184_notes:examples:week5_flux_two_radii

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184_notes:examples:week5_flux_two_radii [2018/07/24 15:02] – curdemma184_notes:examples:week5_flux_two_radii [2021/06/04 00:43] – schram45
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   * $\Phi_e$ for each sphere   * $\Phi_e$ for each sphere
   * $\text{d}\vec{A}$ or $\vec{A}$, if necessary   * $\text{d}\vec{A}$ or $\vec{A}$, if necessary
- 
-===Approximations & Assumptions=== 
-  * There are no other charges that contribute appreciably to the flux calculation. 
-  * There is no background electric field. 
-  * The electric fluxes through the spherical shells are due only to the point charge. 
  
 ===Representations=== ===Representations===
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 $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$
   * We represent the situation with the following diagram. Note that the circles are indeed spherical shells, not rings as they appear.   * We represent the situation with the following diagram. Note that the circles are indeed spherical shells, not rings as they appear.
-{{ 184_notes:5_flux_two_radii.png?300 |Point charge and two spherical shells}}+[{{ 184_notes:5_flux_two_radii.png?300 |Point charge and two spherical shells}}] 
 + 
 +<WRAP TIP> 
 +===Approximations & Assumptions=== 
 +  * There are no other charges that contribute appreciably to the flux calculation. 
 +  * There is no background electric field. 
 +  * The electric fluxes through the spherical shells are due only to the point charge. 
 +  * Perfect spheres. 
 +  * Constant charge for the point charge (no charging/discharging). 
 +</WRAP> 
 ====Solution==== ====Solution====
 Before we dive into calculations, let's consider how we can simplify the problem by thinking about the nature of the electric field due to a point charge and of the $\text{d}\vec{A}$ vector for a spherical shell. The magnitude of the electric field will be constant along the surface of a given sphere, since the surface is a constant distance away from the point charge. Further, $\vec{E}$ will always be parallel to $\text{d}\vec{A}$ on these spherical shells, since both are directed along the radial direction from the point charge. See below for a visual. A more in-depth discussion of these symmetries can be found in the notes of [[184_notes:eflux_curved#Making_Use_of_Symmetry|using symmetry]] to simplify our flux calculation. Before we dive into calculations, let's consider how we can simplify the problem by thinking about the nature of the electric field due to a point charge and of the $\text{d}\vec{A}$ vector for a spherical shell. The magnitude of the electric field will be constant along the surface of a given sphere, since the surface is a constant distance away from the point charge. Further, $\vec{E}$ will always be parallel to $\text{d}\vec{A}$ on these spherical shells, since both are directed along the radial direction from the point charge. See below for a visual. A more in-depth discussion of these symmetries can be found in the notes of [[184_notes:eflux_curved#Making_Use_of_Symmetry|using symmetry]] to simplify our flux calculation.
  
-{{ 184_notes:electricflux4.jpg?400 |Area-vectors and E-field-vectors point in same direction}}+[{{ 184_notes:electricflux4.jpg?400 |Area-vectors and E-field-vectors point in same direction}}]
  
 Since the shell is a fixed distance from the point charge, the electric field has constant magnitude on the shell. Since $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude (on the shell), the dot product simplifies substantially: $\vec{E}\bullet \text{d}\vec{A} = E\text{d}A$. We can now rewrite our flux representation: Since the shell is a fixed distance from the point charge, the electric field has constant magnitude on the shell. Since $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude (on the shell), the dot product simplifies substantially: $\vec{E}\bullet \text{d}\vec{A} = E\text{d}A$. We can now rewrite our flux representation:
  • 184_notes/examples/week5_flux_two_radii.txt
  • Last modified: 2021/06/04 00:47
  • by schram45