Energy density of electric field

AI Thread Summary
The discussion revolves around calculating the electric field magnitude at a distance of 1 meter from a 40W light bulb, which radiates 4W isotropically as light. The energy density formula, 1/2ε₀E², is referenced to derive the electric field strength. Participants explore how to convert the power output into energy per unit area, considering the isotropic distribution of light. The intensity of sunlight at the Earth's surface, approximately 1400W/m², is also mentioned as a comparative measure. The conversation emphasizes understanding the relationship between power, intensity, and electric field calculations in electromagnetic energy.
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Homework Statement



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>

Homework Equations



Energy density = 1/2\epsilon_{}0E^{}2

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.
 
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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:

<br /> I = \frac{1}{\pi}~Wm^{-2}<br />

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

<br /> I = \frac{Energy \times velocity}{Volume}<br />

so.. i guess I've given u enough hint now.. [at what velocity does e.m energy travel?]
 
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