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184_notes:examples:week5_flux_cylinder_line [2017/09/25 15:40] – [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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$$\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ | $$\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ | ||
* We represent the situation with the following diagram. | * We represent the situation with the following diagram. | ||
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====Solution==== | ====Solution==== | ||
First, we evaluate the situation qualitatively. Consider the electric field vectors from the charged line near the surface of the cylinder: | 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 electric field vectors 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 vector points 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 field vectors 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 vector points directly away from and perpendicular to the line of charge, as indicated in the formula for electric field from a line charge. | ||