Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

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1 1 Qualifying Examination Electicity and Magnetism Solutions Januay 12, 2006 PROBLEM EA. a. Fist, we conside a unit length of cylinde to find the elationship between the total chage pe unit length λ and the chage density ρ( = A: λ = R A 2 ddφ = 2πA 2 d = 2πAR 3 /3, o A = o The total chage pe unit length within adius ( R is then Q( = 2πA 3 /3 = λ 3 /R 3. 3λ 2πR 3. It is clea fom symmety that the field has only a adial component, which can be evaluated using a Gaussian pillbox and the integal fom of Gauss law: 4πQ( = E.dS = E dφ = 2πE. Inside the cylinde, R, one has 2πE = 4πλ 3 /R 3 and E = 2λ 2 /R 3. Outside the cylinde, R and Q( = λ, so E = 2λ/. The potential is deived fom E = Φ. Choosing it to be 0 at = R, we find: The figue is below. Inside the cylinde, Φ( = 2λ 3 ( 3 R 3 1. Outside the cylinde, Φ( = 2λ ln( R. FIG. 1: The potential as a function of.

2 2 PROBLEM EB. a. To find the velocity of the c.m. system, we find a boost that makes the two photons have equal enegies and equal but oppositely diected momenta. It is sufficient to use the fist condition: E c.m. = γ(1 βe 1 = γ(1 + βe 2, with E 1 the highe enegy photon, and E 2 the lowe enegy photon. The boost goes in the same diection as the highe enegy photon, 1, so that its c.m. enegy is boosted lowe. This leads to β = E 1 E 2 E 1 + E 2. We evaluate γ = 1 1 β 2 = E 1 + E 2 2 E 1 E 2. Numeically, β is about , and γ b We apply the boost to photon 1: E c.m. = E 1 + E 2 2 E 1 E 2 (1 E 1 E 2 E 1 + E 2 E 1 = 2E 2 2 E 1 E 2 E 1 = E 1 E 2. Numeically, the c.m. photon enegies ae 1 kev. PROBLEM EC1. a. To calculate B in spheical coodinates, ecalling ˆ ˆθ = ˆφ, and that the vecto potential has only a ˆφ tem, A φ = m sin θ, we use the fomula fom the Jackson cheat sheet: 2 [ ] [ B = A 1 = ˆ sin θ θ (sin θa φ + ˆθ 1 ] (A φ B = m [2 cos θ ˆ + sin θ ˆθ ] 3 Pehaps the simplest way to calculate the magnetic field in Catesian coodinates is just to tansfom the spheical coodinate expession using ˆ = sin θ cos φ ˆx + sin θ sin φ ŷ + cos θ ẑ, and ˆθ = cos θ cos φ ˆx + cos θ sin φ ŷ + sin θ ẑ. We obtain B = m 3 [ 3 sin θ cos θ cos φ ˆx + 3 sin θ cos θ sin φ ŷ + ( 2 cos 2 θ sin 2 θ ẑ ] We now keep 2 = x 2 + y 2 + z 2 and use ρ 2 = x 2 + y 2 fo bevity, but eplace cos θ z, sin θ ρ, cos φ x, and sin φ y to obtain: ρ ρ B = m 5 [ 3xz ˆx + 3yz ŷ + ( 2z 2 ρ 2 ẑ ] b. The medium with µ = equies the field to be puely in the ẑ diection at the z = 0 plane, which is satisfied by an image dipole with the same magnitude and diection,

