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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 184_notes:examples:week4_charge_cylinder [2018/02/03 22:37] – tallpaul | 184_notes:examples:week4_charge_cylinder [2021/07/22 18:21] (current) – schram45 | ||
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| ===Goal=== | ===Goal=== | ||
| * Find the electric field at $P$. | * 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=== | ===Representations=== | ||
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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. | ||
| + | |||