Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 184_notes:examples:week5_flux_cube_plane [2017/09/19 13:31] – [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: | =====Example: | ||
| - | Suppose you have a plane of charge with uniform surface charge density of σ=−4μC/m2. What is the electric flux through a cube with side-length 0.5 m that is placed halfway into the plane? Feel free to use the electric field due to an infinite uniform plane of charge: →E=σ2ϵ0(±ˆz) (where ±ˆz 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 σ=−4μC/m2. What is the electric flux through a cube with side-length $l=0.5 \text{ m}thatisplacedhalfwayintotheplane?Feelfreetousetheelectricfieldduetoaninfiniteuniformplaneofcharge:\vec{E} = \frac{\sigma}{2\epsilon_0}(\pm\hat{z})(where\pm\hat{z}$ 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=== | ===Facts=== | ||
| - | * The cube has side-length $q=1 \text{0.5 m}$. | + | * The cube has side-length $l=0.5 \text{ |
| * The cube is halfway into the plane -- presumably this means the plane bisects the cube. | * The cube is halfway into the plane -- presumably this means the plane bisects the cube. | ||
| * The plane has surface charge density σ=−4μC/m2. | * The plane has surface charge density σ=−4μC/m2. | ||
| Line 12: | Line 13: | ||
| ===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=== | ||
| Line 34: | Line 35: | ||
| &= \frac{\sigma l^2}{\epsilon_0} | &= \frac{\sigma l^2}{\epsilon_0} | ||
| \end{align*} | \end{align*} | ||
| - | 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 " | + | When we plug in values for σ, l, and ϵ0, we get $\Phi_{\text{cube}}=1.15\cdot 10^5\text{ Vm}$. |
| - | $$\Phi_{\text{cylinder}}=0$$ | + | |
| - | We gain more confidence when we read the [[184_notes: | + | Notice that in the [[184_notes: |
| - | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | + | $$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$ |
| - | Since the total charge of the dipole | + | 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, |