Problem with understanding polarization

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neworder1
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Suppose we have a ball made of linear dielectric with permittivity [tex]\epsilon[/tex], with some initial homogenous polarization [tex]\vec{P}[/tex] aligned with z axis. Then we know that inside the ball the polarization generates an electric field [tex]\vec{E}=\frac{-1}{3\epsilon_{0}}\vec{P}[/tex] (standard calculation). But we also know that in a linear dielectric we have the relation [tex]\vec{P}=(\epsilon - \epsilon_{0})\vec{E}[/tex], and these two equations lead to contradiction since we have [tex]\vec{E}=\frac{-1}{3\epsilon_{0}}\vec{P} = \frac{-1}{3\epsilon_{0}}(\epsilon - \epsilon_{0})\vec{E}[/tex]. Does it mean that a linear dielectric can't be polarized this way without an external field?
 
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neworder1 said:
Suppose we have a ball made of linear dielectric with permittivity [tex]\epsilon[/tex], with some initial homogenous polarization [tex]\vec{P}[/tex] aligned with z axis. Then we know that inside the ball the polarization generates an electric field [tex]\vec{E}=\frac{-1}{3\epsilon_{0}}\vec{P}[/tex] (standard calculation). But we also know that in a linear dielectric we have the relation [tex]\vec{P}=(\epsilon - \epsilon_{0})\vec{E}[/tex], and these two equations lead to contradiction since we have [tex]\vec{E}=\frac{-1}{3\epsilon_{0}}\vec{P} = \frac{-1}{3\epsilon_{0}}(\epsilon - \epsilon_{0})\vec{E}[/tex]. Does it mean that a linear dielectric can't be polarized this way without an external field?

Yeah, of course! If there is no field, and the object is polarized... it's not a linear material.