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| 184_notes:examples:week4_charge_cylinder [2021/05/26 13:32] – [Solution] schram45 | 184_notes:examples:week4_charge_cylinder [2021/06/29 18:07] – schram45 | ||
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| \vec{E} &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{1}{1-\frac{L}{2z}}-\frac{1}{1+\frac{L}{2z}}\right) \\ | \vec{E} &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{1}{1-\frac{L}{2z}}-\frac{1}{1+\frac{L}{2z}}\right) \\ | ||
| &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{\left(1+\frac{L}{2z}\right)-\left(1-\frac{L}{2z}\right)}{\left(1-\frac{L}{2z}\right)\left(1+\frac{L}{2z}\right)}\right) \\ | &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{\left(1+\frac{L}{2z}\right)-\left(1-\frac{L}{2z}\right)}{\left(1-\frac{L}{2z}\right)\left(1+\frac{L}{2z}\right)}\right) \\ | ||
| - | &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz%5} | + | &= \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{\frac{L}{z}}{1-\frac{L^2}{4z^2}}\right) = \frac{Q\hat{x}}{4\pi\epsilon_0z^2}\left(\frac{1}{1-\frac{L^2}{4z^2}}\right) \\ |
| + | &= \frac{Q\hat{x}}{4\pi\epsilon_0\left(z^2-\frac{L^2}{4}\right)} | ||
| \end{align*} | \end{align*} | ||
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| + | Since $z$ is very large we will once again eliminate any constant terms tied in with it.$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}\hat{x}$$ | ||
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