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Homework Help: Uniform Charge Distribution

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A rod 16.0 cm long is uniformly charged and has a total charge of -21.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 42.0 cm from its center.

    2. Relevant equations

    3. The attempt at a solution

    The equation I am advised to use is, [itex]\vec{E}=k_e\frac{\lambda l}{d(l+d)}[/itex], which becomes [itex]\vec{E}=k_e\frac{q}{d(l+d)}[/itex]. I understand how the numerator is formed, but how the author comes up with the numerator is something I am a little confused about. Could some possible explain how this formula is developed and what precisely it means?
  2. jcsd
  3. Feb 2, 2013 #2
    The numerator in this equation is simply the charge related to the charge density.

    where λ = delta Q / delta L -> delta Q = λ*delta L

    The equation as a whole of course is simply the definition of the electric field from Coulomb's law.
  4. Feb 2, 2013 #3
    Okay, I understand that. But what about the denominator, shouldn't it somehow resemble r^2?
  5. Feb 2, 2013 #4
    It is r^2 if it is a point charge but you have a rod here! You need to select a small part of the rod, find the electric field due to that small part and integrate. You would require a bit of calculus here.
  6. Feb 2, 2013 #5
    Oh, I see. I thought that that was the general pattern for all electric field equations. But as you have mentioned, that is not the case though.
  7. Feb 2, 2013 #6
    The formula you mentioned doesn't seem right to me. Are you sure you wrote down the correct formula?
  8. Feb 2, 2013 #7
    It's correct

    λ = Q/L

    dE = k λ(delta x)/(x^2)

    Integrate from d to d+l (with an added bit of algebra) to obtain the expression given.

    The continuous charge distribution for a uniformly charged rod is usually given as follows, and is maybe the more familiar one, but used in a different physical situation:

    http://www.phy-astr.gsu.edu/cymbalyuk/Lecture13.pdf [Broken]

    EDIT: pdf equation not relevant
    Last edited by a moderator: May 6, 2017
  9. Feb 2, 2013 #8
    Nope, it isn't correct.
    The pdf you linked shows the correct formula and its derivation.

    You can verify the formula mentioned by the OP by simply considering the length of the rod to be much larger than d. If l>>>d then expression becomes kλ/d. This is wrong because electric field due to an infinitely large thin wire is 2kλ/d at a distance d from it.
    Last edited by a moderator: May 6, 2017
  10. Feb 2, 2013 #9
    The equation given by the OP is relevant when the point charge is along the axis of the rod, and the rod is finite. The equation given in the pdf, the more familiar form, is for a different physical situation as shown in the diagrams for the pdf.

    To understand the validity of the OP's equation, consider the point charge to be along the axis of the rod. Define l as being the length of the rod, and d to be the minimum x position of the rod where the point charge is at the origin. Integrate dE from d to d+l, take out the constants from the integrand, then find a common denominator to derive the OP's expression. It is valid. Furthermore, it contains the necessary information for a numerical solution to the OP's original question.
    Last edited: Feb 2, 2013
  11. Feb 2, 2013 #10
    Explain me why you integrate from d to d+l. Your limits are completely wrong. Show me the equation you set up for dE.
  12. Feb 2, 2013 #11


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    Homework Helper

    Before using the formula you need to know the meaning of the notations. What are d and l?

    The picture explains it. Integrate the contribution of the little element dq=λdl from x=d to x=d+l.


    Attached Files:

  13. Feb 2, 2013 #12
    Woops, I read the question wrong. Sorry for the trouble.
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