# The electric field inside and outside of a dielectric

• Pushoam
In summary, when a dielectric is present in a region with an electric field of ##\vec E_0##, the net electric field becomes ##\vec E_{net} = \vec E_0 + \vec E_p##, where ##\vec E_p## is the electric field due to the polarization of the dielectric. For a linear dielectric, this polarization field is 0 outside the dielectric, so the net field remains ##\vec E_{net} = \vec E_0## outside the dielectric. Inside the dielectric, the net field becomes ##\vec E_{net} = \epsilon_r \vec E_0##, where ##\epsilon_r## is the 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?

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.,
$$\frac{\vec{x}}{a} \cdot \epsilon \vec{E}_{<}=\frac{\vec{x}}{a} \epsilon_0 \vec{E}_{>}.$$
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
$$\vec{E}_<=\frac{3 E_0}{2+\epsilon_{\text{rel}}} \vec{e}_3.$$

Pushoam
Thanks.

The statement you made that outside the sphere the field is a superposition of a homogeneous field and an induced dipole is right but I believe you are omitting the x and y components. I do not quite see why you drop them.

Even for the simplest case, (infinite parallel plate capacitor, maybe) the net field decreases and not increases when the dielectric is present. So, you need to divide by the dielectric constant and not multiply by it.

vanhees71 said:
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.,
$$\frac{\vec{x}}{a} \cdot \epsilon \vec{E}_{<}=\frac{\vec{x}}{a} \epsilon_0 \vec{E}_{>}.$$
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
$$\vec{E}_<=\frac{3 E_0}{2+\epsilon_{\text{rel}}} \vec{e}_3.$$

I really have to emphasize that the field outside the sphere has an x and y component, so that this answer is incomplete unless I am interpreting it wrong. If anyone is interested I can post my own answer and take the time to LateX it.

djamil said:
The statement you made that outside the sphere the field is a superposition of a homogeneous field and an induced dipole is right but I believe you are omitting the x and y components. I do not quite see why you drop them.

@vanhees71's result for outside the sphere is
vanhees71 said:
So outside you have a superposition of a homogeneous field and an induced dipole field,
$$\vec{E}_>=E_0 \vec{e}_3 +E_0 a^3 \frac{\epsilon_{\text{rel}}-1}{2+\epsilon_{\text{rel}}} \frac{3 x_3 \vec{x}-r^2 \vec{e}_3}{r^5}$$
Note the presence in the last term of the position vector ##\vec x## which has x and y components.

vanhees71

## 1. What is a dielectric?

A dielectric is a material that does not conduct electricity. It is often used as an insulator in electronic devices.

## 2. How does a dielectric affect the electric field?

When placed in an electric field, a dielectric polarizes and creates an opposing electric field. This reduces the overall electric field strength inside the material.

## 3. What is the difference between the electric field inside and outside of a dielectric?

The electric field inside a dielectric is weaker than the electric field outside due to the polarization of the material. The electric field outside is not affected by the presence of the dielectric.

## 4. How does the dielectric constant affect the electric field?

The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to store electrical energy. A higher dielectric constant means the material can polarize more easily, resulting in a stronger opposing electric field and a weaker overall electric field inside the material.

## 5. Can the electric field inside a dielectric be completely eliminated?

No, the electric field inside a dielectric cannot be completely eliminated. The material will always polarize to some degree and create an opposing electric field. However, the electric field can be significantly reduced depending on the properties of the dielectric and the strength of the external electric field.

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