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--- 184_notes:examples:week5_flux_cube_plane [2017/09/19 12:46] – [Example: Flux through a Cube on a Charged Plane] 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: $\vec{E} = \frac{\sigma}{2\epsilon_0}(\pom\hat{z})$ (where $\pom\hat{z}$ 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 18: / Line 19: @@
 $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A}$$
   * We represent the electric field due to a plane of uniform surface charge density with:
-$$\vec{E} = \frac{\sigma}{2\epsilon_0}(\pom\hat{z})$$
+$$\vec{E} = \frac{\sigma}{2\epsilon_0}(\pm\hat{z})$$
   * We represent the situation with the following diagram.
-{{ 184_notes:5_flux_two_radii.png?300 |Point charge and two spherical shells}}
+{{ 184_notes:5_cube_plane.png?400 |Charged Plane and Cubic Surface}}
 ====Solution====
-Before we dive into calculations, let's consider how we can simplify the problem. Think 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. 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.
+First, we evaluate the situation qualitatively. Consider the electric field lines of the charged plane:
+{{ 184_notes:5_plane_field_lines.png?400 |Negatively Charged Plane -- Electric Field Lines}}
-Since $\vec{E}$ is constant with respect to $\text{d}\vec{A}$ (in this case, it is sufficient that $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude), we can rewrite our flux representation:
+You might notice that we have oriented the cube conveniently. The electric field is parallel to the sides of the cube, so there are no electric field lines entering or exiting from the side of the cube. So the flux through the sides should be $0$. For the top and bottom of the cube, the electric field lines are perpendicular to the surface, which means they are parallel to the area-vectors. These facts will greatly simplify our integral calculation of the flux.
-$$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$
-We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge's radial axis) since it is constant on the surface of a given spherical shell. We use the formula for the electric field from a point charge.
-$$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$
-We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $E=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we have $E=2.5\cdot 10^6 \text{ N/C}$.
-To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by $\text{d}A$. Since we are integrating this little piece over the entire shell, we end up with the area of the shell's surface:
-$$\int\text{d}A=A=4\pi r^2$$
-The last expression, $4\pi r^2$, is just the surface area of a sphere. We can plug in values for $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $A=1.13\cdot 10^{-2} \text{ m}^2$. For the larger shell, we have $A=4.52\cdot 10^{-2} \text{ m}^2$.
-Now, we bring it together to find electric flux, which after all our simplifications can be written as $\Phi_e=EA$. For our two shells:
 \begin{align*}
-\Phi_{\text{small}} &= 1.0\cdot 10^7 \text{ N/C } \cdot 1.13\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^5 \text{ Nm}^2\text{/C} \\
+\Phi_{\text{total}} &= \Phi_{\text{sides}}+\Phi_{\text{top}}+\Phi_{\text{bottom}} \\
-\Phi_{\text{large}} &= 2.5\cdot 10^6 \text{ N/C } \cdot 4.52\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^5 \text{ Nm}^2\text{/C}
+                    &= 0 + \int_{\text{top}}\vec{E}\bullet \text{d}\vec{A} + \int_{\text{bottom}}\vec{E}\bullet \text{d}\vec{A} \\
+                    &= \int_{\text{top}}E\hat{z}\bullet \text{d}A\hat{z} + \int_{\text{bottom}}E(-\hat{z})\bullet \text{d}A(-\hat{z}) \\
+                    &= E\int_{\text{top}}\text{d}A + E\int_{\text{bottom}}\text{d}A \\
+                    &= E\cdot(A_{\text{top}}+A_{\text{bottom}}) \\
+                    &= \frac{\sigma}{2\epsilon_0}(2l^2) \\
+                    &= \frac{\sigma l^2}{\epsilon_0}
 \end{align*}
+When we plug in values for $\sigma$, $l$, and $\epsilon_0$, we get $\Phi_{\text{cube}}=1.15\cdot 10^5\text{ Vm}$.
-We get the same answer for both shells! It turns out that the radius of the shell does not affect the electric flux for this example. You'll see later that electric flux through a closed surface depends //only// on the charge enclosed. The surface can be weirdly shaped, and we can always figure out the flux as long as we know the charge enclosed.
+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
+$$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$
+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.