1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Force on a dielectric slab

  1. Sep 3, 2014 #1
    ...partially inserted between the plates of a parallel plate capacitor.

    I have a few questions about this - it is section 4.4.4 of Griffiths, 3rd edition.

    Imagine a parallel plate capacitor, of plate area l*d, separation d, filled with a linear dielectric of susceptibility X. Now imagine sliding the dielectric out of the plates in the direction that the plates have length l by a distance x. The idea is to find the force on the dielectric.

    You can do this for two cases - constant charge and constant voltage. Both give the force as
    F=-ε0XwV2/2d along the x direction where V is the p.d between the plates. I understand both of these derivations.

    However understanding the results physically is proving to be an issue.

    First of all if we let x=0 so the dielectric slab is perfectly between the plates, why is the force non-zero - apparently there is still a force in the negative x direction which does not make sense at all, from the symmetry of the scenario for starters.

    Secondly, at constant charge, F=-ε0XwV2/2d but V is not a constant and has x dependence - this makes me happy as I would expect the force to vary with x. However at constant voltage, F=-ε0XwV2/2d but now V is a constant, and so is this telling me the force is the same on the slab wherever it is in the universe (provided it remains in the correct 'slot' between the plates). This just wouldn't make sense surely because as x gets very large, the fringe field causing the force should be vanishingly small.

    If anyone can answer these I would be grateful! I have a few more issues about it but not understanding the above probably means I'm not in a position to worry about them just yet.
  2. jcsd
  3. Sep 3, 2014 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    The analysis is based on Griffith's equation (4.62) for the capacitance when the dielectric is displaced ##x## from being fully inserted: $$C = \frac{\varepsilon_0 a}{d}(\varepsilon_r l - \chi_e x)$$
    If you think about how this expression is derived, for what range of ##x## does this expression hold?

    In particular, is it valid for ##x > l ## ?

    Is it valid for ##x << l \,##?

    Consider the assumptions made about a parallel plate capacitor when deriving ##C = \frac{\varepsilon A}{d}##.

    EDIT: Watch a qualitative demo video here .
    Last edited: Sep 3, 2014
  4. Sep 4, 2014 #3
    I can see it's not valid for x>l, as the capacitance should then be C=ε0wl/d. So if the capacitance is constant for x>l, does this mean the force on the slab is zero when the slab is out of the capacitor 'slot'? I think the video would sort of agree with that.

    For x<<l, I'm not too sure - I'm guessing it's because we require that d is much less than the two dimensions of the capacitor (why exactly do we need this though?), and if x<<l, this isn't the case for the part of the capacitor in the vacuum, invalidating the capacitance expression...

    Onto another problem - Griffiths states 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound'.

    However although the force expressions are the same in each case, F=-ε0XwV2/2d if you consider them in their separate contexts, they appear to be different. For constant voltage, if you pick some voltage V, F is the same for all x. In the constant charge case, if you pick some V, you get the same F as in the constant voltage case but because V is not being held constant, you can only get this value of V at a certain value of x such that V=Q/C (as C depends on x). So then the variation of force with x appears to be different, despite the expressions being the same - so doesn't that contradict the above statement, or am I misunderstanding it?
    Last edited: Sep 4, 2014
  5. Sep 4, 2014 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    There will still be some force on the dielectric when it it outside the capacitor. But the force decreases rapidly with distance once its outside. There is a "bowing out" or "fringing" of the field of the plates at the edges of the plates. This field that exists outside of the region between the plates would create some force on the dielectric when it is outside.

    Yes, that's right.

    I agree with what you are saying. How the force varies with x is different for keeping Q constant and keeping V constant. But, as you noted, in both cases you can express the force at any position with the same equation F=-ε0XwV2/2d. Griffiths is right when he says that the force at some position of the dielectric ultimately depends on just the distribution of free and bound charges at that position. For a particular x, the distribution of the charges will generally be different when keeping V constant as compared to keeping Q constant. So, the force at some x will not be the same for the two cases, in general. But in either case, the force can be expressed by the same equation.

    I agree with you that Griffiths statement [ 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound' ] could be interpreted as implying that the force should vary with x in the same way for the two cases. As you noted, that would be incorrect.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted