184_notes:examples:week7_cylindrical_capacitor

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184_notes:examples:week7_cylindrical_capacitor [2017/10/10 14:21] – [Solution] tallpaul184_notes:examples:week7_cylindrical_capacitor [2021/06/15 13:52] (current) – schram45
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 +[[184_notes:cap_in_cir|Return to Capacitors in Circuit]]
 +
 =====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:7_cylindrical_cap.png?200 |Cylindrical Capacitor}}+[{{ 184_notes:7_cylindrical_cap.png?200 |Cylindrical Capacitor}}]
 ====Solution==== ====Solution====
 In order to find capacitance, we need a charge $Q$ and a potential difference $\Delta V$. But the potential difference depends on how much charge is on the cylindrical plates. The idea here is to use the charge that is on the plates and use that charge to then solve for $\Delta V$. When we compute the capacitance, the charge on the plates should cancel out so that the capacitance only depends on the shape/size of the cylinders. For now, we say there is a charge $Q$ on the inner cylinder, and a charge $-Q$ on the outer cylinder. Since the cylinders are conductors (as they would be in any capacitor), the charge is uniformly distributed on the cylindrical shells. In order to find capacitance, we need a charge $Q$ and a potential difference $\Delta V$. But the potential difference depends on how much charge is on the cylindrical plates. The idea here is to use the charge that is on the plates and use that charge to then solve for $\Delta V$. When we compute the capacitance, the charge on the plates should cancel out so that the capacitance only depends on the shape/size of the cylinders. For now, we say there is a charge $Q$ on the inner cylinder, and a charge $-Q$ on the outer cylinder. Since the cylinders are conductors (as they would be in any capacitor), the charge is uniformly distributed on the cylindrical shells.
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 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 $a<s<b$. We also show vectors for area and electric field for the top and wall of the cylinder. 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 $a<s<b$. We also show vectors for area and electric field for the top and wall of the cylinder.
  
-{{ 184_notes:7_cylindrical_cap_gauss.png?200 |Cylindrical Capacitor with Gaussian Surface}}+[{{ 184_notes:7_cylindrical_cap_gauss.png?250 |Cylindrical Capacitor with Gaussian Surface}}]
  
 We have done a [[184_notes:examples:week5_flux_cylinder_line|similar problem]] involving Gauss's Law before. In order to find the electric flux in terms of the electric field, we write $$\Phi=\int\vec{E}\bullet \text{d}\vec{A}$$ We have done a [[184_notes:examples:week5_flux_cylinder_line|similar problem]] involving Gauss's Law before. In order to find the electric flux in terms of the electric field, we write $$\Phi=\int\vec{E}\bullet \text{d}\vec{A}$$
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                                  &= E(s) (2\pi sh)                                  &= E(s) (2\pi sh)
 \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.
 +</WRAP>
  
 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 $a<s<b$). 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 $a<s<b$).
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  • Last modified: 2017/10/10 14:21
  • by tallpaul