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In the image above there are two infinite planes. The infinite plane at $z = a$ has a surface charge $\sigma$ while the infinite plane at $z = 0$ has a surface charge $-\sigma$. The dielectric that has a dielectric constant $\epsilon_1$ is present in the region $ 0 < z < a$, while the dielectric that has a dielectric constant $\epsilon_2$ is present in the regions $ a < z$ and $z < 0$. Using Gauss' law, the Electric field produced by the infinite plane at $z = a$ is: $$ \vec{E}_1 = \frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z>a\quad \text{and} \quad \vec{E}_1 = -\frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z<a \tag{1}. $$ While for the infinite plane at $z = 0$ the E-field is: $$ \vec{E}_2 = -\frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z>0 \quad \text{and} \quad \vec{E}_2 = \frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z<0. \tag{2} $$ Clearly the E-fields in regions $z > a$ and $z < 0$ are zero while the E-field in $ 0 < z < a$ is: $$ \vec{E} = -\frac{2\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad 0<z<a.\tag{3} $$ If $\epsilon_1 = \epsilon_2$ then:

$$ \vec{E} = -\frac{\sigma}{\epsilon_0(\epsilon_1)}\hat{z}\quad 0<z<a.\tag{4} $$ The potential difference between $z =a$ and $z = 0$ is: $$ V = \frac{\sigma}{\epsilon_0\epsilon_1}a.\tag{5} $$ With equation (5) the capacitance is: $$ C = \frac{Q}{V} = \frac{\sigma\cdot A}{\frac{\sigma}{\epsilon_0\epsilon_1}a} = \epsilon_1 \frac{A \epsilon_0}{a} = \epsilon_1 C_{\text{vacuum}} \tag{6} $$ which is expected. However, if we say $\epsilon_1 \neq \epsilon_2$ then the capacitance becomes:

$$ C = \frac{\epsilon_1+\epsilon_2}{2} C_{\text{vacuum}} \tag{7} $$ Somehow the external dielectric has changed the capacitance, if this is true why does this happen? If not where have I made the error?

Edit 1: I assumed that both the dielectrics were linear dielectrics. Furthermore, when calculating the flux I placed a square box that encapsulated a portion of the infinite plane. With this Gauss' law becomes: $$ \iint_{s}\vec{D}\cdot\vec{dA} = \iint\vec{D_{above}}\cdot\vec{dA} + \iint\vec{D_{below}}\cdot\vec{dA} = (D_{top}+D_{below})A = A \sigma, \tag{8} $$ where $A$ is the area of a face, also, The unit normal vector for $\vec{dA}$ always point in the same direction as the displacement field, that is why I added the contributions ($D_{top}+D_{below}$). In equation (8) only the contributions from the faces of the box whose normal is perpendicular to the plane are taken. Since I was looking at linear dielectrics I used: $$ \vec{D} = \epsilon_0\vec{E} + \vec{P} = \epsilon_0\epsilon_r\vec{E}. \tag{9} $$ Using equations (8) and (9) I obtained equations (1) and (2), we get: $$ D_{top}+D_{below} = \epsilon_0\epsilon_2 E+\epsilon_0\epsilon_1 E = \sigma. \tag{10} $$

Edit 2: It turns out that placing the same $E$ in equation (10) is wrong (physically not possible) and this was the source of my problems. To rectify the problem we must realize that the magnitude of $D$ is the same on the top and below. Using equation (8) we get: $$ D = \frac{\sigma}{2}.\tag{11} $$ Using equations (9) and (11) the E-fields for the infinite plane at $z = a$ is: $$ \vec{E} = \frac{\sigma}{2\epsilon_0\epsilon_2}\quad z>a\quad \text{and}\quad \vec{E} = -\frac{\sigma}{2\epsilon_0\epsilon_1}\quad 0<z<a. \tag{12} $$ While for the infinite place at $z = 0$ the E-fields are: $$ \vec{E} = -\frac{\sigma}{2\epsilon_0\epsilon_1}\quad 0<z<a\quad \text{and}\quad \vec{E} = \frac{\sigma}{2\epsilon_0\epsilon_2}\quad z<0.\tag{13} $$ With this the fields in the region where dielectric 2 is present cancel and go to zero while the region where dielectric 1 is present the field is: $$ \vec{E} = -\frac{\sigma}{\epsilon_0\epsilon_1} \rightarrow V = a\frac{\sigma}{\epsilon_0\epsilon_1}\rightarrow C = \epsilon_1 C_{\text{vacuum}}. $$

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  • $\begingroup$ Note that this derivation assumes that the plates are infinite. The result would presumably not hold exactly for a finite-sized capacitor where the regions "inside" and outside" the capacitor had different dielectric constants — though this result would be a very good approximation to the exact result. $\endgroup$ Commented May 5, 2020 at 18:12

2 Answers 2

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Draw a Gaussian surface around both plates and the net enclosed charge is zero. Consequently the net flux is zero and there is no electric field produced outside the capacitor.

The external dielectrics are irrelevant.

Hope this helps

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  • $\begingroup$ Yes, is a very nice way of testing whether my statement was correct. $\endgroup$
    – greatangle
    Commented May 5, 2020 at 11:23
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No, it does not. The problem is in Eqs. (1) and (2), i.e. in your expressions for the electric fields. When you apply Gauss's law in terms of $E$, you should consider all charges, also the ones accumulated in the dielectrics. That is why in this case it is more convenient to apply Gauss's law in terms of the electric displacement field, usually denoted as $D$.

Think also just a moment about it: Eqs. (1) and (2) cannot be right, since the electric field generated by the plane cannot have the same magnitude in two different dielectrics!

Applying Gauss's law in terms of $D$, you should be able to show that the electric field inside the capacitor, hence the capacitance, do not depend on the external dielectric. I can add the derivation, if you want (I was once downvoted for writing too many details).

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  • $\begingroup$ I have used Gauss' law for dielectrics (I added an edit). I completely agree with your second statement, E-field cannot be the same on both sides. I applied Gauss' law in terms of D and then converted to E-field (see edit 1), but I think I can see where my error arises now. It is in equation (10) where I placed both E-fields to be equal. I think more details are always welcomed. $\endgroup$
    – greatangle
    Commented May 5, 2020 at 11:18
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    $\begingroup$ @greatangle Very good. The problem is now in (10). The electric field is not the same above and below the plane. It is actually the displacement field (D) which is the same! So you can calculate D, and from this calculate E above and below the plane. $\endgroup$
    – fra_pero
    Commented May 5, 2020 at 11:24
  • $\begingroup$ When you say the displacement field (D) is the same, above and below you mean to say that the magnitude of D is the same. However, the direction is opposite. $\endgroup$
    – greatangle
    Commented May 5, 2020 at 11:29
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    $\begingroup$ @greatangle I think I commented too fast without reading fully your comment. Or maybe you were editing at the same time. Very good, I am glad you found the error by yourself. $\endgroup$
    – fra_pero
    Commented May 5, 2020 at 11:29
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    $\begingroup$ @greatangle Exact. The magnitude is the same and the direction is opposite. $\endgroup$
    – fra_pero
    Commented May 5, 2020 at 11:29

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