- #1
Blastrix91
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Homework Statement
http://img809.imageshack.us/img809/5676/unavngivetk.png
Additional table info:
The molarmass ##M_{air}## is ##29 * 10^{-3} g/mol = 29*\frac{10^{-3}}{1000} kg/mol##
The mass density ## \rho_{air} ## is ##1.3 kg/m^{3}##
The dielectric constant ## K_{air}## is 1.00059
My teacher told me to consider the air as 1 molecule.
Homework Equations
http://img94.imageshack.us/img94/1667/unavngivet2w.png
The Attempt at a Solution
Here is my approach:
Basically it is only N we need which is molecules per volume:
[tex] N = \frac{Molecules}{V} = \frac{N_a n}{V} = \frac{m}{M_{air}} \frac{N_a}{V} = \frac{\rho_{air}V}{M_{air}} \frac{N_a}{V} = \frac{\rho_{air}N_a}{M_{air}} [/tex]
[tex] \alpha = \frac{3 \varepsilon_0 M_{air}}{\rho_{air}N_a} \frac{K_{air}-1}{K_{air}+1} [/tex]
Inserting the values:[tex]\frac{3(8.854*10^{-12})(29*\frac{10^{-3}}{1000})}{1.3(6.022*10^{23})} * \frac{1.00059 - 1}{1.00059 + 1} = 2.902 * 10^{-43}[/tex]
The answer in the back of my book says it is supposed to be: ##\alpha = 9.7 * 10^{-41}##
Can any of you guys spot where I did wrong? (The table values were some I got from my teacher and if my approach is right, then he might have wrote them wrong)
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