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Circular arc of charge, integration question

  1. Apr 13, 2006 #1
    I have a circular arc of wire centered at the point (0,0). It has a radius of [tex]r[/tex], extends from [tex]\theta = -60[/tex] to [tex]\theta = 60[/tex] and also holds a charge [tex]q[/tex]. For the differential electric field I have the following equation:

    [tex]
    dE = \frac{\lambda ds}{4 \pi \epsilon_0 r^2}
    [/tex]
    Where [tex]ds[/tex] is the length of a differential element of the arc.

    To the find the [tex]x[/tex] component of the electric field I this equation:

    [tex]
    dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} ds
    [/tex]

    To integrate this I have to set [tex]ds = r d\theta[/tex] so that the above equation reads:

    [tex]
    dE_{x} = \frac{\lambda}{4 \pi \epsilon_0 r^2} cos(\theta)} r d\theta
    [/tex]

    Where does the [tex]r[/tex] come from in the statement [tex]ds = r d\theta[/tex]?
     
  2. jcsd
  3. Apr 13, 2006 #2

    nrqed

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    Gold Member


    That is just the relation between an arclength and the subtended angle in a circle. Recall that, as long as the angle is given in radians, we have [itex] s = r \theta [/itex], right? This is true for any arc fo a circle. If you tell me that an arc of a circle subtends [itex] {\pi \over 8} [/itex]radians, for example, and that the circle has a radius of 20 cm, then the length of the arc is simply [itex] {5 \pi \over 2} [/itex] cm.

    For an infinitesimal subtended angle [itex] d \theta [/itex] the relation is obviously that the infinitesima arc length is [itex] ds = r d \theta [/itex].
     
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