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184_notes:examples:week5_flux_cylinder [2017/09/22 15:17] – dmcpadden | 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. If we imagine a cylindrical surface with radius $R$ and height $h$ is situated in the field so that the bases of the cylinder are parallel to the $xz$-plane. What is the electric flux through the cylinder? | A constant electric field $\vec{E}$ is directed along the $x$-axis. If we imagine a cylindrical surface with radius $R$ and height $h$ is situated in the field so that the bases of the cylinder are parallel to the $xz$-plane. What is the electric flux through the cylinder? | ||
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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 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: | ||
$$\Phi_{\text{top}}=\vec{E}\bullet\vec{A}_{\text{top}}=(E\hat{x})\bullet(\pi R^2\hat{y})=0$$ | $$\Phi_{\text{top}}=\vec{E}\bullet\vec{A}_{\text{top}}=(E\hat{x})\bullet(\pi R^2\hat{y})=0$$ | ||
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$$\Phi_{\text{bottom}}=\vec{E}\bullet\vec{A}_{\text{bottom}}=(E\hat{x})\bullet(\pi R^2(-\hat{y}))=0$$ | $$\Phi_{\text{bottom}}=\vec{E}\bullet\vec{A}_{\text{bottom}}=(E\hat{x})\bullet(\pi R^2(-\hat{y}))=0$$ | ||
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. | 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: | + | [{{ 184_notes: |
If we were to calculate the electric flux through the cylinder' | If we were to calculate the electric flux through the cylinder' | ||
$$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$ | $$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$ | ||
- | FIXME I think it'd be worth showing the flux through A1 and the flux through A2 to emphasize that the parallel E bit will make a positive flux and an anti-parallel E bit will make a negative flux so they cancel. I think we're missing | + | Each little piece will have a " |
+ | $$\Phi_{A_1} = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_{A_2}$$ | ||
- | Within the sum, we can match up the little pieces of area with their opposites (like $\vec{A}_1$ and $\vec{A}_2$ in the figure). | + | It shouldn' |
- | $$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} | + | $$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} = 0$$ |
We can continue in the manner for the entire wall, and we will find that | We can continue in the manner for the entire wall, and we will find that | ||
$$\Phi_{\text{wall}}=0$$ | $$\Phi_{\text{wall}}=0$$ | ||
In total, | In total, | ||
$$\Phi_{\text{cylinder}}=\Phi_{\text{top}}+\Phi_{\text{bottom}}+\Phi_{\text{wall}}=0$$ | $$\Phi_{\text{cylinder}}=\Phi_{\text{top}}+\Phi_{\text{bottom}}+\Phi_{\text{wall}}=0$$ |