Energy in an Electric Field of an EM Wave

[bobsalive]

I have been desperately trying to find an equation (or a set of equations where I can derive an equation) that shows a relationship between the energy in an electric field and the wavelength or frequency of an electromagnetic wave.

I am trying to show a relationship between Wavelength and Intensity of an electromagnetic wave, and working backwards, the Intensity is proportional to the Energy Density of the wave, half of which is the electric energy density, which can be calculated by multiply the permittivity of free space with the energy in the electric field squared. But I can't find the equation for energy in the electric field (in relation to wavelength)! Any ideas?

 

[xepma]

E = hf

 

[bobsalive]

That is the energy in the wave, not the energy in the electric field! There is energy in the magnetic field as well isn't there?

 

[lanedance]

have a look at this
http://www.phy.duke.edu/~lee/P54/energy.pdf

In classical electromagnetics I think the energy is related directly to the Intensity (or field strength squared) independent of wavelength, this is because the average of a sinusoid is independent of its period.

However in the quantum limit the energy of a photon is directly related to the wavelength by $$E = \hbar f$$ as mentioned previously

 

[bobsalive]

Thanks lanedance for the link. I read it, and I've already covered the stuff in that page, what I am trying to find is an equation that will tell me what E equals (E as in the electric field strength, NOT the energy of the wave).

A photon does not have a charge, so E = f/q is an inappropriate equation...

Edit: I know that I = uc, and because u = εE^2, I = cεE^2. But what does the E equal?

 

[lanedance]

I'm not totally sure what you're asking...:uhh:

The field E(x,t) is an allowable solution of the wave equation generated from maxwells equations

One allowable example of a plane wave propogating in the x dir'n with frequency $$\omega$$ would be
$$E(x,t) = (0,E_y,0)$$with $$E_y (x,t) = E_0 \cos{(kx -\omega t)}$$
$$B(x,t) = (0,0,B_z)$$with $$B_z (x,t) = B_0 \cos{(kx -\omega t)}$$

with magnetic & electric field strengths related by:
$$B_0 = \frac{E_0}{c}$$

Thanks lanedance for the link. I read it, and I've already covered the stuff in that page, what I am trying to find is an equation that will tell me what E equals (E as in the electric field strength, NOT the energy of the wave).

A photon does not have a charge, so E = f/q is an inappropriate equation...

the E is in fact the electric field.. so if we palced a test charge q in the electric field it would experience a force F = Eq

is this what you mean?

Edit: I know that I = uc, and because u = εE^2, I = cεE^2. But what does the E equal?

 

[bobsalive]

I'm not sure about maxwell's equations, but I think I've sorted out the problem by using E=V/d (Electric Field = Voltage / Distance) which has provided me a suitable answer.

Thanks for all your help!

 

[Wallabyted]

This might help too ... (from
http://www.osha.gov/SLTC/radiofrequ...etic_fieldmemo/electromagnetic.html#section_3 )

Pd = E x H
Watts/meter2 = Volts/meter x Amperes/meter

where
Pd = the power density,
E = the electric field strength in volts per meter,
H = the magnetic field strength in amperes per meter.

The above equation yields units of W/m2. The units of mW/cm2 are more often used when making surveys. One mW/cm2 is the same power density as 10 W/m2 The following equation can be used to obtain these units directly:

Pd = 0.1 x E x H mW/cm2