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The van der Waals force is a well-understood physical phenomena, as are electrostatics. When particles get down to a nanoscale size and separation, van der Waals forces can be very strong and become a dominant force for causing particles to adhere to each other to surfaces or each other.

When particles are charged though, we instead solve for the electrostatic potential field and use that to determine the force on them. But in this case, are van der Waals forces still present? After all, van der Waals forces themselves are electrostatic phenomena - caused by induced dipole-dipole reactions between the two objects.

I know that in the case of DLVO theory that the electrostatic and van der Waals effects are simply added together linearly. However can this always be done?

Let's say that we have two small charged particles near each other, something like 50nm in diameter and separated by 1nm (close enough that van der Waals force is significant even if there is no net charge on either particle). You can solve for the electrostatic force on each particle by solving for the electrostatic potential field $\nabla^2\phi=\frac{\rho}{\varepsilon_0}$ with appropriate boundary conditions. It's of course complicated for anything but the most simple of geometries, but we can always use a finite element or other numerical method to get a numerical solution.

The real question though is, will there still be a separate van der Waals force between these two spheres, or does solving for the electrostatic potential account for that as well? The van der Waals force between these two particles can be simplified when the separation distance $h$ is much smaller than the particle radii $R_!$ and $R_2$:

$$F_{vdw}=\frac{A R_1 R_2}{6(R_1+R_2)h^2}$$

Where $A$ is the Hamaker constant.

So can I just add the $F_{vdw}$ to the electrostatic force $F_{es}$, or is there a more subtle relationship between the two?

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  • $\begingroup$ It depends on how you model the atoms. If you model them as point charges then the van der Waals interaction does not naturally arise and needs to be added on explicitly as a correction. $\endgroup$ – lemon Aug 30 '16 at 16:28
  • $\begingroup$ Depends on the characteristics of your system. If you're dealing with a very large array of particles modelled as points, then you'll get good results using statistical mechanics as if the particles were non-interacting and only later adding the dipole-dipole interactions as a correction (I just noticed @lemon just said that). If you have only two atoms in a molecule and want to study that system, then you'll use quantum mechanics and the dipole-dipole interaction will already be present in your Hamiltonian. $\endgroup$ – QuantumBrick Aug 30 '16 at 16:33
  • $\begingroup$ @lemon, for my example I said 50nm particles that are 1nm apart. Obviously in that case modeling them as point charges is not the best way to model them. $\endgroup$ – Derek Aug 30 '16 at 16:49

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