Why can we model spherical capacitor with two dielectrics as two capacitors in series?

[zenterix]

Homework Statement

Consider a conducting spherical shell with an inner radius $a$ and an outer radius $c$.

Let the space between the two spherical surfaces be filled with two different dielectric materials such that the dielectric constant is $\kappa_1$ between $a$ and $b$ and $\kappa_2$ between $b$ and $c$.

Relevant Equations

Determine the capacitance of the system.

I solved this problem by simply applying the formula for capacitance. The potential difference between a point on the inner shell and the outer shell is computed by considering the electric fields to be $\frac{E_0}{\kappa_1}$ between radius $a$ and $b$ and $\frac{E_0}{\kappa_2}$ between $b$ and $c$, where $E_0$ is the electric field between capacitor plates without the dielectric materials.

After some algebra one reaches the correct answer.

The solution I saw in the book I am reading takes a different route, which I would like to understand.

The book says that the system can be treated as two capacitors connected in series.

From what I gathered, one capacitor is located between $a$ an $b$ and the other between $b$ and $c$.

As far as I know, we model the dielectric as forming a dipole which creates an electric field opposite the capacitor's field. The dipole is such that the positive charge is near the capacitor's negative plate, and the negative charge is near the capacitor's positive plate.

Therefore, I see an inner positively charged shell of the capacitor, then the negative end of the innermost dielectric dipole, then further out at $b$ there is the positive end of the dipole.

Now, this doesn't sound very correct, because at $b$ we would also have the negative pole of the dipole of the outermost dielectric material.

I am confused so let me stop writing. My question remains: why can we model the scenario in this problem as two capacitors in series?

 

[haruspex]

at b we would also have the negative pole of the dipole of the outermost dielectric material.

That is why it works. If we posit equal and opposite charges at the same point then we haven't changed anything.

 

[Delta2]

The easy answer to your question is that as you might already know by treatments of capacitors in series and parallel in textbooks (using Kirchoff's laws ) that

• Capacitors in Parallel have same voltage but different charge each
• Capacitors in Series have same charge but different voltage

We are in the 2nd case here, that is the voltage is different because the respective electric fields and the separation distances are different, but the charge is the same

However there seems to be a crucial difference: We are missing two conducting "plates" that is the inner shell of the outer capacitor and the outer shell of the inner capacitor. I don't have a very good answer here, I think this is countered by a surface charge density that we have at the conceptual spherical shell that is formed by the interface between the two dielectrics.

By reading your post more carefully as you pointed out we should have two thin layers of -Q and +Q charge there that is the total charge there will be zero, so it doesn't matter if those two conducting shells exist or not their total charge would be zero, given of course that the electric field is such as those charges were there.

 

[DaveE]

Another way to think of this is that because of the radial symmetry of the E-field there is an equipotential surface where the dielectrics meet. Inserting a conductor at that surface would make no change to the fields. Then you would literally have two capacitors in series.

 

[Delta2]

Another way to think of this is that because of the radial symmetry of the E-field there is an equipotential surface where the dielectrics meet. Inserting a conductor at that surface would make no change to the fields. Then you would literally have two capacitors in series.

Why didn't I think of this :D