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184_notes:examples:week5_flux_cube_plane [2017/09/19 13:28] – tallpaul | 184_notes:examples:week5_flux_cube_plane [2017/09/19 13:47] – [Solution] tallpaul | ||
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=====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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&= \frac{\sigma}{2\epsilon_0}(2l^2) \\ | &= \frac{\sigma}{2\epsilon_0}(2l^2) \\ | ||
&= \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 $\sigma$, $l$, and $\epsilon_0$, 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, |