# Paralle plate capacitor (with dielectric)

EDIT: I figured it out.

## Homework Statement

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

1. Find $\vec D$, $\vec E$, $$\rho_S [/itex] ================ ------- ---------|---------|--------| ---------|---------|-------V0 ---------|---------|--------| ================-------- NOTE: The dielectric is on the right side (I colored it red) ## Homework Equations Note: Book convention uses $\rho$ as charge density. Boundary Conditions: [tex] B_{1t} = B_{2t}$$
$$D_{1n} - D_{2n} = \rho_S$$

$$\vec E = \frac{\vec D}{\epsilon}$$

Ideal Parallel Plate:
$$\vec E = \hat e \frac{V_{12}}{d}$$

## 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,

$$\vec E_1 = -\hat z \frac{V_{12}}{d} =-\hat z \frac{V_{0}}{d}$$
$$\vec E_2 = -\hat z \frac{V_{0}}{d}$$

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

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 $\vec E_1 = \vec E_t$.

Moving on to the second BC we note:
$$\vec D_1 = \epsilon \vec E_1 = \epsilon_0 \vec E_1$$
$$\vec D_2 = \epsilon \vec E_2 = \epsilon_0 \epsilon_r \vec E_2$$

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

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

So,
$$\vec D_1 - \vec D_2 = -\hat z \frac{\epsilon_0 V_0}{d} (1-\epsilon_r)$$
Now the normal component is $\hat -x$ (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:
$$\hat n \cdot (\vec D_1 - \vec D_2) = \rho_S$$
This implies, $\rho_S = 0$

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

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