νϕ ) Apply the boundary condition at the botto face: Φ(ρ,ϕ=0)=0 0=(a 0 ln ρ)( A 0 )+ (a ν ρ ν +b ν ρ ν )( A ν +B ν ) A 0 =0 and B ν = A ν The solution becoes: Φ(ρ, ϕ)=(a 0 lnρ)ϕ+ (a ν ρ ν +b ν ρ ν ) sin(νϕ) Apply the boundary condition at the top face: Φ(ρ,ϕ=β)=0 0=(a 0 ln ρ)β+ (a ν ρ ν +b ν ρ ν )sin (νβ) a 0 =0 and b 0 =0 and 0=sin(νβ) which leads to The solution becoes: ν= π β where =, 2... Φ(ρ,ϕ)= (a ρ π /β +b ρ π/β )sin ( π ϕ/β) where =, 2... The solution in both regions ust have this for. We ust now look at each region separately to get any farther. In the region close to the origin (I), we ust have a valid solution at the origin, so that b = 0, leading to: Φ I (ρ,ϕ)= a ρ π/β sin ( π ϕ/β) In the region far fro the origin (II), we ust have a valid solution at infinity, so that a = 0, leading to: Φ II (ρ,ϕ)= b ρ π /β sin(π ϕ/β) We now apply boundary conditions to link the two regions, reebering that there is a line charge present at the boundary:

3 (E 2 E ) n= σ [ Φ II ρ + Φ I δ(ϕ ϕ ') = ρ ]at ρ=ρ' ρ' sin( π ϕ/β)[b ρ ' π/ β +a ρ' π/β ]= β π δ(ϕ ϕ') Multiply both side by a sine and integrate with respect to the polar angle: β sin( π ϕ/β)sin(nπϕ/β)d ϕ[b ρ ' π /β +a ρ ' π/β ]= β 0 π sin(nπ ϕ' /β) [b ρ ' π /β +a ρ ' π/β ]= 2 π The other boundary condition is: sin (π ϕ '/β) n E 2 =n E [ Φ I ϕ = Φ II ϕ ]at ρ=ρ' a ρ' π/β π/β =b ρ' Solving the syste of equations in two independent variables as represented by the above equations in boxes, we find: b =ρ' π/β π sin ( π ϕ' /β) a =ρ ' π/ β π sin(π ϕ'/β) The final solution becoes: Φ(ρ, ϕ)= π ρ π/β < ρ π/β > sin(π ϕ/β)sin(π ϕ'/β) If we set =4 π, this potential becoes the Green function: G(ρ,ϕ,ρ ',ϕ ')=4 = ρ π/β < ρ π/β > sin ( π ϕ/β)sin ( π ϕ' /β)

4 (b) Let us try to convert the su into closed for. First note that the sine function is the iaginary part of the coplex exponential, so that the Green function becoes: G(ρ,ϕ,ρ ',ϕ ')=4 = ρ π/β < ρ π/β > I(e i π ϕ/β )I(e i πϕ '/β ) Next use the identity: 2 I(z )I(z 2 )=R[ z z 2 +z * z 2 ] G(ρ,ϕ,ρ ',ϕ ')=R[ 2 = G(ρ,ϕ,ρ ',ϕ ')=R[ 2 = ρ < Z π /β ρ > π /β e i π(ϕ+ϕ ')/ β = Z 2 where Z =ρ < π/ β ρ > π/β e i π (ϕ+ϕ')/β and Z 2 =ρ < π /β ρ > π /β e i π(ϕ ϕ')/β Now use = Z = ln( Z ) G(ρ,ϕ,ρ ',ϕ ')=2R [ ln ( Z ) ln ( Z 2 )] G(ρ,ϕ,ρ ',ϕ ')=2 R[ ln ( Z Z 2)] Use 2 R[ln z]=ln(z z *) : G(ρ,ϕ,ρ ',ϕ ')= ln( + Z 2 2 R(Z ) + Z R(Z 2 )) Insert back in Z and Z 2 : G(ρ,ϕ,ρ ',ϕ ')= ln( +ρ 2π/β 2 < ρ π/β > 2ρ π/β < ρ π/β > cos(π(ϕ+ϕ +ρ 2π/β 2 < ρ π/β > 2ρ π/β < ρ π/β > cos(π(ϕ ϕ ')/β)) G(ρ,ϕ,ρ ', ϕ ')= ln( ρ 2π /β 2 π/ > +ρ β < 2(ρ < ρ > ) π /β cos(π(ϕ+ϕ')/β) 2π ρ /β 2 π/ > +ρ β < 2(ρ < ρ > ) cos(π(ϕ ϕ')/β)) π /β ] = ρ π/ β π < ρ /β i π(ϕ ϕ' > e )/β] Because of the syetry, we can drop the greater-than and less-than subscripts: G(ρ,ϕ,ρ ',ϕ ')= ln( ρ 2π /β +ρ ' 2 π/β 2 (ρ ρ') π/β cos(π (ϕ+ϕ ')/β) ρ 2π /β +ρ ' 2 π/β 2 (ρ ρ') π/β cos(π (ϕ ϕ ')/β))

5 (c) For β = π this reduces to: G(ρ,ϕ,ρ ', ϕ ')= ln ( x x '' ) ln ( x x ' ) 4 π 2 π 2 π This is the potential of a unit line charge located at x' near a grounded conducting plane in the x-z axis, which is effectively the sae as a negative iage line charge at the irror location x''. y (x', y') - (x', -y') x For β = π/2 this reduces to: G(ρ,ϕ,ρ ',ϕ ')= ln( ρ 4 +ρ' 4 2(ρρ ') 2 cos(2(ϕ+ϕ')) ρ 4 +ρ' 4 2(ρρ ') 2 cos(2(ϕ ϕ'))) G(ρ,ϕ,ρ ',ϕ ')= [ ln( ρ 2 +ρ ' 2 2ρρ ' cos(ϕ ϕ ')) ln( ρ 2 +ρ ' 2 +2ρρ ' cos(ϕ+ϕ ')) 4π 2π ln( ρ 2 +ρ' 2 2ρρ 'cos(ϕ+ϕ'))+ln( ρ 2 +ρ ' 2 +2ρρ ' cos(ϕ ϕ'))] This is the potential of a unit line charge located at x' near a right-angle interior edge, which is equivalent to three iage charges, two negative ones offset by plus and inus 90 degrees, and one positive one offset by 80 degrees. This result atches the one found in Proble 2.3. y (-x', y') - (-x', -y') + (x', y') - (x', -y') x

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