184_notes:examples:week4_two_segments

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184_notes:examples:week4_two_segments [2018/02/03 21:21] – [Solution] tallpaul184_notes:examples:week4_two_segments [2021/05/25 14:28] (current) – schram45
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 +[[184_notes:dq|Return to $dQ$]]
 +
 ===== 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 $\vec{E}_P$ at a point $P$ that is located at $\vec{r}_P=r_x\hat{x}+r_y\hat{y}$. 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 $\vec{E}_P$ at a point $P$ that is located at $\vec{r}_P=r_x\hat{x}+r_y\hat{y}$. You don't have to evaluate integrals in the expression.
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   * Use $\lambda$ to write an expression for $\text{d}Q$.   * Use $\lambda$ to write an expression for $\text{d}Q$.
   * Assign a variable location to the $\text{d}Q$ piece, and then use that location to find the separation vector, $\vec{r}$.   * Assign a variable location to the $\text{d}Q$ piece, and then use that location to find the separation vector, $\vec{r}$.
-  * 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 $\text{d}Q$ 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 $\text{d}Q$: $$\text{d}Q=\lambda\text{d}l=\frac{Q\text{d}y}{L}$$
 +
 +<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>
  
 {{ 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}}
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  • Last modified: 2018/02/03 21:21
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