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184_notes:examples:week5_flux_cylinder_line [2017/09/19 14:30] – tallpaul | 184_notes:examples:week5_flux_cylinder_line [2017/09/25 15:40] – [Solution] tallpaul | ||
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=====Example: | =====Example: | ||
- | Suppose you have a line of charge with a uniform linear charge density of $\lambda=15\mu\text{C/ | + | Suppose you have a line of charge with a uniform linear charge density of $\lambda=15\mu\text{C/ |
===Facts=== | ===Facts=== | ||
Line 13: | Line 13: | ||
* 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 cylinder is aligned with respect to the line so that its bases are perpendicular to the line, and its wall is parallel (as described). | * The cylinder is aligned with respect to the line so that its bases are perpendicular to the line, and its wall is parallel (as described). | ||
+ | * Line of charge is very very long | ||
===Representations=== | ===Representations=== | ||
* We represent the electric flux through a surface with: | * We represent the electric flux through a surface with: | ||
$$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A}$$ | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A}$$ | ||
- | * We represent the electric field due to a line of uniform | + | * We represent the electric field due to an infinite |
$$\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ | $$\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ | ||
* We represent the situation with the following diagram. | * We represent the situation with the following diagram. | ||
{{ 184_notes: | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | First, we evaluate the situation qualitatively. Consider the electric field lines of the charged | + | First, we evaluate the situation qualitatively. Consider the electric field vectors from the charged |
- | {{ 184_notes:5_plane_field_lines.png?400 |Negatively | + | {{ 184_notes:5_line_field_lines.png?400 ||Positively |
- | You might notice that we have oriented | + | It's a little tough to demonstrate |
+ | |||
+ | The electric field vectors are parallel to the bases of the cylinder, so $\vec{E}\bullet\text{d}\vec{A}=0$ on the bases. So the flux through the bases should be $0$. For the wall of the cylinder, the electric field vectors | ||
\begin{align*} | \begin{align*} | ||
- | \Phi_{\text{total}} &= \Phi_{\text{sides}}+\Phi_{\text{top}}+\Phi_{\text{bottom}} \\ | + | \Phi_{\text{total}} &= \Phi_{\text{bases}}+\Phi_{\text{wall}} \\ |
- | &= 0 + \int_{\text{top}}\vec{E}\bullet \text{d}\vec{A} + \int_{\text{bottom}}\vec{E}\bullet \text{d}\vec{A} \\ | + | &= 0 + \int_{\text{wall}}\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}) | + | &= \int_{\text{wall}}E\hat{r}\bullet \text{d}A\hat{r} \\ |
- | &= E\int_{\text{top}}\text{d}A + E\int_{\text{bottom}}\text{d}A \\ | + | &= E\int_{\text{wall}}\text{d}A \\ |
- | & | + | & |
- | &= \frac{\sigma}{2\epsilon_0}(2l^2) \\ | + | &= \frac{\lambda}{2\pi R \epsilon_0}\cdot 2\pi R l \\ |
- | &= \frac{\sigma l^2}{\epsilon_0} | + | &= \frac{\lambda |
\end{align*} | \end{align*} | ||
- | When we plug in values for $\sigma$, $l$, and $\epsilon_0$, | + | When we plug in values for $\lambda$, $l$, and $\epsilon_0$, |
- | Notice that in the [[184_notes: | + | Notice that in the [[184_notes: |
- | $$\Phi_{\text{cube}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\sigma l^2}{\epsilon_0}$$ | + | $$\Phi_{\text{cylinder}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\lambda |
- | This is the same result! An alternative question for this example could have been: What is the electric field due to a uniformly charged | + | This is the same result! An alternative question for this example could have been: What is the electric field due to a uniformly charged |