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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
    E=KQ/R^2


    3. The attempt at a solution
    dQ=[tex]\lambda[/tex]ds
    [tex]\lambda[/tex]=Q/[tex]\pi[/tex]R
    dQ=[tex]\lambda[/tex]Rd[tex]\Theta[/tex]
    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

    dynamicsolo

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    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
    thanks
     
  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

    alphysicist

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    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.
     
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