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184_notes:examples:week4_charge_cylinder [2017/09/13 01:32] – [Solution] tallpaul | 184_notes:examples:week4_charge_cylinder [2021/06/29 18:10] – schram45 | ||
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===== Example: Electric Field from a Cylindrical Shell of Charge ===== | ===== Example: Electric Field from a Cylindrical Shell of Charge ===== | ||
=== Note: Super Challenge Problem!! -- This is a beyond the scope of this class (so you won't be expected to solve this kind of problem), but it is a cool example of how to expand from lines to areas of charge if you are interested === | === Note: Super Challenge Problem!! -- This is a beyond the scope of this class (so you won't be expected to solve this kind of problem), but it is a cool example of how to expand from lines to areas of charge if you are interested === | ||
+ | |||
Suppose we have a cylindrical shell with radius $R$ and length $L$ that has a uniform charge distribution with total charge $Q$. The cylinder does not have bases, so the charge is only distributed on the wall that wraps around the cylinder at the radius $R$. What is the electric field at a point $P$, which is a distance $z$ from the center of the cylinder, along the axis that passes through the center of the cylinder and parallel to its wall? What happens to the electric field as $z = 0$? What about for very large $z$? Why? | Suppose we have a cylindrical shell with radius $R$ and length $L$ that has a uniform charge distribution with total charge $Q$. The cylinder does not have bases, so the charge is only distributed on the wall that wraps around the cylinder at the radius $R$. What is the electric field at a point $P$, which is a distance $z$ from the center of the cylinder, along the axis that passes through the center of the cylinder and parallel to its wall? What happens to the electric field as $z = 0$? What about for very large $z$? Why? | ||
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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/ |
- | ===Lacking=== | + | ===Goal=== |
- | * Electric field at $P$. | + | * Find the electric field at $P$. |
- | * $\text{d}Q$ and $\vec{r}$ | + | |
- | + | ||
- | ===Approximations & Assumptions=== | + | |
- | * The electric field at $P$ is due solely to the cylinder. | + | |
- | * The thickness of the cylindrical shell is infinitesimally small, and we can approximate it as 2-dimensional shell. | + | |
===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==== | ||
+ | <WRAP TIP> | ||
+ | === Approximation === | ||
+ | We begin with an approximation, | ||
+ | * The thickness of the cylindrical shell is infinitesimally small, and we can approximate it as 2-dimensional shell. | ||
+ | </ | ||
+ | |||
+ | We also make a plan to tackle the integrating, | ||
+ | |||
+ | <WRAP TIP> | ||
+ | === Plan === | ||
+ | We will use integration to find the electric field from the entire cylindrical shell. We'll go through the following steps. | ||
+ | * Slice the cylindrical shell into thin rings, which we know about from the [[184_notes: | ||
+ | * Write an expression for $\text{d}Q$, | ||
+ | * 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 $\text{d}\vec{E}$. | ||
+ | * Figure out the bounds of the integral, and integrate to find electric field at $P$. | ||
+ | </ | ||
+ | |||
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 |