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--- 184_notes:examples:week5_flux_cube_plane [2017/09/19 13:34] – [Solution] tallpaul
+++ 184_notes:examples:week5_flux_cube_plane [2017/09/22 15:57] (current) – dmcpadden
@@ Line 1: / Line 1: @@
+FIXME - this is more or a homework problem for them. So I'm not sure we want to use this one...
 =====Example: Flux through a Cube on a Charged Plane=====
-Suppose you have a plane of charge with uniform surface charge density of $\sigma=-4\mu\text{C/m}^2$. What is the electric flux through a cube with side-length $0.5 \text{ m}$ that is placed halfway into the plane? Feel free to use the electric field due to an infinite uniform plane of charge: $\vec{E} = \frac{\sigma}{2\epsilon_0}(\pm\hat{z})$ (where $\pm\hat{z}$ points away from plane). Notice that the strength of the electric field does not depend on the distance from the plane -- it is constant apart from a change in direction when you cross over to the other side of the plane.
+Suppose you have a plane of charge with a uniform surface charge density of $\sigma=-4\mu\text{C/m}^2$. What is the electric flux through a cube with side-length $l=0.5 \text{ m}$ that is placed halfway into the plane? Feel free to use the electric field due to an infinite uniform plane of charge: $\vec{E} = \frac{\sigma}{2\epsilon_0}(\pm\hat{z})$ (where $\pm\hat{z}$ points away from plane). Notice that the strength of the electric field does not depend on the distance from the plane -- it is constant apart from a change in direction when you cross over to the other side of the plane.
 ===Facts===
-  * The cube has side-length $q=1 \text{0.5 m}$.
+  * The cube has side-length $l=0.5 \text{ m}$.
   * The cube is halfway into the plane -- presumably this means the plane bisects the cube.
   * The plane has surface charge density $\sigma=-4\mu\text{C/m}^2$.
@@ Line 12: / Line 13: @@
 ===Approximations & Assumptions===
   * 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===
@@ Line 35: / Line 36: @@
 \end{align*}
 When we plug in values for $\sigma$, $l$, and $\epsilon_0$, we get $\Phi_{\text{cube}}=1.15\cdot 10^5\text{ Vm}$.
-You can imagine that if we were able to draw this in three dimensions, we would have just as many field lines entering the cylinder as exiting. Since flux is a measure of the "flow" of electric field through a surface, we could say that the flux is zero: the inward flow cancels with the outward flow. For now, we tentatively write:
-$$\Phi_{\text{cylinder}}=0$$
+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
-We gain more confidence when we read the [[184_notes:gauss_ex|next section of notes]], where we define "Gauss' Law". This law states that the total flux through a close surface is the amount of charge divided by $\epsilon_0$, the [[https://en.wikipedia.org/wiki/Vacuum_permittivity|permittivity of free space]].
+$$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$
-$$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$
+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.
-Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above.