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| 184_notes:examples:week4_two_segments [2018/02/03 21:21] – [Solution] tallpaul | 184_notes:examples:week4_two_segments [2021/05/25 14:28] (current) – schram45 |
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| | [[184_notes:dq|Return to dQ]] |
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| ===== Example: Two Segments of Charge ===== | ===== Example: Two Segments of Charge ===== |
| Suppose we have two segments of uniformly distributed charge, one with total charge +Q, the other with −Q. The two segments each have length L, and lie crossed at their endpoints in the xy-plane. The segment with charge +Q lies along the y-axis, and the segment with charge −Q lies along the x-axis. See below for a diagram of the situation. Create an expression for the electric field →EP at a point P that is located at →rP=rxˆx+ryˆy. You don't have to evaluate integrals in the expression. | Suppose we have two segments of uniformly distributed charge, one with total charge +Q, the other with −Q. The two segments each have length L, and lie crossed at their endpoints in the xy-plane. The segment with charge +Q lies along the y-axis, and the segment with charge −Q lies along the x-axis. See below for a diagram of the situation. Create an expression for the electric field →EP at a point P that is located at →rP=rxˆx+ryˆy. You don't have to evaluate integrals in the expression. |
| * Use λ to write an expression for dQ. | * Use λ to write an expression for dQ. |
| * Assign a variable location to the dQ piece, and then use that location to find the separation vector, →r. | * Assign a variable location to the dQ piece, and then use that location to find the separation vector, →r. |
| * Write an expression for \text{d}\vec{E}. | * Write an expression for $\text{d}\vec{E}$. |
| * Figure out the bounds of the integral, and integrate to find electric field at P. | * Figure out the bounds of the integral, and integrate to find electric field at P. |
| * Repeat the above steps for the other segment of charge. | * Repeat the above steps for the other segment of charge. |
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| Because we know that electric fields add through superposition, we can treat each of the charges separately, find the electric field, then add the fields together at P at the end. We can begin with the electric field due to the segment along the y-axis. We start by finding dQ and →r. The charge is uniformly distributed so we have a simple line charge density of λ=Q/L. The segment extends in the y-direction, so we have dl=dy. This gives us dQ: dQ=λdl=QdyL
| Because we know that electric fields add through superposition, we can treat each of the charges separately, find the electric field, then add the fields together at P at the end. We can begin with the electric field due to the segment along the y-axis. We start by finding dQ and →r. The charge is uniformly distributed so we have a simple line charge density of λ=Q/L. The segment extends in the y-direction, so we have dl=dy. This gives us dQ: dQ=λdl=QdyL
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| | <WRAP TIP> |
| | ===Assumption=== |
| | The charge is evenly distributed along each segment of charge. This allows each little piece of charge to have the same value along each line. |
| | </WRAP> |
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| {{ 184_notes:4_two_segments_pos_dq.png?450 |dQ for Segment on y-axis}} | {{ 184_notes:4_two_segments_pos_dq.png?450 |dQ for Segment on y-axis}} |