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Griffiths problem 2.37

  1. Mar 9, 2008 #1
    [SOLVED] Griffiths problem 2.37

    1. The problem statement, all variables and given/known data
    Two large metal plates (each of area A) are held a distance d apart. Suppose we put a charge Q on each plate; what is the electrostatic pressure on the plates?

    2. Relevant equations

    3. The attempt at a solution
    I am so helplessly lost with this problem. How do I get the electric field just outside of the surfaces of the plates? Its not a capacitor since the charges are not opposite!
  2. jcsd
  3. Mar 9, 2008 #2
    No, it's not a capacitor, but is there any reason why Gauss's law can't still work? What is the electric field between, how about outside? Should take a couple lines of work with good uses of symmetry.
  4. Mar 9, 2008 #3
    What Gaussian surface am I supposed to use? A box that goes through box plates? A box that goes through one plate? A box that goes through neither? A don't see how any of them would be helpful since you cannot evaluate the surface integral.
  5. Mar 9, 2008 #4


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    Is A>>d^2? If not, you can't use Gauss's law, and it is a difficult problem.
    If yes, the E field of one plate is easy to find, and F=QE.
  6. Mar 9, 2008 #5
    Griffiths calling things "large" is always a keyword for, "approximate as infinite." Does this help at all with your predicament of evaluating as a surface integral?
  7. Mar 9, 2008 #6
    I assume A is much greater than d^2. But still I don't see how to find the E-field. Let us draw a Gaussian pillbox that saddles one of the plates. The flux through the sides will be 0 by symmetry because the plates are large. But there is still the top and the bottom that we don't know the flux through. We assume the E-field is constant at each. Knowing the E-field at one gives you the E-field at the other. But how do you get the E-field at either one?
  8. Mar 9, 2008 #7


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    For an isolated plate, E through each face is the same by symmetry.
    Griffith's does this in detail.
  9. Mar 9, 2008 #8
    This is not an isolated plate.
  10. Mar 9, 2008 #9
    Tell me, how would you find the electric fields for a capacitor? Perhaps, let's work from some more familiar ground.
  11. Mar 9, 2008 #10
    Okay. That was a helpful post Mindscrape.

    I see what is going on. We are making the (very unrealistic) assumption that the electric field due to each plate is constant at distances within [itex]\pm d [/itex] of that plate and is equal to [itex]Q/2\epsilon_0 A[/itex] in the normal direction. Then you guys are right, the problem is easy since application of Gauss's Law is trivial.
    Last edited: Mar 10, 2008
  12. Mar 9, 2008 #11
    Actually the problem is not quite solved yet. Equation 2.52 depends on E which is given as the electric field just outside the surface, but in this case, isn't the electric field different on the two sides of the plates i.e. it is 0 between the plates and [itex]Q/A \epsilon_0[/itex] on the other side isn't it. So what do I use for E in eqn 2.72?
  13. Mar 9, 2008 #12


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    To find the force on a plate, you treat one plate as if it is isolated.
    Then use F=QE on the other plate. That's the same as how you find the force between two + point charges. There, too, the total field on a line between the charges is zero.
  14. Mar 9, 2008 #13
    It's actually not so unrealistic. Think about what an infinite plane really means. Picture yourself above an infinite plane, as you back away it will look the same (since there isn't really anything to reference). If you are really close to a surface, or the surface is really long then it has the same effect.
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