184_notes:examples:week4_tilted_segment

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184_notes:examples:week4_tilted_segment [2018/02/05 17:29] – tallpaul184_notes:examples:week4_tilted_segment [2018/06/12 18:49] – [Solution] tallpaul
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 +[[184_notes:dq|Return to $dQ$]]
 +
 ===== A Tilted Segment of Charge ===== ===== A Tilted Segment of Charge =====
 Suppose we have a segment of uniformly distributed charge stretching from the point $\langle 0,0,0 \rangle$ to $\langle 1 \text{ m}, 1 \text{ m}, 0 \rangle$, 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. 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.
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 It will also be helpful to see how the dimensions of $\text{d}Q$ break down. Here is how we choose to label it: It will also be helpful to see how the dimensions of $\text{d}Q$ break down. Here is how we choose to label it:
-{{ 184_notes:4_q.png?500 |Tilted Segment dQ Representation}}+{{ 184_notes:4_dq_dimensions.png?300 |Tilted Segment dQ Representation}}
  
 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 $\lambda=Q/\sqrt{2} \text{ m}$. When we define $\text{d}l$, we want it align with the segment, so we can have $\text{d}l=\sqrt{\text{d}x^2+\text{d}y^2}$. Since $x=y$ along the segment, we can simplify a little bit. $\text{d}l=\sqrt{\text{d}x^2+\text{d}x^2}=\sqrt{2}\text{d}x$. Note, that we chose to express in terms of $\text{d}x$, instead of $\text{d}y$. This is completely arbitrary, and the solution would be just as valid the other way. Now, we can write an expression for $\text{d}Q$: 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 $\lambda=Q/\sqrt{2} \text{ m}$. When we define $\text{d}l$, we want it align with the segment, so we can have $\text{d}l=\sqrt{\text{d}x^2+\text{d}y^2}$. Since $x=y$ along the segment, we can simplify a little bit. $\text{d}l=\sqrt{\text{d}x^2+\text{d}x^2}=\sqrt{2}\text{d}x$. Note, that we chose to express in terms of $\text{d}x$, instead of $\text{d}y$. This is completely arbitrary, and the solution would be just as valid the other way. Now, we can write an expression for $\text{d}Q$:
  • 184_notes/examples/week4_tilted_segment.txt
  • Last modified: 2018/06/12 18:49
  • by tallpaul