3 3 positioned at z = d. The B field fo z > 0 is then the sum of the two dipole fields. We can obtain these most simply with the Catesian coodinate B-field fomula, with the tansfomation z z ± d z ±. Also using 2 ± = x2 + y 2 + (z ± d 2, the esult is: ( B 3x(z m = + + 3x(z ( 3y(z ˆx y(z ( 2(z ŷ ρ 2 + 2(z 2 ρ 2 ẑ c. The foce exeted on the dipole is given by F = ( m B, whee B is the field fom the image dipole, evaluated at x = y = 0, z = d. Since m is in the ẑ diection, only the ẑ component of the field mattes, and thus F = (m 2 2(z + d 2 x 2 y 2 (x 2 + y 2 + (z + d 2 5/2 It is clea fom cylindical symmety that thee is no x o y component to the foce. In the algeba, this is because the x 2 and y 2 in the numeato and denominato lead to the deivatives being popotional to odd powes of x and y espectively, which vanish at x = y = 0. Thus, we only explicitly evaluate the deivative with espect to z: [ F = m 2 4(z + d (x 2 + y 2 + (z + d 2 + ( 5(z + d(2(z + ] d2 x 2 y 2 ẑ 5/2 (x 2 + y 2 + (z + d 2 7/2 F = m 2 9(z + d(x2 + y 2 6(z + d 3 3m 2 ẑ (x 2 + y 2 + (z + d 2 7/2 x,y 0, z d ẑ 8d 4 Finally, as noted by WK in his caveat, a movement of the dipole also implies a movement of the image chage, which leads to a facto of 2 geate foce: F = 3m2 4d 4 The minus sign indicates that the foce is attactive. ẑ

4 4 PROBLEM EC2. Assume the conducting plate is an xy plane at z = 0, and the electic dipole is p = p 0 ŷ at z = d on the z axis. a. The electic dipole induces suface chages so that the electic field is puely nomal to the xy plane at z = 0. This can be epesented by an image electic dipole p image = p 0 ŷ, at position z = d. The vecto potential fo a fee dipole p at location d is A( = ik p eik d d. Since vecto potentials ae additive, the total vecto potential of the two dipoles is: [ ] e A( ik + d eik d = ik p + d d. Thus, A( = ik eik [ p ikd cos θ e 1 + (d/ cos θ e ikd cos θ 1 (d/ cos θ whee we used the appoximation d so ± d ± d cos θ. We now expand, e.g., e ikd cos θ = 1 + ikd cos θ +... and (1 + (d/ cos θ 1 = 1 (d/ cos θ..., and we take to be lage, to obtain A( ik eik p [2ikd cos θ] 2k2 d cos θ eik We find the B field fom B = A. Since is lage, we neglect the ˆθ and ˆφ tems, which lead to exta powes of 1/. We also appoximate e ik ik eik, the leading ode in. Thus, E( = B( n. B 2ik 3 d cos θ eik The Poynting vecto S = c E B = c B 2 = c 8π 8π 2π k6 d 2 cos 2 θ n p 2 / 2. ˆ p With the vectos witten in tems of unit vectos p = p e y, and n = sinθ cosφ e x + sinθ sinφ e y + cosθ e z, we find the adiation fo the electic dipole located a distance d in font of a conducting plate: = dω 2 S = c 2π k6 d 2 cos 2 θ (cos 2 θ + sin 2 θ cos 2 φ. Afte integation ove angles the total powe is P = 4 15 c k6 d 2 p 2. Compae this with the adiation patten fo a fee electic dipole: = µo dω 32π k4 c 3 p 2 (cos 2 θ + sin 2 θ cos 2 φ, and a total powe of P = µo 16π k4 c 3 p p ],

