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Electric field at a point due to an infinite line charge

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Charge is distributed uniformly along an infinite straight line with density λ. Develop the expression for E at the general point P.

    2. Relevant equations

    The electric field at a point P, caused by N point charges Qi, each a distance ri from P, is given by [itex]\mathbf{E} = \sum \limits_{i=1}^N \frac{Q_i}{4\pi \varepsilon_0 {r_i}^2} \hat{r}_i[/itex].

    For continuous charge, the formula becomes an integral: [itex]\mathbf{E} = \int \frac{1}{4\pi \varepsilon_0 {r}^2}\hat{r}\:dQ[/itex], with the unit vector varying along the integral. (This probably isn't quite correct but it should be the right idea.)

    3. The attempt at a solution
    I think I'm really just having big issues setting up the integral. There are a lot of things going into it, and I'm pretty sure I'm doing something wrong when dealing with the r vectors. Here is what I did. I'm not necessarily interested in evaluating the integral, more just setting it up correctly. After that point I could attempt to evaluate it or try it in cylindrical coordinates.

    I want to try doing this using Cartesian coordinates, even if using cylindrical coordinates may be better (the solution I have uses those, but even using those I run into similar problems).

    We can choose any coordinate system we like, so let's have the line of charge run along the z-axis, and have the point P lie in the x,y-plane. The coordinates of P are then (x′, y′, 0) and the line charge is at (x,y) = (0,0).

    If we consider an infinitesimal part of the line charge dz, and say that that piece has charge dQ, then we can make the equation dQ = λ dz.

    Consider an infinitesimal piece of the line at the point z = z′. Its distance from P is [itex]r = \sqrt{(x')^2 + (y')^2 + (z')^2}[/itex] along the vector [itex]\mathbf{r} = (x', y', z')[/itex]. So [itex]\hat{r} = \frac{1}{r}\mathbf{r} = \frac{1}{\sqrt{(x')^2 + (y')^2 + (z')^2}}\hat{x} + \frac{1}{\sqrt{(x')^2 + (y')^2 + (z')^2}}\hat{y} + \frac{1}{\sqrt{(x')^2 + (y')^2 + (z')^2}}\hat{z}[/itex].

    So, this means my integral is [itex]\mathbf{E} = \int_{-\infty}^{\infty} \int_0^{y'} \int_0^{x'} \frac{\lambda}{4\pi \varepsilon_0 ((x')^2 + (y')^2 + (z')^2)}\hat{r} \: dx \, dy \, dz[/itex]. Is this true? I don't see unprimed x, y, or z anywhere in the integral, which worries me. I think they should be there. Reading the solution makes me think that every primed variable I have should actually be unprimed.

    If it is true, I wouldn't even know where to begin in solving it, since there are vectors in there and I'm still not sure what to do with vectors in integrals if the differential is also not a vector that is given a vector operation (meaning, the integral ends with something like · dr or × dr, and yes, I do think that dx dy dz = dr.) Would I split it up component-wise? Does that even work? Where did the dx and dy even come from since dQ = λ dz only? I know they have to be in there but I don't know from where.

    Also, from looking at the solution which uses cylindrical coordinates, I think I'm missing a factor of [itex]\sqrt{(x')^2 + (y')^2 + (z')^2}[/itex] in the integrand's denominator. Am I?

    Thank you very much!

    (Also, I know that the z-components all will cancel due to symmetry, but I'm trying to see if I can do it the "complete" way this time. Also.. I probably wouldn't know exactly where it was permitted to remove z-components and where they had to be left in.)
  2. jcsd
  3. Apr 15, 2014 #2

    Simon Bridge

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    You need only integrate over the volume containing the charge - which is a line in this case.
    So if you are doing a volume integral you probably got confused somewhere.

    Prepare the coordinates:
    Put the line of charge up the z axis.
    Put the point P at position ##\vec r_P = (d,0,0)##

    Now break the charge up into infinitesimals:
    ##dq=\lambda dz##
    treat dq as a point charge - you know the formula for the electric field due to a point charge.

    Now you can think about the electric field due to an arbitrary infinitesimal charge:
    So - what is the electric field due to the charge dq at position ##\vec r = (0,0,z)##
    Compare with the electric field due to dq charge at ##\vec r = (0,0,-z)##

    From there you should be able to get an expression for dE that you can integrate.
  4. Apr 15, 2014 #3


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    I think you'd be way better off solving this using Gauss's law, with a cylindrical surface. You should be able to get the answer from there in like 2 steps...
  5. Apr 15, 2014 #4

    Simon Bridge

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    ... this problem is a common exercise for students before they get to Gauss' law.
  6. Apr 15, 2014 #5
    All right yeah, I did have an issue with using a volume integral, since it seems like I still have to use all coordinates. Also, instead of considering the charge dQ at (0,0,z), I thought you had to use (0,0,z′) or some other variable that was not the same as the axis. But since you're integrating over the entire line, you have to use the same variable as the one that "progresses" across the line during integration!

