Uniform Field & Poisson equation Mismatch?

AI Thread Summary
The discussion centers on the mismatch between the Laplacian of the electrostatic potential in a uniform electric field and the Poisson equation. The potential derived from a uniform field leads to a Laplacian of zero, while the Poisson equation suggests a non-zero charge density. It is clarified that a uniform electric field implies zero charge density, as per Gauss' Law. The conversation highlights that a truly uniform field is unphysical and can only be approximated in specific scenarios, such as between large charged plates. Ultimately, the uniform field condition necessitates an implicit assumption of zero charge density.
Apteronotus
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Hi,

I'm getting some confusing results and can't figure out what is wrong
Suppose we have a uniform field

E=[0,0,E_z] in a dielectric media.

By E=-\nabla\psi we can deduce that \psi(x,y,z)=-z E_z

But, taking the Laplacian
\nabla^2\psi=\frac{\partial^2 (-zE_z)}{\partial z^2}=0
does not match the results of the Poisson equation
\nabla^2\psi=-\frac{\rho}{\epsilon_m \epsilon_o}

what am I missing?
 
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In the region where the field is uniform, the charge density is zero by Gauss' Law:

\mathbf{ \nabla } \cdot \mathbf{E} = \frac{ \partial }{ \partial z }E_z = \frac{ \rho }{ \epsilon_0 }
 
Ahh I see. So \rho=0 is an implicit condition for us to have the uniform field in the first place?
 
Sure, you have one of Maxwell's (microscopic) equations, saying that
\vec{\nabla} \cdot \vec{E}=\rho
(Heaviside-Lorentz units). This means that a homogeneous electric field necessarily leads to 0 charge density.

Of course, in nature there is no such thing as a global homogeneous field. You get such a field only in a quite unphysical situation. To that end consider the electrostatic potential of a point charge Q at rest at the position \vec{a}. That's of course the corresponding Coulomb potential,
\Phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}-\vec{a}|}.
Now let |\vec{a}| \gg \vec{x}. Then you can expand the potential around \vec{x}=0. You find
\Phi(\vec{x})=\frac{Q}{4 \pi |\vec{a}|}+\frac{Q \vec{x} \cdot \vec{a}}{4 \pi |\vec{a}|^3}.
Now you obtain the potential for a homogeneous field, by letting |\vec{a}| \rightarrow \infty in such a way that Q \vec{a}/|\vec{a}|^3=-\vec{E}=\text{const}. The constant first term you can subtract beforehand. Then you get
\Phi(\vec{x}) \rightarrow -\vec{x} \cdot \vec{E}.
As you see you have to use an infinite charge at infinite distance to make a homogeneous electric field everywhere in space. That's a rather unphysical situation.

In practice you get a quite good approximation of a homogenous electric field between two large charged plates at small distance, in the region in the middle between the plates.
 
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