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184_notes:examples:week5_flux_cylinder [2017/09/15 22:31] – [Solution] tallpaul | 184_notes:examples:week5_flux_cylinder [2018/07/24 14:52] (current) – curdemma | ||
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===== Example: Flux through a Closed Cylinder ===== | ===== Example: Flux through a Closed Cylinder ===== | ||
- | A constant electric field $\vec{E}$ is directed along the $x$-axis. | + | A constant electric field $\vec{E}$ is directed along the $x$-axis. |
===Facts=== | ===Facts=== | ||
Line 8: | Line 10: | ||
===Lacking=== | ===Lacking=== | ||
- | * $\Phi_e$ | + | * $\Phi_{\text{cylinder}}$ |
* $\vec{A}$ of cylinder, or $\text{d}\vec{A}$ pieces. | * $\vec{A}$ of cylinder, or $\text{d}\vec{A}$ pieces. | ||
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$$\Phi=\vec{E}\bullet \vec{A}$$ | $$\Phi=\vec{E}\bullet \vec{A}$$ | ||
* We represent the situation with the following diagram: | * We represent the situation with the following diagram: | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | Before we dive into the math, let's reason about the nature of the situation. It will be helpful to visualize how the area vector $\text{d}\vec{A}$ would look for different parts of the cylinder' | + | Before we dive into the math, let's reason about the nature of the situation. It will be helpful to visualize how the area vector $\text{d}\vec{A}$ would look for different parts of the cylinder' |
- | {{ 184_notes: | + | [{{ 184_notes: |
- | Notice that the area vectors on the bases of the cylinder are pointing along the $y$-axis. Since the electric field is aligned with the $x$ axis, there will be no flux through the top and bottom of the cylinder. The math is right here: | + | Notice that the area vectors on the bases of the cylinder are pointing along the $y$-axis. Since the electric field is aligned with the $x$ axis, there will be no flux through the top and bottom of the cylinder. The math is here: |
$$\Phi_{\text{top}}=\vec{E}\bullet\vec{A}_{\text{top}}=(E\hat{x})\bullet(\pi R^2\hat{y})=0$$ | $$\Phi_{\text{top}}=\vec{E}\bullet\vec{A}_{\text{top}}=(E\hat{x})\bullet(\pi R^2\hat{y})=0$$ | ||
The same is true for bottom. For completeness, | The same is true for bottom. For completeness, | ||
- | $$\Phi_{\text{bottom}}=\vec{E}\bullet\vec{A}_{\text{bottom}=(E\hat{x})\bullet(\pi R^2(-\hat{y}))=0$$ | + | $$\Phi_{\text{bottom}}=\vec{E}\bullet\vec{A}_{\text{bottom}}=(E\hat{x})\bullet(\pi R^2(-\hat{y}))=0$$ |
+ | The wall of the cylinder is not so easy. Before we set up some nasty integral, consider the symmetry of the cylinder. For every tiny piece of the wall, there another tiny piece directly opposite, which will have the opposite area-vector. See below for a visual. | ||
+ | [{{ 184_notes: | ||
+ | If we were to calculate the electric flux through the cylinder' | ||
+ | $$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$ | ||
+ | |||
+ | Each little piece will have a " | ||
+ | $$\Phi_{A_1} = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_{A_2}$$ | ||
+ | |||
+ | It shouldn' | ||
+ | $$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} = 0$$ | ||
+ | We can continue in the manner for the entire wall, and we will find that | ||
+ | $$\Phi_{\text{wall}}=0$$ | ||
+ | In total, | ||
+ | $$\Phi_{\text{cylinder}}=\Phi_{\text{top}}+\Phi_{\text{bottom}}+\Phi_{\text{wall}}=0$$ |