1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E-Field at center of arc

  1. Feb 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the field at the center of curvature of an arc of arbitary angle [tex]\alpha[/tex]

    ([tex]\alpha[/tex] is with the x-axis)


    2. Relevant equations
    [tex]E=\frac{kQ}{R^2}\widehat{r}[/tex]

    [tex]S=R\alpha[/tex]

    [tex]\lambda=\frac{Q}{S}[/tex]

    3. The attempt at a solution
    I divide the arc into small pieces ds. [tex]ds=rd\alpha[/tex]

    [tex]\lambda = \frac{dQ}{ds}[/tex]

    This gives me: [tex]dE = \frac{k\lambda d \alpha}{R}\widehat{r}[/tex]
    I would have to integrate this / its components (X and Y)

    the solution I'm given is: [tex]E_y=\frac{2k \lambda sin(\frac{\alpha}{2})}{R}[/tex]

    However, I don't understand the following: Why is there only a solution for Y? As far as I can tell the X components don't cancel since its an arc of arbitrary length, not a half or full circle. Also, shouldn't the Y component be in terms of cos? The angle alpha is with the X axis, I don't understand why E_y would be in terms of sin (since it become cos after integration).

    Thank you!
     
  2. jcsd
  3. Feb 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks to me like they oriented the arc so that it's centered on the y-axis. (The statement that "[itex]\alpha[/itex] is with the x-axis" does seem a bit odd.) Of course the easy way to solve the problem regardless of the orientation of the arc, is to align it with some axis, solve the integral as show, then transform it back to the given axis.
     
  4. Feb 22, 2008 #3
    Well, they give me a picture with the arc starting from the positive x axis, going through an angle alpha which is counterclockwise from the positive x-axis. That's what is confusing me...

    Also, shouldn't there be two components to the elctric field? X and y? Regardless of where the arc is oriented?
     
  5. Feb 22, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That does seem to contradict the answer.

    By symmetry, you know the electric field must be along the line from the center of curvature to the midpoint of the arc. To find the magnitude, just choose an axis along that direction to do your integration.

    Whether the field has x and y components depends on its orientation to the axis. If the arc were centered on the x-axis, it would only have an x-component. For whatever reason, in contradiction to the picture you were given, they chose to align the arc with the y-axis. But according to how you described the picture, it would have both an x and y component.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?