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Electric field exerted at P due to a non-conducting disk

  1. Mar 19, 2012 #1
    A non-conducting disk has radius 1.25cm and charge 6.55nc.
    What is the electric field (magnitude) experienced at a point P - where p = 2cm and is on the x-axis.
    The disk's centre is placed at the origin.


    I approached this problem using the general formula derived:

    k/2Σσ [1 - (1/((R^2/X^2 +1)^1/2))]

    However I have no idea what the surface linear charge density should be due to dQ and dR.

    Thanks guys, really appreciated ... =]
     
  2. jcsd
  3. Mar 19, 2012 #2

    tiny-tim

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    ohh! you mean the linear charge density for the ring of radius r to r+dr ?

    well, the charge on the ring is dQ = Q times area of ring / area of disc

    so the linear charge density is that divided by the length (ie the circumference of the ring) :wink:
     
  4. Mar 21, 2012 #3
    No worries - solved :)
     
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