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How to prove that electric field strengths are the same

  1. Jul 13, 2012 #1
    hi, it's my first post :)
    this is not really a homework problem but it's almost like that so...

    1. The problem statement, all variables and given/known data
    in the xy-plane, suppose that there is a non-uniform charge density distribution between x=0 and x=d. the distribution is dependent only on x so that for any vertical line x = a (0≤a≤d), the charge density anywhere on that line is the same. for x<0 and x>d, the charge density is 0.

    i would like to show that the magnitude of the electric fields at any point (x,y) where x< 0 and at any point (x,y) where x> d are the same.


    3. The attempt at a solution
    I can show that the field strength in both areas are constant.

    if I "zoom out" far enough , the area between x = 0 and x =d would appear like a line charge with constant linear charge density. by symmetry, i then argue that the field strength m units to the left of that line charge is equal to the field strength m units to the right of the line charge.

    because the field in both areas are constant, they must be equal

    is my solution valid? if it were, is there any other more "mathematical" solution?
     
  2. jcsd
  3. Jul 15, 2012 #2

    rude man

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    What's the Poisson equation for x < 0 and x > d?

    What then is the relationship between the potential and the electric field? What 1st order diff. eq. do you wind up with, and what is its solution?
     
  4. Jul 15, 2012 #3
    Thank you for the reply.
    We haven't covered the Poisson equation yet. We've covered Maxwell's equations though, both integral and differential forms and the boundary conditions. So, I was thinking if there's another way without using that Poisson equation...

    But thanks, i'll check that. :smile:
     
  5. Jul 15, 2012 #4

    rude man

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    Actually, I should just have said, use divE = ρ/ε. One of the Maxwell relations, of course.
     
  6. Jul 15, 2012 #5
    yes, i've done that

    ∂Ex(x) / ∂x = ε0ρ(x) for 0<x<d
    and
    ∂Ex(x) / ∂x = 0 for 0>x and x>d

    Ex (x)= ε0∫ρ(x)dx (from 0 to x) + Ex(0) for 0<x<d
    and
    Ex (x)= Ex(0) for x<0
    and
    Ex (x)= Ex(d) for x>d


    my problem is how to arrive at the conclusion that
    Ex(0) = Ex(d) ?
     
  7. Jul 16, 2012 #6

    rude man

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    Oops, I see what you mean. I mistakenly thought you needed to show only that the E fields for x < 0 and x > d are constant, not the same constant.

    The thing here is, we're dealing with infinities, which can be tricky. Consider a thin strip dx situated at x = x0 on 0 < x < d and extending from -∞ < y < ∞. Now, you know that it doesn't matter where along x > d you are, the E field is the same. So the E field must be the same value for x < 0 as for x > d, just by symmetry.

    In other words, say you pick a value x = 3d. I can then say, OK, go to x = -3d + 2x0, which is equidistant on the other side of the strip, so obviously the E fields due to that strip must be the same for x < 0 as for x > d. Then you just argue superposition of the infinite number of strips existing between x=0 and x=d, and you have the result that the field is the same in both areas.

    To sum up, you first show (which you already did) that E(x<0) = c1 and E(x>d) = c2. Then you show that c1 = c2.

    You could go whole-hog and do the double integration & you'd find the same thing (I haven't done it!).

    It sounds goofy: say ρ(x) = kx, 0 < x < d, k > 0. The you'd think the E field for x > d must be greater than for x < 0. Which is why I say that infinities can do strange things!
     
    Last edited: Jul 16, 2012
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