More than one dielectric ibetween a parallel plate capacitor

[rusty009]

Hey, I am having some troubles with this question, the question asks me to calculate the capacitance of a parallel platecapacitor where the plates are separated by 3mm, and in between the are three layers of dielectric 1mm thick with different relative permitivities. Do I find the total by adding the three permitivities together :S?

P.S I am also given the area of capacitor and the values of the permitivities

Thanks in advance

 

[berkeman]

Welcome to the PF. The way to approach this problem is to envision floating metal plates of zero thickness, placed at the 2 interfaces between the 3 dielectrics. Is that enough of a hint?

 

[rusty009]

thanks for the reply, I still don't understand the problem :S.

 

[Hootenanny]

thanks for the reply, I still don't understand the problem :S.

What can you say about the charge distribution on each of these plates placed at the interface?

If your still struggling, try sketching the set up.

 

[rusty009]

Hey, I have set it up but I am still having no luck, I have read through all my notes as well but I can't find anything about it. In reply to Hootenanny's post, I think that the charge will be equally distributed along the plates.

 

[Hootenanny]

Hey, I have set it up but I am still having no luck, I have read through all my notes as well but I can't find anything about it. In reply to Hootenanny's post, I think that the charge will be equally distributed along the plates.

Have you sketched the set-up? If so, what does it remind you of?

 

[rusty009]

Ok, I have been reading up a lot, and have encountered the notion of electric flux across the dielectrics, but the thing is I need to know the charge of the capacitor to work with electric flux and I do not have it. I have sketched it up but I am not familiar with this kind of problem.

 

[Hootenanny]

I have sketched it up but I am not familiar with this kind of problem.

Doesn't it remind you of three parallel plate capacitors?

 

[rusty009]

I thought that a parallel capacitor needed to be between two conductors, but arnt there four capacitors ?

 

[Hootenanny]

I thought that a parallel capacitor needed to be between two conductors, but arnt there four capacitors ?

Metals don't conduct? There are four plates, but only three capacitors since a capacitor is formed by a pair of plates.

 

[rusty009]

Hey, Ok I think I might be starting to understand it, but I have 5 plates, 3 dielectrics between two parallel plates. do I then use C=e0erA/d on each capactor then add them up? Thanks for your help.

 

[rusty009]

but the capacitors would be in series so I would have to add them up with 1/c=1/c1 ... , is tis the right method?

 

[Hootenanny]

Hey, Ok I think I might be starting to understand it, but I have 5 plates, 3 dielectrics between two parallel plates.

No if you have drawn it correctly, you should only have four plates, one plate on the boundary between the three dielectric (this makes two plates) and two plates on the outside.

but the capacitors would be in series so I would have to add them up with 1/c=1/c1 ... , is tis the right method?

Indeed, you can treat the capacitors as if they are in series.

 

[rusty009]

Ok, instead of the previous idea I decided to use a diffeent method and I think it worked. I'm basing this formula on electric flux, and that electric flux will stay constant throughout the capacitor so that Electric flux= e0*er*E , so for three different relative permitivities I have three different electric field stregths so that,

e0*er1*E1=e0*er2*E2=e0*er3*E3 where er1 er2 and er3 are the three different permitiities

the e0's will cancell eachothr out giving,

er1*E1=er2*E2=er3*E3

now with this relationship we can assume that

E1=&/e0*er1 E2=&/e0*er2 E3=&/e0*er3 where & is sigma ( charge per length)

and from this we have a consistent set of equations where we can use the formula

E= V/d to get an equation for the voltage across the capactor:

V=E1*d1+E2*d2+E3*d3= &/e0( d1/er1 + d2/er2 + d3/er3 )

and finally I used the formula C=&A/(d1/er1 + d2/er2 + d3/er3 ) where A is the surface area of the plates on the capacitor.

so this method can provide the capacitance of a parallel plate capacitor with virtually infinite permitivites, assuming you know what they are and you know A. I thought I should share this method withthe forum for any future questions that relate to this one :D