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184_notes:linecharge [2018/06/05 14:00] – curdemma | 184_notes:linecharge [2021/05/26 13:51] – 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: | ||
$$\vec{r}=\vec{r}_{obs}-\vec{r}_{source}$$ | $$\vec{r}=\vec{r}_{obs}-\vec{r}_{source}$$ | ||
- | This is still true and is a general definition. In this case, $\vec{r}_{obs}$ would be the vector that points from the origin to Point A (our observation point). Since Point A is in the +x region, we get the distance to Point A to be $\vec{r}_{obs}=\frac{L}{2} + d \hat{x}$. The source in our case is the " | + | This is still true and is a general definition. In this case, $\vec{r}_{obs}$ would be the vector that points from the origin to Point A (our observation point). Since Point A is in the +x region, we get the distance to Point A to be $\vec{r}_{obs}=(\frac{L}{2} + d) \hat{x}$. The source in our case is the " |
- | $$\vec{r}=\vec{r}_{obs}-\vec{r}_{source}=+\frac{L}{2} + d \hat{x}-(+x)\hat{x}$$ | + | $$\vec{r}=\vec{r}_{obs}-\vec{r}_{source}=(+\frac{L}{2} + d) \hat{x} - (+x)\hat{x}$$ |
- | $$\vec{r}= \langle \frac{L}{2}+d-x, | + | $$\vec{r}= \langle \frac{L}{2}+d-x, |
Because the $\vec{r}$ points in a single direction, the magnitude is pretty simple: | Because the $\vec{r}$ points in a single direction, the magnitude is pretty simple: | ||
$$|\vec{r}|=r=\sqrt{(\frac{L}{2}+d-x)^2+0^2+0^2}=\frac{L}{2}+d-x$$ | $$|\vec{r}|=r=\sqrt{(\frac{L}{2}+d-x)^2+0^2+0^2}=\frac{L}{2}+d-x$$ | ||
- | === 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 | ||
+ | * [[: | ||
+ | {{youtube> | ||
/ | / |