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--- 184_notes:examples:week5_gauss_ball [2021/06/07 13:44] – 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>
 Notice that the above result is the same for both  $r<R$  and  $r>R$ . Next, we need to find the total charge enclosed by the Gaussian surface. Here, it is helpful to determine the charge density in our insulator. There is helpful section in the notes on [[184_notes:dq|determining charge density]]. Since our ball is uniformly charged throughout its volume, we just need to divide the total charge by the total volume. The volume charge density is then:
  $\rho=\frac{Q_{\text{total}}}{V_{\text{total}}}=\frac{Q}{\frac{4}{3}\pi R^3}=\frac{3Q}{4\pi R^3}$
+<WRAP TIP>
+===Assumption===
+Having a constant charge density is only possible if we assume that the ball is evenly charged throughout its volume. This may or may not be the case depending on how our insulator is charged.
+</WRAP>
 The charge enclosed by the Gaussian surface, then, would relate to the charge density. For  $r<R$ , the surface does not enclose the entire ball, so we would just multiply the volume that the Gaussian surface encloses by the charge density to find the charge contained in the Gaussian surface.
  $Q_{\text{enclosed}}(r<R)=\rho \cdot V_{\text{spherical shell}} = \frac{3Q}{4\pi R^3}\cdot \frac{4}{3}\pi r^3=\frac{Qr^3}{R^3}$
@@ Line 72: / 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  $r>R$ .