1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Electric field of a charged arc

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A charge of 18 nC is uniformly distributed along a straight rod of length 4.7 m that is bent into a circular arc with a radius of 2.4 m. What is the magnitude of the electric field at the center of curvature of the arc?

    2. Relevant equations

    3. The attempt at a solution
    magnitude dE=kdQ/R^2=K[tex]\lambda[/tex]Rd[tex]\Theta[/tex]/R^2=KQd[tex]\Theta[/tex]/R^2
    sin[tex]\Theta[/tex]magnitude dE=KQd[tex]\Theta[/tex]sin[tex]\Theta[/tex]/R^2
    magnitude Ey=KQ[-cos[tex]\Theta[/tex]]from a to b/[tex]\pi[/tex]R^2

    the limits of integration would be a=0 and b=4.7(2[tex]\Pi[/tex])/2[tex]\Pi[/tex]R
    I got E=4.31e-1 n/c but it was wrong.
    Last edited: Jun 22, 2008
  2. jcsd
  3. Jun 22, 2008 #2


    User Avatar
    Homework Helper

    EDIT: On looking through all of what you wrote, I can tell you this.

    It is not correct that [tex]\lambda = Q/\pi R[/tex] , but you can just use

    [tex]dQ = \lambda R d\theta [/tex].

    However, in your field calculation, look at what field component cancels out in the integration. (You won't integrate dE sin[tex]\theta[/tex]...).

    Make a picture of the arc and the point at its center. Choose an axis through the center that is symmetrically placed through the arc; this will let you take advantage of a symmetry consideration and also clarify how to set up the field integration.
  4. Jun 23, 2008 #3
  5. Jul 3, 2008 #4
    Why is [tex] dQ = \lambda R d\theta [/tex] ?
    When [tex] \lambda [/tex] has the Unit [tex] [\lambda] = \frac{C}{m} [/tex]

    For a circle I understand it.
    [tex] dQ = Q \left( \frac{dl}{2 \pi r} \right) = \lambda dl [/tex]
  6. Jul 3, 2008 #5
    Make sure to take into account that the electric field at a point is a vector.
  7. Jul 4, 2008 #6
    Don't I do that with [tex] dE_x = dE cos(\theta) [/tex]

    [tex] E = E_x = \int\limits_{-\theta }^{\theta} dE cos(\theta) [/tex]

    I thought [tex] dQ = \lambda R d\theta [/tex] is the charge density, which is a scalar.
  8. Jul 4, 2008 #7
    You need to express dQ as a function of d(theta) so you can integrate from 0 to 2*(pi) (if it's a circle)
    [tex] \lambda = \frac{18nC}{4.7m} [/tex]
  9. Jul 4, 2008 #8
    I think I did not express myself the right way. Sorry

    The only thing I don't understand is why there is a R inside [tex] dQ = \lambda R d\theta
    [/tex]. The first time i tried to solve it I used [tex] dQ = \lambda d\theta [/tex]
    Because as you said, I need a function of theta.
    This is how it worked for the circle with [tex] dQ = \lambda dl [/tex]

    So for the arc it is [tex] \left[ \lambda \right] = \frac{C}{\circ m} [/tex]
    I can't interpret that geometrically, it seems to me just to make it fit right
  10. Jul 4, 2008 #9


    User Avatar
    Homework Helper

    Hi SamTaylor,

    As a first look, we know it can't be [tex] dQ = \lambda d\theta [/tex] because that doesn't have the right units--coulombs on the left, and (coulombs/meter) on the right.

    The linear charge density [itex]\lambda[/itex] is the charge per length, and here the length is along the arc. So if [itex]s[/itex] is the length along the arc, a true statement to begin with would be

    [tex]dQ = \lambda\, ds[/tex]

    If you then think about the relationship between the length of a circular arc s, the radius r, and the angle [itex]\theta[/itex], you'll get the right formula.
  11. Jul 4, 2008 #10
    Thanks a lot.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook