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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}

Last edited: