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Calculating the electric field

  1. Oct 7, 2016 #1

    diredragon

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    IMG_1565.JPG IMG_1564.JPG 1. The problem statement, all variables and given/known data
    Calculate the resultant electric field acting on some point ##x##. The electric field is generated by a long thin rod and the line charge density is given. ##p_l=\frac{dQ}{dL}##
    2. Relevant equations
    3. The attempt at a solution

    I have uploaded two images, one of the problem and one of my solution. My problem is the following. When i calculate the ##dL## part my solutions differs from the one given in the book. Can someone take a look at the?
    just a little edit: in the picture you see ##pldl##it is actually ##p_ldl##
     
    Last edited: Oct 7, 2016
  2. jcsd
  3. Oct 7, 2016 #2

    Simon Bridge

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    You seem to be defining dl to be the length of the line segment between ##z## and ##z+dz## (top diagram) ... but surely that is just ##dz##? Why change the name?
    Looking at your dl then ... actually I think that may work ... you'd get ##dl = x\;d\alpha / \cos^2\alpha## or something...
    Some of the other expressions in the first pic do not work... like ##r\; d\alpha \cos\alpha = dl## is not true, and I don't see how you did ##r=\sqrt{\alpha_0^2+z^2}##

    Why are you trying to do the integration wrt ##\alpha## anyway?

    Go back a bit: at position x, the radial component of the field due to charge ##dq## between ##z## and ##z+dz## is:
    ##dE_x = (dq/r^2)\cos\alpha## (using your diagrams).

    Notice that ##\cos\alpha = x/r##, ##r^2=x^2+z^2##, and ##dq = \lambda\; dz##
    ... where ##\lambda## is the linear charge density. now the calculation is easy to set up.
     
    Last edited: Oct 7, 2016
  4. Oct 7, 2016 #3

    diredragon

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    Oh yeah i see..i was looking at the small segment wrong and thats ##x_0## in the square root, I just wrote it funny, sorry bout that and thanks! :)
     
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