184_notes:examples:week4_tilted_segment

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184_notes:examples:week4_tilted_segment [2018/06/07 13:12] – curdemma184_notes:examples:week4_tilted_segment [2018/06/12 18:49] – [Solution] tallpaul
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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