Homework 6 and 7 Solutions, Electromagnetic Theory II Dr. Christopher S. Baird, Spring 2015 University of Massachusetts Lowell

Transcription

1 Homewok 6 and 7 Solutions, Electomagnetic Theoy II D. Chistophe S. Baid, Sping 2015 Univesity of Massachusetts Lowell Poblem 1 Jackson 9.3 Two halves of a spheical metallic shell of adius R and infinite conductivity ae sepaated by a vey small insulating gap. An altenating potential is applied between the two halves of the sphee so that the potentials ae ±V cos ωt. In the long-wavelength limit, find the adiation fields, the angula distibution of adiated powe, and the total adiated powe fom the sphee. SOLUTION: Two halves of a sphee that ae chaged oppositely should immediately emind us of a dipole so that we can assume the dipole tem is the dominant tem. We just need to find the dipole moment of this configuation and then we can apply the equations in the class notes fo the fields adiated by a dipole. Let us conside the sphee at an instant in time when the voltages ae at thei peak. We peviously found the extenal potential due to such a sphee in tems of Legende polynomials. We found the fist tem to be: =V 3 2 R 2 2 cos We know that the potential due to a electic dipole pointing in the z diection should look like: = 1 p 4 0 cos 2 Setting these equal and solving fo p we end up with p=6 0 V R 2 z Now the potentials on the sphee vay accoding to cos(ωt), which is just the eal pat of the complex exponential signifying hamonic dependence. We aleady found in the class notes the adiation-zone fields due to an electic dipole vaying hamonically in time. We can wite them down immediately and plug in the dipole moment we have hee. B= k 2 p 4 k p ei k t B= 3 V k 2 R 2 2 c i k t e sin

2 E= k 2 p k k p e i k t 4 0 E= 3 2 V k 2 R 2 ei k t sin The time-aveaged angula distibution of adiated powe is: d = 1 2 R[ 2 E H *] d = 1 2 [ R[ V k 2 R 2 ei k t d = 9 8 V 2 k 4 R 4 sin 2 The total adiated powe is: P= 9 8 V 2 k 4 R 4 sin 2 d, P= 3 V 2 k 4 R 4 sin ] [ 1 3 V k 2 R 2 i k t e sin 0 2 c ] ]

3 Poblem 2 A chage +Q is placed at the (x, y, z) point (a, 0, 0) and a chage -Q is placed at the point (-a, 0, 0). These two chages have thei chage oscillate in phase at the fequency ω so that they fom a hamonically oscillating electic dipole. Additionally, a sepaate chage +Q is placed at (0, a, 0) and a sepaate chage -Q is placed at (0, -a, 0). These othe two chages oscillate in phase with each othe at the fequency ω so that they fom anothe hamonically oscillating electic dipole. The oscillation of the second two chages lags behind the oscillation of the fist two chages by a fixed phase diffeence θ 0. (a) Find the electic dipole moment of the fist two chages. Find the electic dipole moment of the othe two chages. (b) Find the total electic field poduced by this system in the fa-field by adding the dipole-adiation electic field fom each dipole. (c) What does the fa-field electic field educe down to along the z axis? Along the x axis? Along the y axis? SOLUTION: (a) The electic dipole moment is defined as: p= x' ρ(x')d x' Fo points chages, this educes down to: p= j Q j x j Applied to the fist two chages, this becomes: p 1 =Q a x+( Q)( a x) p 1 =2 Q a x Explicitly witing out the oscillation in time, this becomes: p 1 =2Q a x e i ω t Fo the othe two chages we find: p 2 =2Q a ŷ e i ωt +i θ 0 (b) The electic field in the fa-field due to the fist two chages is: E 1 = k 2 p 1 k ( k p1 ) ei(k ω t) 4π ϵ 0 E 1 = k 2 Q a e k ( k i (k ω t ) x) 2π ϵ 0

4 The electic field in the fa-field due to the othe two chages is: E 2 = k2 Q a 2π ϵ 0 k ( k ŷ) e i(k ω t+θ0) The total electic field is: E= k 2 Q a e i(k ω t) 2π ϵ 0 [ k ( k x)+ k ( k ŷ)e i θ 0] Since the system is at the oigin, the wavevecto diection is the adial diection: E= k 2 Q a e i(k ω t) 2πϵ 0 [ ( x)+ ( ŷ)e i θ 0 ] Expand out the vectos into ectangula components and evaluate the coss poducts to find: E= k 2 Q a e i(k ω t) [ x( sin 2 θsin 2 ϕ cos 2 θ)+ŷ sin 2 θcosϕ sin ϕ+ẑsin θcosθ cosϕ+ 2π ϵ 0 x(sin 2 θsin ϕcos ϕ)e i θ 0 +ŷ( sin 2 θ cos 2 ϕ cos 2 θ)e i θ 0 +ẑsin θcosθsin ϕ e i θ 0] (c) Along the z axis ( θ=0 ), this becomes: E= k 2 Q a e i (k ω t ) [ x+ŷ e i θ 0 ] 2 π ϵ 0 If the phase diffeence is zeo, this becomes: E= k 2 Q a e i (k ω t ) [ x+ŷ ] 2π ϵ 0 which is a linealy-polaized wave taveling along the z axis polaized 45 degees to the x axis. If the phase diffeence is 180 degees, then the electic field along the z axis becomes: E= k 2 Q a e i (k ω t ) [ x ŷ ] 2π ϵ 0 which is a linealy-polaized wave polaized -45 degees to the x axis. If the phase diffeence is 90 degees, then we have: E= k 2 Q a e i (k ω t ) [ x+i ŷ ] 2π ϵ 0 which is a left-ciculaly polaized-wave. Similaly, a phase diffeence of -90 degees would yield a ight-ciculaly polaized wave. We can theefoe ceate all sots of simple polaizations fo waves adiated along the z axis if we popely choose the phase diffeence between the chage oscillations.

5 Along the x axis, the electic field can be found (plug θ=π/2 and ϕ=0 into the geneal solution): E= k 2 Q a e i (k ω t +θ0) ŷ 2 π ϵ 0 Theefoe, the waves adiated along the x axis ae always linealy-polaized in the y diection. This makes sense if we emembe that electic dipoles cannot adiate long thei axes, and the x axis is aligned with the axis of the fist dipole, so that only the field of the second dipole can contibute. Along the y axis, the electic field can be found (plug θ=π/2 and ϕ=π /2 into the geneal solution): E= k 2 Q a e i (k ω t ) x 2 π ϵ 0 Again, the adiated wave is linealy-polaized because one of the dipoles cannot adiate in this diection.