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Electric field quesiton

  1. Feb 23, 2014 #1
    When doing problems such as finding the electric field of an object, my book often makes substations like: ## dq = \lambda dr = \lambda d\theta\cdot r## (this is in reference to finding the electric field at the center of a rod shaped in a semicircle). See attached for full solution to this.

    My question is, how can you let ##\theta## be the angle from the vertical and still make the ## s = r\theta## substitution?? Wouldn't that give you the length of the semicircle starting from the top, to the angle where you direct it to?
    Rephrasing: Wouldn't theta change with every element dx?

    Attached Files:

  2. jcsd
  3. Feb 23, 2014 #2

    Stephen Tashi

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    Science Advisor

    It looks like the rod is hollow. The problem with interpreting the full solution is that you haven't given the full statement of the problem and defined what the various variables mean.

    I think it would.

    [itex] s = r \theta [/itex] and the solution uses [itex] ds = r d \theta [/itex]. There is no assumption that [itex] \theta [/itex] is constant wtih respect to the [itex] x [/itex] coordinate of end of the radius.
  4. Feb 23, 2014 #3


    Staff: Mentor

    I don't understand your concern. Θ is s/r by definition, so you can always make that substitution.
  5. Feb 23, 2014 #4
    But if you're summing up the whole segment, using tiny bits of dx at a time, how do you just get dx?
    I understand ##s=r\theta##, but wouldn't ##\theta## get really large eventually and overlap previously summed up segments? Sorry it's very confusing to explain
  6. Feb 23, 2014 #5


    Staff: Mentor

    I don't see dx anywhere in those equations so I am not sure what you are talking about.

    Are you perhaps asking a general question about how integration works by summing up an infinite number of infinitesimal segments?
  7. Feb 24, 2014 #6
    Yes, that is what I'm asking. I understand that, but how do you sum up all of the dx segments using s=r*theta? What is the strategy?
  8. Feb 24, 2014 #7


    Staff: Mentor

    Since s = r θ we can take differences of both sides and get Δs = r Δθ. So, for example, if you would break s up into a finite number of non-overlapping segments and sum them up ##\Sigma \; f \; \Delta s## then that is exactly the same as breaking θ up into the same finite number of segments ##\Sigma \; f \; r\Delta \theta##. Then simply change that finite number of segments into an infinite number of segments and you have ##\int f \; r \; d\theta##. There is no overlapping in s and therefore no overlapping in θ.
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