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184_notes:examples:week5_flux_cylinder [2017/09/15 22:41] – [Solution] tallpaul | 184_notes:examples:week5_flux_cylinder [2017/09/24 18:59] – [Solution] tallpaul | ||
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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=== | ||
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===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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{{ 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, | ||
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If we were to calculate the electric flux through the cylinder' | If we were to calculate the electric flux through the cylinder' | ||
$$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$ | $$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$ | ||
- | Within the sum, we can match up the little | + | |
- | $$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_1}$$ | + | Each little |
+ | $$\Phi_{A_1} = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_{A_2}$$ | ||
+ | |||
+ | 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, | ||
+ | $$\Phi_{\text{cylinder}}=\Phi_{\text{top}}+\Phi_{\text{bottom}}+\Phi_{\text{wall}}=0$$ |