Paralle plate capacitor (with dielectric)

[FrogPad]

EDIT: I figured it out.

Homework Statement

A parallel-plate capacitor of width [itex] 2w [/itex], length [itex] L [/itex] and separation [itex] d [/itex] is partially filled with dielectric material of [itex] \epsilon_0 \epsilon_r [/itex]. A voltage [itex] V_0 [/itex] is applied.

1. Find [itex] \vec D [/itex], [itex] \vec E [/itex], [tex] \rho_S [/itex]

================ -------
---------|---------|--------|
---------|---------|-------V0
---------|---------|--------|
================--------
NOTE: The dielectric is on the right side (I colored it red)

Homework Equations

Note: Book convention uses [itex] \rho [/itex] as charge density.

Boundary Conditions:
[tex] B_{1t} = B_{2t} [/tex]
[tex] D_{1n} - D_{2n} = \rho_S [/tex]

[tex] \vec E = \frac{\vec D}{\epsilon} [/tex]

Ideal Parallel Plate:
[tex] \vec E = \hat e \frac{V_{12}}{d} [/tex]

The Attempt at a Solution

We first assume the use of ideal conditions, thus the expressions above hold.

Next we note that the E field is directed from the top plate to the bottom plate. So we have,

[tex] \vec E_1 = -\hat z \frac{V_{12}}{d} =-\hat z \frac{V_{0}}{d}[/tex]
[tex] \vec E_2 = -\hat z \frac{V_{0}}{d} [/tex]

The only component of the E fields are tangential components. ie,
[tex] \vec E_1 = -\hat z E_t + \hat e E_n [/tex]
Where: [itex] \hat e [/itex] is a basis vector in some direction, and [itex] E_n = 0[/itex]

I'm basically trying to say that the E fields consist of no normal components to the interface.

Thus we satisfy our first BC with [itex] \vec E_1 = \vec E_t [/itex].

Moving on to the second BC we note:
[tex] \vec D_1 = \epsilon \vec E_1 = \epsilon_0 \vec E_1 [/tex]
[tex] \vec D_2 = \epsilon \vec E_2 = \epsilon_0 \epsilon_r \vec E_2 [/tex]

[tex] \vec D_1 = \epsilon_0 \left(-\hat z \frac{V_{-}}{d} \right) [/tex]
[tex] \vec D_2 = \epsilon_0 \epsilon_r \left(-\hat z \frac{V_{0}}{d} \right) [/tex]

I'm confused here (and probably above too).

So,
[tex] \vec D_1 - \vec D_2 = -\hat z \frac{\epsilon_0 V_0}{d} (1-\epsilon_r) [/tex]
Now the normal component is [itex] \hat -x [/itex] (is this right?). (The coordinate system I'm using is x is left to right, y is in and out of page, and z is up and down.)

So doing the dot product:
[tex] \hat n \cdot (\vec D_1 - \vec D_2) = \rho_S [/tex]
This implies, [itex] \rho_S = 0 [/itex]

... I'm just confused here. I don't understand this part at all. Any help would be great!

 

[Jhero]

Hello,

Thank you for your post and for sharing your thoughts on the problem. From your attempt at a solution, it seems like you have a good understanding of the boundary conditions and how to apply them to this problem. However, I do see a few areas where you may be getting confused.

Firstly, your expression for \vec{E_1} is correct, but your expression for \vec{E_2} is not. Remember that the electric field inside a dielectric material is reduced by a factor of \epsilon_r compared to the electric field in vacuum. So, the correct expression for \vec{E_2} would be:

\vec{E_2} = -\hat{z}\frac{V_0}{d\epsilon_r}

Next, your expressions for \vec{D_1} and \vec{D_2} are also incorrect. Remember that the electric displacement vector, \vec{D}, is equal to \epsilon_0\vec{E} in vacuum, but in a dielectric material it is equal to \epsilon\vec{E}. So, the correct expressions for \vec{D_1} and \vec{D_2} would be:

\vec{D_1} = \epsilon_0\vec{E_1} = -\epsilon_0\hat{z}\frac{V_0}{d}
\vec{D_2} = \epsilon\vec{E_2} = -\epsilon_0\epsilon_r\hat{z}\frac{V_0}{d\epsilon_r}

Now, when you take the dot product of \hat{n} and (\vec{D_1} - \vec{D_2}), you should get a non-zero surface charge density. This is because the normal component of the electric displacement vector is not zero at the interface between the dielectric material and the vacuum. The normal component of \vec{D_1} is -\epsilon_0\hat{x}\frac{V_0}{d}, while the normal component of \vec{D_2} is -\epsilon_0\epsilon_r\hat{x}\frac{V_0}{d\epsilon_r}. So, the dot product would be:

\hat{n}\cdot(\vec{D_1} - \vec{D_2}) = -\epsilon_0\hat{x}\frac