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Parallel Plates and charge density

  1. Apr 12, 2005 #1
    Three very large square planes of charge are arranged as shown (on edge) in Fig. 21-70. From left to right, the planes have charge densities per unit area of -0.50 µC/m2, +0.20 µC/m2, and -0.30 µC/m2. Find the total electric field (direction and magnitude) at the points A, B, C, and D. Assume the plates are much larger than the distance AD.

    A | B | C | D

    I know that the Efield= charge density/2epsilon but I'm having trouble adding the fields. For example, Point B I would think equals E of the first plate + E of the second plate (due to the positive and negative signs). My answer of 2.82e4 + 1.13e4 = 3.95e4...is this correct?

    Thanks in advance for the help, it is greatly appreciated.
     
  2. jcsd
  3. Apr 12, 2005 #2

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    All plates contribute to the total field in all regions. You just need to give them the appropriate sign depending on which side of the plate you're on. Just remember the field lines of a plate point towards the plate if the charge is negative and away if it's positive.
     
  4. Apr 12, 2005 #3
    ok, but would point B, for example, be influenced by the charge density on the 3rd plate? If so, I am assuming that i should draw the diagram with arrows and add up all arrows in one direction and subtract all arrows in the opposite direction.
     
  5. Apr 12, 2005 #4

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    Yes. Don't forget the superposition principle, that the total field is equal to the sum of the field contributions by all charges.
     
  6. Apr 12, 2005 #5
    I'm not sure if I'm going about this correctly. Just for your reference, Plate 1 has a E of -2.82e4, Plate 2 an E of 1.13e4, and Plate 3 an E of -1.69e4........

    Let's take point B:
    I am considering fields to the left to be negative and to the right to be positive.
    Plate 1 will be a negative 2.82e4
    Plate 2 will be a negative 1.13e4
    Plate 3 will be a positive 1.69e4
    add those all up and the Efield that B experiences should be -2.26e4

    Is that correct?
     
  7. Apr 12, 2005 #6

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    That looks right.
     
  8. Apr 12, 2005 #7
    thanks, everything came out correct.
     
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