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| 184_notes:examples:week7_cylindrical_capacitor [2018/06/19 15:37] – curdemma | 184_notes:examples:week7_cylindrical_capacitor [2021/06/15 13:52] (current) – schram45 | ||
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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=== | ||
| Line 46: | Line 42: | ||
| & | & | ||
| \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< | ||