# Homework Help: Why can multiple dialectrics be treated as capacitors

1. Sep 25, 2015

### Gabe805

1. The problem statement, all variables and given/known data

I have come upon a number of problems where there are two or three different dialectrics inside a capacitor. according to the solutions manual I can often treat the dialectrics as capacitors in series or parallel. My question is why can you treat them as capacitors? Isn't the definition of a capacitor two plates of equal and opposite charge? Are not most dialectrics insulators? how do we know that both ends of the dialectric have equal and opposite charge?

2. Relevant equations
C=qv, v=-∫Eds

3. The attempt at a solution
If there are two dialectrics "in series" between a capacitor I know that the dialectric next to the positve end of the capacitor aquires a slightly negative charge due to induction but the other end that is next to the other dialectric is where I am having difficulty.

2. Sep 25, 2015

### Staff: Mentor

Ýou don't need plates for a capacitor, those are just the easiest cases for analysis.

You can imagine a very thin conducting plate between the dielectrics and parallel to the other plates (or a perfect insulator orthogonal to them), then you have your charged plates (or physical separation). This plate doesn't change the electrical properties at all.
Charge conservation, exactly because they don't conduct their total charge has to stay the same.

3. Sep 25, 2015

### ehild

The metal plates of the capacitor are equipotential surfaces, and all planes parallel with them are also equipotentials. An equipotential surface can be replaced by a thin metal plate, it does not change the electric field. This additional metal plate can be considered as two plates touching each other, but that is the same as if they were connected by a wire, that is, the capacitor can be replaced by parallel capacitors this way.
The electric field moves the electrons in the inserted plate so that one side is negative, the other is positive. When you slice that plate in two, one half gets negative charge, the other gets positive charge.

4. Sep 27, 2015

### rude man

Alternatively, forget about putting thin plates or sheets between the different dielectric regions:
Say you have 3 dielectric layers: d = d1 + d2 + d3 = distance between the two capacitor plates.
C = Q/V = Q/(E1 d1 + E2 d2 + E3 d3)
where E is the electric field in regions 1, 2 and 3 resp.
Fact: D = εE is continuous across all layers. So we can write
D = ε1 E1 = ε2 E2 = ε3 E3 = constant
So we can write C = (Q/D)/(d1/ε1 + d2/ε2 + d3/ε3)
But we know that for any parallel-plate capacitor i, Ci = εiAi/di, Ai = area of plates for that capacitor.
or dii =Ai/Ci
so finally C = (Q/AD)/(1/C1 + 1/C2 + 1/C3) = (1/C1 + 1/C2 + 1/C3)
which of course is three capacitors in series.
(The constant Q/AD is found from Gauss's theorem to be 1).
The extension to 2, 4 or more regions is obvious.