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Energy density of electric field

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Assume that 10% of the energy dissipated by a 40W light bulb is radiated isotropically in the form of light. What is the magnitude of he E-field at a distance of 1m? What is it for sun light on the Earth's surface, given the Sun provides ~ 1400W/m2>

    2. Relevant equations

    Energy density = 1/2[tex]\epsilon[/tex][tex]_{}0[/tex]E[tex]^{}2[/tex]

    3. The attempt at a solution

    Ok so I know that we have 4W available, but how do I convert this into and energy per unit area on the surface or volume? I assume we have to use the equation for the energy denstiy of a magnetic field given above.
    Last edited: Apr 8, 2008
  2. jcsd
  3. Apr 8, 2008 #2
    If it spreads out isotropically in all directions, then the power [4 W] is uniform at any point on the sphere of the radius, and can be given by:

    I = \frac{1}{\pi}~Wm^{-2}

    You have the same thing i.e. Intensity for the 'sun' question. Now, Intensity is simply:

    I = \frac{Energy \times velocity}{Volume}

    so.. i guess i've given u enough hint now.. [at what velocity does e.m energy travel?]
    Last edited: Apr 8, 2008
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