**Required math: calculus**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Chapter 4, Post 3.

In an earlier post, we did a crude estimate of the separation between electron and proton in hydrogen due to polarization in an electric field. In that case, we assumed that the electron cloud was a uniformly charged sphere, and the result was that the induced dipole moment is proportional to the applied electric field.

Although this is also the result experimentally, it is possible to construct artificial cases where this relation isn’t true. For example, suppose that the electron cloud is a sphere where the charge

density is proportional to the distance from the centre out to

a distance . That is

We can find the total charge:

We can rewrite the charge density in terms of the total charge:

If the applied field separates the electron cloud and nucleus by a distance then we can do the same calculation as in the earlier post to get a relation between the dipole moment and applied field. The nucleus must feel an equal and opposite field when it moves out the distance so using Gauss’s law we get

The dipole moment is so we get

In this case .

If we have a radially symmetric charge density, we can work out the condition on so that . In general, from Gauss’s law, we have

Since is very small, we can expand in the region in a Taylor series to get

In order for , the leading term in the integral must contain , which means that we must have . This is true of the uniformly charged sphere and also the exponential distribution predicted by quantum mechanics, but obviously not for the case in the current example.

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EdHi, what do you mean by “the leading term in the integral must containing d^3”. Which integral? And how does it relate with rho(0) not equal to 0?

growescienceIn order for , since we must have where is some constant. From equation 8, the LHS will contain a factor of , so the leading term in the integral must contain .

Using the Taylor series expansion:

so if the leading term is to be proportional to we must have .

EdThanks for the quick reply. You have great explanations here! 🙂