# Homework Help: Capacitance of a parallel plate capacitor with dielectrics

1. Feb 25, 2013

### fluidistic

1. The problem statement, all variables and given/known data
1)The capacitance
2)The electric field
3)The surface charge density
4)The polarization charge at the interface between the dielectrics.
Of 2 set ups. The first one is a parallel plate capacitor which is half filled with a dielectric $\varepsilon _1$ and half filled with a dielectric $\varepsilon _2$. The separation of the 2 dielectrics is a vertical plane.
In the other set up, the separation is a horizontal plane.

2. Relevant equations
C=Q/V.
$\vec D= \varepsilon \vec E$.
$(\vec D _2 - \vec D_1 )\cdot \hat n _{21}= \sigma {free}$.
$(\vec E_ 2 - \vec E_1 ) \times \hat n _{21}=0$.
$\sigma _{\text{pol}}=- (\vec P_2 - \vec P_1 ) \cdot \hat n_{21}$ with $P_i=(\varepsilon _i - \varepsilon _0) \vec E_i$.
3. The attempt at a solution
Strangely I'm asked to find the capacitance before the electric field?
So I go with the first set up. I assume a plate has a charge Q and the other -Q and that they are separated by a distance d.
I'm interested in the difference of potential between both plates, "V".
Now I believe that the D field will be different according the region of the dielectric. However $\vec D_1 = \varepsilon _1 \vec E_1$ where $\vec E _1 = - \vec \nabla \Phi _1$ for the first region and $\vec D_2 = \varepsilon _2 \vec E_2$ where $\vec E _2 = - \vec \nabla \Phi _2$ for the 2nd region. Now when I want to get "V", or in my terms $\Phi (d)- \Phi (0)$, I realize that the result will depend upon which Phi I choose. In other words it seems that "V" depends on the region, which of course is totally false. I don't know what I'm doing wrong.

Last edited: Feb 25, 2013
2. Feb 26, 2013

### ehild

The potential is the same along a capacitor plate. In the first set-up, you can imagine the two parts of the capacitor filled with different dielectrics as two capacitors connected in parallel.

ehild

3. Feb 26, 2013

### fluidistic

Thanks for the help ehild!
Yes I realize this. The problem is that I'm not getting it mathematically.
I see, I had some intuition it would be like that, thanks for the confirmation. Although I'd rather "prove it" mathematically instead of assuming it.

From $\vec D = - \vec \nabla \Phi$ and $\vec D _i = \varepsilon _i \vec E$, I get that $V=\Phi (d) - \Phi (0) =- \int _0 ^d \vec D _i \cdot d \vec l = - \varepsilon _i E_i d$ which depends on the region (and should not).
Hmm I see something's wrong, V cannot be path dependent. Why do I get it to be path dependent?!

4. Feb 28, 2013

### ehild

The electric field strength is negative gradient of potential, E=-gradΦ and D=εE. The potential is the same along a plate, so E is the same everywhere.

ehild

5. Feb 28, 2013

### fluidistic

I see, thank you once again. I start to understand the physics here, I believe.
Over the surface of the bottom plate, $\sigma _{\text { free}}=D_1=\varepsilon _1 E$ for the region 1 and $\sigma _{\text { free}}=D_2=\varepsilon _2 E$ for the region 2.
Where E as you said, is constant. It's worth $4\pi \sigma _{\text{free}}$ and pointing up (if the upper plate has a charge of Q and lower plate -Q) where I use Gaussian units (must divide by 4 pi epsilon_0 to get SI units).
Now at the interface between the dielectrics, I use the boundary condition $\sigma _ {\text{polarization}}=-(\vec P_2 - \vec P_1 ) \cdot \hat n _{21}= (\varepsilon _1 - \varepsilon _2 )E$. Here I'm stupefied because if I understand well, a charge out of nowhere appears at the surface between the dielectrics, INSIDE the capacitor. That goes against my intuition. I'd appreciate if someone can comment on this polarized charge.
Now if I assume that each plate has an area "A", I get that $\sigma _{\text{free}}A=Q \Rightarrow C= \frac{\sigma _{\text{free}}A}{V}$ where $V=Ed$, yielding $C=\frac{A}{4\pi d}$ which is false since it does not depend on the dielectric. I probably used a wrong relation between $\vec E$ and $\sigma _{\text{free}}$. That's really confusing!

Edit: I think I know where my error is. $E=4\pi \sigma _{\text{free}}$ should read $E=4\pi \sigma _{\text{total}}=4\pi (\sigma _{\text{polarized}}+ \sigma _{\text{free}})$.
I'm going to work on this.
Edit 2: Still not working. I get another non sense, $C=\frac{A}{4d \pi (1+ \sigma _{\text{polarized}})}$ which depends on the region.
Edit 3: I'm lost, I don't know what's going on.

Last edited: Feb 28, 2013
6. Mar 3, 2013

### ehild

The capacitor was charged initially and had the charge Q. That charge stays on the plates even after the dielectric was placed in. After inserting the dielectric, the electric field is the same in both parts, the D field is ε1E and ε2E. D is related to the surface charge density: D = σ (In SI units). So you have Q1=(A/2)ε1E charge on one part of capacitor and Q2=(A/2)ε2E on the other part. The charge can rearrange itself along a metal plate.The total charge is Q=Q1+Q2. You get an equation for E. From E you get the potential difference U, and C=Q/U.

ehild

7. Mar 4, 2013

### fluidistic

Thank you once more for your help.
This gave me $C=\frac{A(\varepsilon _1 + \varepsilon _2 ) }{d}$.
I had not realized I had not to take into account the polarized charge, because although it's not a free charge, I think the dielectrics produces a charge just over (at the surface of) each plate. But D is not affected by those charges.
For the 2nd set up I get $C=\frac{2A}{d} \left ( \frac{\varepsilon _1 \varepsilon _2 }{\varepsilon _1+\varepsilon _2} \right )$.

Last edited: Mar 4, 2013
8. Mar 4, 2013

### ehild

You miss a factor 1/2 in the first capacitance. The second expression is correct.

You see, there are two capacitors of half area and different ε-s connected in parallel in the first case, and two capacitors with half the original distance between the plates, connected in series in the second case.

ehild