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The force tha apears betwin the plates of a capacitor

  1. Jan 2, 2007 #1
    helo :smile:

    have a proble here can someone help me out?

    supouse i know the surface of the plates of a capacitor ( that are flat surfaces place one in front of anougthere ) the distance bethwin them and the charge.
    how do i calculate the force that apears betwin them?
     
  2. jcsd
  3. Jan 2, 2007 #2

    berkeman

    User Avatar

    Staff: Mentor

    There is an equation in the beginning of electrostatics that gives you the force on a charge based on the amount of charge and the electric field value. Are you familiar with it? If you don't see it in your text (honest, it's right in the beginning), you can do a quick search on wikipedia.org for electrostatics.

    Welcome to PF, BTW.
     
  4. Jan 2, 2007 #3
    It's slightly ambiguous what "force" you are looking for.When you say "the charge," are you referring to a charge between the plates or the charge distributed on the plates?

    If you are referring to a charged object between the plates, do what berkemann said.

    If you are referring to the charge on the plates, then you can calculate the force that one capacitor plate exerts on the other with the concept of virtual work.
     
  5. Jan 2, 2007 #4
    I think he/she probably meant the electrostatic force that one plate exerts on the other. If the distance of the plate is relatively much smaller than their dimension, one can calculate the electric field resulted by one plate using Gauss's law. Then simply multiply that by the charge on one plate and you get the force.
     
  6. Jan 4, 2007 #5
    Try this link (parallel plate?)
    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html

    Or this Eq....q=(eo)EA

    where (eo)= 8.85pF/m
    E = electric field
    A = area
     
  7. Jan 4, 2007 #6
    It's not that simple, because the plates are infinite.

    You would have to move one plate a differential amount, and calculate the change in energy. Then you would calculate the initial capacitance, and the final capacitance, and plug in to the initial energy and final energy, then you solve for the force (because the differential movement of the plate does work [itex]\textbf{F}\cdot{d\textbf{x}}[/itex]).
     
  8. Jan 4, 2007 #7
    if the plates are infinite, then even better, the electric field calculated from gauss's law is exact (no longer approximation).

    I don't know why one would want to find the work done, then differentiate it to find force (I assume that is what you want by integrating then using [tex]\vec{F}=-\nabla U[/tex])

    The force on the charge by the plate is simple enough. Imagine a Gaussian surface, a cylinder going through the plate. by Gauss's law:
    [tex]\oint \vec{E}\cdot \vec{dA}=\frac{Q}{\epsilon_0}[/tex]

    by symmetry, let the base area of the cylinder be A:

    [tex]2EA=\frac{Q}{\epsilon_0}[/tex]

    assume a 2D charge density of [itex]\sigma[/itex], A cancels:
    [tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

    if you have a charge q whatever distance away from the plate, you have:
    [tex]F=qE[/tex]

    and the force by the charge on the plate is the same but opposite in sign due to Newton's third law (assuming the electric field isn't changing).

    This calculate is useful when you want to find the field strength at a distance very close to the plate, so that the plate seems infinitely big. However, if you get sufficiently far away, you have to do serious integration to find out the field strength, and the result will depend on the shape of the plate.
     
    Last edited: Jan 4, 2007
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