Tips for Homework 5: Solving Electrostatics with Series Expansions and Fourier Analysis - , Assignments of Physics

Download Tips for Homework 5: Solving Electrostatics with Series Expansions and Fourier Analysis - and more Assignments Physics in PDF only on Docsity! Hints for homework 5 Chapters 5,6 of PS 5.7. This is a series expansion problem similar to Example 1 of PS. The difference is that now both the cosh and sinh terms are kept as the problem is neither purely odd or even in the y-direction. The relation between the coefficients of the cosh and sinh terms is found from the condition V (x,−a/2) = 0 and the condition V (x, a/2) = V0 is used for the Fourier analysis. An alternative method is to use superposition where we write the problem as a sum of two problems: one with V0/2,−V0/2 boundary conditions in the y-direction and the other with V0/2, V0/2 boundary conditions in the y-direction. Both have zero boundary conditions in the x-direction. The latter is a little nicer from a physical perspective. Using the first method, we have, V (x, y) = ∞∑ n=0 cos((2n+ 1) πx a )[Ancosh((2n+ 1) πy a ) +Bnsinh((2n+ 1) πy a )] (1) Then, V (x,−a/2) = Ancosh((2n+1) π 2 )−Bnsinh((2n+1) π 2 ) = 0 or An Bn = tanh[(2n+1) π 2 ] (2) Plugging this into Eq. (1) gives, V (x, y) = ∞∑ n=0 cos((2n+1) πx a )Bn[tanh((2n+1) π 2 )cosh((2n+1) πy a )+sinh((2n+1) πy a )] (3) Evaluating this expression at the boundary y = a/2 gives, V (x, a/2) = V0 = ∞∑ n=0 cos((2n+ 1) πx a )Bn[2sinh((2n+ 1) π 2 )] = ∞∑ n=0 B′(n)cos((2n+ 1) πx a (4) where B′(n) = 2B(n)sinh((2n+ 1)π 2 ). The Fourier analysis of this expression is exactly the same as that of Example 1 of PS so we find (see Eq. 5.29 of PS). B(n) = 4V0(−1)n π(2n+ 1) 1 2sinh((2n+ 1)π 2 ) (5) Substituting this expression into Eq. (3) gives, V (x, y) = ∞∑ n=0 2V0(−1)n π(2n+ 1) cos((2n+ 1) πx a )[ cosh((2n+ 1)πy a ) cosh((2n+ 1)π 2 ) + sinh((2n+ 1)πy a ) sinh((2n+ 1)π 2 ) ] (6) Taking a common denominator and using the identities sinh(2A) = 2sinhAcoshA, and sinh(A+B) = sinh(A)cosh(B) + cosh(A)sinh(B) yields the result quoted in PS i.e. V (x, y) = ∞∑ n=0 4V0(−1)n π(2n+ 1) cos((2n+ 1) πx a ) sinh[(2n+ 1)(πy a + π 2 )] sinh((2n+ 1)π) (7) 1 5.8. a) This problem is solved by making a superposition of the solutions to problem 5.7, yielding, V (x, y) = ∞∑ n=0 4(−1)n π(2n+ 1) [V2cos((2n+ 1) πx a ) sinh[(2n+ 1)(πy a + π 2 )] sinh((2n+ 1)π) +V1cos((2n+ 1) πy a ) sinh[(2n+ 1)(πx a + π 2 )] sinh((2n+ 1)π) + V3cos((2n+ 1) πy a ) sinh[(2n+ 1)(−πx a + π 2 )] sinh((2n+ 1)π) ] (8) b) To find the value at the origin, evaluate the result of a) at (0,0), yielding, V (0, 0) = ∞∑ n=0 4(−1)n π(2n+ 1) [(V1 + V2 + V3) sinh[(2n+ 1)(π 2 )] sinh((2n+ 1)π) ] = 1 4 (V1 + V2 + V3) (9) where we used, ∞∑ n=0 4(−1)n (2n+ 1) sinh((2n+ 1)π/2) sinh((2n+ 1)π) = π 4 . (10) 5.9. Only one term is needed here, the l = 0 term in the spherical polar expansion∑ (Alr l +Bl/r l+1)Pl(cosθ), so V (r, θ) = A0 +B0/r. The boundary conditions then give the two equations, V1 = A0 +B0/R1; V2 = A0 +B0/R2 (11) Solving these yields, A0 = V1 − R2(V1 − V2) R2 −R1 ; B0 = R1R2(V1 − V2) R2 −R1 (12) and the potential, V (r, θ) = V1 + R2(V1 − V2) R2 −R1 ( R1 r − 1) (13) 5.10. a) The first non-trivial case is P2(u), as no recursion is required for P0(u) = 1, and P1(u) = u. Using the recurrence relation for