Why is the Electric Field of a Polarized Atom Different in Textbooks?

[Tony Hau]

Homework Statement

A primitive model for an atom consists of point molecules $(+q)$ surrounded by a uniformly charged spherical cloud $(-q)$ of radius $a$ (Fig.1). Calculate the atmoic polarizability of such an atom.

Relevant Equations

$E_{dip}(r,\theta) = \frac {p}{4 \pi \epsilon_o r^{3}}(2cos(\theta)\hat {\mathbf r} + sin(\theta) \hat {\mathbf \theta})$

The question is like this:

The solution is like this:

However, according to the equation for $E_{dip}$ , what I think is that it should be: $$E=\frac {1}{4 \pi \epsilon_o} \frac {qd}{d^3} \hat {\mathbf z} $$, where I take the centre of the sphere in figure 2 as the centre of the coordinate, and positive z-axis towards right.

Actually it doesn't have to be that complicated. The electric field experienced by the positive charge on the left in figure 2 can be simply calculated by the Coulomb's law $E=\frac{1}{4 \pi \epsilon_o}\frac{-q}{r^2}$. Anyway, I don't know why the textbook gives something different.

 

[Abhishek11235]

The electric field experienced by the positive charge on the left in figure 2 can be simply calculated by the Coulomb's law $E=\frac{1}{4 \pi \epsilon_o}\frac{-q}{r^2}$.

Really? It is a uniformly charged sphere and we are calculating the electric field inside the sphere!

 

[Tony Hau]

Really? It is a uniformly charged sphere and we are calculating the electric field inside the sphere!

I suppose we can treat the electron cloud as a point charge, just like what we do for centre of mass? If that's not the case, why would the author draw two points inside the sphere?