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Homework Help: Simple Coulomb's Law problem, struggling with basic calculus method

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A point charge +q is located a distance d from one end of a uniformly charged rod. The rod has total charge +Q and length L. (The rod and the point charge are each held fixed in place.)

    What is the force on the point charge due to the rod? ​

    2. Relevant equations

    [tex]F = \frac{k q Q}{r^2} [/tex] (Coulomb's Law)

    3. The attempt at a solution

    This is a really conceptual sort of question that comes up all the time for me--a simple integration problem of getting a common variable and then integrating. It's the setup that I've never understood.

    So here's what I think we're doing conceptually: we're applying Coulomb's law an infinite number of times to an infinite number of small charges on this line of charge and adding them together.

    If [tex]F = \frac{k q Q}{r^2} [/tex],

    then [tex]dF = \frac{k q dQ}{r^2}[/tex], where [tex]dQ = Q (\frac{dl}{L}) = \frac {Q}{L} dl = \lambda dl [/tex]

    (to find dF, we let dQ be an infinitely small charge and dF will be its infinite contribution. to find dQ, we take the infinite length that this charge possesses, dl, and take the percentage of the entire line that dl takes up. dl / L is like an infinite percentage that we multiply by Q to get dQ.)

    So we have:

    [tex]F = \int \frac{k q dQ}{r^2} = \int \frac{k q Q dl}{L r^2}. [/tex]

    Here's where I have trouble. r is obviously changing for each charge, so I can't just pull it out of the integral. How to I write r, the distance from the point charge to each dQ, in terms of dl, an infinite length of the line, so that I can integrate?

    More generally, how should I go about setting up integrals like this? I'm never sure if I should integrate dl and convert r, or integrate dr and convert dl. Does anyone have any tips on how to think about this?
  2. jcsd
  3. Dec 13, 2009 #2


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    Hi mikey555! :smile:

    (btw, I'd always use x rather than l, because it's easier to write and to read :wink:)

    Your difficulty is that you're using dl, but not l itself.

    That's the tail without the dog!

    Your proof (or your thought process) should begin, not with "consider an infinitesimal length dl", but "consider an infinitesimal length from l to (l + dl)".

    Then you can immediately see that l goes from 0 to L (which gives you your limits of integration), and r = |d - l|. :smile:
    If the point charge were at the end of the rod, then r and l would be the same. It isn't, so r is double-valued over the length 2d … you need a single-valued parameter to integrate over, so it needs to be l, not r. :wink:
  4. Dec 13, 2009 #3
    Thanks for the fast response.

    What does double-valued mean? Where does 2d come from (I thought we were only dealing with the length d!)?

    I think what you said could really help but I don't understand it! If you could rephrase what I quoted above, it would be really helpful.
  5. Dec 13, 2009 #4


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    I meant that r is measured from that point d from the end, so for values of r less than d, there's two sections at distance r, one on the right and one on the left (but for r > d, there's only one section).

    So if you integrated over r, you'd have to integrate from r = 0 to r = d, and then start again and integrate over r = 0 to r = L-d (unless you're willing to have negative values of r, which really doesn't appeal to me), but if you integrate over l, measured from the end of the rod, then l goes from 0 to L.
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