A capacitor with two dielectrics in series

[artis]

Hi,

I would like to know how to calculate the total capacitance for a capacitor that has a certain plate overlap area and two dielectrics in between, one being a solid state dielectric and the other one being air.
This is not a school project, I just thought about it and tried to calculate myself but failed.
So my plate area is 8000 cm2, the total distance between plates is 1mm of which 0.1 is a solid dielectric with K=100 and the rest 0.9mm is air.
If one could show the formula and the explanation with my numbers in it it would be great, that helps me understand,

thanks.

 

[berkeman]

The easiest way to handle two dielectrics in series in a capacitor is to remember that the interface between the two dielectrics will be an equipotential surface, just like if there were a very thin piece of metal at the interface. This divides the capacitor up into two capacitors, each with the same area, but a smaller spacing between the "plates". In the figure below, can you write the equation for C for the right-hand capacitor? Remember how the capacitance for series capacitors adds up...

https://i.ytimg.com/vi/DZJqS6G9pVM/hqdefault.jpg

 

[davenn]

The easiest way to handle two dielectrics in series in a capacitor is to remember that the interface between the two dielectrics will be an equipotential surface, just like if there were a very thin piece of metal at the interface. This divides the capacitor up into two capacitors, each with the same area,

cheers
That sounds familiar, I was thinking such, but wasn't sure that I was right
Old age and senility setting in hahahaDave

 

[artis]

so I basically calculate the capacitance for two different capacitors and then I sum the two different capacitances together and divide them by 2, since two series capacitor capacitance is half the total capacitance, correct?
Something tells me this is not correct.
well I found an online calculator , since I'm dumb

 

[vanhees71]

It's capacitances in series, not in parallel! The correct answer is that you have
$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}.$$
@berkeman is right in #4: It follows from the electrostatic equations and the boundary conditions at the plates and the boundary between the two dielectrics. Note that there are no free charges at the dielectrics and thus also no surface charge there, i.e., the normal components of $\vec{D}=\epsilon \vec{E}$ are continuous across this boundary and thus the normal component of $\vec{E}$ jumps (because of the change in polarization between the two dielectrics). The tangential components of $\vec{E}$ are continuous at every boundary surface. The potentials are thus continuous across all boundaries too. That's sufficient to see that in your configuration the boundaries of different adjacent dielectrics are indeed equipotential lines since the electric field is perpendicular to this boundary, and thus the total capacitance is the capacitance of the series of capacitors discribed in 4.

Note that in general the boundaries of different dielctrica need NOT be equipotential surfaces. Check the case where the two dielectrics are just rotated by 90 degrees and think about the boundary problem and how to get the capacitance in this case!

 

[artis]

ok, thanks, I made a mistake with my formula because in theory I knew that with two different dielectrics in series I essentially have two capacitors in series.

although I used this formula , I calculated the capacitance imagining that I'm calculating for two different capacitors, so i first calculated the first dielectric using it's thickness and constant and then the second, then i put the results in place of the C for each capacitance both 1 and 2.
Somehow my numbers were strange, maybe because I used uF.

 

[sophiecentaur]

Somehow my numbers were strange, maybe because I used uF.

Using μF should make no difference as long as the units are the same for both capacitors. The answer you got could seem strange because the net capacitance will be less than the value of the smaller one. Capactiors in series are treated the same as Resistors in Parallel, remember.