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--- 184_notes:examples:week4_charge_cylinder [2018/02/05 17:33] – tallpaul
+++ 184_notes:examples:week4_charge_cylinder [2021/06/29 18:09] – schram45
@@ Line 70: / Line 70: @@
 \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%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*}
+Since $z$ is very large we will once again eliminate any constant terms tied in with it.$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}\hat{x}$$
+As we can see this is exactly the equation we get for a point charge! This should be expected. When viewing objects from sufficiently far away they will appear as point charges, kinda like looking at a person from a distance.