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184_notes:examples:week7_cylindrical_capacitor [2017/10/06 18:18] – [Solution] tallpaul | 184_notes:examples:week7_cylindrical_capacitor [2021/06/15 13:52] (current) – schram45 | ||
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=====Finding the Capacitance of a Cylindrical Capacitor===== | =====Finding the Capacitance of a Cylindrical Capacitor===== | ||
Find the capacitance of a cylindrical capacitor. The structure of the capacitor is a cylindrical shell inside another cylindrical shell. The two shells become oppositely charged when the capacitor is connected to a power source. The length of the cylinders is $L$, and their radii are $a$ and $b$, with $a<b$. | Find the capacitance of a cylindrical capacitor. The structure of the capacitor is a cylindrical shell inside another cylindrical shell. The two shells become oppositely charged when the capacitor is connected to a power source. The length of the cylinders is $L$, and their radii are $a$ and $b$, with $a<b$. | ||
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===Lacking=== | ===Lacking=== | ||
* Capacitance | * Capacitance | ||
- | |||
- | ===Approximations & Assumptions=== | ||
- | * The cylinders are much longer than they are far from one another, i.e., $L >> a, b$. | ||
===Representations=== | ===Representations=== | ||
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* We represent the situation below. | * We represent the situation below. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | In order to find capacitance, | + | In order to find capacitance, |
- | In order to arrive at potential difference, we will need to go through | + | In order to arrive at potential difference, we will need to find the electric field. Remember from the [[184_notes: |
We will end up integrating in the radial direction (a convenient choice, as this is how the electric field is directed!) from $a$ to $b$, which will give us the potential difference between the two cylinders. | We will end up integrating in the radial direction (a convenient choice, as this is how the electric field is directed!) from $a$ to $b$, which will give us the potential difference between the two cylinders. | ||
- | In order to find the electric field between the cylinders, we will use Gauss' Law. Below, we show a Gaussian surface that is cylindrical and fits inside the capacitor, with a radius $s$, which $a< | + | In order to find the electric field between the cylinders, we will use Gauss' Law. Below, we show a Gaussian surface that is cylindrical and fits inside the capacitor, with a radius $s$, which $a< |
- | {{ 184_notes: | + | [{{ 184_notes: |
We have done a [[184_notes: | We have done a [[184_notes: | ||
- | The integral is over the entire Gaussian surface, upon which $\vec{E}\bullet \text{d}\vec{A}$ takes on different values. On the top and bottom of the Gaussian surface, the electric field is directed radially, whereas the area-vectors point up and down, respectively. Both cases yield $\vec{E}\bullet \text{d}\vec{A}=0$, | + | The integral is over the entire Gaussian surface, upon which $\vec{E}\bullet \text{d}\vec{A}$ takes on different values. On the top and bottom of the Gaussian surface, the electric field is directed radially, whereas the area-vectors point up and down, respectively |
- | We write $E(s)$ because the electric field magnitude depends only on the distance from the central vertical axis. We can rotate the entire system about this axis and it does not change, so $E$ cannot depend on $\phi$. We can shift the entire system along this axis, and it does not change (using the "$L$ is very large" assumption), | + | We write $E(s)$ because the electric field magnitude depends only on the distance from the central vertical axis. Note, this notation is read as " |
At this point, we can apply Gauss' Law. We equate flux to the charge enclosed divided by $\epsilon_0$: | At this point, we can apply Gauss' Law. We equate flux to the charge enclosed divided by $\epsilon_0$: | ||
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& | & | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | <WRAP TIP> | ||
+ | ===Approximations & Assumptions=== | ||
+ | In order to take the electric field term out of the integral there are two assumptions that must be made. | ||
+ | * The charge is uniformly distributed amongst the cylindrical plates: Any charge concentrations would create inconsistencies in the electric field from the charges cylinders. This is a good assumption for highly conductive plate materials. | ||
+ | * The length of the cylinders is much greater than how far they are apart: This allows the electric field to be constant along the length of the cylinder at a given radius so long as the last assumption also holds. | ||
+ | </ | ||
The last thing we need is $Q_{enclosed}$. This is simply the fraction of $Q$ that the Gaussian surface encloses. Since the height of the Gaussian cylinder is $h$, we have $Q_{enclosed}=\frac{h}{L}Q$. We can now write the magnitude of the electric field at a radius $s$ from the central vertical axis (given that $a< | The last thing we need is $Q_{enclosed}$. This is simply the fraction of $Q$ that the Gaussian surface encloses. Since the height of the Gaussian cylinder is $h$, we have $Q_{enclosed}=\frac{h}{L}Q$. We can now write the magnitude of the electric field at a radius $s$ from the central vertical axis (given that $a< | ||
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& | & | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | Finally, we can find the capacitance based on the charge of one cylinder and the potential we just found: | ||
+ | $$C=\frac{Q}{\Delta V} = \frac{2\pi\epsilon_0 L}{\log\left(\frac{b}{a}\right)}$$ | ||
+ | Notice that the capacitance no longer depends on the charge Q, but only on the shape/size of our capacitor ($a$, $b$, and $L$ ) |