Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
184_notes:examples:week4_charge_ring [2017/09/13 01:27] – [Solution] tallpaul | 184_notes:examples:week4_charge_ring [2021/05/25 14:35] – schram45 | ||
---|---|---|---|
Line 1: | Line 1: | ||
+ | [[184_notes: | ||
+ | |||
===== Example: Electric Field from a Ring of Charge ===== | ===== Example: Electric Field from a Ring of Charge ===== | ||
- | Suppose we have a ring with radius $R$ that has a uniform charge distribution with total charge $Q$. What is the electric field at a point $P$, which is a distance $z$ from the center of the ring, along a line perpendicular to the plane of the ring? What happens to the electric field as $z = 0$ (i.e., when $P$ is in the center of the ring rather than above it)? Why? | + | Suppose we have a ring with radius $R$ that has a uniform charge distribution with total charge $Q$. What is the electric field at a point $P$, which is a distance $z$ from the center of the ring, along a line perpendicular to the plane of the ring? What happens to the electric field if $z = 0$ (i.e., when $P$ is in the center of the ring rather than above it)? Why? |
===Facts=== | ===Facts=== | ||
Line 9: | Line 11: | ||
* Electric field at $P$. | * Electric field at $P$. | ||
* $\text{d}Q$ and $\vec{r}$ | * $\text{d}Q$ and $\vec{r}$ | ||
- | + | | |
- | ===Approximations & Assumptions=== | + | |
- | | + | |
- | * The thickness of the ring is infinitesimally small, and we can approximate it as a circle. | + | |
- | * The ring is not discharging -- $Q$ is constant. | + | |
===Representations=== | ===Representations=== | ||
- | * The electric field from the point charge can be written as $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}$$ | + | We can represent the ring and $P$ as follows, with coordinates chosen conveniently. We choose cylindrical coordinates because we will be integrating over the length of the ring, and being able to represent its radius as constant will simplify calculations. |
- | * We can represent the ring and $P$ as follows, with coordinates chosen conveniently. We choose cylindrical coordinates because we will be integrating over the length of the ring, and being able to represent its radius as constant will simplify calculations. | + | |
{{ 184_notes: | {{ 184_notes: | ||
- | * We can represent $\text{d}Q$ and $\vec{r}$ for our ring as follows: | + | |
- | {{ 184_notes: | + | <WRAP TIP> |
+ | ===Approximation=== | ||
+ | The ring is in a perfect circle. | ||
+ | </ | ||
====Solution==== | ====Solution==== | ||
- | Just like before, we want to write our $\text{d}Q$ in terms of the little bit of length; however this time, our $\text{d}Q$ is spread over a little bit of the ring. We could try to write this little bit of length in terms of $\text{d}x$ or $\text{d}y$, | + | <WRAP TIP> |
+ | === Approximation === | ||
+ | We begin with an approximation, | ||
+ | * The thickness of the ring is infinitesimally small, and we can approximate it as a circle. | ||
+ | </ | ||
+ | We also make a plan to tackle the integrating, | ||
+ | |||
+ | <WRAP TIP> | ||
+ | === Plan === | ||
+ | We will use integration to find the electric field from the ring. We'll go through the following steps. | ||
+ | * Reason about the geometry of the ring to find and $\text{d}l$ and $\lambda$. | ||
+ | * 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 $P$. | ||
+ | </ | ||
+ | |||
+ | Just like before, we want to write our $\text{d}Q$ in terms of the little bit of length; however this time, our $\text{d}Q$ is spread over a little bit of the ring. We could try to write this little bit of length in terms of $\text{d}x$ or $\text{d}y$, | ||
+ | {{ 184_notes: | ||
So the $\text{d}Q$ in our representation takes up a small angle out of the whole circle, which we can call $\text{d}\phi$. The length of our $\text{d}Q$ is therefore $R\text{d}\phi$ (which comes from the [[https:// | So the $\text{d}Q$ in our representation takes up a small angle out of the whole circle, which we can call $\text{d}\phi$. The length of our $\text{d}Q$ is therefore $R\text{d}\phi$ (which comes from the [[https:// | ||
$$\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$$ | $$\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$$ | ||
