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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 |
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| [[184_notes:dq|Return to dQ]] |
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===== A Tilted Segment of Charge ===== | ===== A Tilted Segment of Charge ===== |
Suppose we have a segment of uniformly distributed charge stretching from the point ⟨0,0,0⟩ to ⟨1 m,1 m,0⟩, which has total charge Q. We also have a point P=⟨2 m,0,0⟩. Define a convenient dQ for the segment, and →r between a point on the segment to the point P. Also, give appropriate limits on an integration over dQ (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 dQ and →r in a simpler way and redo. | Suppose we have a segment of uniformly distributed charge stretching from the point ⟨0,0,0⟩ to ⟨1 m,1 m,0⟩, which has total charge Q. We also have a point P=⟨2 m,0,0⟩. Define a convenient dQ for the segment, and →r between a point on the segment to the point P. Also, give appropriate limits on an integration over dQ (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 dQ and →r in a simpler way and redo. |
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It will also be helpful to see how the dimensions of dQ break down. Here is how we choose to label it: | It will also be helpful to see how the dimensions of dQ 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}} |
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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 √2 m, so we can define the line charge density λ=Q/√2 m. When we define dl, we want it align with the segment, so we can have dl=√dx2+dy2. Since x=y along the segment, we can simplify a little bit. dl=√dx2+dx2=√2dx. Note, that we chose to express in terms of dx, instead of dy. This is completely arbitrary, and the solution would be just as valid the other way. Now, we can write an expression for dQ: | 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 √2 m, so we can define the line charge density λ=Q/√2 m. When we define dl, we want it align with the segment, so we can have dl=√dx2+dy2. Since x=y along the segment, we can simplify a little bit. dl=√dx2+dx2=√2dx. Note, that we chose to express in terms of dx, instead of dy. This is completely arbitrary, and the solution would be just as valid the other way. Now, we can write an expression for dQ: |