Polarizability – linear charge distribution

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 {R}. That is

\displaystyle  \rho(r)=kr \ \ \ \ \ (1)

We can find the total charge:

\displaystyle   q \displaystyle  = \displaystyle  4\pi\int_{0}^{R}kr\cdot r^{2}dr\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \pi kR^{4} \ \ \ \ \ (3)

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

\displaystyle  \rho(r)=\frac{q}{\pi R^{4}}r \ \ \ \ \ (4)

If the applied field separates the electron cloud and nucleus by a distance {d} 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 {d} so using Gauss’s law we get

\displaystyle   4\pi d^{2}E \displaystyle  = \displaystyle  \frac{4\pi}{\epsilon_{0}}\int_{0}^{d}\frac{q}{\pi R^{4}}r\cdot r^{2}dr\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{qd^{4}}{\epsilon_{0}R^{4}} \ \ \ \ \ (6)

The dipole moment is {p=qd} so we get

\displaystyle  p=R^{2}\sqrt{4\pi\epsilon_{0}qE} \ \ \ \ \ (7)

In this case {p\propto E^{1/2}}.

If we have a radially symmetric charge density, we can work out the condition on {\rho(r)} so that {p\propto E}. In general, from Gauss’s law, we have

\displaystyle  4\pi d^{2}E=\frac{4\pi}{\epsilon_{0}}\int_{0}^{d}\rho(r)r^{2}dr \ \ \ \ \ (8)

Since {d} is very small, we can expand {\rho(r)} in the region {[0,d]} in a Taylor series to get

\displaystyle  \rho(r)=\rho(0)+r\rho'(0)+\ldots  \ \ \ \ \ (9)

In order for {p\propto E}, the leading term in the integral must contain {d^{3}}, which means that we must have {\rho(0)\ne0}. 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.

3 thoughts on “Polarizability – linear charge distribution

  1. Ed

    Hi, 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?

    Reply
    1. growescience

      In order for {ppropto E}, since {p=qd} we must have {E=Ad} where {A} is some constant. From equation 8, the LHS will contain a factor of {d^{3}}, so the leading term in the integral must contain {d^{3}}.
      Using the Taylor series expansion:

      $latex
      displaystyle int_{0}^{d}rho(r)r^{2}dr &fg=000000&bg=e5e4e8$
      displaystyle  = displaystyle  int_{0}^{d}r^{2}left[rho(0)+rrho'(0)+…right]dr
      displaystyle  displaystyle  = displaystyle  frac{d^{3}}{3}rholeft(0right)+frac{d^{4}}{4}rho'left(0right)+…

      so if the leading term is to be proportional to {d^{3}} we must have {rholeft(0right)ne0}.

      Reply

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