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Calculate the energy density of the Earth's atmosphere

 
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Mar17-12, 09:44 AM   #1
 

Calculate the energy density of the Earth's atmosphere


Not really homework, just practice for a midterm, I also have the correct answers; but I guess this is the correct section.

1. The problem statement, all variables and given/known data

In the Earth's atmosphere we have an electric field with a vertical direction down towards the earth. The lower part of the atmosphere has a typical field strength of 100 V/m. The strength of the earth's magnetic field is approximately 50*10^(-6) T.

Find the energy density in each of the two layers.

I assume the "two layers" are the upper and lower layers.

2. Relevant equations

[tex] \mbox{Energy density} = \frac{\mbox{Electric energy}}{\mbox{Volume}} = (1/2)\kappa \epsilon_0 E^2[/tex], where k is the Dielectric constant, e_0 is the permittivity of the space (8.85*10^(-12))

We also have that [tex]\kappa = \frac{E_0}{E}, E = \frac{F}{q_0}[/tex].

3. The attempt at a solution

I assume that I could just obtain the energy density directly from using [tex](1/2)\kappa \epsilon_0 E^2[/tex] directly? I know the correct answers should be [tex]u_1 = 4.4\cdot 10^{-8} J, u_2 = 4.4\cdot 10^{-4} J[/tex] However I don't know how to proceed to obtain the variables.
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Mar17-12, 01:41 PM   #2
 
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hi phystudent515!

i find the question very confusing

also the answer … energy density isn't in J, it's in Pa (pascals)
have you given us the whole question?
Mar17-12, 05:48 PM   #3
 
Quote by tiny-tim View Post
hi phystudent515!

i find the question very confusing

also the answer … energy density isn't in J, it's in Pa (pascals)
have you given us the whole question?
Forgive me, the unit for the correct answers are [tex]J/m^3[/tex], I simply misread. However I don't know if this seems more correct or not.

I did omit some text due to it being translated by hand. I'll try to restate the problem text somewhat better worded (due to translation):

"Due to lightning discharges a separation of charge will be created between the atmosphere and the surface of the Earth. The result of which is a vertical electric field in the atmosphere, pointing down towards the Earth. In the lower part of the atmosphere the field strength is typically 100 V/m. The strength of the Earth's magnetic field is approximately 50*10^(-6) T.

Find the energy density in each of the fields."
That is the entire problem text with nothing omitted.

I have tried the following to obtain the same solutions:

For the electric field: [tex](1/2)\kappa \epsilon_0 E^2[/tex] = 1/2 * 1 * 8.85*10^(-12) * 100^2 = 4.4 * 10^(-8), which is correct.

For the magnetic field: I use [tex](1/2) \epsilon_0 E^2 + \frac{B^2}{2\mu_0}[/tex] = 1/2 * 8.85*10^(-12) * 100^2 + (50*10^(-6))^2/(2*4*pi*10^(-7)) = 9.94*10^(-4), which is not correct. I observe however that it is roughly twice that of the result I'm looking for.
Mar19-12, 05:22 AM   #4
 
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Calculate the energy density of the Earth's atmosphere


hi phystudent515!

yes, that second 4.4 is clearly a misprint …

the typesetter has got bored and typed the number twice!

1/2 B2o is the correct formula

btw, J/m3 and Pa are the same, see eg http://en.wikipedia.org/wiki/Energy_...Energy_density
Energy per unit volume has the same physical units as pressure, and in many circumstances is an exact synonym: for example, the energy density of the magnetic field may be expressed as (and behaves as) a physical pressure …

In short, pressure is a measure of volumetric enthalpy of a system.
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