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--- 184_notes:examples:week4_charge_cylinder [2018/02/03 21:50] – tallpaul
+++ 184_notes:examples:week4_charge_cylinder [2021/07/22 18:21] (current) – schram45
@@ Line 12: / Line 12: @@
 ===Goal===
   * Find the electric field at  $P$ .
+===Assumptions===
+  * This is a perfect cylinder: This simplifies down the geometry of the problem and allows us to use any equations related to the geometry of the cylinder
+  * There is no top or bottom: We make this assumption so that we can break the cylinder up into rings and not have to do anything with the top or bottom sides of the cylinder (essentially we are dealing with a 2D tube).
 ===Representations===
@@ Line 30: / Line 34: @@
   * Slice the cylindrical shell into thin rings, which we know about from the [[184_notes:examples:week4_charge_ring|previous example]].
   * Write an expression for  $\text{d}Q$ , which is the charge of one of the rings.
-  * Decided o
+  * Decide on a consistent way to define the location of the ring, and use this to write an expression for  $\text{d}Q$
   * Assign a variable location to the  $\text{d}Q$  piece, and then use that location to find the separation vector,  $\vec{r}$ .
   * Write an expression for  $\text{d}\vec{E}$ .
@@ Line 42: / Line 46: @@
 Now that we are okay on defining  $\text{d}Q$  and  $\vec{r}$ , we update the representation to reflect these decisions:
 {{ 184_notes:4_cylinder_dq.png?400 |Cylindrical Shell dQ}}
+text etgdsygt fzuhf gfsfzubfuzsuf z xd uyfg cgi.
+kki99ki.
 Trivially, we also have  $r=|\vec{r}|=|z-x|$ . We retain the absolute value notation, so that we can generalize for when  $P$  is inside the cylinder. You see below that the absolute value notation immediately drops out.
@@ Line 67: / Line 74: @@
 \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{\frac{L}{z}}{1-\frac{L^2}{4z^2}}\right) \\
+        &= \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 charged objects from far away they can be approximated as points, kinda like looking at a person from a distance.