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Homework Help: Electrical Field between plate

  1. Sep 3, 2011 #1
    Problem:

    Two square metal plates are placed parallel to each other, separated by a distance d= 1.56 cm. The plates have sides of length L = 0.560 m. One of the plates has charge Q= + 1.74×10-6 C, while the other plate has charge -Q. Calculate the magnitude of the electric field between the plates, not close to the edge, i.e., assume a uniform surface charge distribution

    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. Would I use E=sigma/epsilon OR E=sigma/(2*epsilon) here for the correct answer?

    epsilon is just 8.85 here not 8.85E-12 im 99% but I dont really know why.

    How in the WORLD do I find the area to find sigma though?
     
  2. jcsd
  3. Sep 3, 2011 #2

    Doc Al

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    Staff: Mentor

    You want the total field, of course.

    You're wrong here: It's 8.85E-12.

    The plates are square. Find the area.
     
  4. Sep 5, 2011 #3
    Im still not getting it right...

    I used 1.74×10-6 C as my charge and found the area of the one plate...do I need to find the total area (ie. add the two areas?)

    I then got sigma using the above and plugged it into E= sigma / (2* epsilon)

    I tried using 8.85 and 8.85E-12 and both incorrect.

    Help?
     
  5. Sep 5, 2011 #4

    Doc Al

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    Good. What did you get for the area.

    No.

    What did you get for sigma?

    No point in plugging the wrong value.
     
  6. Sep 5, 2011 #5
    area came out to: .3136 m^2
    sigma came out to be: 5.548E-6

    Plugged that into E = sigma / (2* epsilon) gave me: 313472.85 N/C and it was incorrect.
     
  7. Sep 5, 2011 #6

    Doc Al

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    You want the total field, not just that from one side. Double your answer.
     
  8. Sep 5, 2011 #7
    I thought since we were doing 2*epsilon it was for both plates...aka total field.

    I have the answer right now but dont really get why we had to double as I thought it was as stated above.

    Thanks for the help!
     
  9. Sep 5, 2011 #8

    Doc Al

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    That factor of 2 is in the denominator, so that version is only half the field. (See my response in post #2.)
     
  10. Sep 5, 2011 #9
    Thank you very much...I posted one more question if you dont mind.

    Much appreciated!
     
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