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Calculate the energy density of the Earth's atmosphere

by phystudent515
Tags: atmosphere, density, earth, energy
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phystudent515
#1
Mar17-12, 09:44 AM
P: 4
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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tiny-tim
#2
Mar17-12, 01:41 PM
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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?
phystudent515
#3
Mar17-12, 05:48 PM
P: 4
Quote 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.

tiny-tim
#4
Mar19-12, 05:22 AM
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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 B2/o 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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