I have recently become curious about modeling the repulsion of everyday objects in contact with one another. By repulsion I mean as you attempt to walk through a wall, the pain in your nose suddenly alerts you to the fact that it won't be possible. I've come up with this exercise:

What is the mutual pressure exerted by two dipole sheets of finite extent, each area $A$, separated by a distance $D$? Assume $D \gg$ d where the dipole moment $p = ed$ and $e = 1.6\times10^{-19}$. There are $N$ dipoles per unit area. Make an order of magnitude estimate for $N$ to represent a density one would find in ordinary matter. Make a numerical plot for $10^{-10} < D < 10^{-3}$ meters.

This is a crude electrostatic model of two electrically neutral materials brought close together. The oppositely oriented dipole sheets are generated by the electronic cloud repulsion and deformation as the two interfaces approach. It is meant to serve as a quantitative estimate of the maximum distance at which Coulomb repulsion could possibly stop one body from passing through another. Interface physics is pretty rich, but I wonder how this model does in predicting such pressures.

If anyone has related references please post them, especially any experimental studies.

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    $\begingroup$ This sounds an awful lot like a homework problem... if it is, could you add the homework tag? (Homework questions are fine here, we just want them to be identified as such) And if it isn't a homework problem, I'd suggest rewording it so that it doesn't give that impression. $\endgroup$
    – David Z
    Commented Dec 8, 2010 at 5:44
  • $\begingroup$ This is supposed to be a reaction to my discussion with Pete about net electrostatic force between two lattices. My toy model was a one-atom thick layer of $+$ and $-$ charges arranged in checkerboard pattern (which is quite crude but should still give us some information; you can add more layers though). But this question seems to be even cruder, so it's irrelevant to the above. The reason is that the answer in my model depends on how exactly you align $+$ and $-$ of the two lattices. $\endgroup$
    – Marek
    Commented Dec 8, 2010 at 11:39
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    $\begingroup$ For $D$ much larger than the typical distance between dipoles ($1/\sqrt{N}) we can consider the two dipole sheets to be made of a continuous 'dipole density', that is, two sheets of opposite charge per sheet of dipoles. Then there should be no force at all, because these sheets are like capacitor plates (no field outside). Correct? Shorter distances are of course more complicated and depend on details. $\endgroup$
    – Greg P
    Commented Dec 8, 2010 at 20:42
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    $\begingroup$ Even for infinite sheets, there is a nonzero force at small distances of the order of 1/sqrt(N). Because instead of a sheet of dipole density, you might model it (as a very first guess) as a surface with sinusoidally varying dipole density. Then there will indeed be a field outside the sheet, but it will decay within a distance of order 1/sqrt(N) from the surface. I have the calculation somewhere...of course, all of this is just the electrostatic contribution. It is a very complicated problem at small scales! $\endgroup$
    – Greg P
    Commented Dec 8, 2010 at 23:41
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    $\begingroup$ I see, Greg, the effect you are speaking of arises from the granularity of the atoms/molecules which indeed would remain for infinite sheets $\endgroup$
    – Pete
    Commented Dec 9, 2010 at 1:10

1 Answer 1


I'll take a shot at it with a few assumptions. Since you're talking about contact between macroscopic objects, and the area of the sheets represents the area of contact, we can assume $R>>D>>d$. As has been mentioned in the comments, we cannot extrapolate to infinite sheets or there won't be an electric field.

The method here will be to keep the first sheet fixed, and estimate the potential energy of the second sheet as a function of $D$. The derivative $\frac{dU}{dD}$ will give us the force suffered by it.

Placing the fixed sheet on the x-y plane, with its center on the origin, we have to start by calculating its electric field. We'll restrict ourselves to calculating the electric field it generates on a point that lies on the $z$ axis, a distance $D$ from its center (justified later). For that, we'll assume the sheet is disk-shaped, and treat this dipole disk as a pair of charged disks separated by a distance $d$ that coincide with the planes $z=\pm d/2$. The electric field generated by a charged disk is easy to find on textbooks (and easy to integrate as well), being

$$ E(z) = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{z}{\sqrt{R^2+z^2}}\right).$$

Rewriting it to suit our dipole:

$$ E_T(z) = E(z+d') - E(z-d') = \frac{\sigma}{2\epsilon_0}\left(- \frac{z+d'}{\sqrt{R^2+(z+d')^2}} + \frac{z-d'}{\sqrt{R^2+(z-d')^2}}\right),$$ where $d'=d/2$ and $\sigma = eN$. We could use a couple approximations on the above equation, but they're unnecessary right now. The point is that we know what is the electric field that the fixed sheet generates on the center of the second sheet.

