184_notes:examples:week5_gauss_ball

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184_notes:examples:week5_gauss_ball [2021/06/07 13:51] – schram45184_notes:examples:week5_gauss_ball [2021/06/07 14:02] (current) – schram45
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   * The "Gaussian surface" we will use for our application of Gauss' Law.   * The "Gaussian surface" we will use for our application of Gauss' Law.
   * $\vec{E}(\vec{r})$   * $\vec{E}(\vec{r})$
- 
-===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=== ===Representations===
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   * 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.   * 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}}] [{{ 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)==== ====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. 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.
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 <WRAP TIP> <WRAP TIP>
 ===Approximation=== ===Approximation===
-In order to take the electric field term out of the integral we must assume our ball is 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.+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.
 </WRAP> </WRAP>
  
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          \end{cases}          \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)==== ====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 $r>R$. 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 $r>R$.
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  • Last modified: 2021/06/07 13:51
  • by schram45