    This is where I got:
    For the charge at (0,0,z), r = (−d,0,z), so [itex]r = \sqrt{d^2 + z^2}
    [/itex] and [itex] \hat{r} = (\frac{-d}{r}, 0, \frac{z}{r})[/itex].
    For the charge at (0,0,−z), r = (−d,0,−z), so [itex]r = \sqrt{d^2 + z^2}[/itex] and [itex] \hat{r} = (\frac{-d}{r}, 0, \frac{-z}{r})[/itex].
    You can add these together and then perform the integral from z = 0 to z = ∞, and every positive value of z will include the result from the negative value of z in it.
    What you are left with is:
    [tex] \mathbf{E} = \int_0^\infty \left(\frac{\lambda}{4 \pi \varepsilon_0 r^2} \cdot \frac{1}{r}(-d\hat{x} + z \hat{z}) + \frac{\lambda}{4 \pi \varepsilon_0 r^2} \cdot \frac{1}{r}(-d\hat{x} - z \hat{z})\right)\: dz[/tex]
    [tex] \mathbf{E} = \int_0^\infty \frac{-2 \lambda d \hat{x}}{4 \pi \varepsilon_0 (d^2 + z^2)^{\frac{3}{2}}} \: dz = \frac{-2 \lambda d \hat{x}}{4 \pi \varepsilon_0} \int_0^\infty (d^2 + z^2)^{- \frac{3}{2}} \: dz[/tex]
    (It's a little confusing with d being the location of the point P as well as the differential operator.)

    Is this the correct integral? I think it's close but not quite there. It can be solved using trigonometric substitution, so it's doable. Here are issues I'm having with it:
    1. It "double-counts" the charge dQ at z = 0.
    2. I don't think I used the [itex]\hat{r}[/itex] in the integral correctly.
    3. What if the two charges at z and −z didn't cancel, and you had a vector with two components in the integrand? How would you perform that integral?

    Thanks a lot for your help!

    And yes, it's been a bit over 2 years since I took my electricity and magnetism course, and longer since I took vector calculus, so I'm going through some books now to relearn them since I'm going to need them, and I'm just beginning so I haven't quite made it to Gauss's Law again (but I do know how it works!).
  7. Apr 15, 2014 #6

    Simon Bridge

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    The primed notation is a convenience - you'd normally use it where the limits in the integration include an axis-label variable. Otherwise you'd get clumsy looking notations like: ##\int_1^x x\; dx## in some cases. That's not so much a problem here.

    You can change the d to something else - it is more usual to use r or R (and use cylindrical-polar coordinates).
    Some people typeset the differential d differently so you can have ##\int dx\;\text{d}x## but the big thing with yours is when you named the unit vectors after the axis.

    (note ##(x,y,z) = x\hat \imath + y\hat\jmath + z\hat k## ... could help with the confusion.)

    What I particularly wanted you to notice was that the components of E pointing along the z axis will cancel out to leave only the component pointing in the x direction ... so you can do the integral for the magnitudes.

    1. but the contribution from z=0 is infinitesimal ... tend to zero. It's the sum that counts.
    2. so long as you got the distance correct.
    3. a. correct 2. carefully...

    You'll find lots of derivations online t compare your's with.

    Consider - the charge element is at point Q, the origin is at point O and the point we want the E-field is P. This makes a right-angles triangle OPQ - so we can use trigonometry.

    From the symmetry you just did you can write that ##\vec E=E_x\hat\imath##

    If I put θ=∠QPO, then the x-component of the E field will be:$$E_x = E\cos\theta = \frac{d}{\sqrt{z^2+d^2}}E$$

    $$dE_x = \frac{1}{4\pi\epsilon_0}\frac{dq}{z^2+d^2}\frac{d}{\sqrt{z^2+d^2}} \\ \implies E_x = \frac{\lambda d}{2\pi\epsilon_0}\int_0^\infty \frac{dz}{(z^2+d^2)^{3/2}}$$
  8. Apr 15, 2014 #7


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    Oh...in that case, sorry for my intrusion.
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