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--- 184_notes:examples:week5_flux_cylinder [2017/09/24 18:57] – [Solution] tallpaul
+++ 184_notes:examples:week5_flux_cylinder [2017/09/24 18:59] – [Solution] tallpaul
@@ Line 34: / Line 34: @@
 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:
-$$\Phi_1 = \vec{E}\bullet\vec{A}_1 = E\hat{x}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2)$$
+$$\Phi_{A_1} = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_{A_2}$$
-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't be too surprising at this point to see that everything cancels out. When we match up the terms, we get something like:
+It shouldn't be too surprising at this point to see that everything cancels out. When we match up the terms, we get:
-$$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} = \vec{E}\bullet\vec{A}_1+\vec{E}\bullet\vec{A}_2 = \vec{E}\bullet(\vec{A}_1+\vec{A}_2)=\vec{E}\bullet(0)=0$$
+$$\Phi_{\text{matching pieces}}=\Phi_{A_1}+\Phi_{A_2} = 0$$
 We can continue in the manner for the entire wall, and we will find that
 $$\Phi_{\text{wall}}=0$$
 In total,
 $$\Phi_{\text{cylinder}}=\Phi_{\text{top}}+\Phi_{\text{bottom}}+\Phi_{\text{wall}}=0$$