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Electric Field above a Quarter Disk

  1. Sep 21, 2013 #1

    • 1. Problem Statement.
    Find the Electrostatic Field at a distance z above the xy-plane and along the z-axis due to a constant surface charge σ confined to the region a < √(x2+y2) < b, 0 < α < pi/2.
    In other words, find the Electric Field a distance z above a quarter disk occupying the region between the positive x and y axes from r = a to r = b.

    • 2. Known Equations
    E = kq/r2
    dq = σdA
    dA = pi/2 * r' *dr'
    r = √(r'2+z2)

    • 3. Attempt
    The lack of symmetry is what loses me in this question. I know I need to set up tiny quarter rings and then integrate from A to B but I don't know what to do about the vector notation.
     
    Last edited: Sep 21, 2013
  2. jcsd
  3. Sep 21, 2013 #2
    You can break it into x, y, z coordinates. r^2 = (x^2 + y^2 + z^2).
    Write it as a double integral over dx and dy.
     
  4. Sep 22, 2013 #3

    jtbell

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    Staff: Mentor

    Note to the Homework Helpers: This was originally posted in the Classical Physics forum. I moved it here instead of deleting it because the OP showed some thought and posted in a format that sort of resembles the homework forums template. Carry on!
     
  5. Sep 22, 2013 #4
    Firstly, here I think you must write [itex]\displaystyle{dA=rdrd\varphi }[/itex] ([itex]\displaystyle{\varphi }[/itex] is the azimuth). Note that if you have two charges at the same distance from the origin but in different angles they don't create the same field at the point along the z-axis. The two fields have the same magnitude but different direction! So you can not neglect the angle at the integration as you integrate vector to find the field.

    You can solve it in such a way, but you have to find the field of the ring at first. So I think it's easier to use double integral in cylindrical coordinates (that's from where the [itex]\displaystyle{dA }[/itex] I write above comes from). It's almost the same method with yours but you can find immediately the total field of the disk.

    To take into account that [itex]\displaystyle{\vec{E}}[/itex] is a vector it may help you to write Coulomb's law like that:
    [tex]\displaystyle{d\vec{E}=\frac{dq}{4\pi \varepsilon _0}\frac{\vec{z}-\vec{r}}{\left|\vec{z}-\vec{r} \right|^3}=\frac{dq}{4\pi \varepsilon _0}\frac{\vec{z}-\vec{r}}{(r^2+z^2)^{3/2}}}[/tex]
    where [itex]\displaystyle{\vec{z}}[/itex] is the point where you want to find the field position vector and [itex]\displaystyle{\vec{r}}[/itex] is [itex]\displaystyle{dq}[/itex] position vector.

    You can analyze these vectors in the Cartesian unit vectors and then use a double integral in cylindrical. Note that it's difficult to integrate with cylindrical unit vectors because they are not constant!
     
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