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3-dimensional charge density for a finite thin wire

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Express the 3D charge density [tex]\rho[/tex] for a thin wire with length Z and uniform linear charge density [tex]\lambda[/tex] along the z-axis in terms of a two-dimensional dirac-delta function.


    2. Relevant equations

    The three dimensional charge density is the total charge over a volume.


    3. The attempt at a solution

    I am not sure how to proceed with this question. Last night I attempted to calculate the electric field components, which I believe I did correctly, but was heavy in the algebra. I had intended to use Gauss law to calculate the charge density from the Electric field, hoping that a delta function would pop out somewhere. I talked to my professor today and he told me I was using the wrong approach, and that the solution to this problem is much simpler. Unfortunately I'm not sure I entirely understand how to use the dirac-delta function, and I feel stuck. I'm not sure how to start this problem. All I know is that

    [tex]Q=\int^{Z}_{0}\lambda dz[/tex]

    This integral just evaluates to [tex]\lambda Z[/tex].

    I have no idea how to proceed. Any help would be appreciated.

    Thanks.
     
  2. jcsd
  3. Sep 23, 2009 #2

    gabbagabbahey

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    Well, you have [itex]Q=\lambda Z[/itex], but by definition of volume charge desity, you also have:

    [tex]Q=\lambda Z=\int_{\text{all space}} \rho(\textbf{x})d^3x[/itex]

    Where is [itex]\rho[/itex] zero? Where is it undefined?

    P.S....why not just use John Wall as your screen name instead?:wink:
     
  4. Sep 23, 2009 #3
    Thanks for the reply.

    I can't seem to grasp the concept of the delta function, despite reading the text and going through and even seeming to understand examples involving the delta function in the text. I know what it does and what its properties are, but as far as actually using it it breaks apart for me. Are there any resources you could point me to where I may get a better understanding of the delta function?

    Your equation makes sense to me. Rho should be zero anywhere where we're not at x=0, y=0, 0<z<Z. I wouldn't think that it's undefined anywhere as we could express it with the dirac-delta function even at the line of charge. That said, from my feeble understanding of the delta function, your equation makes sense if we have a point charge. I can't see how to make that a two-dimensional delta function.

    As for this user name, I believe I created it before his name got big. That said, I still wouldn't compare him to the G.O.A.T. At least not yet :-p

    Thanks again for your help.
     
  5. Sep 23, 2009 #4

    gabbagabbahey

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    Right.

    But the delta function is undefined at specific locations...For example, [itex]\delta(x)[/itex] is zero for [itex]x\neq0[/itex] and undefined at [itex]x=0[/itex]...Its the integral of [itex]\delta(x)[/itex] over some interval that will never be undefined.

    You understand that I am using [itex]\rho(\textbf{x})[/itex] to represent the localized charge density right? If so, the equation had better make sense to you, it is the definition of localized charge density!

    Anyways, using that definition, look at a point an the wire... [itex]d^3 x=dxdydz[/itex] in Cartesian coordinates....what will [itex]dx[/itex] and [itex]dy[/itex] be for an infinitesimally thin wire? What will that make the charge density [itex]\rho=\frac{dQ}{d^3 x}[/itex] on the wire?
     
  6. Sep 23, 2009 #5
    I'm sorry, I feel really stupid. You make this sound so easy, but I just don't seem to get it.

    Here's my train of thought right now.

    For an infinitesimal point charge

    [tex]\lambda dz => 4\pi \delta^{3}(r)=4\pi\delta(x)\delta(y)\delta(z-k)=\rho[/tex]

    So for the wire would I just integrate over z?

    dx and dy should be 0 for an infinitesimally thin wire, but I'm not sure how to answer your last expression.

    I'm sorry I don't understand... if it is easier I can just turn it in without answering this particular question and wait for the solution to be handed out. I just don't think I get this particular concept. I'm not sure why.

    Thanks for your help.
     
  7. Sep 23, 2009 #6

    gabbagabbahey

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    I'm not sure what you are trying to say here...

    A point charge at the origin has a charge density [itex]\rho=q\delta^3(\textbf{r})=\delta(x)\delta(y)\delta(z)[/itex]..

    The reason is that the charge density everywhere expect the origin is zero (since there's no charge there), the charge density at the origin is undefined (since [itex]\rho=\frac{q}{dxdydz}[/itex] and [itex]dx=dy=dz=0[/itex] for an infinitesimal point), and despite this, the total charge is finite [itex]q=\int_{\text{all space}}\rho dx dy dz[/itex]...These are exactly the properties that define the 3D dirac delta function (zero everywhere, except at a single point, where it is undefined, yet its integral is finite), so you know [itex]\rho[/itex] must be of the form [itex]\rho(\textbf{r})=f(\textbf{r})\delta^3(\textbf{r})[/itex]

    Using this form you have [tex]q=\int_{\text{all space}}f(\textbf{r})\delta^3(\textbf{r}) dx dy dz=f(0)[/tex], so choosing [itex]f(\textbf{r})=q[/itex] is appropriate.

    Use the same reasoning as above....if [itex]dx=dy=0[/itex] for a thin wire, the charge density will be undefined on the wire...You also know that it is zero everywhere else, and its integral is finite (since the total charge is just [itex]\lambda Z[/itex]....so, the expected form of the charge density will be [itex]\rho(\teftbf{r})=f(\textbf{r})\delta(x)\delta(y)[/itex]...right?
     
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