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184_notes:examples:week5_flux_cube_plane [2017/09/19 12:47] – [Example: Flux through a Cube on a Charged Plane] tallpaul | 184_notes:examples:week5_flux_cube_plane [2017/09/22 15:57] (current) – dmcpadden | ||
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+ | 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 $\sigma=-4\mu\text{C/ | + | Suppose you have a plane of charge with a uniform surface charge density of $\sigma=-4\mu\text{C/ |
===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 $\sigma=-4\mu\text{C/ | * The plane has surface charge density $\sigma=-4\mu\text{C/ | ||
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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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$$\vec{E} = \frac{\sigma}{2\epsilon_0}(\pm\hat{z})$$ | $$\vec{E} = \frac{\sigma}{2\epsilon_0}(\pm\hat{z})$$ | ||
* We represent the situation with the following diagram. | * We represent the situation with the following diagram. | ||
- | {{ 184_notes:5_flux_two_radii.png?300 |Point charge | + | {{ 184_notes:5_cube_plane.png?400 |Charged Plane and Cubic Surface}} |
====Solution==== | ====Solution==== | ||
- | Before we dive into calculations, let's consider how we can simplify | + | First, we evaluate |
- | + | {{ 184_notes: | |
- | 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 |
- | + | ||
- | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ | + | |
- | + | ||
- | We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge' | + | |
- | + | ||
- | $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ | + | |
- | + | ||
- | 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' | + | |
- | $$\int\text{d}A=A=4\pi r^2$$ | + | |
- | The last expression, $4\pi r^2$, is just the surface | + | |
- | + | ||
- | Now, we bring it together to find electric | + | |
\begin{align*} | \begin{align*} | ||
- | \Phi_{\text{small}} & | + | \Phi_{\text{total}} &= \Phi_{\text{sides}}+\Phi_{\text{top}}+\Phi_{\text{bottom}} \\ |
- | \Phi_{\text{large}} & | + | &= 0 + \int_{\text{top}}\vec{E}\bullet |
+ | | ||
+ | &= 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*} | \end{align*} | ||
+ | When we plug in values for $\sigma$, $l$, and $\epsilon_0$, | ||
- | We get the same answer for both shells! It turns out that the radius | + | Notice |
+ | $$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$ | ||
+ | This is the same result! An alternative question for this example could have been: What is the electric |