Physics 505 Fall 2005 Practice Midterm Solutions. The midterm will be a 120 minute open book, open notes exam. Do all three problems.

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1 Physics 55 Fll 25 Pctice Midtem Solutions The midtem will e 2 minute open ook, open notes exm. Do ll thee polems.. A two-dimensionl polem is defined y semi-cicul wedge with φ nd ρ. Fo the Diichlet polem, it is possile to expnd the Geen s function s Gρ, φ; ρ, φ = m= mπφ mπφ g m ρ, ρ sin sin Wite down the ppopite diffeentil eqution tht g m ρ, ρ must stisfy. Note tht the expnsion in tems of sinmπφ/ is designed to stisfy Diichlet oundy counditions on the stight segments of the wedge. The Geen s function eqution we wish to solve is 2 x Gρ, φ; ρ, φ = 4πδ 2 x x = 4π ρ δρ ρ δφ φ By completeness, we hve mπφ mπφ sin sin m= = 2 δφ φ Hence 2 x Gρ, φ; ρ, φ = 8π ρ δρ ρ mπφ mπφ sin sin m= Using the pol coodinte expession fo the Lplcin, we find 2 x Gρ, φ; ρ, φ = ρ m= ρ ρ ρ 2 mπ ρ 2 g m ρ, ρ mπφ mπφ sin sin

2 Comping this with yields the ODE ρ ρ ρ ρ 2 mπ ρ 2 g m ρ, ρ = 8π ρ δρ ρ which my e conveted into Stum-Liouville fom y multiplying y ρ ρ ρ ρ 2 mπ ρ g m ρ, ρ = 8π δρ ρ Solve the Geen s function eqution fo g m ρ, ρ suject to Diichlet oundy conditions nd wite down the esult fo Gρ, φ; ρ, φ. The Diichlet oundy conditions e tht g m ρ, ρ vnish when ρ = o, nmely g m ρ, = g m ρ, =. Fo these homogeneous oundy conditions, the Geen s function tkes the fom g m ρ, ρ = 8π A uv whee uρ nd vρ e solutions to the homogeneous eqution stisfying oundy conditions u = v =, nd A is elted to the Wonskin y W u, v = A/ρ. Noting tht the solution to the homogeneous dil eqution hs the fom g m ρ, ρ ρ ±mπ/ it is esy to wite down the ppopite uρ nd vρ u = ρ mπ 2mπ ρ v = ρ mπ Computing the Wonskin yields As esult W = uv vu = mπ 2m ρ ρ g m ρ, ρ = 4 m = 4 m mπ 2m ρ ρ = ρ 2mπ mπ 2mπ ρ< 2mπ mπ 2mπ 2mπ ρ + 2mπ 2mπ ρ + 2mπ 2mπ 2mπ 2mπ ρ 2mπ ρ 2mπ ρ 2mπ 2mπ

3 Comining this with the ngul functions yields the finl esult Gρ, φ; ρ, φ = m= 4 m ρ< mπ 2mπ mπφ sin 2mπ sin Note tht this hs the expected ehvio s eithe o. 2mπ mπφ 2. A conducting spheicl shell of inne dius is held t zeo potentil. The inteio of the shell is filled with electic chge of volume density ρ = ρ 2 sin 2 θ Find the potentil eveywhee inside the shell. To otin the potentil, we mke use of the Geen s function fo the inteio of conducting sphee G x, x = 4π l,m l < > 2l + > l+ 2l+ Y lmω Y lm Ω Actully, ecuse of zimuthl symmety, we only need the m = components of the Geen s function expnsion G x, x = l < l > > l+ 2l+ P l cos θ P l cos θ + m Although the chge density is specified in tems of sin 2 θ, this cn e conveted into Legende polynomils. Since sin 2 θ = cos 2 θ, nd since P l cos θ is of degee cos θ l, we see tht sin 2 θ hs to e line comintion of P nd P 2. It is not too hd to see tht sin 2 θ = 2 P cos θ P 2 cos θ We now note tht since the sufce is held t zeo potentil the solution in the inteio is given y Φ x = ρ x G x, x d x = ρ 2 2 P l cos θ 4πɛ P cos θ P 2 cos θ l P l cos θ l < > 2l+ > l+ d dφ dcos θ

4 By othogonlity of Legende polynomils, this ecomes P cos θ d Φ x = 2ρ 2 ɛ > > 5 P 2 < > 5 2cos θ > d d > 5 P 2cos θ < 2 > 2 > 5 d d + d 5 P 2cos θ d = 2ρ 2 P cos θ ɛ = 2ρ 2 P cos θ ɛ = 2ρ 2 P cos θ ln ɛ 6 P 2cos θ 2 5 Inseting the expessions fo Legende polynomils, this ecomes Φ x = 2ρ 2 ln ɛ 2 cos 2 2 θ Wht is the sufce chge density on the inside sufce of the shell? The sufce chge density is given y Φ σ = ɛ = 2ρ 2 = = 2ρ 2 cos2 θ = cos2 θ d Note tht only the l = tem contiutes to the totl chge induced on the shell. This is simply Q shell = 2ρ 4π2 = 8πρ This is the negtive of the chge contined in the inteio Q inside = ρ x d x = 2πρ sin 2 θ dcos θ = 8πρ. A thin disk of dius lies in the x-y plne with its cente t the coodinte oigin. The disk is unifomly chged with sufce density σ.

5 Clculte the multipole moments of the chge distiution. Mke sue to indicte which moments e non-vnishing. The volume chge density fo the disk cn e witten s ρ x = σ δcos θ povided <. Note tht the fcto of / ensues unifom sufce chge density since dρ = ρ x d x = σ δcos θ2 d dφ dcos θ = σ d dφ θ=π/2 nd d dφ is the stndd e element in pol coodintes. The multipole moments e then given y q lm = l YlmΩ x d x = σ l+ Y lm θ, φδcos θ d dφ dcos θ By zimuthl symmety, only the m = moments e non-vnishing. Integting the φ nd θ ngles gives q l, = 2πσY l, π 2l + 2, l+ d = 2πσ 4π P l l+2 l + 2 Note tht Y l, is independent of φ. Since the Legende polynomils e even nd odd depending on l, we see tht only even l moments e non-vnishing 4k + πpl q 2k, = σ 2k+2 = k 4k + Γk + 2 σ 2k+2 2k + 2 2k +! Since the disk is unifomly chged, the totl chge is simply q = σπ 2. This llows us to wite q 2k, = k 4k + Γk + 2 q 2k 2πk +! The fist two non-vnishing moments e q = 4π q q 2 = 5 4 4π q2 Wite down the multipole expnsion fo the potentil in explicit fom up to the fist two non-vnishing tems. The multipole expnsion yields Φ x = Y Ω 4π q = q 4π 4π = q + 5 q Y 2 Ω 2 4π P 2cos θ π 4π P 2cos θ + = q 2 8 cos2 θ +