# Calculate the energy density of the Earth's atmosphere

by phystudent515
Tags: atmosphere, density, earth, energy
 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 $$\mbox{Energy density} = \frac{\mbox{Electric energy}}{\mbox{Volume}} = (1/2)\kappa \epsilon_0 E^2$$, where k is the Dielectric constant, e_0 is the permittivity of the space (8.85*10^(-12)) We also have that $$\kappa = \frac{E_0}{E}, E = \frac{F}{q_0}$$. 3. The attempt at a solution I assume that I could just obtain the energy density directly from using $$(1/2)\kappa \epsilon_0 E^2$$ directly? I know the correct answers should be $$u_1 = 4.4\cdot 10^{-8} J, u_2 = 4.4\cdot 10^{-4} J$$ However I don't know how to proceed to obtain the variables.
 Sci Advisor HW Helper Thanks P: 26,157 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?
P: 4
 Quote by tiny-tim 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 $$J/m^3$$, 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: $$(1/2)\kappa \epsilon_0 E^2$$ = 1/2 * 1 * 8.85*10^(-12) * 100^2 = 4.4 * 10^(-8), which is correct.

For the magnetic field: I use $$(1/2) \epsilon_0 E^2 + \frac{B^2}{2\mu_0}$$ = 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.

HW Helper
Thanks
P: 26,157
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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