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E-field between insulator and conductor

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data

    I have a slab of conductor on the ground with thickness, a, and it extends in all directions infinitely. Then I have a slab of insulator above the conductor of thickness, a, and it also extends infinitely in all directions. The two slabs are a distance, 2a, from each other. The conductor is grounded. The insulator has some charge density p. It induces a charge density of -pa on the top surface of the conductor. The bottom surface of the conductor has zero charge density. The question assumes the E-field is always in the z-direction, and asks for the E-field for all values of z. Note that the situation is in electrostatic equilibrium.

    2. Relevant equations

    none really..

    3. The attempt at a solution

    I decided to split up the question into two parts: find the E-field due to the conductor for all z values, assuming the insulator isn't there.. then find the E-field due to the insulator, assuming the conductor isn't there.. then by the principle of super-position, add them up, and have the answer.

    For conductors I know that eE = D, where e is epsilon, E is the E-field strength, and D is the surface charge density. So above the conductor, the E-field is E = (-pa)/e pointing up. The E-field within and below the conductor is zero.

    For the insulator, I found the E-field above it to be E = (pa)/(2e) pointing up, below the insulator to be E = (-pa)/(2e), and within the insulator to be E = (p(z - 3.5a))/(e) pointing up.

    My main question is in regards to below the conductor, because when I add up the E-fields generated by both the insulator and conductor, I get an E-field below the conductor, which I know is not possible. Is there some concept I'm missing here? Does the principle of superposition not work here? Maybe it's because the conductor is grounded? If possible, could you start me down the correct path for sovling this..? or just correct where I went wrong?

    Thanks a bunch.
     
  2. jcsd
  3. Feb 14, 2008 #2
    Anyone know?
     
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