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| 184_notes:examples:week4_two_segments [2018/06/07 13:12] – curdemma | 184_notes:examples:week4_two_segments [2021/05/25 14:28] (current) – schram45 |
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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 $\text{d}Q$ and $\vec{r}$. The charge is uniformly distributed so we have a simple line charge density of $\lambda=Q/L$. The segment extends in the $y$-direction, so we have $\text{d}l=\text{d}y$. This gives us $\text{d}Q$: $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}y}{L}$$ | 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 $\text{d}Q$ and $\vec{r}$. The charge is uniformly distributed so we have a simple line charge density of $\lambda=Q/L$. The segment extends in the $y$-direction, so we have $\text{d}l=\text{d}y$. This gives us $\text{d}Q$: $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}y}{L}$$ |
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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}} |