Differences
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Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
184_notes:linecharge [2020/08/20 16:02] – dmcpadden | 184_notes:linecharge [2021/02/13 19:13] – [Putting it together] bartonmo | ||
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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 $\frac{L}{2}$, |
$$\vec{E}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ | $$\vec{E}=\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\frac{dx}{(\frac{L}{2}+d-x)^2}\hat{x}$$ | ||