184_notes:examples:week4_charge_ring

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184_notes:examples:week4_charge_ring [2018/02/03 21:31] – [Solution] tallpaul184_notes:examples:week4_charge_ring [2021/05/25 14:38] (current) – schram45
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 +[[184_notes:linecharge|Return to line of charge]]
 +
 ===== 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 if $z = 0$ (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?
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 We begin with an approximation, which will make our calculations simpler, and makes sense based on our representation: We begin with an approximation, which will make our calculations simpler, and makes sense based on our representation:
   * 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.
 </WRAP> </WRAP>
 We also make a plan to tackle the integrating, which is a little tougher in cylindrical coordinates. We also make a plan to tackle the integrating, which is a little tougher in cylindrical coordinates.
- 
-This example is complicated enough that it's worthwhile to make a plan. 
  
 <WRAP TIP> <WRAP TIP>
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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://en.wikipedia.org/wiki/Arc_length|arc length]] formula). Remember from the notes on [[:184_notes:dq#dQ_-_Chunks_of_Charge|line charges]] that we can write $\text{d}Q=\lambda\text{d}l$. Since the charge $Q$ is uniformly distributed on the ring, we use the length of the ring or circumference ($2\pi R$) to write the line charge density $\lambda=Q/2\pi R$. Now, we have an expression for $\text{d}Q$: 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://en.wikipedia.org/wiki/Arc_length|arc length]] formula). Remember from the notes on [[:184_notes:dq#dQ_-_Chunks_of_Charge|line charges]] that we can write $\text{d}Q=\lambda\text{d}l$. Since the charge $Q$ is uniformly distributed on the ring, we use the length of the ring or circumference ($2\pi R$) to write the line charge density $\lambda=Q/2\pi R$. Now, we have an expression for $\text{d}Q$:
 $$\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$$ $$\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$$
 +
 +<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.
 +</WRAP>
  
 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.
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