# What happens when a dielectric-ball is placed in the uniform electric field?

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, $$E\ =\ |\textbf{E}|\ ,\ P=|\textbf{P}|\tag{Q1-2}$$

Fig. 1: Dielectric-ball

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.

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)

2. The vacuum isn't polarized, $$P$$ is the polarization of the ball, not the vacuum.
• @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. – Ofek Gillon Dec 25 '19 at 15:01