Physics - Electric Fields

Discussion in 'School Work Help' started by khaotic, Jan 12, 2009.

  1. khaotic

    khaotic Fobulous

    A rod of 14.0 cm long is uniformly charged and has a total charge of -22µC. Determine the magnitude and direction of the electric field along the axis of the rod at point 36.0 cm from its center.

    No interested in final answer. Only the solution.
  2. iiimj4everiii

    iiimj4everiii Well-Known Member

    Not sure if this is right. If the answer is wrong, don't read below xD. Answer I got is around -1,506,787.07 N/C and the direction of the electric field lines are toward the rod.


    A uniform rod 14cm or .14m with a total charge Q of -22
    µC. First things first, since the rod is -ve, the electric field lines are pointing toward the rod.

    Now you need to draw a picture to have a better understanding. It's a rod of 14cm or .14m with a test charge 36cm or .36m above the center of the rod. You should picture a straight rod on the bottom and a point test charge on the top. The radius between the test charge and rod changes as it goes from the center of the rod to the edge of the rod. I'll try to draw a picture and upload it here when I'm done going through the steps.

    In general:

    F = keq1q2/r²

    For Electric field, we use the second charge as a infinitesimally charge, a test charge q0

    F = keqq0/r²

    F/q0 = E = keq/r²

    where ke is a constant, q is the charge of the source, and r is the distance between the charge of the source and the test charge q0

    dE = kedq/r²

    E = ke

    λ = Q/X. dq = λdx

    E = ke

    x = ytan(Θ)

    dx = ysec²(Θ)dΘ = ydΘ/cos²(Θ)

    E =

    y/r = cos(Θ)

    1/r = cos(Θ)/y

    1/r² = cos²(Θ)/y²

    E =
    keλ∫ydΘcos²(Θ)/y²cos²(Θ) = keλ∫dΘ/y

    E = keQ/(xy)

    .07/.36 = tan(Θ)


    -11π/180 <
    Θ < 11π/180

    E = KeQ/(xy)[Θ from
    -11π/180 to 11π/180]

    E = keQ/(xy)[11π/180 + 11π/180]

    E = (8.99 x 10^9)(-22 x 10^-6)(11π/90)/(.14 * .36)
    ≈ -1,506,787.07

    Attached Files:

    #2 iiimj4everiii, Jan 31, 2010
    Last edited: Jan 31, 2010