1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electrical Field between plate
Loading...