# Displacement vector in parallel plate capactor

This ought to be simple, but I'm running into some questions... Let's say we have a parallel plate cap with some linear homogeneous dielectric media between the plates. The plates are distance $a$ apart with the top plate holding charge $\sigma$ and the bottom plate $-\sigma$.

I understand that I need to employ Gauss' law in the form

$$\int \vec{D} \cdot d\hat{a} = Q_{free}$$ $$D \pi r^2 = \pi r^2 \sigma$$ $$D = \sigma$$

And I understand that the displacement vector must point in the same direction as the electric field emanating from $+\sigma$, let's say down -$\hat{z}$

$$\vec{D} = -\sigma\hat{z}$$

This apparently is my answer. But I don't get why I stop here. What about the bottom plate with $-\sigma$? Wouldn't Gauss's law on the bottom surface give

$$D \pi r^2 = \pi r^2 (-\sigma)$$ $$D = -\sigma$$ $$\vec{D} = (-\sigma)(-\hat{z}) = \sigma\hat{z} ?$$

Here I'm compelled the add the vectors giving $\vec{D} = \vec{0}$, which I know is wrong.

I'm hoping someone can spot the flaw in my thinking and give a few words of guidance.

When using Gauss's law to calculate the flux through a closed surface we take the field component in the normal direction of the surface. The normal direction always points outward for the closed surface.

Going by the math, it seems like you're making a Gaussian cylinder that encloses one of the plates. The difference between making a cylinder that encloses the top plate with charge $\sigma$ and the bottom plate with charge $-\sigma$ is that the surface normal of the face inside the capacitor will point in different directions.

Say the top plate is $\sigma$ and the bottom one is $-\sigma$. If you enclose the top plate, then the normal direction of the surface inside the capacitor will be in the $-\hat{z}$ direction. If you enclose the bottom plate, the normal direction of the surface inside the capacitor will be in the $\hat{z}$ direction. This difference in the normal direction of your closed surface will change the sign of the flux integral.

• Got it. That explains why using either the top or bottom plate gives $\vec{D}=-\sigma\hat{z}$. Could you explain why I don't want to sum this result from both the top and bottom plate to get $\vec{D}=-2\sigma\hat{z}$? – suneater Apr 9 '15 at 6:52
• Summing is not necessary because although the Gaussian surface you make only encloses one side of the capacitor, it takes into account the electric field created by both plates. Only by taking into account both plates can we say that the $D$ outside the capacitor is zero. If you summed $D$ while assuming it's zero outside the capacitor you're double counting the contribution. – MonkeysUncle Apr 10 '15 at 0:42
• Only in the case where you treat each plate individually would you add the results together. It's like how a single charged plate with distribution $\sigma$ gives a symmetric field with strength $\frac{\sigma}{2\epsilon_0}$, and then two plates with $\pm \sigma$ will cancel outside and give $\frac{\sigma}{\epsilon_0}$ inside. – MonkeysUncle Apr 10 '15 at 0:53