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Uniformly charged wire

  1. Feb 2, 2005 #1
    A uniformly charged wire with a charge density of 4 microCoulombs/meter lies on the x-axis between x=1m and x=3m. What is the y-component of the corresponding electric field at y=3m on the y-axis?

    I'm not really sure where to go with this. I want to treat the rod as an infinite number of point charges but I'm not sure how to calculate (y-component of) the electric field caused by each of these points.
     
  2. jcsd
  3. Feb 2, 2005 #2
    suppose there is a point charge on the point y=3 , find the electric fireld there, actually the electric field there for suppose due to the charge on the X=1 m can be broken up into two components one along -x and another along +y axis..they can be computed separately by integrating ..in this case i think you only need to compute for the y aixs one...for the integration take elemental lengths dx for the wire...

    glad to be of help,
    Arpan Roy
    royarpan@hotmail.com
     
  4. Feb 2, 2005 #3

    HallsofIvy

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    Note that, for each "dx" on one side of the point, there is a corresponding "dx" the same distance on the other side. The horizontal components of force of those will cancel but the vertical components will add.
     
  5. Feb 2, 2005 #4
    Since this problem is not symmetrical, the horizontal components of the [itex]\vec{E}[/itex] do not cancel. The easiest approach is probably to calculate the vertical ( [itex]\vec{E_y}[/itex] ) component seperately.

    Draw a diagram of the situation with the given axis, and choose an arbitrary piece of charge [itex]dq[/itex] of the wire.

    Come up with an equation for the corresponding electric field [itex]\vec{dE}[/itex] due to [itex]dq[/itex] at the point (0,3).

    Figure a way to represent [itex]dq[/itex] in terms of [itex]dx[/itex] so you can integrate with respect to x.
    (hint: it involves the linear charge density )

    Break the equation into the vertical component of [itex]\vec{E}[/itex] and integrate with respect to x
    (hint: [itex]\vec{dE_y} = \vec{dE} sin \theta[/itex] where theta is the angle between a line parallel to the x axis and [itex]\vec{E}[/itex] )
    (another hint: you will have to come up with an equation for [itex]sin \theta[/itex] in terms of x so you can integrate )

    good luck :smile:
    -MS
     
    Last edited: Feb 2, 2005
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