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| 184_notes:examples:week4_charge_ring [2021/05/25 14:32] – schram45 | 184_notes:examples:week4_charge_ring [2021/05/25 14:38] (current) – schram45 | ||
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| 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: | ||
| - | |||
| - | <WRAP TIP> | ||
| - | ===Approximation=== | ||
| - | The ring is in a perfect circle. | ||
| - | </ | ||
| ====Solution==== | ====Solution==== | ||
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| We begin with an approximation, | We begin with an approximation, | ||
| * The thickness of the ring is infinitesimally small, and we can approximate it as a circle. | * The thickness of the ring is infinitesimally small, and we can approximate it as a circle. | ||
| + | * The ring is in a perfect circle. | ||
| </ | </ | ||
| We also make a plan to tackle the integrating, | We also make a plan to tackle the integrating, | ||
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| 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:// | ||
| $$\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$$ | $$\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$$ | ||
| + | |||
| + | <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. If this were not true, the charge density along the ring would not be constant. | ||
| + | </ | ||
| 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 here, because as you will soon see, all terms containing $\hat{s}$ will disappear. | 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 here, because as you will soon see, all terms containing $\hat{s}$ will disappear. | ||