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Electric field in insulating slab

  1. Feb 5, 2014 #1
    A large, thin, insulating slab [itex]2[/itex] m x [itex]2[/itex] m x [itex]5[/itex] mm has a charge of [itex]2 [/itex] x [itex] 10^{-10}[/itex] C distributed uniformly throughout its volume. determine the electric field at observation point P, which is located within the slab, beneath its centre, [itex]0.5[/itex] mm from the top face.

    udI1k.png

    I started off by making two Gaussian surfaces in the shape of cylinders. The first ranges from the bottom of the slab to point P. The second ranges from point P to the top of the slab. I then calculated the net electric field on point P due to the electric fields traveling out of the ends of the two Gaussian surfaces, using Gauss' Law.

    [itex]E = E_{1} - E_{2} = \frac{Q_{1}}{2A\epsilon_{o}} - \frac{Q_{2}}{2A\epsilon_{o}}[/itex]

    where [itex]E[/itex] is the net electric field acting on point P, [itex]E_{1}[/itex] is the electric field from the first Gaussian surface, and [itex]E_{2}[/itex] is the electric field from the second Gaussian surface.

    The charge density is as follows:

    [itex]\rho = Q/V = 2\cdot 10^{-10}/(2\cdot2\cdot0.005) = 1\cdot10^{-8}[/itex]

    [itex]Q_{1}=\rho V = \rho A \cdot 0.0045 = 4.5\cdot 10^{-11}A[/itex]

    [itex]Q_{2}=\rho V = \rho A \cdot 0.0005 = 5\cdot 10^{-12}A[/itex]

    [itex]E = \frac{4.5\cdot 10^{-11}A}{2A\epsilon_{o}} - \frac{5\cdot 10^{-12}A}{2A\epsilon_{o}}[/itex]

    [itex]E = 2.26[/itex] N/C

    My final answer is 2.26 N/C. Is my process correct?
     
  2. jcsd
  3. Feb 5, 2014 #2

    collinsmark

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    That looks correct to me. :approve:

    Oh, except for one thing. You said, "Gaussian surfaces in the shape of cylinders." I think you meant pillboxes. :smile:
     
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