I The electric field inside and outside of a dielectric

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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?
 

vanhees71

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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
$$\vec{n} \times \vec{E}_{<}=-A_{<} \frac{\vec{x}}{a} \times \vec{e}_3=\vec{n} \times \vec{E}_>= \frac{\vec{x}}{a} \times \vec{e}_3\left (E_0-\frac{B_>}{a^3} \right).$$
This leads to
$$\frac{B_>}{a^3}-E_0=A_{<}.$$
Since further ##\sigma_{\text{f}}=0##, also the normal components of ##\vec{D}## must be continuous, i.e.,
$$\frac{\vec{x}}{a} \cdot \epsilon \vec{E}_{<}=\frac{\vec{x}}{a} \epsilon_0 \vec{E}_{>}.$$
Plugging this in leads to
$$-A_{<} \epsilon \frac{x_3}{a}=\frac{x_3}{a} \epsilon_0 \left (E_0+\frac{2 B_{>}}{a^3} \right)$$
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
$$\vec{E}_<=\frac{3 E_0}{2+\epsilon_{\text{rel}}} \vec{e}_3.$$
 
890
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Thanks.
 

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