- #1
sergiokapone
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Let we have a dielectric with field ##E## inside and with a little hole. I have problem. I get a two different answers on this problem, and I try to understand which one of them correct.
As mentioned in http://www.feynmanlectures.caltech.edu/II_11.html#Ch11-S4 (11.25), the electric field in cavity (in SGS)
\begin{equation}
E_{hole} = E + \frac{4\pi}{3}P
\end{equation}
with (11.8) ##P = \frac{1}{4\pi}(\epsilon-1)E## (in SGS), the field inside hole:
\begin{equation}
E_{hole} = \frac{\epsilon + 2}{3} E
\end{equation}
Another solution based directly on the superposition pinciple and boundary conditions for the electric field.
Lets look to the sphere's equatorial plane. The field on every point consist of field ##E## - field inside dielectric and a field of dipole ##\frac{p}{r^3}##:
\begin{equation}
E_{equator} = E + \frac{p}{\epsilon r^3}
\end{equation}
Boundary conditions for the tangential fields
\begin{equation}
E_{hole} = E_{equator}
\end{equation}
For one of the pole point
\begin{equation}
E_{pole} = E - \frac{2p}{\epsilon r^3}
\end{equation}
Boundary conditions for the normal field on pole
\begin{equation}
E_{hole} = \epsilon E_{pole}
\end{equation}
Thus I have two equations
\begin{align}
E_{hole} = E + \frac{p}{\epsilon r^3} \\
E_{hole}/\epsilon = E - \frac{2p}{\epsilon r^3}
\end{align}
This two equations give me the different answer:
\begin{equation}
E_{hole} = \frac{3\epsilon}{2\epsilon + 1} E
\end{equation}
As mentioned in http://www.feynmanlectures.caltech.edu/II_11.html#Ch11-S4 (11.25), the electric field in cavity (in SGS)
\begin{equation}
E_{hole} = E + \frac{4\pi}{3}P
\end{equation}
with (11.8) ##P = \frac{1}{4\pi}(\epsilon-1)E## (in SGS), the field inside hole:
\begin{equation}
E_{hole} = \frac{\epsilon + 2}{3} E
\end{equation}
Another solution based directly on the superposition pinciple and boundary conditions for the electric field.
Lets look to the sphere's equatorial plane. The field on every point consist of field ##E## - field inside dielectric and a field of dipole ##\frac{p}{r^3}##:
\begin{equation}
E_{equator} = E + \frac{p}{\epsilon r^3}
\end{equation}
Boundary conditions for the tangential fields
\begin{equation}
E_{hole} = E_{equator}
\end{equation}
For one of the pole point
\begin{equation}
E_{pole} = E - \frac{2p}{\epsilon r^3}
\end{equation}
Boundary conditions for the normal field on pole
\begin{equation}
E_{hole} = \epsilon E_{pole}
\end{equation}
Thus I have two equations
\begin{align}
E_{hole} = E + \frac{p}{\epsilon r^3} \\
E_{hole}/\epsilon = E - \frac{2p}{\epsilon r^3}
\end{align}
This two equations give me the different answer:
\begin{equation}
E_{hole} = \frac{3\epsilon}{2\epsilon + 1} E
\end{equation}
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