- | To find an expression for $\vec{r}$, we can also consult the representation. $\vec{r}$ points from the location of $\text{d}Q$ to the point $P$. The location of $\text{d}Q$ is $\vec{r}_{\text{d}Q}=R\hat{s}$. This unit vector $\hat{s}$ may be unfamiliar, since we are used to working in Cartesian coordinates. $\hat{s}$ is the unit vector that points along the radius of a cylinder centered on the $z$-axis in our cylindrical coordinate system. In fact, $\hat{s}$ actually depends on $\phi$, and is more appropriately written as a function in terms of $\phi$, or $\hat{s}(\phi)$. We do not acknowledge the $\phi$-dependence in our solution | + | <WRAP TIP> |
+ | ===Assumption=== | ||
+ | The charge is evenly distributed along the ring. This also assumes the ring is a perfect conductor where charges will distribute evenly along the conductor. | ||
+ | </ | ||
+ | |||
+ | To find an expression for $\vec{r}$, we can also consult the representation. $\vec{r}$ points from the location of $\text{d}Q$ to the point $P$. The location of $\text{d}Q$ is $\vec{r}_{\text{d}Q}=R\hat{s}$. This unit vector $\hat{s}$ may be unfamiliar, since we are used to working in Cartesian coordinates. $\hat{s}$ is the unit vector that points along the radius of a cylinder centered on the $z$-axis in our cylindrical coordinate system. In fact, $\hat{s}$ actually depends on $\phi$, and is more appropriately written as a function in terms of $\phi$, or $\hat{s}(\phi)$. We do not acknowledge the $\phi$-dependence in some of our expressions | ||
The location of $P$ is $\vec{r}_P=z\hat{z}$. Then we have $\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=z\hat{z}-R\hat{s}$. This makes sense looking at the direction $\vec{r}$ points in our representation -- it points in towards the center ($-\hat{s}$ bit) and points up toward P ($+\hat{z}$ bit). We also need to know the magnitude $|\vec{r}|$, | The location of $P$ is $\vec{r}_P=z\hat{z}$. Then we have $\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=z\hat{z}-R\hat{s}$. This makes sense looking at the direction $\vec{r}$ points in our representation -- it points in towards the center ($-\hat{s}$ bit) and points up toward P ($+\hat{z}$ bit). We also need to know the magnitude $|\vec{r}|$, | ||
Line 39: | Line 62: | ||
There are still a couple issues to sort out before we proceed. First, what are the limits of integration? | There are still a couple issues to sort out before we proceed. First, what are the limits of integration? | ||
+ | |||
+ | Before we dive into the second issue, it's worth updated the plan, since this ended up a bit more complicated than anticipated. | ||
+ | |||
+ | <WRAP TIP> | ||
+ | === Plan === | ||
+ | We need to update the plan. Here are the steps we will take. We just now finished setting up the integral. | ||
+ | * We are not sure how to take an integral when $\hat{s}$ is involved. | ||
+ | * Zoom out. Try to figure out what the electric field might look like qualitatively. | ||
+ | * Try to simplify the integral to match our expectations for what the result will be. | ||
+ | </ | ||
The second issue has to do with the $\hat{s}$ terms, since $\hat{s}$ depends on $\phi$. Before we proceed, let's split up the integral for clarity: | The second issue has to do with the $\hat{s}$ terms, since $\hat{s}$ depends on $\phi$. Before we proceed, let's split up the integral for clarity: | ||
$$\vec{E}=\frac{Q}{8\pi^2\epsilon_0}\int_0^{2\pi}\frac{\text{d}\phi}{(R^2+z^2)^{3/ | $$\vec{E}=\frac{Q}{8\pi^2\epsilon_0}\int_0^{2\pi}\frac{\text{d}\phi}{(R^2+z^2)^{3/ | ||
- | Since $\hat{s}$ depends on $\phi$, $\hat{s}$ is not constant over the path of our integration. We can actually use the symmetry of the example to get rid of this issue: Notice that the ring and the point $P$ do not change when we rotate them about $z$-axis. We call this " | + | Since $\hat{s}$ depends on $\phi$, $\hat{s}$ is not constant over the path of our integration. We can actually use the symmetry of the example to get rid of this issue. Notice that electric field vectors from opposite points in the ring would have the same radial components, which will cancel out, since the radial components have opposite |
{{ 184_notes: | {{ 184_notes: | ||
- | Now, we can write the electric field vector | + | Now, we can write the electric field vector |
\begin{align*} | \begin{align*} | ||
- | \vec{E} & | + | \vec{E} &= \frac{Q}{8\pi^2\epsilon_0}\int_0^{2\pi}\frac{z\text{d}\phi}{(R^2+z^2)^{3/ |
&= \frac{1}{8\pi^2\epsilon_0}\frac{Qz\hat{z}}{(R^2+z^2)^{3/ | &= \frac{1}{8\pi^2\epsilon_0}\frac{Qz\hat{z}}{(R^2+z^2)^{3/ | ||
&= \frac{1}{8\pi^2\epsilon_0}\frac{Qz\hat{z}}{(R^2+z^2)^{3/ | &= \frac{1}{8\pi^2\epsilon_0}\frac{Qz\hat{z}}{(R^2+z^2)^{3/ |