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184_notes:examples:week4_charge_cylinder [2018/02/03 22:37] – tallpaul | 184_notes:examples:week4_charge_cylinder [2021/06/29 18:10] – schram45 | ||
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* Slice the cylindrical shell into thin rings, which we know about from the [[184_notes: | * 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 $\text{d}Q$, | ||
- | * 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, $\vec{r}$. | * 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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Now that we are okay on defining $\text{d}Q$ and $\vec{r}$, we update the representation to reflect these decisions: | Now that we are okay on defining $\text{d}Q$ and $\vec{r}$, we update the representation to reflect these decisions: | ||
{{ 184_notes: | {{ 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)} | ||
\end{align*} | \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. | ||
+ |