Why Capacitors in Parallels vs. Series: Coaxial Capacitor Case

[PhysicsRock]

Homework Statement

A cylindrical capacitor is made off of two coaxial metal tubes. Here, $r_1$ refers to the outer radius of the inner tube and $r_2$ the inner radius of the outer tube. Both metal pieces have a length of $l$. Between the two pipes, a glass tube is inserted from one side, a distance $a$ ($0 \leq a \leq l$) into the capacitor (filling the gap entirely). It's relative permittivity is $\varepsilon_r > 1$. Calculate the capacitance of the contraption as a function of $a$.

Relevant Equations

Capacitance of a cylindrial capacitor $\displaystyle C = \frac{2 \pi \varepsilon_0 L}{\displaystyle \ln\left( \frac{r_2}{r_1} \right)}$.

So my idea was to separate the capacitor into two individual ones, one of length $l - a$ filled with a vacuum and one of length $a$ filled with the glass tube. The capacitances then are

$$
C_0 = \frac{2 \pi \varepsilon_0 (l-a)}{\displaystyle \ln\left( \frac{r_2}{r_1} \right)}
$$

for the vacuum capacitor, and

$$
C_1 = \frac{2 \pi \varepsilon_0 \varepsilon_r a}{\displaystyle \ln\left( \frac{r_2}{r_1} \right)}
$$

for the capacitor with the dielectric. Originally, I thought they must be in series, however, doing the math, the overall capacitance for that case would be

$$
C = \frac{2 \pi \varepsilon_0 \varepsilon_r (l-a) a}{l + a (\varepsilon_r - 1)} \frac{1}{\displaystyle \ln\left( \frac{r_2}{r_1} \right)}.
$$

This, however, doesn't make any sense. For example, when plugging in $a = 0$, what one would expect is that the capacitance is equal to that of one cylindrical capacitor of length $l$ filled entirely with a vacuum. According to the above expression though, it would be zero.

So I tried calculating the capacitance for them being in parallel and I get

$$
C = \frac{2 \pi \varepsilon_0}{\displaystyle \ln\left( \frac{r_2}{r_1} \right)} [ l + a (\varepsilon_r - 1) ]
$$

which does satisfy all expectations, for example for the scenario discussed above. This leads to the conclusion that the capacitors must in fact be placed in parallel. However, I don't understand why, since typically for such problems the separated capacitors are always in series. Can any of you explain to me why this is the case here?

Thank you.

 

[kuruman]

Each of the metal tubes is an equipotential. This means that the potential difference across the vacuum capacitor is the same as the potential difference across the glass capacitor. Two capacitors that have the same potential difference across them form a parallel combination.

 

[PhysicsRock]

Each of the metal tubes is an equipotential. This means that the potential difference across the vacuum capacitor is the same as the potential difference across the glass capacitor. Two capacitors that have the same potential difference across them form a parallel combination.

Thank you. That makes total sense.