I am thinking about the classic problem of a uniformly polarized sphere, within which the polarization is in $z$ direction.

I've been trying to find the electric field inside the uniformly polarized sphere using the electric displacement $\vec{D} = \varepsilon_{0}\vec{E} + \vec{P}$. By using the "Gauss Law" for electric displacement $\oint \vec{D} \cdot d\vec{A} = Q_{\text{free}}$, $\vec{D}$ is $0$ because there is no free charge inside the material. Hence, I have $$ \varepsilon_0\vec{E} + \vec{P}=0$$ from the above equation the electric field inside the sphere is $\vec{E}= -\frac{\vec{P}}{\varepsilon_0}$. But the correct answer is $\vec{E}=-\frac{\vec{P}}{3\varepsilon_0}$. Can anyone please explain this inconsistency?

  • $\begingroup$ You aren't considering the fact that you have a sphere. One way to approach this would be to calculate the bound charge, which would be non-zero at the surface. Then calculate the electric field within that shell of charge (and I'm guessing you've done a similar problem before). Also, the policy on this site is to not solve homework-like problems, so this question may get closed. A conceptual question such as "what's an approach to calculating the electric field within a polarized volume" might be better received. $\endgroup$
    – Gilbert
    Nov 17, 2023 at 18:25
  • $\begingroup$ I calculated the electric field inside the sphere by considering the surface-bound charge density, but I wanted to try a different approach by first calculating the Displacement vector first and then finding the electric field. $\endgroup$ Nov 18, 2023 at 17:52

2 Answers 2

  1. You don't need to specify the $z$ direction.

  2. To get a uniform polarization in a linear dielectric, you need to immerse it in a uniform external field, $E$, where $E$ is the value far from the field.

When the sphere is present, there is polarization so that $E$ in the sphere is modified, and:

$$ E_{in} = E + E_P $$

That should fix your problem.

You can also just calculate $E_{in}$ from the non-free charges with:

$$ \rho(\vec r) \propto \nabla \cdot \vec P $$

and at the surface of the sphere

$$ \nabla \cdot \vec P \propto \vec z \propto Y_1^0(\theta, \phi) $$

which implies an interior dipole field (which is uniform) and external dipole field superimposed on $E$


The macroscopic Gauss's Law for $\vec{D}$ is not enough to solve this problem. In a (static) system with the divergence equation $\vec{\nabla}\cdot\vec{E}=0$, you can conclude that $\vec{E}$ is a constant, since the curl equation $\vec{\nabla}\times\vec{E}=0$ is also known. Hence, by Helmholtz's Theorem, $\vec{E}$ is zero if if vanishes at spatial infinity. However, an analogous argument does not hold for $\vec{D}$, since $\vec{D}$ is not a curl-free (that is, conservative) vector field. In general, $\vec{\nabla}\times\vec{D}=\vec{\nabla}\times\vec{P}$, and while $\vec{\nabla}\times\vec{P}=0$ inside the sphere, it is singular (meaning it is a $\delta$-function) at the surface of the sphere.

Because, in general, $\vec{\nabla}\times\vec{D}\neq0$, using $\vec{D}$ to calculate the fields is not useful in most situations in which there is not a planar, cylindrical, or sphereical symmetry (in which cases, the symmetry ensures $\vec{\nabla}\times\vec{D}=0$). It is more useful to work with the conservative field $\vec{E}=-\vec{\nabla}\Phi$, solving for the scalar potential $\Phi$ using a method such as separation of variables. To solve the problem that way, rather than by introducing $\vec{D}$, you should calculated the bound charges $\rho_{b}$ and $\sigma_{b}$ from the polarization $\vec{P}$. In this case, $\rho_{b}=0$ and $\sigma_{b}=P\cos\theta$, which makes for an easy separation of variables solution (with only $\ell=0$ terms).

To see concretely why this more general method is necessary, note any constant $\vec{E}=E_{0}\hat{n}$ inside the sphere would satisfy $\vec{\nabla}\cdot\vec{E}=0$ (or equivalently $\oint d\vec{S}\cdot\vec{E}=0$) inside the sphere. So Gauss's Law cannot possibly be enough to fix $\vec{E}$ (or $\vec{D}$).


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