- #1
Ocirne94
- 8
- 0
Hi all,
I'm stuck on this incompatibility within the differential form of Gauss' thearem (or Maxwell's first equation) with dielectrics.
[itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}+\rho_{bound}}{\epsilon_{0}}[/itex]
[itex]\rho_{bound}=-\vec{\nabla}\cdot\vec{P}[/itex]
But with a linear, homogeneous, isotropic dielectric we have
[itex]-\vec{\nabla}\cdot\vec{P}=0=\rho_{bound}[/itex]
And therefore we get
[itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}}{\epsilon_{0}}[/itex] (1)
But using the general formula we have
[itex]\vec{\nabla}\cdot\epsilon_{0}\vec{E} + \vec{\nabla}\cdot\vec{P}=\rho_{free}[/itex]
[itex]\vec{P}=\epsilon_0\cdot\chi_0\vec{E}[/itex]
So ([itex]1+\chi_e=\epsilon_r[/itex], [itex]\epsilon_0\cdot\epsilon_r=\epsilon[/itex])
[itex]\vec{\nabla}\cdot\epsilon\vec{E}=\rho_{free}[/itex],
which means
[itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}}{\epsilon}[/itex]
and is incompatible with (1).
Where is the mistake?
Thank you in advance,
Ocirne
I'm stuck on this incompatibility within the differential form of Gauss' thearem (or Maxwell's first equation) with dielectrics.
[itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}+\rho_{bound}}{\epsilon_{0}}[/itex]
[itex]\rho_{bound}=-\vec{\nabla}\cdot\vec{P}[/itex]
But with a linear, homogeneous, isotropic dielectric we have
[itex]-\vec{\nabla}\cdot\vec{P}=0=\rho_{bound}[/itex]
And therefore we get
[itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}}{\epsilon_{0}}[/itex] (1)
But using the general formula we have
[itex]\vec{\nabla}\cdot\epsilon_{0}\vec{E} + \vec{\nabla}\cdot\vec{P}=\rho_{free}[/itex]
[itex]\vec{P}=\epsilon_0\cdot\chi_0\vec{E}[/itex]
So ([itex]1+\chi_e=\epsilon_r[/itex], [itex]\epsilon_0\cdot\epsilon_r=\epsilon[/itex])
[itex]\vec{\nabla}\cdot\epsilon\vec{E}=\rho_{free}[/itex],
which means
[itex]\vec{\nabla}\cdot\vec{E}=\frac{\rho_{free}}{\epsilon}[/itex]
and is incompatible with (1).
Where is the mistake?
Thank you in advance,
Ocirne