184_notes:examples:week5_flux_two_radii

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184_notes:examples:week5_flux_two_radii [2017/09/25 13:11] – [Solution] tallpaul184_notes:examples:week5_flux_two_radii [2021/06/04 00:47] (current) – schram45
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 +[[184_notes:eflux_curved|Return to Electric Flux through Curved Surfaces notes]]
 +
 =====Example: Flux through Two Spherical Shells===== =====Example: Flux through Two Spherical Shells=====
 Suppose you have a point charge with value 1μC. What are the fluxes through two spherical shells centered at the point charge, one with radius 3 cm and the other with radius 6 cm? Suppose you have a point charge with value 1μC. What are the fluxes through two spherical shells centered at the point charge, one with radius 3 cm and the other with radius 6 cm?
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   * Φe for each sphere   * Φe for each sphere
   * d→A or →A, if necessary   * d→A or →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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 →E=14πϵ0qr2ˆr →E=14πϵ0qr2ˆr
   * 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 a few approximations and assumptions we should make in order to simplify our model. 
 +  * 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. 
 +The first three assumptions ensure that there is nothing else contributing or affecting the flux through our spheres in the model. 
 +  * Perfect spheres: This will simplify our area vectors and allows us to use geometric equations for spheres in our calculations. 
 +  * Constant charge for the point charge: Ensures that the point charge is not charging or discharging with time. 
 +</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 d→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, →E will always be parallel to d→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 d→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, →E will always be parallel to d→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|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}}]
  
-FIXME (Quick something about why the dot product simplifies) Since →E is constant with respect to d→A (in this case, it is sufficient that $\vec{E}isparallelto\text{d}\vec{A}$ and has constant magnitude), we can 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 →E is parallel to d→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:
  
 Φe=∫→E∙d→A=E∫dA Φe=∫→E∙d→A=E∫dA
  
-FIXME (You could use this to justify why E is constant on area) We can rewrite E  since it is constant on the surface of a given spherical shell. (Where E is scalar value representing →E as a magnitude, with a sign indicating direction along point charge's radial axis.) We use the formula for the electric field from a point charge.+Note that E is a scalar value representing →E as a magnitude, which includes a sign indicating direction along the point charge's radial axis. We can rewrite E using the formula for the electric field from a point charge.
  
 E=14πϵ0qr2|ˆr|=14πϵ0qr2 E=14πϵ0qr2|ˆr|=14πϵ0qr2
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