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I am reading a solved exercise about a parallel plate capacitor in which states that the electric flux density between the 2 plates is:

$$D=p_{s}$$

where $p_{s}$ is the surface current density of one plate.

My question is why is this correct? Isn't the previous relationship a boundary condition which is true only in the surface of the plate? Why is this correct for between the plates also?

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    $\begingroup$ Puzzled why (at the time of writing this) two people flagged to close as "homework", yet nobody tagged it as such. I think this question is clearly asking about a principle and falls in the scope of "on topic" questions. $\endgroup$ – Floris Jan 15 '16 at 13:53
  • $\begingroup$ @Floris I can't see the close votes but I can't understand why anyone would vote this as a homework question when I really don't ask anything about homework. $\endgroup$ – Adam Jan 15 '16 at 14:06
  • $\begingroup$ @Floris: Note that it is not required that a VTC as HW on a post have the HW tag in order to do so. $\endgroup$ – Kyle Kanos Jan 15 '16 at 23:56
  • $\begingroup$ @KyleKanos I appreciate that - but wouldn't it be a reasonable thing to do? $\endgroup$ – Floris Jan 16 '16 at 0:01
  • $\begingroup$ @Floris: probably, but sometimes we don't have time beyond a few mouse clicks...or are lazy. $\endgroup$ – Kyle Kanos Jan 16 '16 at 0:11
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In a parallel plate capacitor we assume that the electric field between the plates is uniform, i.e. it doesn't spread out. Given that it doesn't spread out, the electric flux at the surface is the same as the electric flux in between.

In general the electric field is given by $E = \frac{Q}{\epsilon A}$. For a point particle, its electric field spreads out into a sphere, so $A = 4\pi r^2$. Given that $A$ depends on $r$, then the electric flux changes with distance.

However in the case of a uniform field $A$ is constant and for a parallel plate capacitor equal to the area of the capacitor plates. So $E$ doesn't vary with distance between the plates.

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$D$ is the flux on any surface $S$ intersected by one of the plates, divided by the intersected area on the plate $A$. Therefore, the actual flux $\Phi = DA$. In addition, from the Gauss Law, $\Phi = Q$, being $Q$ the net charged enclosed by the surface $S$, which is $p_s A$. Finally $D = p_s$ for any surface $S$ (of course, assuming an uniformly distributed charge density

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