Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Field - Charged Rod Question

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A charged rod with uniform charge per length and total charge Q is placed along the z-axis with one end at the origin. The rod is located on the positive z-axis according the the diagram accompanying the problem (not shown). Find the electric field E(x,0,0) at any point along the x-axis.

    2. Relevant equations
    1) I have been given the integral x dx / (x^2 + a^2)^3/2.
    2) I have been given a second integral dx / (x^2 + a^2)^3/2

    3. The attempt at a solution

    Since I know I need to use those two integrals, I solved them first:

    1) = - 1 / square root of (x^2 + a^2)
    2) = x / a * square root of (x^2 + a^2)

    I realize that the total field will be the vector sum of all the segments of the rod (and so I imagine that even though I'm solving for only the x component that it has more than just that). I believe that I need to use the equation E = ke integral of dq / r^2 unit vector r. In this, r is the distance from the charge element to a point and the unit vector r is directed "from the element toward the point." I've played around with this a bit, but I'm at a loss. I realize that I will need to use the two integrals (otherwise they wouldn't have been given in the problem), but I'm not sure where exactly I need to utilize them. I'm not sure where to go with the equation for E that I expect to need, either. Any help would be very much appreciated.
  2. jcsd
  3. Jan 17, 2007 #2
    You take a small element dz at a distance z from the origin whose charge is [tex]dq=\lambda dz[/tex].
    Then the electric field at a distance x on the x axis is [tex]dE(x)=\frac{Kdq}{(x^2+z^2)}[/tex]. Integrate this from 0 to [tex]\frac{Q}{\lambda}[/tex] and that I think is your answer. Here, [tex]\lambda[/tex] is charge per unit length.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook