- #1

PhysicsRock

- 114

- 18

- 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.

$$

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.