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--- 184_notes:examples:week4_tilted_segment [2018/02/03 20:41] – tallpaul
+++ 184_notes:examples:week4_tilted_segment [2018/06/12 18:49] (current) – [Solution] tallpaul
@@ Line 1: / Line 1: @@
+[[184_notes:dq|Return to  $dQ$ ]]
 ===== 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.
@@ Line 13: / Line 15: @@
   * 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 representation:
 {{ 184_notes:4_tilted_segment.png?350 |Axes with Tilted Segment}}
-  * We can represent  $\text{d}Q$  and  $\vec{r}$  for our line as follows:
-{{ 184_notes:4_tilted_segment_dq.png?500 |Tilted Segment dQ Representation}}
   * 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. We could have chosen the  $x$ -axis, and arrived at a very similar answer. Whichever you like is fine!
 {{ 184_notes:4_tilted_segment_rotated.png?800 |Tilted Segment with New Axes}}
@@ Line 25: / Line 25: @@
 </WRAP>
-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  $\lambda=Q/\sqrt{2} \text{ m}$ . 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  $\text{d}Q$ :
+We know how to draw  $\text{d}Q$  and  $\vec{r}$ , so we can start with an update to the representation.
+{{ 184_notes:4_tilted_segment_dq.png?500 |Tilted Segment dQ Representation}}
+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_dq_dimensions.png?200 |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  $\text{d}Q$ :
  $\text{d}Q=\lambda\text{d}l=\frac{\sqrt{2}}{\sqrt{2} \text{ m}}Q\text{d}x=\frac{Q}{1 \text{ m}}\text{d}x$
@@ Line 34: / Line 40: @@
 ----
-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  $\lambda=Q/\sqrt{2} \text{ m}$ . When we define  $\text{d}l$ , we want it align with the segment, which is much simpler this time:  $\text{d}l=\text{d}y$ . Now, we can write an expression for  $\text{d}Q$ :
+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  $\lambda=Q/\sqrt{2} \text{ m}$ . When we define  $\text{d}l$ , we want it align with the segment, which is much simpler this time:  $\text{d}l=\text{d}y$ . Now, we can
- $\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}y}{\sqrt{2} \text{ m}}$
-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  $\vec{r}_P=\langle \sqrt{2} \text{ m},\sqrt{2} \text{ m},0 \rangle$ , and  $\vec{r}_{\text{d}Q}=\langle 0, y, 0 \rangle$ . We have enough to write  $\vec{r}$ :
- $\vec{r}=\vec{r}_P-\vec{r}_{\text{d}Q}=\langle \sqrt{2} \text{ m}, \sqrt{2} \text{ m}-y, 0 \rangle$
-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 over  $y$ . This means that our limits of integration also have to match the total length that we want to add up //in terms of the  $y$  variable//, which goes from  $0$  to  $\sqrt{2} \text{ m}$ . So our limits of integration would be from  $0$  to  $\sqrt{2} \text{ m}$ .