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184_notes:examples:week5_flux_cylinder_line [2017/09/22 16:04] – dmcpadden | 184_notes:examples:week5_flux_cylinder_line [2017/09/25 15:39] – [Solution] tallpaul | ||
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====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: |
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- | 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 (but the magnitude of the arrows are all the same). 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 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. |
\begin{align*} | \begin{align*} | ||
\Phi_{\text{total}} &= \Phi_{\text{bases}}+\Phi_{\text{wall}} \\ | \Phi_{\text{total}} &= \Phi_{\text{bases}}+\Phi_{\text{wall}} \\ |