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Force between the plates of a plane capacitor

  1. Aug 8, 2011 #1
    I found this problem while self-studying Electricity and Magnetism, and I want to know if my solution is rigorous.

    1. The problem statement, all variables and given/known data
    Show that the plates of a plane capacitor attract each other mutually with a force equal to
    [tex]F=\frac{q^2}{2\epsilon_0 A}[/tex]
    Obtain this result by calculating the work required to increase the separation between the plates from x to x + dx.

    2. Relevant equations
    Energy stored in a capacitor with electric field E, plate area A and distance x between the plates:
    [tex]U=\frac{1}{2}\epsilon_0 E^2 Ax[/tex]

    3. The attempt at a solution
    The energy U stored in a capacitor can be viewed as the work that needs to be done by an external agent to separate its plates by a distance x:
    [tex]U=\frac{1}{2}\epsilon_0 E^2 Ax[/tex]
    The electric field E can be rewritten as [itex]q/(\epsilon_0 A)[/itex], where q is the magnitude of the charge on each plate. Thus, the work can be rewritten as:
    [tex]U=\frac{1}{2}\frac{q^2 x}{\epsilon_0 A}[/tex]
    The work dU done by the external agent to separate the plates from a separation x to a separation x + dx is:
    [tex]dU=\vec{F}\cdot d\vec{x}[/tex]
    where [itex]\vec{F}[/itex] is the force applied by the external agent.
    Because in this situation force and displacement are in the same direction, it may be rewritten as:
    [tex]dU=Fdx[/tex]
    Isolating the force:
    [tex]F=\frac{dU}{dx}[/tex]
    Thus:
    [tex]F=\frac{d}{dx}\frac{1}{2}\frac{q^2 x}{\epsilon_0 A}[/tex]
    [tex]F=\frac{1}{2}\frac{q^2}{\epsilon_0 A}[/tex]

    Thank you in advance.
     
    Last edited: Aug 9, 2011
  2. jcsd
  3. Aug 8, 2011 #2

    Delphi51

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    Homework Helper

    Looks like an error substituting in the q/(ϵ0A). Shouldn't you end up with an A on the bottom instead of on the top?
     
  4. Aug 9, 2011 #3
    Yes, there was an error; I've just corrected it.
     
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