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--- 184_notes:examples:week5_gauss_ball [2021/06/07 13:48] – schram45
+++ 184_notes:examples:week5_gauss_ball [2021/06/07 14:02] (current) – schram45
@@ Line 12: / Line 12: @@
   * The "Gaussian surface" we will use for our application of Gauss' Law.
   *  $\vec{E}(\vec{r})$
-===Approximations & Assumptions===
-  * The ball is a perfect sphere.
-  * There are no other charges that affect our calculations.
-  * The ball is not discharging.
-  * The ball is uniformly charged throughout its volume.
-  * For the sake of representation, we assume  $Q>0$  (we don't make this assumption for the calculation, it just helps to visualize little charges as little  $+$  signs).
 ===Representations===
@@ Line 27: / Line 20: @@
   * We represent the situation with the following diagram. The insulating case (part A) is shown on the left, and the conductor (part B) is on the right.
 [{{ 184_notes:5_ball_insulator_conductor.png?400 |Charged Ball as an Insulator or Conductor}}]
+<WRAP TIP>
+===Assumptions===
+There are a few assumptions that can be made to simplify down our model before starting any calculations.
+  * There are no other charges that affect our calculations.
+  * The ball is not discharging.
+  * For the sake of representation, we assume  $Q>0$  (we don't make this assumption for the calculation, it just helps to visualize little charges as little  $+$  signs).
+</WRAP>
 ====Solution (Part A)====
 A step-by-step approach to using Gauss' Law is shown in the [[184_notes:gauss_ex|notes]] for a line of charge. We will take a similar approach here.
@@ Line 46: / Line 48: @@
 <WRAP TIP>
 ===Approximation===
-In order to take the electric field term out of the integral an approximation must be made.
+In order to take the electric field term out of the integral we must approximate our ball as a perfect sphere. Having a perfect sphere ensures all the area vectors are parallel to the electric field vectors. This also ensures that the electric field is constant through our Gaussian surface at a given radius.
-  * The ball must be a perfect sphere: This ensures the area vectors are parallel to the electric field vectors and gets rid of the dot product. We also know that the electric field should be constant at a given radius, so having a perfect sphere allows the electric field term to be pulled out of the integral.
 </WRAP>
@@ Line 78: / Line 79: @@
          \end{cases}
 \]
-Outside the ball, the electric field exists as if the ball were a point charge!
+Outside the ball, the electric field exists as if the ball were a point charge! It is also important to note that the relationships seen in our above equations also agree with what we already know about insulating spheres of charge.
 ====Solution (Part B)====
 We repeat the process above for the case that the ball is a conductor. Notice that much of the reasoning is the exact same. We still have spherical symmetry, and we choose the same Gaussian surface. It is pictured below for both  $r<R$  and  $r>R$ .