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--- 184_notes:examples:week4_tilted_segment [2018/02/05 17:29] – 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 27: / Line 29: @@
 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?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  $\text{d}Q$ :