# I The electric field inside and outside of a dielectric

#### Pushoam

I bring a dielectric in a region with electric field $\vec E_0$. Net electric field $\vec E_{net} = \vec E_0 + \vec E_p$ , where $\vec E_p$ is electric field due to polarization of dielectric.
For linear dielectric, $\vec E_p$ is 0 outside the dielectric. So,
$\vec E_{net} = \vec E_0$ outside the dielectric ,
$\vec E_{net} = \epsilon_r \vec E_0$ inside the dielectric , $\epsilon_r$ is dielectric constant.

Is this correct?

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#### vanhees71

Gold Member
It's not that easy. You have to solve the corresponding electrostatic boundary problem, i.e.,
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=\rho_{\text{f}}$$
with
$$\vec{t} \cdot (\vec{E}_>-\vec{E}_<)=0, \quad \vec{n} \cdot (\vec{D}_{>}-\vec{D}_{<})=\sigma_\text{f}.$$
Here $\vec{t}$ is any tangent vector along the surface, $\vec{n}$ the normal vector along the surface, and $\rho_{\text{f}}$ and $\sigma_{\text{f}}$ are the free charge density and surface-charge density along the surface, respectively. The $\vec{E}_0$ is not simply undisturbed outside the dielectric, but due to the polarization of the dielectric it contributes also to the field outside.

As an example take a dielectric sphere of radius $a$ in a field that becomes asymptotically homogeneous, i.e.,
$$\vec{E}(\vec{x}) \rightarrow_{|\vec{x}| \rightarrow \infty} E_0 \vec{e}_3.$$
Here we have $\rho_{\text{f}}=0$ and $\sigma_{\text{f}}=0$.

We'll work in spherical coordinates. Because $\vec{E}$ is vortex free, there's a potential
$$\vec{E}=-\vec{\nabla} \Phi.$$
From symmetry it's clear that in the multipole expansion we have only $m=0$ contributions (nothing depends on $\varphi$). The asymptotics tells us that
$$\Phi(\vec{x}) \rightarrow_{|\vec{x}| \rightarrow \infty} =-E_0 x_3=-E_0 r \cos \vartheta.$$

Now this suggests that we can get along with the $\ell=1$ (dipole), $m=0$ fields. Since the Legendre polynomial $P_1(\cos \vartheta)=\cos \vartheta$ we have the ansatz
$$\Phi(\vec{x})=\left (A r + \frac{B}{r^2} \right) \cos \vartheta$$
with constants $A$ and $B$ that are different insight the sphere, i.e., for $r<a$ (permittivity $\epsilon=\epsilon_{\text{rel}} \epsilon_0$) and outside the sphere (permittivity $\epsilon_0$).

Inside the sphere there's no singularity at $r=0$. This means we have
$$\Phi_{<}(\vec{x})=A_{<} r \cos \vartheta=A_{<} x_3.$$
and
$$\Phi_{>}(\vec{x})=\left (A_{>} r + \frac{B_{>}}{r^2} \right) \cos \vartheta = A_{>} x_3 + \frac{B_{>} x_3}{r^3}.$$
Now we have to fulfill the boundary conditions. From the asymptotics at infinity we have $A_{>}=-E_0$. It's easier to work in Cartesian coordinates from now on, which is why I already expressed the potential in terms of the Cartesian coordinates (using $r=\sqrt{x_1^2+x_2^2+x_3^2}$).

The electric field is given by
$$\vec{E}_{<}(\vec{x})=-\vec{\nabla} \Phi_{>}(\vec{x})=-A_{<} \vec{e}_3,\\ \vec{E}_{>}(\vec{x}) = E_0 \vec{e}_3 + \frac{B_{>} (-r^2 \vec{e}_3 +3 x_3 \vec{x})}{r^5}.$$
Now all tangential components of $\vec{E}$ must be continuous along the boundary $r=a$. The normal unit vector along the sphere is $\vec{n}=\vec{x}/a$ and we must have $\vec{n} \times (\vec{E}_{>}-\vec{E}_{<})=0$, which leads to
$$\vec{n} \times \vec{E}_{<}=-A_{<} \frac{\vec{x}}{a} \times \vec{e}_3=\vec{n} \times \vec{E}_>= \frac{\vec{x}}{a} \times \vec{e}_3\left (E_0-\frac{B_>}{a^3} \right).$$
$$\frac{B_>}{a^3}-E_0=A_{<}.$$
Since further $\sigma_{\text{f}}=0$, also the normal components of $\vec{D}$ must be continuous, i.e.,
$$\frac{\vec{x}}{a} \cdot \epsilon \vec{E}_{<}=\frac{\vec{x}}{a} \epsilon_0 \vec{E}_{>}.$$
$$-A_{<} \epsilon \frac{x_3}{a}=\frac{x_3}{a} \epsilon_0 \left (E_0+\frac{2 B_{>}}{a^3} \right)$$
or
$$E_0 + \frac{2 B_{>}}{a^3}=-\epsilon_{\text{rel}} A_{<}.$$
Solving the system of linear equations for $A_<$ and $B_>$ finally yields
$$A_<=-\frac{3 E_0}{2+\epsilon_{\text{rel}}}, \quad B_>=E_0 a^3 \frac{\epsilon_{\text{rel}}-1}{2+\epsilon_{\text{rel}}}.$$
So outside you have a superposition of a homogeneous field and an induced dipole field,
$$\vec{E}_>=E_0 \vec{e}_3 +E_0 a^3 \frac{\epsilon_{\text{rel}}-1}{2+\epsilon_{\text{rel}}} \frac{3 x_3 \vec{x}-r^2 \vec{e}_3}{r^5}$$
and inside the homogenous field
$$\vec{E}_<=\frac{3 E_0}{2+\epsilon_{\text{rel}}} \vec{e}_3.$$

#### Pushoam

Thanks.

"The electric field inside and outside of a dielectric"

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