Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic Fields & Forces

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    The drawing shows a parallel plate capacitor that is moving with a speed of 31 m/s through a 4.0 T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 170 N/C and each plate has an area of 7.5 multiplied by 10 x 10^-4 m^2. What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?


    2. Relevant equations
    F = Bqvsin(theta) --- sin() isn't important in this case because it's perpendicular

    Then there's also E = F/q and C = KEoA/d, but I can't for the life of me figure out how to use it in this case.

    3. The attempt at a solution
    Well, I've established that I'm going to need to find the charge somehow from the electric field and the area of the plates, since all the other info for F = Bvq is given. I'm at a loss to figure out how to get the charge though... thanks in advance for any help/advice.

    Attached Files:

  2. jcsd
  3. Feb 25, 2009 #2
    You're completely correct. As it turns out, if you draw a Guassian cylinder or cube that extends halfway into one of the (conducting) capacitor plates and apply Gauss's Law, you'll find that the electric field from one plate is always [tex]\sigma/2\epsilon_0[/tex], where [tex]\sigma[/tex] is the charge density of the plate (charge over area). If you add to this value the field from the other plate, you'll find that inside a parallel plate capacitor, the electric field always has a value of [tex]\sigma/\epsilon_0[/tex]. So given your value for the electric field and the areas of the plates, you can now calculate the charge on the positive plate.
  4. Feb 25, 2009 #3
    Thanks for your help, swuster. I'm still having a bit of trouble though.

    I've done the following now:

    E = (q/A)/2Eo
    170 = (q/7.5e-4)/(2 * 8.85e-12)
    q = 2.25e-12 C

    Then I substitute in and solve for F:
    F = qvB = (2.25e-12)(31)(4) = 2.79e-10 N, which is wrong, unfortunately.

    I feel like I missed a step when following your instructions, and I didn't quite follow this bit:
    Thanks again for the help, I appreciate it greatly! :)
  5. Feb 25, 2009 #4
    Should be:

    E = (q/A)/Eo

  6. Feb 25, 2009 #5
    Ah, OK, thanks a bunch! Worked like a charm after that!
  7. Feb 25, 2009 #6
    Yep - between the capacitor plates, the electric field adds!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook