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Gauss Law

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data
    I am to find the electric field for a charge distribution of
    $$ \rho(x)= e^{-\kappa \sqrt{x^2}} $$


    2. Relevant equations

    I know that gauss law is $$ \int E \cdot da = \frac{q_{enc}}{\epsilon_0} $$

    3. The attempt at a solution

    I am not sure what the charge distribution looks like. Is this saying that there is only charge along the x axis? or is the charge everywhere? I am also no sure what kind of surface I should be integrating over. Should I be integrating over a circle and then finding the total charge enclosed within?
     
  2. jcsd
  3. Sep 17, 2013 #2
    Is [itex] x [/itex] a vector? If not, assume one dimension. Your surface area will most likely be of a sphere. Also, recall that [itex] q_{enc} [/itex] is the total charge. Can you think of another (more formal) way to write [itex] q_{enc} [/itex]?
     
  4. Sep 17, 2013 #3
    [itex] x [/itex] appears to be a scalar. Does this mean that the charge only exists along the x axis? Or is it also distributed through the y-z plane? And the [itex] q_{enc} [/itex] can be written as [itex]\int \rho(x)[/itex] I believe. So I should be able to just integrate my charge distribution from [itex] -x [/itex] to [itex] x [/itex] and consider the area a sphere of radius [itex]x[/itex]? That doesn't seem quite right to me for some reason since I have an x symmetry should I be using a cylinder? similar to a line of charge along the x axis?
     
    Last edited: Sep 17, 2013
  5. Sep 17, 2013 #4
    Yes, really what we have is a point charge in one dimension, where we only consider the charge density along the x-axis. I suppose a cylinder would be fitting for Gauss's Law. Yes, you are correct about integrating along the x-axis.
     
  6. Sep 17, 2013 #5
    Using a cylinder seems to give me a dependence on both x and y. I feel like there should be a simpler choice of surface, but I cannot seem to think of it. I have also tried a sphere centered at the origin. I am not sure how I would apply a plane.
     
  7. Sep 17, 2013 #6
    Perhaps, we can treat this similar to the case for an infinite wire? Are we finding the E-field at some point say on the y-axis, or some point on the x-axis?

    Must we use Gauss's Law?
     
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