Why Capacitors in Parallels vs. Series: Coaxial Capacitor Case

In summary, the two individual capacitors, one with vacuum and one with a glass dielectric, must be placed in parallel in order to satisfy all expectations and form an equipotential system. This is due to the fact that the potential difference across both capacitors is the same, making them suitable for a parallel combination.
  • #1
PhysicsRock
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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.
 
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  • #2
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.
 
  • #3
kuruman said:
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.
 

1. Why use capacitors in parallel instead of series in a coaxial capacitor case?

In a coaxial capacitor case, using capacitors in parallel allows for a higher capacitance value to be achieved, as the total capacitance is the sum of the individual capacitors. This is beneficial for applications that require a larger capacitance value.

2. What is the difference between using capacitors in parallel and series in a coaxial capacitor case?

The main difference is in the total capacitance achieved. In parallel, the total capacitance is the sum of the individual capacitances, while in series, the total capacitance is less than the individual capacitances. Additionally, in parallel, the voltage across each capacitor is the same, while in series, the voltage is divided among the capacitors.

3. Can I mix capacitors in parallel and series in a coaxial capacitor case?

Yes, it is possible to mix capacitors in parallel and series in a coaxial capacitor case. This can be useful in applications where a specific capacitance value is needed and can only be achieved by combining capacitors in different configurations.

4. What are the advantages of using capacitors in parallel in a coaxial capacitor case?

One advantage is the increased capacitance value that can be achieved. Additionally, using capacitors in parallel can also improve the overall stability and reliability of the circuit, as the load is distributed among multiple capacitors.

5. Are there any drawbacks to using capacitors in parallel in a coaxial capacitor case?

One potential drawback is the increased cost, as using multiple capacitors can be more expensive than using a single larger capacitor. Additionally, using capacitors in parallel can also increase the overall size and complexity of the circuit.

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