Is the following statement correct for a line charge distribution $λ(x)$?

$$ρ(\mathbf r)=λ(x)δ(y)δ(z)$$

If yes - what does it say?

  • 1
    $\begingroup$ What do you mean If yes - what does it say? If you are correct, then it's a charge distribution for a line charge, what more would you want? $\endgroup$
    – Kyle Kanos
    Sep 30, 2015 at 11:44
  • $\begingroup$ Yes, this is true. From this you can go on and calculate it's potential (and thus electric field) by integration. $\endgroup$ Sep 30, 2015 at 12:26

2 Answers 2


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describes a charge density in the form of a (possibly infinite, depends on what your allowed x values are in the system) line in 3D space, where $\lambda(x)$ is the linear charge density as a function of x. The delta functions indicates the charge density is concentrated at one point in the yz plane, but extended in the x axis

For a 1D description of the above, you will simply use


  • $\begingroup$ Many thanks. Just to have things clear. t in your formula is not the time, right? I think t should be replaced by r? $\endgroup$
    – Sweetheart
    Sep 30, 2015 at 13:23
  • $\begingroup$ @Torro I put t, which is time because that's what I saw initially in the unedited question, but it seems to be a typo and Qmechanic has now fixed it. So let me update my answer... $\endgroup$
    – Secret
    Sep 30, 2015 at 13:55

The factor δ(y) indicates that the charge distribtution is non-zero only for y=0, i.e. on the zx plane; likewise, δ(z) that it is non-zero only on the xy plane. Thefore, the product δ(y)δ(x) Indiactes that the charge distribution in non-zero on the intersection of zx and xy planes, i.e. the x-axis. Then, the function λ(x) defines the actual form of the charge density on the x-axis. Not that, since, δ functions are distributions in the mathematical sence, this charge distrubtion makes sence only under 2-D integrals over y and z for a given x. Such a proccedure will reduce ρ(r) to λ(x).


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