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Homework Help: Force on a dipole in a non-uniform electric field

  1. Aug 24, 2014 #1
    I'm going to explain my understanding about a bit of a contradiction I can't resolve, and I was hoping somebody could help me understand it. Sorry about the bombardment of vector components!

    There appears to be two methods we can use to calculate the force (effectively the same in fact, but one is a bit of a shortcut):

    1. Find the potential energy of the dipole using
    Then simply use F=-U to get the force. This comes out as
    Now this method has given me correct answers when I have started with a specific vector function E, found its corresponding U and then got F

    2. The 'shortcut'. We can use a vector calculus identity so that
    F=-U=-(-p.E)=(p.E)=(p.)E. All this means is we can find F without the intermediate stage of needing U. All we need is E. However this in component form gives

    So the same vector F comes out in two different ways. Now, I can't find any reason that makes them equal, but I recall using the first correctly for a specific field, and the second method I'm assuming would have given me the correct answer too, so why aren't they agreeing with one another for a general dipole in a general field in component form? Thanks for any help!
  2. jcsd
  3. Aug 25, 2014 #2
    Anybody? :)
  4. Aug 25, 2014 #3
    Your electrostatic field has zero curl, so the following holds:
    \frac{\partial E_x}{\partial z} = \frac{\partial E_z}{\partial x}, \quad \frac{\partial E_y}{\partial x} = \frac{\partial E_x}{\partial y}, \quad \frac{\partial E_z}{\partial y} = \frac{\partial E_y}{\partial z}
    It comes directly from the definition of curl:

    Just a reminder that the identity you're using in (2) is:
    \nabla(\mathbf{u}\cdot\mathbf{v}) = (\mathbf{u}\cdot\nabla)\mathbf{v} + (\mathbf{v}\cdot\nabla)\mathbf{u} + \mathbf{u}\times(\nabla\times\mathbf{v}) + \mathbf{v}\times(\nabla\times\mathbf{u})
    If you're not working with an electrostatic field, then things get a bit more complicated.
    Last edited: Aug 25, 2014
  5. Aug 25, 2014 #4

    rude man

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    ∇(p.E) = (p.∇)E + (E.∇)p + p x (∇ x E) + E x (∇ x p)

    so in you last step you seem to have omitted no less than 3 extra terms!
  6. Aug 25, 2014 #5
    That's fine, though, for a permanent dipole moment and electrostatic field.
  7. Aug 26, 2014 #6
    Ah so simple! I've been trying to get my head around this for ages - thanks :)

    Edit: A further thought - I'm not sure if I am going to be very clear here so if it doesn't make sense let me know.

    Suppose we have a dielectric medium in which the dipole moment varies with position. Then although we have shown the two formulae for the force to be equivalent, in this case the F=(p.E) approach implies we must differentiate the dipole moment components, whilst F=(p.)E approach does not. Is this not a contradiction (my book seems to suggest it is)?

    The only way I can think of resolving this is: in deriving both of these expressions we assumed the dipole moment was constant - i.e I took the dipole moment components from out of the derivatives when applying the gradient to p.E and when using the vector calculus identity for the second approach, we assumed the divergence and curl of the dipole moment were zero. Therefore we should in fact not be relying on any of these expressions for this situation. My book however states that the second approach would be valid here - it seems somewhere that may have 'got lucky' possibly. This is a bit sketchy but just all I can think of.
    Last edited: Aug 26, 2014
  8. Aug 27, 2014 #7
    What book are you reading?

    My suggestion is to stick with ##\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}##, since it's the more general form. IIRC, it's strictly only valid for dipoles with very short displacement vectors (since its derivation uses a first-order Taylor expansion of the electric field), but it works for induced (nonconservative) electric fields as well.
  9. Aug 28, 2014 #8
    It's Griffiths, in a footnote on page 165.

    I will quote it:
    'In the present context Eq. 4.5 could be written more conveniently as F=(p.E). However it is safer to stick with (p.)E because we will be applying the formula to materials in which the dipole moment (per unit volume) is itself a function of position and this second expression would imply (incorrectly) that p too is to be differentiated.

    Eq. 4.5 is F=(p.)E.

    I don't really understand why there seems to be an inconsistency between the two formulae given as we have shown them to be the same (assuming constant p) but it is quite evident there is for non-constant p.
  10. Aug 28, 2014 #9
    To be honest, I've never really given the form ##\mathbf{F} = \nabla(\mathbf{p}\cdot\mathbf{E})## much thought. You'd have to start out with a potential field, so that precludes induced electrical fields, which is why I think most tend to avoid it. In the third edition of Griffiths, he also takes the approach of using a first-order Taylor expansion of ##\mathbf{E}## to derive ##\mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E}## on page 164.

    Say you had a neutral atom in an electrical field such that it's polarized and you get an induced dipole with a moment proportional to the electrical field ##\mathbf{p} = \alpha \mathbf{E}##, as shown in Griffiths on page 161. That makes it seem like ##\mathbf{p}## itself is a vector field, but as far as I can tell, it's understood that ##\mathbf{p} = \alpha \mathbf{E}(\mathbf{r}_0)##, where ##\mathbf{r}_0## is the position vector of the dipole at some instant in time. As such, you shouldn't include it as a vector field in the coordinates ##(x,y,z)## when determining ##\nabla(\mathbf{p}\cdot\mathbf{E})##, i.e. ##\mathbf{p}## is assigned a value depending on its current position in the electrical field.

    I think that's what Griffiths means in the footnote - at least, that's all that makes sense to me.
    Last edited: Aug 28, 2014
  11. Aug 29, 2014 #10
    Ah I see, it's not that there's an inconsistency then, it's just that one of them can trick you into wanting to differentiate p. Thankyou :)
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