We'll now have to use the approximations. This result is the electric field that's felt by a single dipole on the second sheet. If the sheets were infinite, symmetry would imply that all dipoles on the second sheet are under the same (constant) electric field, and the potential energy per unit area of the second sheet would be $U=\vec{E}_T\cdot\vec{p}\,N = E_T(D)p\,N$. Since $R>>D$, the border effects won't be too relevant, and we'll assume that $E_T$ is constant along the entire second sheet (much like one does inside capacitors). That leads to the total potential energy of the second sheet $$U(D)=\frac{(pN)^2}{2\epsilon_0d}\left(- \frac{D+d'}{\sqrt{R^2+(D+d')^2}} + \frac{D-d'}{\sqrt{R^2+(D-d')^2}}\right)\pi R^2,$$ as well as the potential energy per unit area $$u(D)=\frac{(pN)^2}{2\epsilon_0d}\left(- \frac{D+d'}{\sqrt{R^2+(D+d')^2}} + \frac{D-d'}{\sqrt{R^2+(D-d')^2}}\right).$$

Taking the first derivative with respect to $D$, will give us the force and the pressure (respectively) that the fixed sheet applies on the second sheet.


Here go the answers:

$$F=\frac{\pi(RpN)^2}{2\epsilon_0d}(G_- - G_+)$$

$$P=\frac{(pN)^2}{2\epsilon_0d}(G_- - G_+)$$

$$G_\pm = (R^2 + (D\pm d/2)^2)^{-.5} - (D\pm d/2)^2(R^2 + (D\pm d/2)^2)^{-1.5}.$$

From what I can tell, these are the Force and Pressure that the question asks for. Below are a couple of plots for $G_--G_+$ (measured in $1/m$), just multiply their $y$ values by $\frac{(pN)^2}{2\epsilon_0d}$ to get the pressure. The $x$ axis is $D$ measured in $m$.

This plot uses $R=1mm$. Green line is $d=1nm$, Red line is $d=10nm$. Plot 1

This plot uses $R=1cm$. Green line is $d=1nm$, Red line is $d=10nm$. Plot 2

  • $\begingroup$ Thanks Bruce, good thinking on use of the charged disk. I look forward to your numerical estimates. I don't see how the R dependence is at all problematic. Since that is the source of the field, of course there will be less pressure with less source. The question is, how much pressure does this model give for a square mm or square cm of source. I'm thinking p should be something like the value for water, and then an order of magnitude greater and less (3 cases). $\endgroup$
    – Pete
    Commented Dec 17, 2010 at 15:38
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    $\begingroup$ @Pete: The only problem I have is that pressure will decrease as $R$ increases. The only think I need to do the plots is an estimate for $d$ or for $p$, I have no idea what the value for water is. =) $\endgroup$
    – Malabarba
    Commented Dec 17, 2010 at 16:19
  • $\begingroup$ @Pete: Plots done. $\endgroup$
    – Malabarba
    Commented Dec 17, 2010 at 17:18
  • $\begingroup$ What are y-axis units? I see what you were saying about R now. This is what Greg P was pointing out, that the granularity of matter must be taken into account else things will look like the infinite case when you get too close. But, interesting the smaller R produced a higher (x100) peak pressure with this model. $\endgroup$
    – Pete
    Commented Dec 17, 2010 at 17:52
  • $\begingroup$ @Pete: Indeed. The graph uses SI on both axis, so just multiply it by $\frac{(pN)^2}{2\epsilon_0 d}$ in SI units as well and you'll get the pressure in $N/m^2$. The $x$ axis is in meters, and the $y$ axis is in $m^{-1}$ (inverse meters). $\endgroup$
    – Malabarba
    Commented Dec 17, 2010 at 18:12

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