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B Question about the math involved in E-field calculations

  1. Jun 7, 2016 #1
    When calculating the electric field from a point above a line of charge using coulomb's law, the integral that comes up is of the form [itex]\int { \frac { dx }{ { x }^{ 2 }+{ a }^{ 2 } } } [/itex]. But if the point we were asked for is right in the middle, the horizontal (cosine) components cancel out, leaving only the vertical components, so we'd have to throw in a sine term to make the expression correct. Why, if we add the sine and cosine components, the result is not equal to the expression without sine and cosines?
    Namely,
    [itex]\int { \frac { dx }{ { x }^{ 2 }+{ a }^{ 2 } } } \neq { \left( \int { \frac { dx }{ { x }^{ 2 }+{ a }^{ 2 } } } sin\theta \right) }^{ 2 }\quad +\quad { \left( \int { \frac { dx }{ { x }^{ 2 }+{ a }^{ 2 } } } cos\theta \right) }^{ 2 } [/itex]
     
    Last edited: Jun 7, 2016
  2. jcsd
  3. Jun 7, 2016 #2

    Dale

    Staff: Mentor

    You probably should post your reference for this so we can all be clear. This expression seems too simple to me, so I suspect that they are making some important assumptions.
     
  4. Jun 7, 2016 #3
    Here is an image I found on the web
    SymmetricLine.gif
    The calculation (if I replace d with a, and the charges are taken to be positive ones)
    [itex]dE\quad =\quad k\lambda \frac { dx }{ { r }^{ 2 } } \\ E\quad =\quad k\lambda \int _{ -a }^{ a }{ \frac { dx }{ { x }^{ 2 }+{ a }^{ 2 } } } sin\theta \\ sin\theta \quad =\quad \frac { a }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } [/itex]
     
  5. Jun 7, 2016 #4

    Dale

    Staff: Mentor

    Ok, so you are assuming the point which is on the line bisecting the charge segment and at a distance equal to half the length of the segment.
     
  6. Jun 7, 2016 #5
    Oh I see, sorry. What I mean is a = -a, but I didn't mean for the distance of the point away from the line to be a too. It doesn't matter though does it?
     
  7. Jun 7, 2016 #6

    Dale

    Staff: Mentor

    Well, if d is different from a then d will need to show up in your expression.
     
  8. Jun 7, 2016 #7
    Ok, here is what I have to make it more clear (according to the picture)
    [itex]dE\quad =\quad k\lambda \frac { dx }{ { r }^{ 2 } } \\ E\quad =\quad k\lambda \int _{ -a }^{ a }{ \frac { dx }{ { x }^{ 2 }+{ d }^{ 2 } } } sin\theta \\ sin\theta \quad =\quad \frac { d }{ \sqrt { { x }^{ 2 }+{ d }^{ 2 } } } [/itex]
     
  9. Jun 7, 2016 #8

    Dale

    Staff: Mentor

    Yes, that all looks right. The d is inside the integral and the a shows up only in the limits.
     
  10. Jun 7, 2016 #9
    I guess my question is: If the sine component measures the vertical contribution, and the cosine measures the horizontal, what does [itex]\int { \frac { dx }{ { x }^{ 2 }+{ a }^{ 2 } } } [/itex] do?
     
  11. Jun 7, 2016 #10
    It measures the total contribution. When you do the integral, only the vertical contribution survives.
     
    Last edited: Jun 7, 2016
  12. Jun 7, 2016 #11
    That integral comes out as [itex]\frac { 1 }{ a } arctan(1)\quad -\quad \frac { 1 }{ a } arctan(-1)\quad =\quad \frac { \pi }{ 2a } [/itex], which is not equal to the one with sine theta factored in
     
  13. Jun 7, 2016 #12

    Dale

    Staff: Mentor

    The integrand measures the magnitude of the E field from one small piece of the charge, but the integral is not physically meaningful.

    If you have two force vectors (3,0) and (0,4) then the resultant force is (3,4). The magnitude of the resultant force is 5, not 7. You can write down the sum of the magnitudes, and you can calculate it, but it isn't useful as far as I know.
     
  14. Jun 7, 2016 #13
    Thanks, that cleared some things up. That integral is not physically meaningful because the fields are not all in the same direction.
     
  15. Jun 7, 2016 #14

    Dale

    Staff: Mentor

    Yes, exactly
     
  16. Jun 8, 2016 #15
    I see now what you were asking. Yes, the integrand is the magnitude of the E-field, including horizontal and vertical components, from a small element on the line. The integral would be adding those contributions without taking the direction of the components into account. An easy way to see this is If d = 0. We'd expect a zero net field at the point of interest, because of equal contributions from the left and right side, but the integral would give a nonzero result, so not meaningful as already mentioned.
     
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