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relativespeak
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Homework Statement
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}
Homework Equations
\vec{D}_{0}=\epsilon_{0}\vec{E} when \vec{P}=0 (inside the cavity).
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}.
The answer is right, but I'm not sure about my work/reasoning.
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