# Field and Displacement Inside Carvities Within A Dielectric

1. Oct 30, 2012

### relativespeak

1. The problem statement, all variables and given/known data

Suppose the field inside a large piece of dielectric is $\vec{E}_{0}$, so that the electric displacement is $\vec{D}_{0}=\epsilon_{0}\vec{E}+\vec{P}$.

a. Now a small spherical cavity is hollowed out of the material. Find the field at the center of the cavity in terms of $\vec{E}_{0}$ and $\vec{P}$. Also find the displacement at the center of the cavity in terms of $\vec{D}_{0}$ and $\vec{P}$.

b. Do the same for a long needle-shaped cavity running parallel to $\vec{P}$.

c. Do the same for a thin wafer shaped cavity perpendicular to $\vec{P}$

2. Relevant equations

$\vec{D}_{0}=\epsilon_{0}\vec{E}$ when $\vec{P}=0$ (inside the cavity).

3. The attempt at a solution

For each part, I found the E of the cavity as if it were a solid piece, and then subtracted that from $\vec{E}_{0}$ to account for the change in $\vec{E}$. Then, because $\vec{D}_{0}=\epsilon_{0}\vec{E}$ inside the cavity, $\vec{D}_{0}=\epsilon_{0}\vec{E}_{new}$.

a. This part was simple. Inside a polarized sphere, $\vec{E}=\frac{-1}{3\epsilon_{0}} \vec{P}$, so $\vec{E}_{new}=\vec{E}_{0}+\frac{1}{3\epsilon_{0}} \vec{P}$, and $\vec{D}=\vec{D}_{0}-\frac{2}{3}\vec{P}$.

b. Here it gets more tricky. The length of the needle is parallel to $\vec{P}$, so the $\vec{E}$ is pointed into and outside the ends. There would be a - charge and + charge at opposite ends, creating a field between them. But the field is negligible because the distance between them is large compared to the dipole? In that case, $\vec{E}_{new}=\vec{E}_{0}$. Once we neglect the field inside, I can move forward, but can anyone explain better why exactly we ignore that field.

c. Now we have a thin wafer-shaped cavity. I tried to use potential to calculate the electric field within the wafer.

$V=\frac{1}{4\pi\epsilon_{0}}\oint_{S}\frac{1}{r} \vec{P}\cdot d\vec{a}$.

(I did not use the second half of this eqn because we are assuming uniform polarization, which means the gradient of polarization is 0.)

From here I find
$\vec{E}=-\vec{\triangledown}V=\frac{-1}{2\epsilon_{0}}\vec{P}$.

I then multiplied this by 2 to account for the top and bottom sides of the wafer and found

$\vec{E}_{new}=\vec{E}+\frac{1}{\epsilon_{0}}\vec{P}$.