Relationship between Electric and Magnetic Field Amplitudes in Light Waves

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Homework Statement

I'm not sure if this goes in introductory physics for not, but anyways...

A light wave is traveling in glass of index 1.50. If the electric field amplitude of the wave is known to be 100 \frac{V}{m}, find (a) the amplitude of the magnetic field and (b) the average magnitude of the Poynting vector.

Homework Equations

E_{e} = \frac{power}{area}
E_{e} = \frac{1}{2}ε_{0}cE_{0}^{2}
n = \frac{c}{\upsilon}
E_{0} = cB_{0}
S = ε_{0}c^{2}E_{0}B_{0}
P = IV

Where
E_{e} is the irradiance
c ≈ 2.998 X 10^{8} \frac{m}{s} is the speed of light
E_{0} is the magnitude of the magnetic field
ε_{0} ≈ 8.8542 X 10^{-12} \frac{(C s)^{2}}{kg m^{3}} is the permittivity of vacuum
n is the refractive index of a material
\upsilon is the velocity of light through the material
B_{0} is the magnitude of the magnetic field
S is the magnitude of the Poynting vector
P is power
V is voltage
I is current

The Attempt at a Solution

For part (a)
I seem to be having some issues processing the given information. I know that irradiance E_{e} is power P divided by areaA. I have 100 \frac{V}{m}, which isn't the irradiance E_{e}. Without being able to find the irradiance E_{e} I'm not sure how to proceed. I'm unsure how to apply the knowledge of the refraction index n. I can solve for the speed of the light through the material \upsilon but I'm not sure what good that really does.

n = \frac{c}{\upsilon}
\upsilon = \frac{c}{n} = \frac{2.998 X 10^{8} \frac{m}{s}}{1.5} ≈ 1.999 X 10^{8} \frac{m}{s}

Once I find the irradiance I can solve for amplitude of the electric field
E_{e} = \frac{1}{2}ε_{0}cE_{0}^{2}
E_{0} = \sqrt{\frac{2E_{e}}{ε_{0}c}} = \sqrt{\frac{2E_{e}}{(2.998 X 10^{8} \frac{m}{s})(8.8542 X 10^{-12} \frac{(C s)^{2}}{kg m^{3}})}}

Once I get this value I can solve for the amplitude of the magnetic field
E_{0} = cB_{0}
B_{0} = \frac{E_{0}}{c}

For part (b)
Once I solve part A I can solve for the average magnitude of the Poynting vector rather easily

S = ε_{0}c^{2}E_{0}B_{0} = (8.8542 X 10^{-12} \frac{(C s)^2}{kg m^{3}})(2.998 X 10^{8} \frac{m}{s})E_{0}B_{0}

Thanks for any help that anyone can provide me in solving this problem.

 

[Simon Bridge]

I have 100 \frac{V}{m}, which isn't the irradiance E_{e}.

You are told what it is in the problem statement -

If the electric field amplitude of the wave is known to be 100V/m

... what is the relationship between the amplitude of the electric and magnetic fields?