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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
@@ Line 16: / Line 16: @@
 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:4_ring_charge.png?350 |Ring Charge Representation}}
-<WRAP TIP>
-===Approximation===
-The ring is in a perfect circle.
-</WRAP>
 ====Solution====
@@ Line 27: / Line 22: @@
 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 ring is in a perfect circle.
 </WRAP>
 We also make a plan to tackle the integrating, which is a little tougher in cylindrical coordinates.
@@ Line 45: / Line 41: @@
 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  $\text{d}Q$ :
  $\text{d}Q=\lambda\text{d}l=\frac{Q}{2\pi R}R\text{d}\phi=\frac{Q\text{d}\phi}{2\pi}$
+<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.