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Determine the electric charge "in the earth"

  1. Sep 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose that the electric field in the Earth's atmosphere is E = 92 N/C, pointing downward. Determine the electric charge in the Earth.

    2. Relevant equations



    3. The attempt at a solution
    I suppose that I'm supposed to construct a Gaussian around the earth and find the flux in.

    After a whole bunch of steps I ended up with and equation that looks like this

    [itex]E=\frac{k_{e} Q_{in}}{a^{3}}r[/itex]

    valid for r<a.

    Since E is negative, the charge of the earth must be negative.

    Now I simply said that a=6430km and r=6400km and tried solving for Q. Apparently that's wrong.
     
  2. jcsd
  3. Sep 5, 2014 #2

    BvU

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    I don't understand the a3 ? isn't there just one radius involved? What's your gaussian surface ?
     
  4. Sep 5, 2014 #3

    gneill

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    Can you show/describe your "whole bunch of steps"? Where does that a variable in the denominator come from? How did you come up with the numerical value for a?
     
  5. Sep 5, 2014 #4

    andrevdh

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    Shouldn't one just consider the earth as a conductive sphere?
     
  6. Sep 5, 2014 #5

    BvU

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    Doesn't matter, conducting or not conducting.
    At least it does not matter outside of the sphere. That's the nice thing about gauss law.
    Of course the charge must be uniformly distributed in the angular coordinates, but that follows from the 92 N/C " everywhere".

    Both helpers try to point out that the 6430 as opposed to the 6400 is falling out of the sky :smile:. Why not have one radius of the earth ?
     
  7. Sep 5, 2014 #6

    ehild

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    The atmosphere is quite wide, at what height is the electric field 92 N/C? If that field is caused by the charge of the Earth alone, it should depend of height.

    Anyway, the atmosphere is outside the Earth's crust, and the formula in the OP refers to the electric field inside a homogeneously charged sphere.

    ehild
     
  8. Sep 5, 2014 #7
    If you recall, if you are outside a spherically symmetric charge (or mass) distribution, then you can model the distribution as a point charge (or mass). Since you are measuring a field outside of the Earth, what does that tell you about how to model the Earth's charge?
     
  9. Sep 5, 2014 #8
    This is all the information that was given in the question. I know it's very vague.

    the a^3 comes from the integral. I said that the Gaussian is inside of the earth. Apparently the formula is correct.

    There are the "hints" given:

    Imagine a spherical surface just above Earth’s surface to enclose the entire planet.

    Assume the field is uniform. Calculate the total electric flux through the entire surface, taking care to use the correct sign.

    Write Gauss’s law relating the flux to the charge enclosed within the surface, and use it to find the enclosed charge.
     
  10. Sep 5, 2014 #9
    That doesn't work.

    E=ke Q/r^2 gives the incorrect answer.
     
  11. Sep 5, 2014 #10

    ehild

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    It must be correct. Show your calculation. What did you use for r?

    ehild
     
  12. Sep 5, 2014 #11
    One thing about problem solving, is that before you start cranking the calculus, or doing the heavy math, you should have an idea of where you are going and an estimate of what the answer will be. This is developed by understanding the physics concepts. One important concept is symmetry. (Look up the spherical cow)

    I am sure you have read and studied Gauss's law which states that the flux through a surface is proportional to the charge enclosed. You also were probably informed that symmetry plays a big part in simplifying Gauss's law. What is more simple than a sphere? In this case, the symmetry really does reduce this problem to that of calculating the electric field a given distance from a point charge. That distance is the Earth's radius.

    Start with that, and make sure you are using the proper units, and correct values for all variables.
    Solve the equation first before putting in the numbers, and carefully check your work.
    then post it here and we can help you check it.

    But yes, the concept will work, unless physics has changed drastically and I heard nothing of it. ;)
     
  13. Sep 6, 2014 #12
    I think you probably trying to make this harder than they intended.
    With your using both and a and an r and have r<a and a^3 and r^1 it seems like you are trying to come up with what fraction of the total charge on the Earth is contained within a shell of radius a or something.

    Or maybe you were trying to start some sort of proof of a simple way to treat a uniformly charged sphere or something.

    Anyway, they give you E in the atmosphere, so we are dealing with the entire Earth contained so you don't need to do any within the radius stuff. And the atmosphere is thin and the bumps and valleys on the Earth are small, it's all insignificant compared to the radius of the Earth so just forget all that and all you need to do is just use radius of the Earth and assume that covers it all, again all the bumps and a few thousand meters here or there doesn't matter or the height of the measurement in the atmosphere, etc. whatever, forget it, so just forget about the a and all that stuff.

    The Earth is spherical, very symmetric so think what that means and how you can treat the charge throughout a solid sphere to make things very simple and the basic rule they surely flat out told you about Gauss Law for such a case. (they probably are not asking you to prove this for this question either)

    And then apply Gauss law around that. You are given radius of the Earth and the magnitude of the electric field and the constant so in the end you should have nothing left in bare variable but the charge that they want and it should be just a simple algebra of multiplying a few numbers to get the answer.
     
    Last edited: Sep 6, 2014
  14. Sep 6, 2014 #13
    Hmm are you mixing wrong units? Using the wrong type of constant (for E&M there are all sorts of different unit sets people use and the constants and factors change depending)? Did you just use the radius of the Earth for r? (make sure to not accidentally use the diameter or something)
     
  15. Sep 6, 2014 #14
    I was using the wrong radius. I looked up the radius from the back of the book and it worked. Thanks.

    This problem was kind of misleading as the first part had me deriving the initial formula that I posted. The last part was finding this charge so I assumed I had to use the formula that I just derived. Anyway, thanks for all the help guys.
     
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