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184_notes:examples:week7_cylindrical_capacitor [2017/10/09 16:05] – dmcpadden | 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., $L >> a, b$. | ||
- | * Cylinders are uniformly charged. | ||
===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, | ||
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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< |
- | FIXME Make the gaussian cylinder a different color and add area/ | + | [{{ 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. Note, this notation is read as " | 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 " | ||
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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< |