5 5 The pesent poblem diffes in paticula: In the y z plane (cosφ = 0 the angula distibution of the adiation is as cos 4 θ, while fo the fee dipole it is like cos 2 θ. In the x z plane (cosφ = 1 the angula distibution of the adiation is as cos 2 θ, while fo the fee dipole it is independent of θ. In geneal (plate + dipole = dω 4k2 d 2 cos 2 θ (fee dipole. dω b. If p is pependicula to the conducting plate, image is + p at location d. Because of the same diection of the two dipoles, the total vecto potential is twice that of a fee dipole. As a esult the adiation patten is multiplied by a facto 4, compaed to the fee electic dipole. dω (plate + dipole = 4 dω (fee dipole. PROBLEM ED1. The total chage on the top plate is Q = σab. a. Students might ecall the field in the capacito is σ/ɛ 0. Othewise, they use E o = ρ ɛ o E o = σ ɛ o ẑ. The Poynting vecto is along the +x axis: S = E o H o = 1 µ o Eo B o = 1 µ o The momentum density is along +x axis, σ ɛ o ẑ B o ŷ = 1 ɛ o µ o σb oˆx = c 2 σb oˆx. p = 1 E c 2 o H o = σb oˆx, and the total momentum is then the volume integal P total = pdv = σb o abhˆx = QB o hˆx. b. Now tun off B o in time t. Since B o = B o ŷ, conside a closed loop in an xz plane, at some fixed y, of the same aea as a coss section though the capacito, A = ah. We have an initial magnetic flux though the loop of Bo ds = B o ahŷ. Thee is an induced E field E.d l = t B o.ds = B oah. t With h a, and the induced field opposing the decease in the magnetic field, we see that E.d l = 2aE ind E ind = B oh 2 t with diection +ˆx at the uppe (+σ plate, but ˆx at the lowe ( σ plate. The foces on both plates ae then equal and in the same +ˆx diection, F = Q E = QB oh 2 t ˆx.

6 6 Integating ove time, p = F dt, gives a total momentum tansfe to each plate of p = (QB o h/2ˆx. Thus, in tuning off the magnetic field, each plate eceives half the momentum stoed in the field. PROBLEM ED2. a. The vecto potential given descibes an electomagnetic wave taveling along the +x axis. The E and B fields ae calculated using E = 1 c A t = 1 f c t ŷ and B = A = f x ẑ. Since the function f depends on u = x ct, we see that f t on the electon is F = q( E + 1 v B, so c F x = e 1 c v yb z = e 1 c v ye y, = c f x, so B z = E y. The foce F y = ee y + e 1 c v xb z = ee y + e 1 c v xe y, and F z = 0. Thus the electon moves in the xy plane. Recalling the hint above, we wite the equations of motion as: v x = e mc v ye y, v y = e m E y + e mc v xe y. Assuming the light wave stats to inteact with the electon at t = 0, the initial conditions ae v x (t = 0 = 0 and v y (t = 0 = 0. b. Following the hint, we fist evaluate d ( v 2 dt x + vy 2 e = 2vx v x + 2v y v y = 2v x mc v ye y + 2v y ( e m E y + e mc v xe y = 2e m v ye y = 2c v x. The tick was to substitute in the equations of motion, twice. One sees that v 2 x + v2 y = 2cv x, and v x thus must be positive. (This is the answe fo pat c. c. Thus, the electon is pushed in the +x diection in the xy plane. The ultimate sign of the y component depends on the details of f(x vt. d. We also see fom the above elations that v y and v x ae not independent, and we may eliminate one of them. The tick is to use v 2 y = 2cv x v 2 x = c 2 (c v x 2, c v x = c 2 v 2 y (note the useful integal to eliminate v x fom the second ( v y equation of motion: v y = e mc E y (c v x v y c2 v 2 y = e mc E y = e mc 2 f t. The integal of the left side ove time is the useful integal: v y 1 ( vy dt = dv c2 vy 2 c2 vy 2 y = acsin. c

7 7 The integal of the ight side ove time is e f mc 2 t dt = e (f(t f( = e mc2 mc 2 f(t. Thus ( e v y (t = c sin mc f(x vt, 2 and ( e v x = c c 2 vy 2 = c c cos mc f(x vt. 2 This seems to me to be a potentially difficult poblem, much moe so than ED1, since it equies seveal math ticks (seveal ticks bette technique? to solve.

Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b.

Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b. Solutions. Plum Pudding Model (a) Find the coesponding electostatic potential inside and outside the atom. Fo R The solution can be found by integating twice, 2 V in = ρ 0 ε 0. V in = ρ 0 6ε 0 2 + a 2