• Support PF! Buy your school textbooks, materials and every day products Here!

More than one dielectric ibetween a parallel plate capacitor

  • Thread starter rusty009
  • Start date
70
0
Hey, Im having some troubles with this question, the question asks me to calculate the capacitance of a parallel platecapacitor where the plates are seperated 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
 

Answers and Replies

berkeman
Mentor
56,097
6,142
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?
 
70
0
thanks for the reply, I still dont understand the problem :S.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
thanks for the reply, I still dont 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.
 
70
0
Hey, I have set it up but I am still having no luck, I have read through all my notes aswell but I cant find anything about it. In reply to Hootenanny's post, I think that the charge will be equally distributed along the plates.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Hey, I have set it up but I am still having no luck, I have read through all my notes aswell but I cant 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?
 
70
0
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
Staff Emeritus
Science Advisor
Gold Member
9,598
6
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?
 
70
0
I thought that a parallel capacitor needed to be between two conductors, but arnt there four capacitors ?
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
I thought that a parallel capacitor needed to be between two conductors, but arnt there four capacitors ?
Metals don't conduct? :confused: There are four plates, but only three capacitors since a capacitor is formed by a pair of plates.
 
70
0
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.
 
70
0
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
Staff Emeritus
Science Advisor
Gold Member
9,598
6
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.
 
70
0
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 teh 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
 

Related Threads for: More than one dielectric ibetween a parallel plate capacitor

Replies
10
Views
6K
Replies
3
Views
610
Replies
2
Views
6K
Replies
4
Views
23K
Replies
8
Views
4K
Replies
6
Views
751
Replies
1
Views
3K
Top