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184_notes:examples:week4_two_segments [2018/02/03 21:12] – [Solution] tallpaul | 184_notes:examples:week4_two_segments [2021/05/25 14:28] (current) – schram45 | ||
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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 $\vec{E}_P$ 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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We begin with an approximation, | We begin with an approximation, | ||
* The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments. | * The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments. | ||
+ | </ | ||
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+ | This example is complicated enough that it's worthwhile to make a plan. | ||
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+ | <WRAP TIP> | ||
+ | === Plan === | ||
+ | We will use integration to find the electric field from each segment, and then add the electric fields together using superposition. We'll go through the following steps. | ||
+ | * For the first segment, find the linear charge density, $\lambda$. | ||
+ | * 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, $\vec{r}$. | ||
+ | * Write an expression for $\text{d}\vec{E}$. | ||
+ | * 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. | ||
+ | * Add the two fields together to find the total electric field at $P$. | ||
</ | </ | ||
Because we know that electric fields add through superposition, | Because we know that electric fields add through superposition, | ||
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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. | ||
+ | </ | ||
{{ 184_notes: | {{ 184_notes: |