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Parallel Plate Capacitor Force

  1. May 6, 2007 #1
    The problem is: Consider a parallel-plate capacitor with plates of area A and with separation d. Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor. Express your answer in terms of given quantities and [tex]\epsilon_0[/tex].

    2. The equation I figured I needed to use for this was one for the potential energy stored in a capacitor. [tex]U_c=\frac{\epsilon_0}{2}(Ad)E^2, E=\frac{\Delta V}{d}[/tex]

    Well, I figured that because force is the derivative of potential energy, I could just take the derivative of the potential energy equation and get what I needed:

    [tex] \frac{d}{dV}(\frac{\epsilon_0}{2}(Ad)(\frac{\Delta V}{d})^2)[/tex]. From this I get [tex]\frac{\epsilon_0 AV}{d}[/tex], having done the proper substitution for E in terms of my given variables (A,d, and V.) I am using "Masteringphysics" is anyone is familiar with that. I also attempted another means, by which I used the formula F=EQ, and substituted [tex]E=\frac{Q}{\epsilon_0 A}[/tex] and [tex]Q=\frac{\epsilon_0 A}{d}*\Delta V_C[/tex]

    After doing this, the answer I got was [tex]\frac{V^2 \epsilon_0 A}{d^2}[/tex]. Apparently I am off by some "multiplicative factor" (I hate it when the program tells me this...) but I haven't the damndest clue as to what it is.
    Last edited: May 6, 2007
  2. jcsd
  3. Feb 1, 2008 #2
    Multiplicative Error

    This worked for me...
  4. Feb 1, 2008 #3

    Shooting Star

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    Homework Helper

    I’ll give you a different derivation, which you may find “simpler”.

    When a voltage V is applied across the plates, suppose the plates receive charges +Q and -Q. The surface charge density on the plates is +s and –s, where s=Q/A. If the plates were infinite in extent, then each would produce an electric field of magnitude s/2e, where,

    s/(2e) = Q/(2Ae) --(1), (I’m writing e for epsilon_nought).

    The sum of the fields of both the plates E = Q/(Ae) and V=E*d
    => Q = AeV/d --(2).

    So, force on one plate due to field of other = Q*field = Q*Q/(2Ae) (from 1)
    = eAV^2/(2d^2) (from 2).

    Note that the electric field of each plate has been calculated for the case when each plate is infinite. In practice, some correction factor is introduced.
    Last edited: Feb 2, 2008
  5. Nov 1, 2009 #4
    The force is the derivative wrt to distance not potential, i.e=F=dU/dx. If derived correctly you should get [tex]\frac{-V^2 \epsilon_0 A}{2 d^2}[/tex].
  6. Oct 25, 2011 #5
    I just came upon this thread and thought it would be worth pointing out that when doing such derivations, one must remember that for the principle of virtual work to be valid, the system must be closed to other forms of energy input. As such, you can't do the derivative with respect to displacement at constant voltage, as charge can flow, changing the energy of the system. It has to be done at constant charge, i.e. open circuit. As Feynman clearly pointed out in FLP, if you make this mistake in the parallel plate capacitor case, you get the negative of the correct answer, so your force is in the wrong direction! I have seen this mistake in a major MEMS textbook, and in physics lab class notes on the internet from a good university, so it's a pretty widespread mistake.
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