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Homework Help: EM Radiation oscillating charged mass

  1. Aug 7, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m and charge q is attached to a spring with force constant k, hanging from the ceiling. Its equilibrium position is a distance h above the floor. It is pulled down a distance d below equilibrium and released, at time t = 0;

    Under the usual assumptions (d << lambda << h) calculate the intensity of the radiation hitting the floor as a function of the distance R from the point directly below q.

    2. Relevant equations

    3. The attempt at a solution
    I see that this is a harmonic oscillator that could be described by x(t) = d cos(wt)
    and a = -w*w*dcos(wt)

    I would like to use Larmor's formula: P = u_0 q^2 a^2 / ( 6 pi m c) but I believe that I may need to revert back the to Poynting vector because we are only trying to find intensity [W/area] so: S = u_0 q^2 a^2 / ( 32 pi m c).

    I think I'm having a hard time determining how to draw the picture to visualize this situation. How do I handle finding the intensity on the floor? I see that the distance from the mass to the floor is (R^2 + (h+delta)^2)^1/2.

    Does anyone have any tips on how to get started with this? Do I need to rewrite the electric and magnetic fields and recalculate the Poynting vector from scratch (re-derive the electric dipole equations essentially?)
  2. jcsd
  3. Aug 7, 2011 #2
    Draw the charge at a height h above a plane. Draw a ring of radius R and thickness dR in said plane centered on the charge. Draw the angles theta and theta + d(theta) that the ring makes.

    From http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node146.html

    We find the differential power, dP, into a solid angle d(omega) goes as:

    dP/d(omega) is proportional to a^2*sin^2(theta) eq. 18.35

    d(omega) = sin(theta)*d(theta)*d(phi) See solid angle:


    Convince your self that the power that goes through the ring is proportional to:


    There is a geometric relationship between dR and d(theta) that you should be able to find from your drawing.

    You know dP/d(omega) but you want dP/dA where dA is the area of the ring.

    But dP/dA = dP/d(omega)*d(omega)/dA

    All you need is the relationship between d(omega) and dA.

    Corrections welcome, hope this helps!
  4. Aug 7, 2011 #3
    Hello Spinnor,

    Thanks for your excellent response, I've almost figured it out.

    From the picture I see that: dR2 = 2(h2 + R2) + dR2 + 2RdR - 2 sqrt(h2 +R2) sqrt( h2 + (R+dR)2) cos( d(theta) )

    I decided that this was a mess but we could probably simplify it to where (R+dR)2 ~= R2 so I get: dR = (h2 + R2) d(theta).

    Then using the solid angle d(omega) = sin(theta) d(theta) d(phi) I re-arrange and plug it in.

    d(omega) = sin(theta) dR d(phi) / sqrt(h2 + R2)

    Now there is where I've struggled. I can't seem to translate that into a useful d(omega)/dA expression. Is my approximation for dR correct? Or am I missing something else?
  5. Aug 7, 2011 #4
    I think we want dA = R*d(phi)*dR

    d(omega)/dA =[sin(theta)*d(theta)*d(phi)]/ R*d(phi)*dR

    dR is related to d(theta)

    [h^2+R^2]*d(theta)/h = dR , so

    d(omega)/dA = [sin(theta)*d(theta)]/R*dR



    So it looks like dP/dA goes as


    We want to get sin(theta) in terms of R. Let straight down be theta = 0, then R/[h^2+R^2]^.5 = sin(theta)

    This was pretty sloppy. Better next time, hope it helps.
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