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Electric field in two dimensions

  1. Jan 29, 2004 #1
    I have this problem which im stuck on: Find an expression for the electric field at point P as in this figure
    P
    |
    |
    |
    |
    |
    |
    +++++++++++++++++++++++++++++++++++++++++++++++++++++++
    <----------------------L-------------------------------->
    Where the distance from P to the positive charged rod is Y.

    All i know is that you need to break the positive charged rod into little differentials such as dL and integrate someting like this -
    Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq
    However i am not sure about the X components and how taht is calculated, please help.
     
  2. jcsd
  3. Jan 30, 2004 #2
    Why dont u apply Gauss Th
     
  4. Jan 31, 2004 #3
    Well this one can be tricky. Now with your picture there and then your claim I am confused, is the point located perpendicular to the end of the line of charge?
    If that is the case then integrating along the length from 0 to L might be easier...
    anyway
    if you look at the picture, as you move from the left to the right, and angle is formed with the line extending from P along L. as you move down L this angle changes.
    As the angle becomes larger the radius changes as well.
    dEx=dEcos(theta)
    dEy=dEsin(theta)
    Now dE=Ko*dq/(r^2)
    substituting the charge density lambda*dx for dq we get
    dE=Ko*Lambda*dx/(r^2)
    if you replace r with the varibles we do know you get
    r=sqrt(L^2+d^2)
    where d is the vertical distance from the line to the point.
    now making the necessary trigonometric changes you can find the the integral for each component.

    Hope that helps
     
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