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--- 184_notes:examples:week5_flux_cylinder [2017/09/15 22:48] – [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 =====
-A constant electric field $\vec{E}$ is directed along the $x$-axis. A cylinder 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?
 ===Facts===
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 If we were to calculate the electric flux through the cylinder's wall, we could set it up like a sum over all the little pieces of area that add up to the entire wall.
 $$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$
-Within the sum, we can match up the little pieces of area with their opposites (like $\vec{A}_1$ and $\vec{A}_2$ in the figure). It shouldn't be too surprising at this point to see that everything cancels out. When we match up the terms, we get something like:
-$$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} = \vec{E}\bullet\vec{A}_1+\vec{E}\bullet\vec{A}_2 = \vec{E}\bullet(\vec{A}_1+\vec{A}_2)=\vec{E}\bullet(0)=0$$
+Each little piece will have a "matching piece", like $\vec{A}_1$ and $\vec{A}_2$ in the figure. Since area-vectors point out of the surface, matching pieces will be opposites: $\vec{A}_1 = - \vec{A}_2$. Since the electric field is constant everywhere, we find that the fluxes are also opposites:
+$$\Phi_{A_1} = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_{A_2}$$
+It shouldn't be too surprising at this point to see that everything cancels out. When we match up the terms, we get:
+$$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} = 0$$
 We can continue in the manner for the entire wall, and we will find that
 $$\Phi_{\text{wall}}=0$$
 In total,
 $$\Phi_{\text{cylinder}}=\Phi_{\text{top}}+\Phi_{\text{bottom}}+\Phi_{\text{wall}}=0$$