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--- 184_notes:examples:week4_charge_cylinder [2018/02/03 21:40] – [Example: Electric Field from a Cylindrical Shell of Charge] 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 27: / Line 31: @@
 <WRAP TIP>
 === Plan ===
-We will use integration to find the electric field from the ring. We'll go through the following steps.
+We will use integration to find the electric field from the entire cylindrical shell. We'll go through the following steps.
-  * Reason about the geometry of the ring to find and $\text{d}l $and$ \lambda$.
+  * 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  $\text{d}Q$ .
+  * Write an expression for $\text{d}Q$, which is the charge of one of the rings.
+  * 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 37: / Line 42: @@
 Notice that our  $\text{d}Q$  is different than other  $\text{d}Q$ s we have used so far. But using a whole ring as our  $\text{d}Q$  makes sense. The cylindrical shell is 2-dimensional, which means that if our  $\text{d}Q$  is a small patch of the surface, then we will have a 2-dimensional integral, which is doable but more complicated than necessary. Instead, we can take a thin slice of the entire cylinder, which gives us a ring. We only need an integral for traversing along the //length// of the cylinder, and we are able to account for the entire surface of the cylinder while travelling in only one dimension. We will represent our  $\text{d}Q$  as a fraction of the total  $Q$  based on the thickness of our ring (we set our coordinates such that  $+x$ -direction is to the right, and the center of the cylinder is at the origin):  $\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}x}{L}$
-We choose  $\vec{r}$  based on what we know about the electric field from a ring. In the [[:184_notes:examples:Week4_charge_ring|previous example]], we determined an electric field vector that depends on  $z$ . Notice that the  $z$  from that example is the same  $z$  here. The vector also was aligned with the  $z$ -axis, which in this example is along the axis that passes through the cylinder and through the point  $P$ . As always,  $\vec{r}$  is directed from source to point of observation:  $\vec{r}=z\hat{x}-x\hat{x}=(z-x)\hat{x}$
+We choose  $\vec{r}$  based on what we know about the electric field from a ring. In the [[:184_notes:examples:Week4_charge_ring|previous example]], we determined an electric field vector that depends on  $z$ . Notice that the  $z$  from that example is the same  $z$  here. The vector also was aligned with the  $z$ -axis, which in this example is along the axis that passes through the cylinder and through the point  $P$ . The tricky part is determining the location of the "source", since a ring occupies many points in space. As long as we are consistent, though, we will be okay. Based on the result previous example, it makes sense to have the source location just be the center of the ring. As always,  $\vec{r}$  is directed from source to point of observation:  $\vec{r}=z\hat{x}-x\hat{x}=(z-x)\hat{x}$
+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 63: / 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) \\
-        &\approx \frac{Q\hat{x}}{4\pi\epsilon_0Lz}\left(\frac{L}{z}\right) \\
+        &= \frac{Q\hat{x}}{4\pi\epsilon_0\left(z^2-\frac{L^2}{4}\right)}
-        &= \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}\hat{x}
 \end{align*}
-So, for large  $z$ , the cylindrical shell looks like a point charge! So we when we are very far away, we can approximate the charge distribution as a point charge -- they both produce the same electric field. This seems to make sense, and is reassuring. We often approximate strange objects as point charges when they are far away, and this is another confirmation that that is an accurate assumption.
+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.