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Electric potential due to finite rod

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    A rod of length L has a total charge Q uniformly distributed along its length. The rod lies along the x axis with its center at the origin.What is the electric potential as a function of position alon gthe x axis for x>L/2


    2. Relevant equations
    V=∫kdq/r = ∫kQ/r dx I'll add limits of integration at the end to make it easier
    V=kQlnr integrated from x+L/2 to x-L/2 which gives me kQln((x-L/2)/(x+L/2)) but the answer was kQ/L * ln((x+L/2)/(x-L/2))

    I don't understand where they got the L from and how my log was the opposite


    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2011 #2
    Ughhhh I did it wrong, but can someone still help me please.
    I got the right answer but that's without taking the limits.
    I did E=kλ(1/(x-L/2) - 1/(x+L/2)) and then integrated the negative of that to get electric potential and got V=kQ/L * ln((x+L/2)/(x-L/2)) <<<< That answer is without taking the limits of integration tho, which I have no idea what they are.
    The book's answer matched mine. Do we have to integrate it over it's limits of integration for this problem? If not, shouldn't there be a +C in the answer???
     
  4. Sep 15, 2011 #3
    can someone please help me..
     
  5. Sep 15, 2011 #4

    SammyS

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    Ask yourself, "What does r represent in the integral ∫kQ/r dx ?"

    Also, for what situation is V=kQlnr valid?
     
  6. Sep 15, 2011 #5
    So the limits of integration are for when x>0 so it would be from 0 to x?
    That doesn't make sense to me tho, the question asked for the electric potential at x>L/2 so why wouldn't it be L/2 to x? wow never mind L/2=0 in this problem. But for future problems if L/2 isn't at the origin you'd integrate from L/2 instead of 0 right? Also, when you integrate it's lnlxl right? THanks.
     
  7. Sep 15, 2011 #6

    SammyS

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    I asked those questions so that you might realize that you can't simply apply V=∫kdq/r = ∫kQ/r dx or V=kQlnr .

    It appears that you arrived at these by trying to use the results for an infinitely long linear charge distribution at a distance, r, from the line of charge. (This distance is measured in a direction perpendicular to the line charge.)

    The problem you posted has a line charge of finite length and you are to find the electric potential at a point that along the line on which the charge lies, at a position beyond where the charge lies.
     
  8. Sep 15, 2011 #7

    SammyS

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    Well, I should have said [itex]\displaystyle V=\int\frac{k\,dq}{r}[/itex] could be OK. The others are wrong.

    The way the problem is worded, you might be led into the error of using the variable, x, in two different (and mutually incompatible) ways. I suggest finding the electric potential at position
    x = a along the x-axis, where a > L/2. Use the variable, x, to integrate over the rod.

    For a piece of the rod at position x, what is the distance of that piece of rod from position, a, on the x-axis? (This distance is represented by the variable, r, in the above integral.)

    How much charge is there between position x and x + dx along the rod? In other words, what is dq at position x along the rod?
    Use the fact that a total charge of, Q, is distributed uniformly along the rod of length, L.​
     
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