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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 184_notes:linecharge [2018/09/12 15:44] – [Horizontal Line of Charge] dmcpadden | 184_notes:linecharge [2021/07/22 18:17] (current) – schram45 | ||
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| Sections 15.1-15.2 in Matter and Interactions (4th edition) | Sections 15.1-15.2 in Matter and Interactions (4th edition) | ||
| - | [[184_notes: | + | /*[[184_notes: |
| - | [[184_notes: | + | [[184_notes: |
| ===== Lines of Charge Examples ===== | ===== Lines of Charge Examples ===== | ||
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| Now we just need to fill in the pieces of this equation. | Now we just need to fill in the pieces of this equation. | ||
| - | === Finding dQ === | + | ==== Finding dQ ==== |
| To find dQ, we want to split this line of charge into little chunks of charge. Since this is a linear charge distribution, | To find dQ, we want to split this line of charge into little chunks of charge. Since this is a linear charge distribution, | ||
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| Now we have the little bit of charge represented in terms of the little bit of length. | Now we have the little bit of charge represented in terms of the little bit of length. | ||
| - | === Finding $\vec{r}$ === | + | ==== Finding $\vec{r}$ |
| To find the $\vec{r}$, we need to write the distance from a general dQ to Point A, which in this case will only be in the $\hat{x}$ direction. Remember that we can write the separation vector $\vec{r}$ as: | To find the $\vec{r}$, we need to write the distance from a general dQ to Point A, which in this case will only be in the $\hat{x}$ direction. Remember that we can write the separation vector $\vec{r}$ as: | ||
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| - | === Putting it together === | + | ==== Putting it together |
| Now, we can fit all the pieces together to find the electric field by plugging in what we found for dQ, $\hat{r}$ and r: | Now, we can fit all the pieces together to find the electric field by plugging in what we found for dQ, $\hat{r}$ and r: | ||
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| $$\vec{E}=\int\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ | $$\vec{E}=\int\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ | ||
| - | The final piece that we need to add is limits to the integral. Since the piece of tape stretches from $-\frac{L}{2}$ to $\frac{L}{2}$, | + | The final piece that we need to add is limits to the integral. Since the piece of tape stretches from $-\frac{L}{2}$ to $\frac{L}{2}$, |
| $$\vec{E}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ | $$\vec{E}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ | ||
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| ==== Examples ==== | ==== Examples ==== | ||
| - | [[: | + | * [[: |
| + | * Video Example: Electric Field from a Ring of Charge | ||
| + | * [[: | ||
| + | * Video Example: Electric Field from a Cylinder of Charge | ||
| + | {{youtube> | ||
| + | {{youtube> | ||
| / | / | ||