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Homework Help: Force on capacitor plates.

  1. Sep 22, 2010 #1
    As I have understood it, it is a classical problem to show that the force between to capacitorplates is [tex] \frac{1}{2}\epsilon _0 E^2 A[/tex], which is at first counterintuitive since one should think that it is [tex]QE = \epsilon_0 E^2 A[/tex]. I am pondering that question a bit now and i saw the MIT lecture http://ocw.mit.edu/courses/physics/...magnetism-spring-2002/video-lectures/embed07/ where it is explained, but I dont quite get the explanation. When is it ever assumed that the conductor has any with? That is.. is it not only a plane in which it does not have a volume and therefore it is absurd to talk about the electric field inside the conductor?

    And why would we ever have to average the two anayway? The efield inside doesnt apply any force. The only force on the plate is from the efield outside and that is certainly E.

    Can anyone explain this to me?
  2. jcsd
  3. Sep 23, 2010 #2
    You would most certainly agree that the electric field outside the plate is due to both plates. You logic is wrong because a source (even extended one like a plate) can not apply force on itself due to its own field.One simple way to look at this is to assume that the plates have an individual field of E1 an E2. Then inside the plate E1-E2 =0 and outside E1+E2=σ/ε. Solve them and use the field of the other plate to calculate the force on the plate in question.
    This result,valid for an infinite plate,is also used to calculate force on the points on a charged conductor due to obvious reasons.
  4. Sep 24, 2010 #3
    Ah! So what you're really saying is that it's all solved with the fact that the plates can only apply forces to eachother?

    I have indeed assumed that the total field inside the capacitor exerts a force on a plate, but half of this field is ofcourse due to one plate itself... so ofcourse a particle moving inside that field would experience a force equal to qE, but the plates them selves does ofcourse only experience 0.5QE from eachother. Right?
  5. Sep 24, 2010 #4
    Yes right.
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