Physics 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel. Problem Set 3. 2 (x x ) 2 + (y y ) 2 + (z + z ) 2

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1 Physics 505 Electricity an Magnetism Fall 003 Prof. G. Raithel Problem Set 3 Problem.7 5 Points a): Green s function: Using cartesian coorinates x = (x, y, z), it is G(x, x ) = 1 (x x ) + (y y ) + (z z ) 1. (1) (x x ) + (y y ) + (z + z ) Note the require symmetry in x an x. A coorinate-free form is G(x, x ) = 1 x x 1 x x + ẑ(ẑ x ). () b): It is Using z G(x, x ) z =0 = Φ(x) = 1 x = y = 4π x = y = z G(x, x ) z=0 V (x )x y. (3) z (x x ) +(y y ) +z 3 an re-writing into cylinrical coorinates, x = ρ cos φ x = ρ cos φ y = ρ sin φ y = ρ sin φ (4) yiels z G(x, x ) z=0 = = z z + ρ + ρ ρρ (cos φ cos φ + sin φ sin φ ) 3 z z + ρ + ρ ρρ cos(φ φ ) 3. (5) The potential thus is Φ(ρ, φ, z) = V z π a ρ ρ φ π φ =0 ρ =0 z + ρ + ρ ρρ cos(φ φ ) 3. (6) c): On the z-axis it is ρ = 0, an

2 a Φ(0,, z) = V z ρ =0 [ ] a ρ ρ φ = V z 1 = V (1 z + ρ 3 z + ρ 0 z z + a ). (7) ): We write an expan Φ(ρ, φ, z) = V z π a ( 1 + ρ ρρ cos(φ φ ) 3/ ) π(ρ + z ) 3/ φ =0 ρ =0 ρ + z ρ ρ φ. (8) (1 + ɛ) 3/ = }{{} ɛ + 15 }{{} 8 ɛ }{{} A B C..., (9) where ɛ = ρ ρρ cos(φ φ ) ρ +z 1. Integration term by term yiels A = π a φ =0 ρ =0 ρ ρ φ = πa (10) B = 3 π φ =0 a ρ =0 ρ ρρ cos(φ φ ) ρ + z ρ ρ φ = 3πa4 4(ρ + z ), (11) where the cos-term integrates to zero. C = 15 8 π φ =0 a ρ =0 ρ 4 4ρρ 3 cos(φ φ ) + 4ρ ρ cos (φ φ ) (ρ + z ) ρ ρ φ 5πa 6 = 8(ρ + z ) + 15πρ a 4 8(ρ + z ), (1) where we have use that π φ =0 cos(φ φ )φ = 0 an π φ =0 cos (φ φ )φ = π. Collecting the terms into Eq. 8, it is foun Φ(ρ, z) = V a ( z 3a 1 (ρ + z ) 3/ 4(ρ + z ) + 5a 4 8(ρ + z ) + 15a ρ ) 8(ρ + z ) +..., q.e.. (13) On-axis, the expression reuces to Φ(ρ = 0, z) = V a z ) (1 3a 4z + 5a4 8z 4..., (14) while the expansion of the result of part c) for large z is Φ(0, z) = V (1 [ z ( a ) ] ) 1/ (1 z + a ) = V 1 + V ) (1 a z z 3a 4z + 5a4 8z 4. (15) The results in Eq. 14 an Eq. 15 agree, as expecte.

3 Problem.8 5 Points a): It is to be shown that the equipotential surfaces of two parallel line charges of equal magnitue an opposite polarity are cyliners. Using the variables ientifie in the figure, we claim that for any potential V there exists a cyliner with raius r an axis location ientifie by x such that for all values of γ the potential on the cyliner is V. We prove the claim by fining a unique solution for x an r. X r r1 r R x (for case V < 0 ) x (for case V > 0 ) Figure 1: Equipotential surfaces of two parallel line charges of equal magnitue an opposite polarity. First we note that for symmetry the axis of the cyliner can only be locate on a straight line through the two line charges. Since the potential at a istance ρ from a line charge is πɛ 0 ln(ρ), at the location ientifie in the figure by γ an r the potential is Φ = πɛ 0 ln r r 1, an an using the law of cosines it is ( ) r πɛ0 V = exp =: α, (16) r 1 α = (R + x) + r (R + x)r cos γ x + r xr cos γ. (17) For that equation to hol for all γ, the ratio of the coefficients of cos γ an the ratio of the terms without cos γ must both be equal, an The secon equation yiels x = α = exp ( πɛ 0V ). α = (R + x) + r x + r r = α x (x R) 1 α α (R + x)r =. (18) xr R α 1, which, when inserte into the first equation, yiels r = αr α 1 with

4 It is note that for V > 0 the value of x is positive. For V it is r 0 an x 0. For V < 0 the value of x is negative an < R. For V it is r 0 an x R. The figure shows both a case of negative an positive V. These finings are important in the next part of the problem. b): To etermine the capacitance, we nee to place two line charges with opposite polarities such that two cyliners with the specifie raii a an b an center-to-center separation are equipotential surfaces of the system. The voltage ifference between these equipotential surfaces will then allow us to calculate the capacitance. First we note that for two cyliners outsie of each other, i.e. > a + b, we are seeking a solution of the type shown in the figure: one circle with positive potential V + aroun the line charge > 0, an one one circle with negative potential V aroun the line charge. From the given answer we suspect that it will be useful to calculate ( a b )/(). Defining ( ) πɛ0 V + α + = exp an ( ) πɛ0 V α = exp (19) an using a = α + R α + 1 > 0 b = α R α 1 > 0 = x + x = α +R α+ 1 α R α 1 (0) it is foun that a b = R α + + α (α + 1)(α 1) > 0 a b = R α + α (α + 1)(α 1) > 0 = α + + α α + α = 1 ( α + α + α α + ) = cosh ( ) πɛ0 (V + V ) (1) Thus, ( cosh 1 a b ) = (V + V ) πɛ 0 = (V + V ) πɛ 0L Q = πɛ 0L C, () an the capacitance per length is c): For a + b, it is a b C L = πɛ 0 cosh 1 ( ), q.e.. (3) a b 1. Since cosh 1 y ln(y) for y 1, it then is

5 ( cosh 1 a b ) ( ln ab [1 a + b ]) ( = ln )+ln ab (1 a + b ) ( ) ln a + b ab (4) an C L = πɛ 0 cosh 1 ( πɛ ) ( ) 0 a b ln ab ( πɛ 0 πɛ ( ) 0 a + b ) + ( ( )) ln ab ln ab a +b (5) The result exhibits the correct behavior for a +b 0, an the lowest-orer correction in a +b. ): For cyliners insie each other, choose voltages V + an V of same polarity. Without loss of generality, we can choose them both positive, an repeat the calculation of b): a = α + R α+ 1 > 0 b = α R α 1 > 0 = x + x = α +R α+ 1 α R α 1 (6) It is foun that a + b = R α + + α (α + 1)(α 1) > 0 a + b = R α + α (α + 1)(α 1) > 0 = α + + α α + α = 1 ( α + α + α α + ) = cosh ( ) πɛ0 (V + V ) (7) an the capacitance per length becomes C L = πɛ 0 cosh 1 ( ). (8) a +b ( ) For = 0, it is cosh 1 a +b = ln ( ) a b. You can show this by application of cosh on both sies an evaluation of the expression on the right. Thus, for = 0 it is, as expecte, C L = πɛ 0 ln ( ) a. (9) b

6 Problem.9 5 Points a): Accoring to Eq..15 in the textbook, it is σ = 3ɛ 0 E 0 cos θ. The electrostatic pressure is P = σ ɛ 0 ˆr. Due to symmetry, only the z-components of the resultant force will integrate to 0. Integration over one hemisphere yiels F 0 = F z = π 1 φ=0 cos θ=0 9 ɛ 0E 0a cos 3 θ cos θφ = ɛ 0 π ( ) 3E0 a. (30) The force is repulsive. b): The aitional charge spreas evenly over the full sphere, yieling an aitional σ 0 = repulsive force ue to aitional electrostatic pressure is Q 4a π. An aitional F 1 = π 1 φ=0 cos θ=0 σ 0 ɛ 0 a cos θ cos θφ = πσ 0 ɛ 0 = Q 3a ɛ 0 π. (31) Also, there is an aitional force F = E 0 Q/ on half the net charge locate on each hemisphere. Noting that F points in the same irection for both hemispheres, while F 0 an F 1 point in opposite irections, the force require to hol the spheres together is just F 0 + F 1, i.e. ( ) 3E0 a F = ɛ 0 π + Q 3a ɛ 0 π. (3) Note that this force oes not epen on the sign of Q. The net force acting on the whole assembly of both hemispheres is F = E 0 Q, as expecte, an epens on the sign of Q.

