184_notes:examples:week5_flux_cube_plane

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184_notes:examples:week5_flux_cube_plane [2017/09/19 13:44] – [Example: Flux through a Cube on a Charged Plane] tallpaul184_notes:examples:week5_flux_cube_plane [2017/09/19 13:47] – [Solution] tallpaul
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 ===Approximations & Assumptions=== ===Approximations & Assumptions===
   * There are no other charges that contribute appreciably to the flux calculation.   * There are no other charges that contribute appreciably to the flux calculation.
-  * The cube is aligned with respect to the plane so that all its faces are either parallel or perpendicular to the plane.+  * The cube is aligned with respect to the plane so that all of its faces are either parallel or perpendicular to the plane.
  
 ===Representations=== ===Representations===
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 When we plug in values for $\sigma$, $l$, and $\epsilon_0$, we get $\Phi_{\text{cube}}=1.15\cdot 10^5\text{ Vm}$. When we plug in values for $\sigma$, $l$, and $\epsilon_0$, we get $\Phi_{\text{cube}}=1.15\cdot 10^5\text{ Vm}$.
  
-Notice that in the [[184_notes:gauss_ex|next section of notes]], we define "Gauss' Law", which states that the total flux through a close surface is the amount of charge divided by $\epsilon_0$. A quick check for this example shows us that the charge enclosed by the sphere covers an area of $l^2$, which means the charge is $\sigma\cdot l^2$. If we had used Gauss' Law, we would have quickly found that+Notice that in the [[184_notes:gauss_ex|next section of notes]], we define "Gauss' Law", which states that the total flux through a close surface is the amount of charge enclosed divided by $\epsilon_0$. A quick check for this example shows us that the charge enclosed by the sphere covers an area of $l^2$, which means the charge is $\sigma\cdot l^2$. If we had used Gauss' Law, we would have quickly found that
 $$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$ $$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$
 This is the same result! An alternative question for this example could have been: What is the electric field due to a uniformly charged plane? If we were not given the electric field at the beginning, we could have used symmetry arguments and Gauss' Law to work backwards, starting with the charge enclosed, and then using the integral formula for electric flux to solve for the electric field. There are sometimes electric fields that we do not know off-hand, and Gauss' Law is often the best tool to find them. This is the same result! An alternative question for this example could have been: What is the electric field due to a uniformly charged plane? If we were not given the electric field at the beginning, we could have used symmetry arguments and Gauss' Law to work backwards, starting with the charge enclosed, and then using the integral formula for electric flux to solve for the electric field. There are sometimes electric fields that we do not know off-hand, and Gauss' Law is often the best tool to find them.
  • 184_notes/examples/week5_flux_cube_plane.txt
  • Last modified: 2017/09/22 15:57
  • by dmcpadden