P2(u), we find that C2 = −3C0, and P2(u) = C0 − 3C0u2. The normalization condition P2(1) = 1 implies that C0 = −1/2, so we find P2(u) = (3u 2 − 1)/2. b) Using Mathematica, we get, P10(u) = 1 256 (−63 + 3465u2 − 30030u4 + 90090u6 − 109395u8 + 46189u10); (14) and P11(u) = 1 256 (−693u+ 15015u3 − 90090u5 + 218790u7 − 230945u9 + 88179u11) (15) 2 Evaluating the LHS gives, σ0 0m [sin( mπ 2 )− sin(3πm 2 ) + sin( mπ 2 )− sin(3πm 2 )] = 2σ0 0m [sin( mπ 2 )− sin(3πm 2 )] (32) The RHS is zero when m is even, which we could have deduced from the outset. When m is odd we write m = (2j + 1) so that sin( (2j+1)π 2 )− sin(3π(2j+1) 2 ) = 2(−1)j. We then find, aj = 4σ0(−1)j 2π0(2j + 1)2Rj−1 , so that Vint = ∑ j 2σ0R(−1)j π0(2j + 1)2 ( r R )2j+1cos((2j + 1)φ) (33) The exterior solution is found in the same way, yielding the same solution other than r/R→ R/r in the above equation. 5.32. This is a spherical co-ordinate problem where only the l = 1 term is needed. It is easy to show that, V (r, θ) = 1 4π0 (− pr R3 + p r2 )cosθ (34) satisfies the boundary conditions, namely a dipole at the origin and zero potential at the conducting sphere, i.e. V (R, θ) = 0. 5.34. a) This problem can be simplified using superposition to write the boundary con- dition as a sum of V0/2 over the whole circle plus −V0/2 on the upper part and V0/2 on the lower part. The latter problem is odd in φ, so the exterior solution is of the form, Vext(r, φ) = V0 2 + ∞∑ n=1 bn rn sin(nφ) (35) The Fourier analysis then gives, − ∫ π 0 V0 2 sin(mφ)dφ+ ∫ 2π π V0 2 sin(mφ) = πbm Rm (36) The LHS gives V0[cos(mπ)− cos(0)]/m which is zero for m even and −2V0/m for m odd, so that, Vext(r, φ) = V0 2 − ∞∑ n odd 2V0 πn ( R r )nsin(nφ) (37) The interior solution is found by making the substitution R/r → r/R in this expression. b) At the origin, V = V0/2. c) The electric field at is given by, Er = ∞∑ n odd 2V0 πn n r ( r R )nsin(nφ); Eφ = 1 r ∞∑ n odd 2V0 πn n( r R )ncos(nφ) (38) 5 At the origin, we then find Er(0, φ) = 2V0 π 1 r ( r R )sin(φ) = 2V0 πR sin(φ) (39) and Eφ(0, φ) = 2V0 π 1 r ( r R )cos(φ) = 2V0 πR cos(φ) (40) Using ĵ = sinφr̂ + cosφφ̂, the electric field reduces to ~E = 2V0ĵ/(πR) 6.3. a) Plot < pz >= p[− 1 a + coth(a)] (41) where a = pE0 kBT = (3.34× 10−30Cm)(106V/m) (1.38 ∗ 1023J/K)T = 0.242K T (42) where T is in Kelvin as a function of T . Also plot the polarizability α = d < pz > dE0 = ( 1 a2 − 1 [sinh(a)]2 ) p kBT (43) as a function of T. A good range to plot over is 0.01K < T < 2K 6.5 a) The bound charge at the surface found from the polarization at the surface, using σb = n̂ · ~P . At the top surface we have σb = P , while at the bottom surface σb = −P . The electric field lines look like those coming from a set of dipoles. b) The electric field inside the slab is like that between two equal and opposite sheets of charge at the slab surfaces, with sheet charge density P at the top and −P at the bottom. The electric field in the center is ~E = −(P/0)k̂. c) The electric field on the axis of the slab, at large distances from the slab, is found using the formula for a dipole field with the dipole moment equal to the polarization times the volume of the slab, V = π(10h)2h = 100πh3. We are asked to find the field at ~r = 100hî, so that, 0 ~E = 3V (~P · r̂)r̂ − V ~P 4πr3 = −V P k̂ 4π(100h)3 = −0.25× 10−4P k̂ (44) 6