Electrostatics > The electric field in the presence of a dielectric > Problem 6.6.12

Statement

$6.6.12.$ The space between the plates of a parallel‑plate capacitor is filled with two layers of different dielectrics of thicknesses $d_1$ and $d_2$. The dielectric permittivities of the dielectrics are $\varepsilon_1$ and $\varepsilon_2$. The plate area is $S$. Find the capacitance of the capacitor. What charge will be induced at the interface between the dielectrics if a charge $\pm q$ is placed on the capacitor plates?

Solution

Our capacitor can be represented as two capacitors connected in series; for them the capacitance is
$$
\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\tag{1}
$$
where
$$
C_i=\frac{\varepsilon_i\varepsilon_0S}{d_i}\tag{2}
$$
$(2)\to(1):$
$$
\frac{1}{C}=\frac{1}{\varepsilon_0S}\left(\frac{d_1}{\varepsilon_1}+\frac{d_2}{\varepsilon_2}\right)
$$
$$
C=\frac{\varepsilon_0\varepsilon_1\varepsilon_2S}{\varepsilon_2d_1+\varepsilon_1d_2}\tag{3}
$$

Now recall that at the interface between dielectrics the normal component of the field undergoes a jump. The dielectrics are isotropic, the field is normal to the interface, and there are no free charges at the interface, so
$$
\varepsilon_2E_2=\varepsilon_1E_1\tag{4}
$$
For the voltage across the capacitor after placing the charge:
$$
V=E_1d_1+E_2d_2=\frac{q}{C}\tag{5}
$$
The surface density of bound (induced) charges at the interface is determined by the jump in the normal component of the polarisation vector:
$$
\sigma'=P_{n1}-P_{n2}\tag{6}
$$
For an isotropic dielectric
$$
\vec P=(\varepsilon-1)\varepsilon_0\vec E
$$
Therefore
$$
\sigma'=(\varepsilon_1-1)\varepsilon_0E_1-(\varepsilon_2-1)\varepsilon_0E_2
=\varepsilon_0[(\varepsilon_1E_1-\varepsilon_2E_2)+E_2-E_1]
$$
Taking $(4)$ into account, we finally obtain a nice expression:
$$
\sigma'=\varepsilon_0(E_2-E_1)\tag{7}
$$

Solve the system $(4), (5), (7)$:
$$
E_1=\frac{q\varepsilon_2}{C(d_1\varepsilon_2+d_2\varepsilon_1)}, \quad
E_2=\frac{\varepsilon_1}{\varepsilon_2}E_1=\frac{q\varepsilon_1}{C(d_1\varepsilon_2+d_2\varepsilon_1)}
$$
$$
\sigma'=\varepsilon_0q\frac{\varepsilon_1-\varepsilon_2}{C(d_1\varepsilon_2+d_2\varepsilon_1)}
=\varepsilon_0q\frac{\varepsilon_2d_1+\varepsilon_1d_2}{\varepsilon_0\varepsilon_1\varepsilon_2S}\, \frac{\varepsilon_1-\varepsilon_2}{d_1\varepsilon_2+d_2\varepsilon_1}
$$
$$
\sigma'=\frac{q'}{S}=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}\frac{q}{S}
$$
$$
q'=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}q
$$

Answer

$$
C=\frac{\varepsilon_0\varepsilon_1\varepsilon_2S}{\varepsilon_2d_1+\varepsilon_1d_2}
$$
$$
q'=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}q
$$