Physics 505 Fall Homework Assignment #4 Solutions

Transcription

1 Physics 505 Fa 2005 Homework Assignment #4 Soutions Textbook probems: Ch. 3: 3.4, 3.6, 3.9, The surface of a hoow conducting sphere of inner radius a is divided into an even number of equa segments by a set of panes; their common ine of intersection is the z axis and they are distributed uniformy in the ange φ. (The segments are ike the skin on wedges of an appe, or the earth s surface between successive meridians of ongitude. The segments are kept at fixed potentias ±V, aternatey. a Set up a series representation for the potentia inside the sphere for the genera case of 2n segments, and carry the cacuation of the coefficients in the series far enough to determine exacty which coefficients are different from zero. For the nonvanishing terms, exhibit the coefficients as an integra over cos θ. The genera spherica harmonic expansion for the potentia inside a sphere of radius a is Φ(r, θ, φ =,m α m a Ym (θ, φ where α m = V (θ, φy m(θ, φdω In this probem, V (θ, φ = ±V is independent of θ, but depends on the azimutha ange φ. It can in fact be thought of as a square wave in φ n =4 V 2 π ϕ V This has a famiiar Fourier expansion V (φ = 4V π [(2k + nφ 2k + This is aready enough to demonstrate that the m vaues in the spherica harmonic expansion can ony take on the vaues ±(2k+n. In terms of associated egendre

2 poynomias, the expansion coefficients are 2 + α m = 4π = 4V π = 4iV ( m! 2± V (φe imφ dφ ( + m! ( m! 4π ( + m! 2 + ( m! 4π ( + m! P m (x dx 2π [(2k + nφe imφ dφ 2k + 0 P m (x dx δ m,(2k+n δ m, (2k+n 2k + P m (x dx Ug P m (x = ( m [( m!/( + m!p m (x, we may write the non-vanishing coefficients as α, (2k+n = ( n+ α,(2k+n = 4iV 2 + ( (2k + n! 2k + 4π ( + (2k + n! P (2k+n (x dx ( for k = 0,, 2,.... Since (2k + n, we see that the first non-vanishing term enters at order = n. Making note of the parity of associated egendre poynomias, P m ( x = ( +m P m (x, we see that the non-vanishing coefficients are given by the sequence α n,n, α n+2,n, α n+4,n, α n+6,n,... α 3n,3n, α 3n+2,3n, α 3n+4,3n,... α 5n,5n, α 5n+2,5n, α 5n+4,5n,.... b For the specia case of n = (two hemispheres determine expicity the potentia up to an incuding a terms with = 3. By a coordinate transformation verify that this reduces to resut (3.36 of Section 3.3. For n =, expicit computation shows that P (x dx =, Inserting this in to ( yieds α, = α, = iv α 3, = α 3, = iv P 3 (x dx = 3π 6, P 3 3 (x dx = 45π 8 3π 2 2π 35π 256, α 3, 3 = α 3,3 = iv 256

3 Hence iv [ 3π a 2 (Y, + Y, + a 3 ( 2π 256 (Y 3, + Y 3, + 35π 256 (Y 3,3 + Y 3, 3 35π 256 Y 3,3 [ 3π ( 3 2π = 2V I a 2 Y, + a 256 Y 3, + [ 3 = 2V I θ eiφ a 4 ( a 28 θ(5 cos2 θ e iφ θ e 3iφ [ 3 = V θ φ a ( 3 θ(5 cos 2 θ φ θ 3 φ a 28 (2 To reate this to the previous resut, we note that the way we have set up the wedges corresponds to taking the top of the +V hemisphere to point aong the ŷ axis. This may be rotated to the ẑ axis by a 90 rotation aong the ˆx axis. Expicity, we take ŷ = ẑ, ẑ = ŷ, ˆx = ˆx or θ φ = cos θ, cos θ = θ φ, θ cos φ = θ cos φ Noting that 3φ = 3 φ + 3 φ cos 2 φ, the ast ine of (2 transforms into [ V a 2 cos ( θ + 3 cos θ (5 2 θ 2 φ a ( cos 3 θ + 3 cos θ 2 θ 2 φ [ 3 = V cos θ a 8 a 2 (5 cos3 θ 3 cos θ [ 3 = V P (cos θ 7 3 P3 (cos θ 2 a 8 a which reproduces the resut (3.36.

4 3.6 Two point charges q and q are ocated on the z axis at z = +a and z = a, respectivey. a Find the eectrostatic potentia as an expansion in spherica harmonics and powers of r for both r > a and r < a. The potentia is ceary q 4πɛ 0 ( x a x + a where a = aẑ points from the origin to the positive charge. Ug the spherica harmonic expansion x x = 4π,m as we as a = aẑ, we obtain q ɛ 0,m 2 + r< r> r< r> + Y m(ˆx Y m (ˆx [Y m(0, φ Y m(π, φ Y m (θ, φ (3 Noting that Y m (0, φ P m ( and that P m ( = δ m,0 we see that ony terms with m = 0 contribute. This is aso obvious from symmetry. Since Y 0 (0, φ = ( Y 0 (π, φ = 2 + 4π the potentia (3 becomes q 4πɛ 0 = q 2πɛ 0 [ ( r < r + =0 > odd r< r> + P (cos θ 4π 2 + Y 0(θ, φ (4 b Keeping the product qa = p/2 constant, take the imit of a 0 and find the potentia for r 0. This is by definition a dipoe aong the z axis and its potentia. Since we wi take a 0, we have r < = a and r > = r. This yieds an expansion of (4 qa ( a 2k P2k+ 2πɛ 0 r 2 (cos θ r

