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184_notes:examples:week4_tilted_segment [2017/09/06 12:56] – [Example: A Tilted Segment of Charge] tallpaul | 184_notes:examples:week4_tilted_segment [2017/09/12 23:46] – [Solution] tallpaul | ||
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===== Example: A Tilted Segment of Charge ===== | ===== Example: A Tilted Segment of Charge ===== | ||
- | Suppose we have a segment of uniformly distributed charge stretching from the point $\langle 0,0,0 \rangle$ to $\langle 1, 1, 0 \rangle$, which has total charge $Q$. We also have a point $P=\langle 2,0,0 \rangle$. Define a convenient $\text{d}Q$ for the segment, and $\vec{r}$ between a point on the segment to the point $P$. Also, give appropriate limits on an integration over $\text{d}Q$ (you don't have to write any integrals, just give appropriate start and end points). First, do this for the given coordinate axes. Second, define a new set of coordinate axes to represent $\text{d}Q$ and $\vec{r}$ in a simpler way and redo. | + | Suppose we have a segment of uniformly distributed charge stretching from the point $\langle 0,0,0 \rangle$ to $\langle 1 \text{ m}, 1 \text{ m}, 0 \rangle$, which has total charge $Q$. We also have a point $P=\langle 2 \text{ m},0,0 \rangle$. Define a convenient $\text{d}Q$ for the segment, and $\vec{r}$ between a point on the segment to the point $P$. Also, give appropriate limits on an integration over $\text{d}Q$ (you don't have to write any integrals, just give appropriate start and end points). First, do this for the given coordinate axes. Second, define a new set of coordinate axes to represent $\text{d}Q$ and $\vec{r}$ in a simpler way and redo. |
===Facts=== | ===Facts=== | ||
- | * The segment stretches from $\langle 0,0,0 \rangle$ to $\langle 1, 1, 0 \rangle$. | + | * The segment stretches from $\langle 0,0,0 \rangle$ to $\langle 1 \text{ m}, 1 \text{ m}, 0 \rangle$. |
* The segment has a charge $Q$, which is uniformly distributed. | * The segment has a charge $Q$, which is uniformly distributed. | ||
- | * $P=\langle 2,0,0 \rangle$. | + | * $P=\langle 2 \text{ m},0,0 \rangle$. |
===Lacking=== | ===Lacking=== | ||
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===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
* The thickness of the segment is infinitesimally small, and we can approximate it as a line segment. | * The thickness of the segment is infinitesimally small, and we can approximate it as a line segment. | ||
+ | * The total charge is a constant - not discharging. | ||
===Representations=== | ===Representations=== | ||
- | * We can draw a set of coordinate axes using what we already know. The first part of the example involves the following orientation: | + | * For the first part, we can draw a set of coordinate axes using what we already know. The first part of the example involves the following orientation: |
{{ 184_notes: | {{ 184_notes: | ||
- | * We can represent $\text{d}Q$ and $\vec{r}$ for our ring as follows: | + | * We can represent $\text{d}Q$ and $\vec{r}$ for our line as follows: |
{{ 184_notes: | {{ 184_notes: | ||
- | * When we define a new set of coordinate axes, it makes sense to line up the segment along an axis. We choose the $y$-axis. | + | * For the second part when we define a new set of coordinate axes, it makes sense to line up the segment along an axis. We choose the $y$-axis. |
{{ 184_notes: | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | In the first set of axes, the segment extends in the $x$ and $y$ directions. A simple calculation of the Pythagorean theorem tells us the length of the segment is $\sqrt{2}$, so we can define the line charge density $\lambda=Q/ | + | In the first set of axes, the segment extends in the $x$ and $y$ directions. A simple calculation of the Pythagorean theorem tells us the total length of the segment is $\sqrt{2} \text{ m}$, so we can define the line charge density $\lambda=Q/ |
- | $$\text{d}Q=\lambda\text{d}l=Q\text{d}x$$ | + | $$\text{d}Q=\lambda\text{d}l=\frac{\sqrt{2}}{\sqrt{2} \text{ m}}Q\text{d}x=\frac{Q}{1 \text{ m}}\text{d}x$$ |
- | The units here might look a little weird, since distance was defined without dimensions in the example statement. | + | |
- | $$\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=\langle 2-x, -x, 0 \rangle$$ | + | Next, we need $\vec{r}$. We will put it in terms of $x$, not $y$, just as we did for $\text{d}Q$. A choice of $y$ instead of $x$ here would be valid had we chosen to express $\text{d}Q$ in terms of $\text{d}y$ earlier. We know $\vec{r}_P=\langle 2 \text{ m},0,0 \rangle$, and $\vec{r}_{\text{d}Q}=\langle x, y, 0 \rangle$. Again, $x=y$, so we can rewrite $\vec{r}_{\text{d}Q}=\langle x, x, 0 \rangle$. We now have enough to write $\vec{r}$: |
- | An integration would occur over $x$, with goes from $0$ to $1$. These would be our limits of integration. | + | $$\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=\langle 2 \text{ m}-x, -x, 0 \rangle$$ |
+ | Because we picked $\text{d}x$ and $x$ as our variable, we are all set up to integrate | ||
---- | ---- | ||
- | In the second set of axes, the segment extends only in the $y$ direction. The length of the segment is still $\sqrt{2}$, so we can define the line charge density $\lambda=Q/ | + | In the second set of axes, the segment extends only in the $y$ direction. This problem is now very similar to the examples in the notes. The length of the segment is still $\sqrt{2} \text{ m}$, so we can define the line charge density $\lambda=Q/ |
- | $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}y}{\sqrt{2}}$$ | + | $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}y}{\sqrt{2} \text{ m}}$$ |
- | Next, we need $\vec{r}$. We will put it in terms of $y$, just as we did for $\text{d}Q$. The location of $P$ is a little different with this new set of axes. Now, we have $\vec{r}_P=\langle \sqrt{2}, | + | Next, we need $\vec{r}$. We will put it in terms of $y$, just as we did for $\text{d}Q$. The location of $P$ is a little different with this new set of axes. Now, we have $\vec{r}_P=\langle \sqrt{2} \text{ m},\sqrt{2} \text{ m},0 \rangle$, and $\vec{r}_{\text{d}Q}=\langle 0, y, 0 \rangle$. We have enough to write $\vec{r}$: |
- | $$\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=\langle \sqrt{2}, \sqrt{2}-y, 0 \rangle$$ | + | $$\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=\langle \sqrt{2} \text{ m}, \sqrt{2} \text{ m}-y, 0 \rangle$$ |
- | An integration would occur over $y$, with goes from $0$ to $\sqrt{2}$. | + | Because we picked $\text{d}y$ and $y$ as our variable, which was the natural choice based on how we chose to set up our coordinate axes, we are all set up to integrate |