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184_notes:examples:week7_cylindrical_capacitor [2017/10/06 17:10] – [Finding the Capacitance of a Cylindrical Capacitor] tallpaul | 184_notes:examples:week7_cylindrical_capacitor [2021/06/15 13:52] (current) – schram45 | ||
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+ | [[184_notes: | ||
+ | |||
=====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==== | ||
- | Let's start with node $A$. Incoming current is $I_1$, | + | In order to find capacitance, |
- | $$I_{A\rightarrow B} = I_{out}-I_2 = I_{in}-I_2 = I_1-I_2 = 5 \text{ A}$$ | + | |
- | We do a similar analysis for node $B$. Incoming current is $I_{A\rightarrow B}$, and outgoing current is $I_3$. Since $I_{A\rightarrow B}=5 \text{ A}$ and $I_3=4 \text{ A}$, we need $I_{B\rightarrow D}$ to be outgoing to balance. To satisfy | + | In order to arrive at potential difference, we will need to find the electric field. Remember from the [[184_notes: |
- | $$I_{B\rightarrow D} = I_{out}-I_3 = I_{in}-I_3 = I_{A\rightarrow B}-I_3 = 1 \text{ | + | |
- | For node $C$, incoming current | + | We will end up integrating in the radial direction (a convenient choice, as this is how the electric field is directed!) from $a$ to $b$, which will give us the potential difference between the two cylinders. |
- | $$I_{C\rightarrow D} = I_{out} = I_{in} = I_2+I_3 = 7 \text{ 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 | ||
+ | |||
+ | [{{ 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 (as shown above). Both cases yield $\vec{E}\bullet \text{d}\vec{A}=0$, so we can safely ignore this part of the integral. On the wall of the Gaussian surface, the electric field and the area-vectors are //both// directed radially, so their dot product simplifies | ||
+ | |||
+ | 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$: | ||
+ | \begin{align*} | ||
+ | \frac{Q_{enclosed}}{\epsilon_0} &= \int\vec{E}\bullet \text{d}\vec{A} \\ | ||
+ | &= \int E(s)\text{d}A \\ | ||
+ | &= E(s) \int \text{d}A \\ | ||
+ | & | ||
+ | & | ||
+ | \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. | ||
+ | </ | ||
- | Lastly, | + | 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 |
- | $$I_{D\rightarrow battery} = I_{out} = I_{in} = I_{B\rightarrow D}+I_{B\rightarrow D} = 8 \text{ A}$$ | + | $$E(s)=\frac{Q_{enclosed}}{\epsilon_0}\frac{1}{2\pi sh} = \frac{Q}{2\pi\epsilon_0 L s}$$ |
- | Notice that $I_{D\rightarrow battery}=I_1$. | + | This is the magnitude |
+ | \begin{align*} | ||
+ | |\Delta V| &= \int_{r_i}^{r_f} \vec{E}\bullet \text{d}\vec{r} \\ | ||
+ | & | ||
+ | & | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
- | {{ 184_notes: | + | Finally, we can find the capacitance based on the charge of one cylinder and the potential we just found: |
+ | $$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 $L$ ) |