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Electric Field Between Plates

  1. Feb 5, 2005 #1
    Alright, here's the problem. You're told that there are two square metal plates with side length L and a distance d away from each other. One has charge +Q, the other -Q. Then they ask for the magnitude for the charge between the plates, not close to the edge.

    Here's what I have so far: The electric field between them obviously isn't 0.

    I've got sigma = (magnitude of charge)/area.
    The electric field for one plate is E = sigma/(2 * epsilon).
    Since the fields from both plates in between them point in the same direction, the total field would be E = sigma/epsilon.

    This was going great until I realized the I still had an unused variable, distance d. I have no idea of what to do with it. I tried multiplying E by d, then dividing it by d, but I don't like the way those numbers look. Please help!
     
  2. jcsd
  3. Feb 5, 2005 #2

    Andrew Mason

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    Separation distance d is immaterial to the electric field if d is small compared to L. Does the question go on to ask about the capacitance?

    AM
     
  4. Feb 5, 2005 #3
    Nope. Correct me if I'm wrong, but this is what I have pictured (since there is no picture to go with the problem):

    | - ---> +
    | - ---> +
    | - ---> +
    L - <-d-> +
    | - ---> +
    | - ---> +
    | - ---> +

    For the record, L = 0.82m and d = 0.022m

    I'm not sure if that would be considered a significant difference, but they gave me a value, so I'm scared to ignore it.
     
    Last edited: Feb 5, 2005
  5. Feb 5, 2005 #4
    bu-du-bump!
     
  6. Feb 5, 2005 #5

    dextercioby

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    I'm sorry,but to me the problem does not make too much sense.What's inbetween the plates??Vacuum,a conductive medium,a dielectric,what??
    Besides,how would u get the charge?

    Daniel.
     
  7. Feb 5, 2005 #6
    Ok, I'll put it verbatim.

    Two square metal plates are placed parallel to each other, separated by a distance d = 2.20 cm. The plates have sides of length L = 0.820 m. One of the plates has charge Q = + 2.70 x 10^-3 C, while the other plate has charge -Q. What is the magnitude of the electric field between the plates, not close to the edge?
     
  8. Feb 5, 2005 #7
    lalalabump
     
  9. Feb 5, 2005 #8

    Doc Al

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    Staff: Mentor

    Better listen to Andrew and ignore that separation distance. Note that the equation you are using for the field:
    "The electric field for one plate is E = sigma/(2 * epsilon)." ​
    is only accurate for distances that are small compared to L. Use it. That's what they want.
     
  10. Feb 5, 2005 #9

    dextercioby

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    That's something totally different.Here's something from your first post: "Then they ask for the magnitude for the charge between the plates, not close to the edge. "... :rolleyes:

    Daniel.
     
  11. Feb 5, 2005 #10
    Hahaha, I was doing it right the whole time. The only thing was that I kept putting 8.55 E -12 for epsilon instead of 8.85. Thanks guys!
     
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