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Electric Field of Finite, Diagonal Line-Charge

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the triangular-shaped wire in the picture. The base is of length 2a and is along the y-axis, and the two sides are of length 3a each. The wire is uniformly charged with charge density (per line) [itex]\lambda[/itex].

    Find the electrostatic field at all points [itex]\vec{x}[/itex] = x[itex]\hat{i}[/itex], where x [itex]\in[/itex] [a; 2a]. There is no need to evaluate any integral. Your final answer must specify the components of the electric field through integrals such that (assuming the numerical values of the integrals are known) the final answer depends only on [itex]\lambda[/itex], a and the constant [itex]\epsilon _0[/itex].

    The associated picture is:

    Problem_zps033b7b86.png


    2. Relevant equations
    [tex]
    \vec{E}(x) = \frac{\vec{F}(x)}{q} \\
    \lambda = \frac{q}{8a}
    [/tex]
    3. The attempt at a solution

    Most of this problem came pretty easy to me. Some symmetry arguments made, etc.—I found the e-field contribution of the base (w/ length 2a), and proceeded to try to find the e-field contribution of one of the sides (I know the [itex]\hat{j}[/itex] components cancel, and the [itex]\hat{i}[/itex] component is just double of the contribution of one of the sides.

    The issue comes up when trying to generalize the distance and direction from some point between x = a and x = 2a and a point on the diagonal wire.

    My process begins like this ([itex]\hat{r}[/itex] is the direction from a point on the wire to the x-point, and r is the distance):

    [tex]
    \vec{E}(x) = \sum{\vec{dE}(x)} = \frac{1}{q} \sum{\vec{dF}(x)}\\
    {\vec{dF}(x)} = \frac{1}{4\pi\epsilon _0} \frac{(q)(dq)}{r^2} \hat{r} \\
    dq = \lambda 8ada - \text{Where "da" is an infinitesimal portion of the wire}
    [/tex]
    Now, I've got to find r and [itex]\hat{r}[/itex]—and relate da to that. This is where I get snagged up.
    x is a fixed point between x = a and x = 2a, and xw is the x-coordinate of a point on the wire.

    [itex]\hat{r}[/itex], the direction from some point on the x-axis and a point on the wire, is:
    [tex]
    r = \frac{(x, 0) - (x_w, y)}{\sqrt{(x-x_w)^2 + y^2}} = \frac{<x - x_w, -y>}{\sqrt{(x-x_w)^2 + y^2}} = \frac{(x-x_w)\hat{i}}{\sqrt{(x-x_w)^2 + y^2}} + \frac{-y\hat{j}}{\sqrt{(x-x_w)^2 + y^2}} = \frac{(x-x_w)\hat{i}}{\sqrt{(x-x_w)^2 + y^2}}
    [/tex]

    The last step is due to the fact that the [itex]\hat{j}[/itex] components cancel one another. First, is this... correct? Does it make sense to do that, and, eventually, have to integrate with respect to xw? Or am I over-complicating things?

    Also, I can't figure out for the life of me what da is. I know it's some infinitesimal amount of wire, but how can I express that mathematically? I end up doing something like [itex]\sqrt{dx^2 + dy^2}[/itex] which I'm pretty sure is wrong.

    Maybe... use the angle made with the vector from the point to the wire and the x-axis, and use that angle to express the infinitesimal length of the wire (angle multiplied by arclength)? I feel like I'm missing something very simple here (there usually is -_-), but I've been mulling over this problem for a while and can't figure it out.

    (I figured out the e-field due to the base simply enough; Pythagoras made that one easy for me. :D)
     
    Last edited: Apr 6, 2013
  2. jcsd
  3. Apr 6, 2013 #2

    haruspex

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    It's not a good idea to use da. a is a constant. Pick some other variable for distance along the wire (or x-coordinate of an element of wire) and work in terms of increments in that. E.g. take (u, v) to be the coordinates and find the relationship between u and v, etc.
     
  4. Apr 6, 2013 #3
    That's what I tried doing, my man (the [itex]\sqrt{dx^2 + dy^2}[/itex] comment).

    Don't worry, this isn't a homework problem (though, because of the way it's presented, I thought it best to post it here). Hints won't help me much at this point. :P I'm studying for an upcoming exam by doing last year's, and have spent way too much time on this problem.

    I tried going the route of rotating the coordinate system, but... uh... that didn't work well.
     
    Last edited: Apr 6, 2013
  5. Apr 7, 2013 #4

    haruspex

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    Maybe, but it was very hard to follow because of the way you've assigned da, x, y, xw. Looks easiest to me to take elements that have coordinates (u, ±v), measuring u from the right hand vertex. I'll measure x from there too. v = αu, where α=√2/4. Length of element is 3αdu. Field in x direction from pair of elements (one above, one below) is ## \frac{6\lambda \alpha du}{4\pi\epsilon _0 ((x-u)^2+\alpha^2 u^2)} ##
     
  6. Apr 7, 2013 #5
    How did you get the [itex]\alpha = \frac{\sqrt{2}}{4}[/itex] and [itex]v = \alpha u[/itex]?

    I keep ending up with [itex]v = -u + 2\sqrt{2}a[/itex]—which makes my calculation of an infinitesimal length of wire really... uh... ugly.
     
  7. Apr 7, 2013 #6

    haruspex

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    As I said, I took the origin as the pointy end of the triangle for simplicity, and with +ve u as being to the left. What (in (u,v) terms) would the equation of the line of the upper wire be?
     
  8. Apr 7, 2013 #7
    Whoa. There's that [itex]\frac{\sqrt{2}}{2}[/itex]

    Geeze. That... that makes the problem so much easier. And one could use two different coordinate systems for the different parts because the value of the integral should come out to be the same anyway.

    Also, approaching the problem as I was (w/o altering the coordinate system) I should be able to express [itex]dl[/itex] (changed it to [itex]dl[/itex] from [itex]da[/itex], since that's easier to follow) in terms of the ratio between the total length of the wire and its total x-coordinate (without having to change coordinate systems or anything). That is, [itex]L = \frac{3}{2\sqrt{2}}u[/itex].

    Which means that [itex]dL = \frac{3}{2\sqrt{4}}du[/itex] (in the old coordinate system).

    Your way simplifies the integral, though. I need to hammer methods like that into my mind.

    This problem has taken so much time, man. Thanks so much for your help. I wish I could upvote you, or give you tokens, or a good recommendation to your boss or something. :P
     
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