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Gauss's Law and E field problem

  1. Jul 19, 2005 #1
    I have two plates, one on top of the other, that have a charge densities +s1 of the top of the top plate, -s2 on the bottom of the top plate, s2 on the top of the bottom plate, and -s1 on the bottom of the bottom plate (s1 and s2 are both positive numbers). I have to find the electric field on top of the plates, inbetween, and below the plates. Using gauss's law I drew a cylinder through the top plate and found the electric field to be s1/e (e=permittivity of free space). On the bottom by symmetry the electric field would be -s1/e. Now my problem is finding the E field inbetween. The E field vector goes away from positive charge and goes toward negative charges. So If i make a cylinder for that goes through the top plate and find the amount of E field going through the bottom part of the cylinder I have E=-s2/e. Now for the bottom plate the E filed points away from the top part of the bottom plate so by using a cylinder and Gauss's law again I have E=s2/e. Thus the E field inbetween should be -s2/e + s2/e=0. But I don't see how this is the case since the E field vector from the bottom plate is in the same direction as the E field vector of the top plate since the bottom is positively charged and the top is negatively charged. Shouldn't I get E=2s2/e ?

    What am I doing wrong? (I hope I didn't get you confused, but I don't know how to do latex.)
     
  2. jcsd
  3. Jul 19, 2005 #2

    OlderDan

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    If I am reading it right, you have 4 planes of charge such that a Gaussian cylinder cutting all 4 planes encloses zero charge. By Gauss' law the integral of the normal component of the field over the surface will be zero. Since the field from any single plane of charge has reflection symmetry, and is independent of the distance from the plane, the fields outside the plates will be zero. Between the plates you will have contributions from all four planes, with a downward contribution from the s1 surfaces and upward from the s2 surfaces.
     
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