# Potential of a planar capacitor with embedded electret

I want to calculate the potential between the two plates of a planar capacitor with embedded electret as shown below. The electret has a surface charge density $\sigma$, dielectric constant $\varepsilon$ and a thickness $d$. The air gap has a thickness $g$.

UPDATE 1: the problem is equivalent to a negative charge density $\sigma$ is placed on the surface of the dielectric as illustrated as negative signs in the figure.

If the electric field is assumed uniform, the potential difference between the two plates are:

$V = -E_g g + E_d d$

Applying Gauss's law for the electret and air inside the capacitor at the boundary of the dielectric and the air gives:

$\oint_SD.da=Q_{free} \to (-D_d+D_g)A=Q_{free}$

Assume that the dielectric is homogenous and isotropic: $D_d=\varepsilon\varepsilon_0 E_d$, while $D_g=\varepsilon_0E_g$. This leads to

$-\varepsilon E_d + E_g = \frac{\sigma}{\varepsilon_0}$

I need one more condition/equation to solve for $E_d$ and $E_g$ to calculate the potential between the two plates. Could anyone please help?

• How did you apply Gauss' Law for electric field with bound surface charges? – GeeJay Aug 29 '16 at 12:48
• I applied Gauss's law for the region consisting the dielectret and air. Was it incorrect? – Cuong Aug 29 '16 at 12:51
• $E_d$ should not be multiplied by $\epsilon$. – GeeJay Aug 29 '16 at 12:57
• $E_d$ is the electric field inside the dielectric (which is the electret) with dielectric constant $\varepsilon$. According to Gauss's law, $E_d$ must be multipled with $\varepsilon$. Could you please explain why? – Cuong Aug 29 '16 at 13:06
• For one, the dimensions don't check out. Permittivity times electric field does not give units of N/C. – GeeJay Aug 29 '16 at 13:12

Remember, $E_d$ and $E_g$ are the net fields in the d and g regions, after the effects of both bound and free surface charges.

Also, note throughout this answer I will use $\epsilon$ to mean relative permittivity (dielectric constant) and $\epsilon*\epsilon_0$ to mean absolute permittivity.

Applying regular Gauss' Law ($\oint_SE\cdot da=\frac{q}{\epsilon_0}$) by using a Gaussian pillbox (with face area A) that extends in both $E_d$ and $E_g$ regions, you get:

$E_g*A-E_d*A=\frac{\sigma*A}{\epsilon_0}$

$\Rightarrow E_g-E_d=\frac{\sigma}{\epsilon_0}$

The following is where I believe you were confused, and for a while, I was too!

Use Gauss' Law in a dielectric medium (you tried to use it) inside this pillbox:

$\oint_SD\cdot da=\oint_S\epsilon\epsilon_0E\cdot da=Q_{free}$

The absolute permittivity of 'd' region is $\epsilon\epsilon_0$ and 'g', ie-air is $\epsilon_0$

$\Rightarrow E_g\epsilon_0-E_d\epsilon\epsilon_0=0$

Because there is no free charge in the dielectric or in air. The only free charge resides on the capacitor plates, and they don't come inside the pillbox.

If the dielectret medium is homogeneous and isotropic,

$P=\sigma=\chi_e\epsilon_0E=(\epsilon-1)\epsilon_0E_d$

Note to OP: I'm afraid I misunderstood what you were trying to say. I assumed $\epsilon$ meant absolute permittivity whereas it is the relative permittivity according to the problem. Now, as to what you were really doing wrong--you used an in-between version of Gauss' Law in a dielectric and just "plain and regular" Gauss' Law.

If you work through it, you will see that all three equations in this answer relating $E_d$ and $E_g$ and $\sigma$ agree with each other.

• I am a little confused. Your statement "The only free charge resides on the capacitor plates, and they don't come inside the pillbox." is contrary to the statement made by @Cuong "Yes, the charge is assumed on the surface of the dielectric material, which is illustrated as negative signs in the figure."? – Farcher Aug 30 '16 at 7:51
• @Farcher Please refer to the image added to the answer. There is charge on the dielectric, but it is bound, not free, and the capacitor plates, which must have charge on them have free charge. – GeeJay Aug 30 '16 at 9:23
• I interpreted the problem as follows. A surface charge density of $\sigma$ is placed on the bottom of the dielectric and this negative charge induces positive charges on the top and bottom plates of the arrangement. This would means that the induced charge density on the top and bottom plates would not be the same? – Farcher Aug 30 '16 at 9:59
• No, the problem is this: there's a capacitor like any other, with opposite signed charge on either plate. A dielectric of thickness d is inserted into it. The electric field of the capacitor induces polarization in the dielectric, so that there is a layer of surface charge on opposite surfaces of the dielectric. Watch the first ten minutes of this – GeeJay Aug 30 '16 at 10:34
• I have seen the Lewin video before. As you will realise I interpreted the question in a different was and assumed that the negative charges on the dielectric were free charges. Thanks for your comments. – Farcher Aug 30 '16 at 12:23

After doing more research, I have found the answer. In my question, either top or bottom plate is grounded, while the other is floating. In my understanding, it is only 1 Dirichlet boundary condition (grounding), while electrostatics is a second order partial differential equation which requires 2 Dirichlet boundary conditions to obtain a unique solution (please correct me if I am wrong). Therefore, there is no answer for my question. However, the correlation between $V, E_d$ and $E_g$ are still held.

Below is a mathematical explanation based on Poisson's equation.

$\nabla.\textbf{D}=\nabla.\varepsilon\textbf{E}=\rho_v \qquad$ and $\qquad \textbf{E}=-\nabla\textbf{V} \qquad \to \nabla^2\textbf{V}=-\frac{\rho}{\varepsilon}$

The system is assumed to be 1D electric field and there is no free charge in the enclosured volume. Hence:

$\nabla^2\textbf{V}=\frac{d^2V}{dy^2}=0 \to V=ay+b \quad$ with assumption that $y=0$ at the bottom and upward.

The potentials in the dielectric medium and air are: $$\begin{cases} V_g(y)=A_gy+B_y & \mathrm{if} ~0 \le y<g\\ V_d(y)=A_dy+B_y & \mathrm{if} ~g \le y \le g+d \end{cases}$$

At the dielectric-air interface: $$V_g(g)=V_d(g)\\ -\varepsilon E_d + E_g = \sigma$$ In my problem, either $V(0)=0$ or $V(g+d)=0$ is another equation. We lack one more to be able to solve.

If $V(0)$ and $V(g+d)$ are clearly specified, the solution can be obtained easily.