# Capacitors in series

1. Feb 28, 2014

### oneplusone

I read about an example in which you had two metal plates, and in between them one third of the distance from the top plate downwards (towards the bottom plate) was made up of κ, and the rest was air. THe problem proceeded to calculate the capacitance of them in series.
I don't get how the top layer acts as a capacitor though. There's no metal plate? Can someone please help me understand what's going on?

2. Feb 28, 2014

### Staff: Mentor

What is "k"?

3. Feb 28, 2014

### UltrafastPED

A picture would make this a bit clearer ...

4. Feb 28, 2014

### 256bits

5. Feb 28, 2014

### 256bits

If I understand you correctly, you have 2 metal plates seperated by a distance d, with an insulator of k > 1 of thickness d/3 and air k =1 of thickness 2d/3 between the plates.

How was the calculation performed in the example?
Normally it would be 1/C = 1/C1 + 1/C2 for series capactitances.
If one would put a metal layer on the material, can you see that the metal layer would have a certain voltage, so you have 2 capacitors connected in parrallel - one with the insulator and one with air as the dielectric.

Or were they calculating the electric field between the plates?

6. Mar 1, 2014

### oneplusone

Sorry for being unclear.

@above you are correct; that's what we did. The problem i was having was understanding why they are two different compactors.

7. Mar 1, 2014

### dauto

They are treated as separate capacitors as a calculation technique. The metal plate in between isn't necessary (no electric charge would move if you added an infinitesimally thin metal plate between the insulator and the air, so it won't affect the calculation.