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Potential of a finite rod

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi all,
    I need to find the potential of a positively charged rod with charge Q. Assuming to the right as positive,the center of the rod is on the origin, and it extends to -L/2 in the negative x direction, and L/2 in the positive x direction.
    There is a point at distance x, located to the right of the rod, on the x axis.
    I have to express the answer in terms of the variables Q, L, x, and appropriate constants.


    2. Relevant equations
    V= ∫K*dQ/R
    λ=Q/L

    3. The attempt at a solution
    I know this question has been asked here before, but I wasn't able to come to a conclusion about it myself based on these posts...so here we go:
    I started by choosing an arbitrary point on the rod, called dQ, that is a distance dx away from the point.
    λ=dQ/dx
    dQ=λ*dx
    I have my equation
    V=∫K*dQ/R
    sub in for dQ
    V=∫K*λ*dx/R
    I know I need to integrate over the length of the rod. So i decided to use the bounds as 0 to L/2, and then multiply the integral by 2.
    This is where I am stuck. I don't know what the value for R is...I am leaning towards x+L..but I am not sure. Also I seem to have gotten rid of my Q variable..so im not sure I should have subbed in λ*dx for it...
    Any suggestions would be greatly appreciated!
     
  2. jcsd
  3. Feb 5, 2015 #2

    rude man

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    x is a finite distance away from the end of the rod. What you say here makes no sense to me.
    Consider various elements of charge dQ = λ dx along the rod. The element of dQ closest to x will be at what distance from x? The element of charge dQ farthest from x will be at what distance from x? So what should be your limits of integration?
     
  4. Feb 5, 2015 #3
    The closest element of dQ to x will be...L/2 + x...I'm not sure about the furthest...I am thinking x+L?
    I am still not sure about the radius R though, because the small section of charge being summed can be at any distance on the rod away from the set point.
     
  5. Feb 5, 2015 #4

    BvU

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    If you already have an x axis, it's better to have point at position x, not at distance x (that's confusing). Make a drawing to find out what is meant with the variables involved. Also you want to clearly distinguish between the integration variable (I see a dx) and the ('fixed' *) x coordinate of the point where you want to calculate the potential.

    *) 'during' the integration

    Not wise. The integrand isn't the same for the other interval
     
  6. Feb 5, 2015 #5
    Thanks everyone for the replies
    After contemplating this further, I think I didn't do a very good job of conceptualizing the situation
    So starting from the beginning:
    Let's call the distance the point is away(so from the origin to the point) x.
    Rod is the same -L/2 to L/2
    The distance that my charge, dQ, is at, from the origin, is D.
    λ=dQ/dD=Q/L, so λdD=(Q/L*dD)
    I know that the distance from the point to my dQ is x-D, so R=x-D
    I want to integrate from -L/2 to L/2
    so:
    K∫λdD/R
    Kλ∫dD/(x-D)
    Integrate:
    Kλ [-ln (x-D)] from -L/2 to L/2
    Kλ[-ln (x- L/2) + ln(x+L/2)]
    using log rules:
    Kλ[ln((x+L/2)/(x-L/2))
    to express in terms of Q, just sub in for λ=Q/L
    I think this is more accurate
     
  7. Feb 5, 2015 #6

    rude man

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    This is correct. Good!
     
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