# Electric field produced by a uniformly polarized sphere

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?

• 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. Nov 17, 2023 at 18:25
• 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. Nov 18, 2023 at 17:52

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:

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

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 $$\vec{D}$$).