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| 184_notes:examples:week5_flux_cylinder [2017/09/24 18:59] – [Solution] tallpaul | 184_notes:examples:week5_flux_cylinder [2017/09/24 18:59] – [Solution] tallpaul |
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| 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: | 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: |
| $$\Phi_1 = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_2$$ | $$\Phi_{A_1} = \vec{E}\bullet\vec{A}_1 = E\cdot(\hat{x}\bullet\vec{A}_1) = E\cdot(\hat{x}\bullet-\vec{A}_2) = -\vec{E}\bullet\vec{A}_2 = -\Phi_{A_2}$$ |
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| It shouldn't be too surprising at this point to see that everything cancels out. When we match up the terms, we get: | It shouldn't be too surprising at this point to see that everything cancels out. When we match up the terms, we get: |