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Let the space filled with a dielectric having a dielectric constant $\varepsilon_1$, there are a uniform electric filed $\textbf{E}_1$ in the space. We set a dielectric-ball in this space.

My text has the following 【Practice problems】and 【Answer of Practice problems】 (originally written in Japanese)(See ref.1). I read them but them but, I'm not sure of the points described in 【My Questions】, below.

【My Questions】

  • Can I understand the $\textbf{P}$ of the【Answer of Practice problems】 is the polarization of Dielectric-ball? If so, Is the vacuum polarized in the setting of the (3) of 【Practice problems】?
  • It seems that the answer assumed the $\textbf{E}$ is uniform in the Dielectric-ball. But, How can I justify this?

【Practice problems】:
Rough translation of the Practice problems, quoted from (ref.1) with minor modification.

Let the space filled with a dielectric having a dielectric constant $\varepsilon_1$, there are a uniform electric filed $\textbf{E}_1$ in the space. We set a dielectric-ball in this space. The dielectric constant and the radius of spherical dielectric is $\varepsilon_2$ and a respectively.

  • (1). Then, find the electric field inside of the dielectric-ball.
  • (2). Draw the Electric flux-lines schematically for the case of $\varepsilon_2\ >\varepsilon_1 $ and, $\varepsilon_2 <\varepsilon_1 $ .
  • (3). In particular, what happens if the dielectric-ball is a void ? (That means, the dielectric-ball is the vacuum therefore, $\varepsilon_2=\varepsilon_0$)

Here, we can suppose the electric polarizability is represented as follows in our case. $$\chi=\ {\varepsilon_2/\varepsilon}_1\ -\ 1\tag{Q1-1}$$ The length of vector are represented by $|\ |$ for example, $$E\ =\ |\textbf{E}|\ ,\ P=|\textbf{P}|\tag{Q1-2}$$

enter image description here
Fig. 1: Dielectric-ball

【Answer of Practice problems】:
Rough translation of the answer of the 【Practice problems】, quoted from (ref.1) with minor modification.

The ring shape area-element $dS$ on the surface of the ball, shown in the figure is expressed as follows. $$dS=2\pi a^{2}sin\theta d\theta \tag{A 1-1}$$ We denote $\textbf{P}$ as the dielectric polarization, then, the polarization charge generated in $dS$ is $$P\cos\theta dS=2P\pi a^2\ (\cos\theta)(\sin\theta)d\theta \tag{A 1-2}$$


The electric field created by this polarization electrification at the center of the sphere is expressed by Coulomb's law as follows. $$dE_0=\left(\frac{Pcos\theta dS}{4\pi\varepsilon_0a^2}\right)cos\theta\ =\frac{P\left(cos^2\theta\right)\ (sin\theta)d\theta}{2\varepsilon_0}\ \tag{A 1-3}$$
Sum up the $dE_0$ over the entire spherical surface, the electric field at the origin O is obtained. This is obtained by integrating the above equation from 0 to π with respect to as follows. $$E_0\ =\ \frac{\textbf{P}}{2\varepsilon_0}\ \int_{0}^{\pi}{\left(cos^2\theta\right)\ (sin\theta)d\theta\ }\ =\frac{\textbf{P}}{3\varepsilon_0} \tag{A 1-4}$$
From the supposition (see (Q1-1)), the electric polarizability is represented as follows in our case, $$\chi=\ {\varepsilon_2/\varepsilon}_1\ -\ 1\tag{A 1-5}$$
The $\textbf{P}$ satisfies $$\mathbf{P}=\ \varepsilon_0\left({\varepsilon_2/\varepsilon}_1\ -\ 1\right)\textbf{E}\tag{A 1-6}$$
On the other hand, the sign of the surface polarization charge is opposite to that of the external electric field. Therefore, the electric field $\textbf{E}$ at the $\textbf{O}$ is $$\textbf{E}=\textbf{E}_1-\textbf{E}_0=\textbf{E}_1-\frac{\textbf{P}}{3\varepsilon_0}\tag{A 1-7}$$
Therefore, $$\textbf{E}=\textbf{E}_1-\frac{\textbf{P}}{3\varepsilon_0}\ =\textbf{E}_1+\ \frac{1}{3}\left(\ 1-{\varepsilon_2/\varepsilon}_1\ \right)\textbf{E} \tag{A 1-8}$$
we get the answer of (1) is as follows. $$\textbf{E}=\frac{\mathbf{3}}{\left(\ 1+{\varepsilon_2/\varepsilon}_1\ \right)}\textbf{E}_\mathbf{1}\tag{A 1-9 }$$ Therefore, the answer of (2) as shown in Fig.2. enter image description here
Fig.2 The answer of (2).

If the dielectric-ball is a void, such that $$\varepsilon_2=\varepsilon_0 \tag{A 1-10}$$ Then, the answer of (3) is as follows. $$\textbf{E}=\frac{\mathbf{3}\varepsilon_s}{\left(\ 2\varepsilon_s\ +\ 1\ \right)}\textbf{E}_\mathbf{1}\tag{A 1-11}$$ Here, $$\varepsilon_1=\varepsilon_0\tag{A 1-12}$$

【Reference】
(ref.1) Hitoshi Ookubo, et.al.;" 電気磁気学 (Electromagnetism)" Shou-Ko-Do,Tokyo (1993/10).P.P.66-67(Written in Japanese)

P.S. I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions. I welcome any corrections and English review. (You can edit my question and description to improve them)

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I'll start with the second question and then the first:

  1. If I remember correctly, the electric field in this particular scenario in the dielectric ball does come out to be uniform, but I have no trivial way to explain why that is so. Instead, I suggest you solve Laplace's equation in and out of the sphere with the correct boundary conditions (this is how I solved it at the time...).

  2. The vacuum isn't polarized, $P$ is the polarization of the ball, not the vacuum.

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  • $\begingroup$ Thank you for your answer. For 2, my textbook use similar manner even when the dielectric ball is the vacant. At that time, I think it is strange to use this P. (See the (3) of 【Practice problems】) How can we justify it? $\endgroup$ – Blue Various Dec 25 '19 at 14:03
  • $\begingroup$ For 1. Is this an analogy to Method of image charges?en.wikipedia.org/wiki/Method_of_image_charges $\endgroup$ – Blue Various Dec 25 '19 at 14:11
  • $\begingroup$ @BlueVarious , for 1: well, this method is allowed due to the uniqueness theorem of the equation, but it isn't an analogy - you really need to solve it. 2. For vacuum, $\epsilon = \epsilon_0 , \chi =0 $ so it isn't polarized. Notice, however, that they define $\chi$ differently, so they pretend the outside is "vacuum" and the inside has a different permitivity compared to it. Meaning although the surface charge will actually rise from the surroundings, they solve the problem in a different way, pretending they come from the relative permitivity of the ball. $\endgroup$ – Ofek Gillon Dec 25 '19 at 15:01
  • $\begingroup$ Thank you for your comment. Oops!! Assuming that there is polarization, but it is zero after all? $\endgroup$ – Blue Various Dec 25 '19 at 15:05

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