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Uniform Field & Poisson equation Mismatch?

  1. Jul 18, 2012 #1
    Hi,

    I'm getting some confusing results and cant figure out what is wrong
    Suppose we have a uniform field

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

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

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

    what am I missing?
     
  2. jcsd
  3. Jul 18, 2012 #2

    gabbagabbahey

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    In the region where the field is uniform, the charge density is zero by Gauss' Law:

    [tex]\mathbf{ \nabla } \cdot \mathbf{E} = \frac{ \partial }{ \partial z }E_z = \frac{ \rho }{ \epsilon_0 }[/tex]
     
  4. Jul 19, 2012 #3
    Ahh I see. So [itex]\rho=0[/itex] is an implicit condition for us to have the uniform field in the first place?
     
  5. Jul 19, 2012 #4

    vanhees71

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    Sure, you have one of Maxwell's (microscopic) equations, saying that
    [tex]\vec{\nabla} \cdot \vec{E}=\rho[/tex]
    (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 [itex]\vec{a}[/itex]. That's of course the corresponding Coulomb potential,
    [tex]\Phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}-\vec{a}|}.[/tex]
    Now let [itex]|\vec{a}| \gg \vec{x}[/itex]. Then you can expand the potential around [itex]\vec{x}=0[/itex]. You find
    [tex]\Phi(\vec{x})=\frac{Q}{4 \pi |\vec{a}|}+\frac{Q \vec{x} \cdot \vec{a}}{4 \pi |\vec{a}|^3}.[/tex]
    Now you obtain the potential for a homogeneous field, by letting [itex]|\vec{a}| \rightarrow \infty[/itex] in such a way that [itex]Q \vec{a}/|\vec{a}|^3=-\vec{E}=\text{const}[/itex]. The constant first term you can subtract beforehand. Then you get
    [tex]\Phi(\vec{x}) \rightarrow -\vec{x} \cdot \vec{E}.[/tex]
    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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