What is the formula for gauss law in dielectric equation?
2 Answers
In the context of dielectrics, Gauss's Law can be expressed in terms of the electric displacement field \(\mathbf{D}\), which accounts for the presence of dielectric materials. The formula for Gauss's Law in a dielectric medium is:
\[ \nabla \cdot \mathbf{D} = \rho_f \]
where:
- \(\mathbf{D}\) is the electric displacement field.
- \(\rho_f\) is the free charge density.
The electric displacement field \(\mathbf{D}\) is related to the electric field \(\mathbf{E}\) and the electric permittivity \(\epsilon\) of the dielectric material by:
\[ \mathbf{D} = \epsilon \mathbf{E} \]
where \(\epsilon\) can be expressed as:
\[ \epsilon = \epsilon_0 \epsilon_r \]
with:
- \(\epsilon_0\) being the permittivity of free space.
- \(\epsilon_r\) being the relative permittivity (dielectric constant) of the material.
So, substituting \(\mathbf{D} = \epsilon \mathbf{E}\) into Gauss's Law gives:
\[ \nabla \cdot (\epsilon \mathbf{E}) = \rho_f \]
This equation describes how the electric displacement field relates to free charge density in the presence of a dielectric material.
\[ \nabla \cdot \mathbf{D} = \rho_f \]
where:
- \(\mathbf{D}\) is the electric displacement field.
- \(\rho_f\) is the free charge density.
The electric displacement field \(\mathbf{D}\) is related to the electric field \(\mathbf{E}\) and the electric permittivity \(\epsilon\) of the dielectric material by:
\[ \mathbf{D} = \epsilon \mathbf{E} \]
where \(\epsilon\) can be expressed as:
\[ \epsilon = \epsilon_0 \epsilon_r \]
with:
- \(\epsilon_0\) being the permittivity of free space.
- \(\epsilon_r\) being the relative permittivity (dielectric constant) of the material.
So, substituting \(\mathbf{D} = \epsilon \mathbf{E}\) into Gauss's Law gives:
\[ \nabla \cdot (\epsilon \mathbf{E}) = \rho_f \]
This equation describes how the electric displacement field relates to free charge density in the presence of a dielectric material.
Gauss's Law in the context of dielectrics can be expressed in terms of the electric displacement field \(\mathbf{D}\) and the electric field \(\mathbf{E}\). The formulation of Gauss's Law for dielectrics incorporates the effects of dielectric materials on the electric field and displacement field.
The formula for Gauss's Law in dielectrics is:
\[ \oint_{\partial V} \mathbf{D} \cdot d\mathbf{A} = Q_{\text{free}} \]
where:
- \(\mathbf{D}\) is the electric displacement field.
- \(d\mathbf{A}\) is the differential area vector on the closed surface \(\partial V\).
- \(Q_{\text{free}}\) is the free charge enclosed within the volume \(V\).
In a dielectric material, \(\mathbf{D}\) is related to the electric field \(\mathbf{E}\) and the polarization \(\mathbf{P}\) of the material by:
\[ \mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} \]
where:
- \(\epsilon_0\) is the permittivity of free space.
- \(\mathbf{P}\) is the polarization field of the dielectric.
Thus, the Gauss's Law can also be expressed in terms of \(\mathbf{E}\) and \(\mathbf{P}\):
\[ \oint_{\partial V} (\epsilon_0 \mathbf{E} + \mathbf{P}) \cdot d\mathbf{A} = Q_{\text{free}} \]
In cases where the dielectric is linear and isotropic, \(\mathbf{P}\) is proportional to \(\mathbf{E}\) by:
\[ \mathbf{P} = \epsilon_0 \chi_e \mathbf{E} \]
where \(\chi_e\) is the electric susceptibility of the dielectric. Thus, the electric displacement field \(\mathbf{D}\) can also be written as:
\[ \mathbf{D} = \epsilon \mathbf{E} \]
where \(\epsilon\) is the permittivity of the dielectric material, \(\epsilon = \epsilon_0 (1 + \chi_e)\).
So, in summary:
- For dielectrics: \(\oint_{\partial V} \mathbf{D} \cdot d\mathbf{A} = Q_{\text{free}}\)
- With linear dielectrics: \(\mathbf{D} = \epsilon_0 \mathbf{E} + \epsilon_0 \chi_e \mathbf{E}\) or \(\mathbf{D} = \epsilon \mathbf{E}\)
The formula for Gauss's Law in dielectrics is:
\[ \oint_{\partial V} \mathbf{D} \cdot d\mathbf{A} = Q_{\text{free}} \]
where:
- \(\mathbf{D}\) is the electric displacement field.
- \(d\mathbf{A}\) is the differential area vector on the closed surface \(\partial V\).
- \(Q_{\text{free}}\) is the free charge enclosed within the volume \(V\).
In a dielectric material, \(\mathbf{D}\) is related to the electric field \(\mathbf{E}\) and the polarization \(\mathbf{P}\) of the material by:
\[ \mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} \]
where:
- \(\epsilon_0\) is the permittivity of free space.
- \(\mathbf{P}\) is the polarization field of the dielectric.
Thus, the Gauss's Law can also be expressed in terms of \(\mathbf{E}\) and \(\mathbf{P}\):
\[ \oint_{\partial V} (\epsilon_0 \mathbf{E} + \mathbf{P}) \cdot d\mathbf{A} = Q_{\text{free}} \]
In cases where the dielectric is linear and isotropic, \(\mathbf{P}\) is proportional to \(\mathbf{E}\) by:
\[ \mathbf{P} = \epsilon_0 \chi_e \mathbf{E} \]
where \(\chi_e\) is the electric susceptibility of the dielectric. Thus, the electric displacement field \(\mathbf{D}\) can also be written as:
\[ \mathbf{D} = \epsilon \mathbf{E} \]
where \(\epsilon\) is the permittivity of the dielectric material, \(\epsilon = \epsilon_0 (1 + \chi_e)\).
So, in summary:
- For dielectrics: \(\oint_{\partial V} \mathbf{D} \cdot d\mathbf{A} = Q_{\text{free}}\)
- With linear dielectrics: \(\mathbf{D} = \epsilon_0 \mathbf{E} + \epsilon_0 \chi_e \mathbf{E}\) or \(\mathbf{D} = \epsilon \mathbf{E}\)