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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 [2021/06/29 18:09] – schram45 | ||
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<WRAP TIP> | <WRAP TIP> | ||
=== Plan === | === 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 | + | * Slice the cylindrical shell into thin rings, which we know about from the [[184_notes: |
- | * 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, $\vec{r}$. | * 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 $\text{d}\vec{E}$. | * Write an expression for $\text{d}\vec{E}$. | ||
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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, | 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, | ||
- | We choose $\vec{r}$ based on what we know about the electric field from a ring. In the [[: | + | We choose $\vec{r}$ based on what we know about the electric field from a ring. In the [[: |
+ | |||
+ | Now that we are okay on defining $\text{d}Q$ and $\vec{r}$, we update the representation to reflect these decisions: | ||
+ | {{ 184_notes: | ||
+ | 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. | 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. | ||
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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}\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_0\left(z^2-\frac{L^2}{4}\right)} |
- | &= \frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}\hat{x} | + | |
\end{align*} | \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 | + | |
+ | 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 |