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--- 184_notes:examples:week4_charge_ring [2018/02/03 21:37] – [Solution] tallpaul
+++ 184_notes:examples:week4_charge_ring [2021/05/25 14:38] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:linecharge|Return to line 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?
@@ Line 20: / 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 38: / 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.