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Electric Field due to a line of charge

  1. Sep 21, 2006 #1
    Have a look at the diagram

    Find the elctric field a distance z above one end of a astraight line segment of lenght L which carries uniform line charge lambda. Check that your formula is consistent with what you would expect for the case z >> L

    [tex] \lambda= q/L [/tex]
    [tex] dq = \lambda dx [/tex]

    For the electric Field in the horizontal (points to the left and is negative)

    [tex] dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{dq}{(z^2 + x^2)} \sin \theta [/tex]

    Subsituting what we know about sin theta and dq

    [tex] dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{(z^2 + x^2)} \frac{x}{\sqrt{z^2 + x^2}} [/tex]

    integrating x = 0 to x = L

    [tex] E_{x} = \int dE_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \int_{x=0}^{x=L} \frac{xdx}{(z^2+x^2)^{\frac{3}{2}}} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{-1}{\sqrt{z^2 + x^2}} \right]_{x=0}^{x=L} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{-1}{\sqrt{z^2 + L^2}} + \frac{1}{z} \right] [/tex]

    [tex] E_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right] [/tex]

    ok so suppose z >> L then the electric field is zero?? Shouldnt it reduce to that of a point charge and not zero?? this should be regardless of whether i solve for x or z right??

    for the Z direction i got
    [tex] E_{z} = \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}} [/tex]
    unlike the last one this one does reduce to the equatiopn for a point charge but is off my a factor of 1/z ...

    combining the two yields

    [tex] \vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right] \hat{x} + \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}} \hat{z} [/tex]

    is there soemthing wrong with the calculationg ofr hte X Horizontal Direction?? Please help

    Thank you in advance for your help and advice!

    Attached Files:

    Last edited: Sep 21, 2006
  2. jcsd
  3. Sep 22, 2006 #2

    Doc Al

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    Staff: Mentor

    If you get far enough away, the field from anything (finite) goes to zero! :wink: But you need to understand how it approaches zero. In this case, think about it: You are very far from a point charge--What's the direction of the field?

    It's not off. You just need to examine its behavior as z goes to infinity. Use a binomial expansion:
    [tex](L^2 + z^2)^{-1/2} \approx (1/z)(1 - \frac{L^2}{2z^2})[/tex]
  4. May 10, 2011 #3
    I'm studying from Griffiths too :) and i've a question about this exercize.

    The electric potential:

    [tex]V = \frac{ \lambda }{4\epsilon_0 \pi} ln\left(\frac{L+\sqrt{z^2+L^2}}{z}\right) [/tex]

    Calculating the electric field from the V's gradient, it has x component always at zero.
    Where the problem?

    I'm sorry for my poor english, good evening mate!
  5. May 13, 2011 #4
    Yeah, Griffith's book is good.

    From your calculation for [tex]\vec{E}[/tex] (yes i think it's correct for both direction), for z >> L, the electric field of component - x will dissapear (since [tex]\frac{1}{z} - \frac{1}{z} = 0)[/tex] and leaves for us the z - component as below:

    after simplifying, [tex]\vec{E} = \frac{\lambda L}{4 \pi \epsilon_{0}z^2}\hat{z}[/tex] for z>>L.

    the case is the same as above example from griffith's book (that is for E at a distance z above midpoint of line L). I quoted it : From far away the line

    "looks" like a point charge [tex]q = 2\lambda L[/tex].

    I think x and y component will be gone since you do grad [tex]\nabla[/tex]-operation to potential (the potential only depend on z). so,

    [tex]\vec{E} = -\nabla V = -\frac{\lambda}{4\pi \epsilon_{0}}\frac{\partial}{\partial z}\ln (\frac{L + \sqrt{z^2 + L^2}}{z})[/tex].

    suppose [tex]k = \sqrt{z^2 + L^2}[/tex],

    it will give [tex]\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}}(\frac{z}{k(L + k)} - \frac{1}{z})\hat{z}[/tex]

    btw, i cant find this problem in griffith's book....
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