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Calculate the volume charge density of the atmosphere

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data

    In the atmosphere and at an altitude of 250 m, you measure the electric field to be 150 N/C directed downward, and you measure the electric field to be 170N/C directed downward at an altitude of 400 m. Calculate the volume charge density of the atmosphere in the region between altitudes of 250 m and 400m, assuming it to be uniform.

    2. Relevant equations



    3. The attempt at a solution

    I just want to know. When I substitute the values in E = (sigma/epsilon 0) and then sigma = Q/Area.. So I get Q = 150 (epsilon 0) A. Now this I will substitute in the formula rho = Q / V.

    Now my question is what will be Area? What value of r should I write in 4 pi r^2 ?
    Will it be radius of earth or the altitude of 250 m? If it's radius of earth, where will I write 250m then?
     
  2. jcsd
  3. Feb 14, 2010 #2

    kuruman

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    This is a Gauss's Law application. If you can picture the Gaussian surface that you should use, you will know what to do with r and what its value(s) ought to be. So what kind of Gaussian surface should you use to "calculate the volume charge density of the atmosphere in the region between altitudes of 250 m and 400m, assuming it to be uniform"?
     
  4. Feb 14, 2010 #3
    Hmm may be cylinder? But how in the world do I get the idea of gaussian surface?!
     
  5. Feb 14, 2010 #4

    kuruman

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    Choose a Gaussian surface that reflects the symmetry of the problem. The Earth is more closely approximated to a sphere than a cylinder. Also, the expression E = σ/ε0 applies only in cases where there is a surface charge. There is no surface charge in the atmosphere at heights of 250 m or 400 m or any height. There is only volume charge density in the atmosphere.
     
    Last edited: Feb 14, 2010
  6. Feb 14, 2010 #5
    So what it should be E = rho / epsilon 0?
    And I got what you are saying. I got the charge in the sphere, Q inside as = 4 E pi epsilon 0 r^(2)

    Is that right? If that is right, then r=radius of the earth only
     
  7. Feb 14, 2010 #6
    Ok I tried it once again.

    This time I took rho = Q / V

    And then substituted Q inside = 4 E pi epsilon 0 r^(2)
    with rho V = 4 E pi epsilon 0 r^(2)

    => rho 4/3 pi r^3 = 4 E pi epsilon 0 r^(2)

    Is it correct?? In that way I can find rho, but then I am still confused about the 'r' s
     
  8. Feb 14, 2010 #7
    Wait the r will be radius of earth plus 250m ?
     
  9. Feb 14, 2010 #8

    kuruman

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    You know that the E field at RE+400 m is 170 N/C and that the E field at RE+250 m is 150 N/C. Both fields are radially in.

    Can you calculate the electric flux through the area of a sphere of radius RE+400 m and then through a sphere of radius RE+250 m?

    The sum of the two would be the total flux through a shell of inner radius RE+250 m and outer radius RE+400 m. What should this total flux be equal to according to Gauss's law?
     
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