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184_notes:linecharge [2018/01/20 17:05] – 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) | ||
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+ | [[184_notes: | ||
===== Lines of Charge Examples ===== | ===== Lines of Charge Examples ===== | ||
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{{youtube> | {{youtube> | ||
==== Horizontal Line of Charge ==== | ==== Horizontal Line of Charge ==== | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
- | Say that we have a horizontal line of charge (this could be like a piece of tape, a plastic pen, a glass rod, etc.) that has a total length of L and a total charge of Q, and we are interested in finding the electric field at Point A that is a distance d away from the end of tape. First thing that we need to do is choose a frame of reference - lets pick x=0 to be in the middle of the piece of tape, with +x direction to the right and -x direction to the left. In this frame of reference, the piece of tape stretches then from −L2 to L2. Note that this choice of reference frame is completely arbitrary. You can pick whatever reference frame you would like (but you can be strategic in your choice - some reference frames will be easier to handle mathematically than others). | + | Say that we have a horizontal line of charge (this could be like a piece of tape, a plastic pen, a glass rod, etc.) that has a total length of $L$ and a total charge of $Q$, and we are interested in finding the electric field at Point $A$ that is a distance |
Before we talked about how to find the electric field using the general equation: | Before we talked about how to find the electric field using the general equation: | ||
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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 →r === | + | ==== Finding →r |
To find the →r, we need to write the distance from a general dQ to Point A, which in this case will only be in the ˆx direction. Remember that we can write the separation vector →r as: | To find the →r, we need to write the distance from a general dQ to Point A, which in this case will only be in the ˆx direction. Remember that we can write the separation vector →r as: | ||
→r=→robs−→rsource | →r=→robs−→rsource | ||
- | This is still true and is a general definition. In this case, →robs 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 →robs=L2+dˆx. The source in our case is the " | + | This is still true and is a general definition. In this case, →robs 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 $\vec{r}_{obs}=(\frac{L}{2} + d) \hat{x}.Thesourceinourcaseisthe"littlebitofcharge"−sotofind\vec{r}_{source}weneedawaytowritewherethedQisrelativetotheorigin.Sincethedistancefrom0todQcanchangedependingonwhereyouputyourdQ,wewillcallthissomevariable,namely"x"tomatchwhatwechosefordQ.So\vec{r}_{source}=(+x)\hat{x}.∗∗Notethatthisdistanceisa"big"xandnotadx(or"littlebitofx")becausethe\vec{r}$ represents a large distance and not an infinitesimal distance**. So the total separation vector here is given by: |
- | →r=→robs−→rsource=+L2+dˆx−(+x)ˆx | + | $$\vec{r}=\vec{r}_{obs}-\vec{r}_{source}=(+\frac{L}{2} + d) \hat{x} - (+x)\hat{x}$$ |
- | →r=⟨L2+d−x,0,0⟩=L2+d−xˆx | + | $$\vec{r}= \langle \frac{L}{2}+d-x, |
Because the →r points in a single direction, the magnitude is pretty simple: | Because the →r points in a single direction, the magnitude is pretty simple: | ||
|→r|=r=√(L2+d−x)2+02+02=L2+d−x | |→r|=r=√(L2+d−x)2+02+02=L2+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, ˆr and r: | Now, we can fit all the pieces together to find the electric field by plugging in what we found for dQ, ˆr and r: | ||
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→E=∫14πϵ0QLdx(L2+d−x)2ˆx | →E=∫14πϵ0QLdx(L2+d−x)2ˆx | ||
- | The final piece that we need to add is limits to the integral. Since the piece of tape stretches from −L2 to L2, | + | The final piece that we need to add is limits to the integral. Since the piece of tape stretches from −L2 to L2, |
→E=∫L2−L214πϵ0QLdx(L2+d−x)2ˆx | →E=∫L2−L214πϵ0QLdx(L2+d−x)2ˆx | ||
- | At this point, you can either calculate the integral on your own or plug it into [[https:// | + | At this point, you can either calculate the integral on your own or plug it into [[https:// |
==== Examples ==== | ==== Examples ==== | ||
- | [[: | + | * [[: |
+ | * Video Example: Electric Field from a Ring of Charge | ||
+ | * [[: | ||
+ | * Video Example: Electric Field from a Cylinder of Charge | ||
+ | {{youtube> | ||
+ | {{youtube> | ||
- | [[: | + | /*[[: |