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Electric Field due to Continuous Line Charge

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the electric field a distance z above the midpoint of a straight line segment of length
    2L, which carries a uniform line charge λ.

    2. Relevant equations
    1) my textbook says :
    E(r) = 1/4πεoV λ(r')/r2 r dl'

    2) and this also works? :
    E(r) = 1/4πεoV λ(r')/r3 r dl'

    both where r is the unit vector from the charge to the point we are caluclating the field at

    3. The attempt at a solution
    I can integrate once I get the setup down just fine. I just wanted to know when and why this two are the same. For example, in my book it says for the setup for this problem :
    E(r) = 2 * 1/4πεoV(λdx/r2)cosθz

    For one thing, how did that cos appear and since it equals z/r couldn't 2) have just been used from the start? And if 2) can be used could I apply it to surface and volume integrals?
  2. jcsd
  3. Dec 10, 2009 #2
    General formula for E is:

    dE = 1/4πεo (dq/r^2)[tex]\hat{r}[/tex]

    However, r = r.[tex]\hat{r}[/tex] (r_hat is unit vector)
    So you can write"

    dE = 1/4πεo (dq/r^3).r

    Electric field of a point above distance Z from a line is a cumulative Electric field caused by every charge in straight line.

    Lets say an angle formed by E caused by one charge and a vertical axis is theta. From there, you can get:
    E_x = E * sin(theta)
    E_y = E * cos(theta)

    And because E in x-axis will automatically blow up, you just need to find E in y-axis.

    Hope this helps
  4. Dec 10, 2009 #3
    Oh now I see. I wasn't even thinking that because r_hat wasn't pointing vertically that the z component wasn't all of the electric field. I was thinking the x-fields cancelled so the field in the z direction was exactly equal to the electric field, but its actually only cos(theta) of it. Thanks a bunch
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