What is the capacitance of a capacitor constructed using foil and paper?

[Number1Ballar]

Homework Statement

Suppose you constructed a capacitor using two 5cm*25cm sheets of foil and a 0.3mm thick piece of paper (k=3.7). What is its capacitance? Calculate to 1 decimal place. Use units of "nF". (1nF=1*10^-9F)

Homework Equations

where d is the distance between the 2 sheets, A is the area of 1 sheet.

The Attempt at a Solution

The equation above is the only equation I could find that involved area of a capacitor. I found A = (0.25m)(0.05m) = 0.0125m². I don't know where the paper comes in, nor do I understand what 'k' is and what formula I am supposed to use for it.

 

[WarPhalange]

I'm not sure but I think k is the dielectric constant of paper.

See, the way a capacitor works is that you have two plates next to each other, charged oppositely. This creates an electric field between them, making the electrons want to jump across to the positive side.

In order to get more out of these plates, you put something in between, which lowers the electric field in between the plates. This means that you now have to pile up more charge on the plates to have the same kind of electric field as you had before, giving your capacitor more storage space, i.e. capacitance.

So in this case you are putting paper in between the two plates, and I *believe* that that is what "k" is and you just plug that in for epsilon.

 

[learningstill]

If you want to use that equation (and I would suggest that) then you need to figure out what your values of \epsilon and \it{d} are.

It would help to picture this capacitor that you're trying to build. From wikipedia:

The simplest practical capacitor consists of two wide, flat, parallel plates separated by a thin dielectric layer.

You need to investigate what a dielectric is and how that relates \epsilon and \it{k}. I'll give you a hint and tell you that \epsilon_{0} is known as the permittivity of free space.

Hope this helps.

 

[Number1Ballar]

hmmm okay I'm starting to see how a capacitor works, but am still a little confused on how to do this problem.

I see that if the paper was in between these 2 sheets, then 'd' would equal 0.0003m, 'A' would equal 0.0125m², and for \epsilon I would plug in the value of k given?

so..

C = (3.7)(0.0125m²) / (0.0003m) = 154.17 ...which seems quite high for the answer...I think I'm still not getting something

 

[learningstill]

You need to better understand the differences between \epsilon, \epsilon_{0} and \it{k}. Try http://en.wikipedia.org/wiki/Dielectric_constant" where \epsilon_{s} as given in their equation is what your equation calls \epsilon.

 

[Number1Ballar]

actually I think I figured it out..

C = (3.7)(8.85 X 10^-12 F/m)(0.0125m²) / (0.0003m) = 1.36 X 10^-9F

= 1.36 nF