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184_notes:examples:week4_charge_cylinder [2018/02/03 21:38] – tallpaul | 184_notes:examples:week4_charge_cylinder [2021/06/29 18:05] – schram45 | ||
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* The cylinder has a charge $Q$, which is uniformly distributed. | * The cylinder has a charge $Q$, which is uniformly distributed. | ||
* The cylinder has length $L$ and radius $R$. | * The cylinder has length $L$ and radius $R$. | ||
- | * The electric field due to a ring with the same dimensions in the $xy$-plane at a point along the $z$-axis is $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/ | + | * The electric field due to a ring with the same dimensions in the $xy$-plane at a point along the $z$-axis is $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/ |
===Goal=== | ===Goal=== | ||
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===Representations=== | ===Representations=== | ||
- | * We can represent the cylindrical shell and $P$ as follows: | ||
{{ 184_notes: | {{ 184_notes: | ||
- | * We can represent $\text{d}Q$ and $\vec{r}$ for our cylinder as follows (since we already know the electric field from a ring along such an axis): | ||
- | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
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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_0z^2}\left(\frac{1}{1-\frac{L^2}{4z^2}}\right) \\ |
- | &\approx | + | &= \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. | ||
+ | |||
+ | \begin{align*} | ||
+ | \vec{E} | ||
\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 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. |