5 Setting qa = p/2 and taking a 0, ony the k = 0 term survives in the sum. The resut is p 4πɛ 0 r 2 P (cos θ = p cos θ 4πɛ 0 r 2 (5 which is the potentia due to a dipoe. c Suppose now that the dipoe of part b is surrounded by a grounded spherica she of radius b concentric with the origin. By inear superposition find the potentia everywhere inside the she. To account for the spherica she, we add to (5 a soution to the (homogeneous apace s equation. For an inside soution, we have [ p 4πɛ 0 r 2 P (cos θ + A r P (cos θ The boundary condition Φ(r = b = 0 corresponds to having =0 A b +2 P (cos θ = P (cos θ =0 Since the egendre poynomias form an orthonorma set, the ony term that can show up on the eft hand side is the = term. We then take A = /b +2, and the resuting soution is p ( 4πɛ 0 r 2 r b 3 cos θ 3.9 A hoow right circuar cyinder of radius b has its axis coincident with the z axis and its ends at z = 0 and z =. The potentia on the end faces is zero, whie the potentia on the cyindrica surface is given as V (φ, z. Ug the appropriate separation of variabes in cyindrica coordinates, find a series soution for the potentia anywhere inside the cyinder. The genera soution obtained by separation of variabes has the form Φ(ρ, φ, z = {J m (kρ or N m (kρ }{e ±imφ}{ e ±kz} However, ce the potentia vanishes on the endcaps, it is natura to take k ik so that the z function obeying boundary conditions is (nπz/. The resut is to use the modified Besse functions I m (kρ and K m (kρ instead. However, for the soution to be reguar at ρ = 0 we discard the K ν (kρ functions, which bow up at vanishing argument. The resuting series expression for the potentia is Φ(ρ, φ, z = m=0 n= ( nπ (amn I m ρ mφ + b mn cos mφ ( nπ z (6

6 In order to satisfy the boundary conditions on the cyindrica surface, we need to have V (φ, z = ( nπb (amn I m mφ + b mn cos mφ ( nπ z m,n This is a doube Fourier series in φ and z. As a resut, the Fourier coefficients are { amn b mn } ( nπb I m = 2π dφ 2 { } ( mφ nπ dz V (φ, z π 0 0 cos mφ z with the caveat that b 0,n must be divided by two. This can be rewritten as { amn b mn } = 2 2π { } ( mφ nπ dφ dz V (φ, z πi m (nπb/ 0 0 cos mφ z (7 (where b 0,n has to be divided by two. 3.0 For the cyinder in Probem 3.9 the cyindrica surface is made of two equa hafcyinders, one at potentia V and the other at potentia V, so that { V for π/2 < φ < π/2 V (φ, z = V for π/2 < φ < 3π/2 a Find the potentia inside the cyinder. To obtain the potentia, we want to find the coefficients a mn and b mn of the expansion (6 and (7. Noting first that V (φ, z = V (φ = ±V is an even function of φ, we see that a the a mn coefficients vanish. We are eft with b mn = = [ 2V π/2 πi m (nπb/ π/2 3π/2 π/2 2V 4 (mπ/2 ( n I m (nπb/ m n dφ cos mφ dz nπz 0 (m 0 This is non-vanishing ony when both m and n are odd. Introducing m = 2k + and n = 2 +, we have b 2k+,2+ = Inserting this into (6 yieds 6V ( k I 2k+ ((2 + πb/ (2k + (2 + 6V =0 ( k I 2k+ ( (2+πρ (2k + (2 + I 2k+ ( (2+πb cos(2k+φ (2 + πz (8

7 b Assuming b, consider the potentia at z = /2 as a function of ρ and φ and compare it with two-dimensiona Probem 2.3. For b both ρ/ and b/ are much ess than one. This aows us to use a sma argument expansion of the modified Besse function I ν (x In addition, for z = /2 we have ( x ν Γ(ν + 2 (2 + πz = ( + 2 π = ( Hence in this imit (8 becomes 6V = 6V k, [ ( k 2k + ( 2 + ( ( ρ 2k+ cos(2k + φ 2 + b [ ( k ( ρ R 2k + b eiφ 2k+ k Noting the Tayor series expansion for arctan tan z = n ( n 2n + z2n+ we arrive at 6V tan (R tan ( ρ b eiφ = 4V π R tan ( ρ b eiφ To cacuate R tan z we reca that Hence For z = (ρ/be iφ we find tan a + tan b = tan a + b ab R tan z = 2 (tan z + tan z = 2 tan z + z z 2 2V π tan 2(ρ/b cos φ (ρ/b 2 = 2V π tan 2bρ cos φ b 2 ρ 2 which reproduces the answer to Probem 2.3 (where V = V 2 = V.