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Charged Square Contour (E Field)

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A line charge of uniform charge density Q' is distributed along a square contour a on a side. The medium is aie. Find the electric field intensity vector at a point that is at a distance a from each of the square vertices.

    2. Relevant equations



    3. The attempt at a solution

    I've got everything figured out expect for the line integral which will give me the electric field vector for one of the sides of the square contour.

    In the figure attached (my work), I've denoted this as, [itex]\vec{E_{1}}[/itex].

    How do I write out this integral?

    I know that the radius in the calculation of the electric field is going to vary from a to h then back to a as we move along one side of our square contour, but how do I represent this knowledge in an integral that will evaluate the changing r for infinitesimly small lengths (dl) along one side of the square contour?
     

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  3. Sep 21, 2011 #2

    rude man

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    I would go with calculating the potential at (0,0,z) and then E = -dV/dz. Integrating the force vectors looks daunting, especially when you have a lousy sense of spatial relations like I do.

    Tale a 1/8 section of the square: (a/2, y = 0 to a/2). Write r(y,z) as the distance from a point on this section to (0,0,a). Integrate dV[r(y,z)]) at (0,0,a) from y = 0 to y = a/2, giving you V[r(z)]. Double this amount by considering the same V contribution from the opposite 1/8 section (-a/2, y = 0 to - a/2). The result is now 1/4 the total potential at (0,0,a). E due to these two sections combined is in the +z direction only. The other three section pairs similarly contribute no x or y component along the z axis.
    Then, E = -dV(z)/dz.

    How's that sound? Haven't done it myself so don't know if there are any stumbling blocks, but looks straightforward to me.
     
    Last edited: Sep 21, 2011
  4. Sep 21, 2011 #3
    We haven't learnt about the potential yet, we are suppose to do it using a line integral.

    How can I set the integral up in order to account for the varying radii as I move a small length dl along the line charge?
     
  5. Sep 21, 2011 #4

    rude man

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    Ok, then I would try as follows:
    1. take 1/8 section, say the one defined by (x=0 to a/2, a/2). Draw a line from a typical element of this section dL to (0,0,a). Come up with |r|, the distance from dL to (0,0,a). This is a function of x (and a of course) only. Come up with the equation for
    r = r1 i + r2 j + r3 k. Again, r1, r2 and r3 are f(x) only. Solve for cos(theta) from
    r dot k = |r|cos(theta). Then dE in the z direction at (0,0,a) due to this one section is
    cos(theta)dE(x). theta will itself be a function of x. Then integrate dE(x) from x = 0 to a/2, multiply by 8, and get your answer.

    Rough way to travel.
     
    Last edited: Sep 21, 2011
  6. Sep 21, 2011 #5
    You can't draw it to (0,0,a), that's not where the reference point is required to be located in space.

    In the question it states it must be a distance a from the vertices of the contour.

    This geometry forces the value of our reference point on the z-axis to be,

    [itex](0,0, \frac{a}{\sqrt{2}})[/itex]

    You lost me on the rest of you're explanation, if you could show me that'd make things alot more clear.
     
  7. Sep 22, 2011 #6

    rude man

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    You're absolutely right, the point of observation is not (0,0,a). Sorry about that It's obviously [0,0,a/√2].

    Next we have to determine the vector r pointing from an element dx located at (x,a/2,0) on the chosen segment (0 < x< a/2, a/2, 0),
    to [(0,0,a/√2].

    Fact: given two points P1(x1,y1,z1) and P2(x2,y2,z2) the vector pointing from P1 to P2 is given by

    r = (x2-x1) i + (y2 - y1) j + (z2 - z1) k. Since P1 for us is (x, a/2, 0), r = r(x). P2 of course is (0,0,a/√2).

    Next, we know that E along the z axis is directed upward along that axis, so we need to find the component of dE along z due to the element of wire dx. (The x and y components of E cancel). But dE is directed along r, so we need to find the angle between r and the z axis. To do that, use r dot k = |r|cosθ . You can see that θ = θ(x).

    Then, dE(x) at (0,0, a/√2) will be Kρdx*cosθ/(|r|^2) k,

    K = 9e9 SI units and both |r| and θ are functions of x, ρ = Q'/4a and k is unit vector in the z direction. Integrate dE from x = 0 to a/2. Multiply by 8 & you're home free.

    This is a real bear of a problem they've handed you, lots of work left for you. Don't hesitate to ask more questions. When you get to potentials you'll see it's much easier that way. Potentials are scalars, not vectors. And we already knew the direction of E before we even started.
     
    Last edited: Sep 22, 2011
  8. Sep 22, 2011 #7

    rude man

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    Oh, oh, I just saw you had the observation point at (0,0, a/√2) and I finally see that that's correct. I made the correction in my final post.
     
    Last edited: Sep 22, 2011
  9. Sep 22, 2011 #8

    You're saying that, [itex]\theta[/itex] is a function of x, is this what it is defined as,

    [itex]\theta(x) = cos^{-1}\left(\frac{\vec{r} \cdot \hat{k}}{|r|} \right)[/itex]

    That's going to be ridiculous to try and integrate, isn't it?

    Or can I just change the limits and integrate with respect to [itex]\theta[/itex]?
     
  10. Sep 22, 2011 #9

    rude man

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    I just noticed that cosθ/(|r|^2) = (r•k)/|r|^3 so you can avoid θ altogether.

    Now dE = Kρdx(r•k)/(|r|^3) k
     
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