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--- 184_notes:examples:week4_charge_ring [2018/02/03 21:27] – 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?
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 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.
-This example is complicated enough that it's worthwhile to make a plan.
 <WRAP TIP>
@@ Line 40: / 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.
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 There are still a couple issues to sort out before we proceed. First, what are the limits of integration? We are integrating over  $\text{d}\phi$ , which represents a small slice of the ring, as shown in our representation. We want to integrate over all of the slices of the ring, so we set our limits of integration to cover all of the angles in the circle:  $0$  to  $2\pi$  (which are the starting and ending values for  $\phi$ ).
+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.
+</WRAP>
 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: