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184_notes:examples:week5_flux_cylinder_line [2017/09/19 14:45] – [Solution] tallpaul | 184_notes:examples:week5_flux_cylinder_line [2018/07/24 15:20] – curdemma | ||
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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/ | ||
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* 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 charge density with: | + | * 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 line: | + | First, we evaluate the situation qualitatively. Consider the electric field vectors from the charged line near the surface of the cylinder: |
- | {{ 184_notes: | + | [{{ 184_notes: |
- | It's a little tough to demonstrate the field lines with only two dimensions to draw on, but you can imagine that the thicker arrows point out of the page more, and the thinner arrows point into the page more. In essence, each arrow should point directly away from and perpendicular to the line of charge, as indicated in the formula for electric field from a line charge. | + | It's a little tough to demonstrate the electric |
- | The electric field is parallel to the bases of the cylinder, so there are no electric field lines entering or exiting from the bases. So the flux through the bases should be $0$. For the wall of the cylinder, 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. | + | 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{bases}}+\Phi_{\text{wall}} \\ | \Phi_{\text{total}} &= \Phi_{\text{bases}}+\Phi_{\text{wall}} \\ |