7 Problem.10 5 Points a): One may expect Eq..14 of the textbook, Φ = E 0 (r a3 r ) cos θ to be the solution. This suspicion turns into certainty by reiterating that Φ = 0 in the volume of interest, an by verification of the bounary conitions. The latter are that on the boss surface r = a an on the plane θ = π/ the potential must be a constant, an that for r it must be E = Φ = E 0 ẑ. The potential Φ = E 0 (r a3 r ) cos θ accors with both bounary conitions. It is σ = ɛ 0 nφ, with ˆn being the normal vector from the volume of interest into the conuctor. Thus, on the boss it is an on the surface σ = ɛ 0 r Φ r=a = ɛ 0 E 0 (1 + a3 r 3 ) cos θ = 3ɛ 0E 0 cos θ, (33) σ = ɛ 0 z Φ θ=π/ = +ɛ 0 rθ Φ θ=π/ = ɛ 0 E 0 (1 a3 r 3 ). (34) 0 E 0 0 E 0 infty z -a a z infty Figure : Sketch of σ. b): The charge on the boss is πa 1 0 3ɛ 0E 0 cos θ cos θ = 3πɛ 0 E 0 a. c): The problem is solve in analogy with an image charge problem of two charges locate outsie a conucting sphere with zero net charge. The original charges q an q are locate at ẑ an ẑ, respectively. The image charges q = q a an q a a a are locate at ẑ an ẑ. The bounary conitions of the image charge problem coincie with the ones in the problem. Using Eq..5 of the textbook an the superposition principle, it is then, on the boss, σ = q ) a (1 4πa a a a 3 cos θ a + a 3 cos θ The charge Q on the boss is πa 1 a σ cos θ, an with z = cos θ, u = 1 + 0, an v = a. (35) it is

8 ) Q = πa q a (1 { 1 } 4πa a z 1 3 z 3 0 u vz 0 u + vz = q ) a (1 { [ ] 1 [ ] } 1 a + v u vz 0 v u + vz 0 = q ) a (1 a 1 a a 1 a 1 a ) = q (1 a a a { a } = q a a + a 1 = q {1 a }, q.e.. (36) + a

9 Problem.30 5 Points Aitional information: You may rea Problem 1.1, 1.4,.15,.16, an use any relevant information given in these problems. In the comparison part of the problem, it is only require to compare the result of the finite-element metho with the exact result. h= Figure 3: Finite element metho for D square problem. For symmetry, it is sufficient to write own linear equations for the three encircle gri points of the figure, k Ψ k φ i φ k xy = ρ i ɛ 0 φ i xy i = 0, 1, V V }{{}}{{} k Ψ k A ki = ρi ɛ 0 h i = 0, 1, (37) The first equation is Eq..80 in the textbook, except that for simplicity we use only one inex k instea of k an l, an we use i instea of i an j. The A ki are 8/3, -1/3 or 0 accoring to the rules explaine in the textbook (see, for instance, Problem.9). The sum over k inclues, in principle, the bounary noes. However, since the bounary is on zero potential, in the present problem the sums o not explicitly show bounary points. Also, when writing own the sums we employ the fact that ue to symmetry the potentials on many noes are equal (see figure). Further, ρ i is constant an equals ρ. We fin 8 3 Ψ Ψ Ψ = h ρ ɛ Ψ Ψ 1 3 Ψ = h ρ ɛ Ψ 0 3 Ψ Ψ = h ρ ɛ 0 8Ψ 0 4Ψ 1 4Ψ = a Ψ 0 + 6Ψ 1 Ψ = a Ψ 0 Ψ 1 + 8Ψ = a (38)

10 where a = 3h ρ ɛ 0. The solution is Ψ 0 = 9 70 a, Ψ 1 = 9 8 a an Ψ = 9 35a. For a charge ensity of unity an h = 0.5, we fin the following numerical values liste uner F.E.M.: F.E.M. Exact 4πɛ 0 Ψ πɛ 0 Ψ πɛ 0 Ψ The exact values are taken from Problem 1.4c). Consiering the roughness of the gri, the accuracy achieve by the finite element metho (F.E.M.) is goo. In contrast to the iteration methos, which require many iterations to converge to a final result, the F.E.M. yiels its final result after only a single calculation. In larger problems, the complexity of the F.E.M. calculation rapily increases with the number of gri points, while iteration methos remain very simple. Total 5 Points Please report typos.