184_notes:examples:week5_flux_cylinder

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184_notes:examples:week5_flux_cylinder [2017/09/22 15:17] – dmcpadden184_notes:examples:week5_flux_cylinder [2017/09/24 18:57] – [Solution] tallpaul
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 $$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$ $$\Phi_{\text{wall}}=\sum_{\text{entire wall}}\Phi_{\text{little piece}}=\sum_{\text{entire wall}}\vec{E}\bullet\vec{A}_{\text{little piece}}$$
  
-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 that step.+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: 
 +$$\Phi_1 = \vec{E}\bullet\vec{A}_1 = E\hat{x}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2)$$
  
 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: 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:
  • 184_notes/examples/week5_flux_cylinder.txt
  • Last modified: 2018/07/24 14:52
  • by curdemma