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184_notes:examples:week4_two_segments [2017/09/13 00:40] – [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. | ||
- | |||
- | {{ 184_notes: | ||
===Facts=== | ===Facts=== | ||
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* The other segment lies on the $x$-axis stretching from $0$ to $L$, with charge $-Q$ uniformly distributed. | * The other segment lies on the $x$-axis stretching from $0$ to $L$, with charge $-Q$ uniformly distributed. | ||
* The point $P$ is at the arbitrary location $\vec{r}_P=r_x\hat{x}+r_y\hat{y}$ | * The point $P$ is at the arbitrary location $\vec{r}_P=r_x\hat{x}+r_y\hat{y}$ | ||
+ | * The electric field due to a point charge is $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}$$ | ||
+ | * The electric field at $P$ is the superposition of contributions from the two segments: $$\vec{E}_P = \vec{E}_{+Q} +\vec{E}_{-Q}$$ | ||
- | ===Lacking=== | + | ===Goal=== |
- | * $\vec{E}_P$ | + | * Find $\vec{E}_P$. |
- | + | ||
- | ===Approximations & Assumptions=== | + | |
- | * The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments. | + | |
- | * $\vec{E}_P$ is made up of contributions from the line segments and nothing else. | + | |
===Representations=== | ===Representations=== | ||
- | * We represent the situation using the diagram shown in the example statement. | + | {{ 184_notes:4_two_segments.png? |
- | * We represent the electric field due to a point charge as $$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}$$ | + | |
- | * We represent the electric field at $P$ as the sum of contributions from the two segments: $$\vec{E}_P = \vec{E}_{+Q} +\vec{E}_{-Q}$$ | + | |
====Solution==== | ====Solution==== | ||
- | Because we know that electric fields add through superposition, | + | <WRAP TIP> |
+ | === Approximation === | ||
+ | We begin with an approximation, | ||
+ | * The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments. | ||
+ | </ | ||
+ | |||
+ | This example is complicated enough that it's worthwhile to make a plan. | ||
+ | |||
+ | <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, | ||
+ | |||
+ | <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: | ||
- | The separation vector $\vec{r}$ points from the source of the electric field to the observation point. The source is $\text{d}Q$, | + | The separation vector $\vec{r}$ points from the source of the electric field to the observation point. The source is $\text{d}Q$, |
Now, we have enough to define the electric field from the small piece ($\text{d}Q$) of the segment - plugging the $\text{d}Q$ and $\vec{r}$ we just found: $$\text{d}\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{\text{d}Q}{r^3}\vec{r}=\frac{1}{4\pi\epsilon_0}\frac{Q\text{d}y}{L\cdot|r_x\hat{x}+(r_y-y)\hat{y}|^3}(r_x\hat{x}+(r_y-y)\hat{y})$$ | Now, we have enough to define the electric field from the small piece ($\text{d}Q$) of the segment - plugging the $\text{d}Q$ and $\vec{r}$ we just found: $$\text{d}\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{\text{d}Q}{r^3}\vec{r}=\frac{1}{4\pi\epsilon_0}\frac{Q\text{d}y}{L\cdot|r_x\hat{x}+(r_y-y)\hat{y}|^3}(r_x\hat{x}+(r_y-y)\hat{y})$$ | ||
Next, we integrate over the entire segment to find an expression for its contribution to the electric field vector at $P$. The limits of our integral are based on the variable of integration, | Next, we integrate over the entire segment to find an expression for its contribution to the electric field vector at $P$. The limits of our integral are based on the variable of integration, | ||
- | $$\vec{E}_{+Q}=\int_0^L\frac{1}{4\pi\epsilon_0}\frac{Q\text{d}y}{L\cdot|\vec{r}_P-y\hat{y}|^3}(\vec{r}_P-y\hat{y})$$ | + | $$\vec{E}_{+Q}=\int_0^L\frac{1}{4\pi\epsilon_0}\frac{Q\text{d}y}{L\cdot|r_x\hat{x}+(r_y-y)\hat{y}|^3}(r_x\hat{x}+(r_y-y)\hat{y})$$ |
Next, we can do a similar analysis to find the electric field vector contribution from the segment that lies along the $x$-axis. See below for a visual of $\text{d}Q$ and $\vec{r}$. See if you can convince yourself that for the segment along the $x$-axis, $\text{d}Q=\frac{-Q\text{d}x}{L}$, | Next, we can do a similar analysis to find the electric field vector contribution from the segment that lies along the $x$-axis. See below for a visual of $\text{d}Q$ and $\vec{r}$. See if you can convince yourself that for the segment along the $x$-axis, $\text{d}Q=\frac{-Q\text{d}x}{L}$, | ||
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$$\vec{E}_{-Q}=\int_0^L\frac{1}{4\pi\epsilon_0}\frac{-Q\text{d}x}{L\cdot|(r_x-x)\hat{x}+r_y\hat{y}|^3}((r_x-x)\hat{x}+r_y\hat{y})$$ | $$\vec{E}_{-Q}=\int_0^L\frac{1}{4\pi\epsilon_0}\frac{-Q\text{d}x}{L\cdot|(r_x-x)\hat{x}+r_y\hat{y}|^3}((r_x-x)\hat{x}+r_y\hat{y})$$ | ||
- | Then the final electric field vector at $P$ is the sum of the two contributions because of superpostion. (You can pull out the constants to simplify the integral if you want.) | + | Then the final electric field vector at $P$ is the sum of the two contributions, because of vector superposition. (You can pull out the constants to simplify the integral if you want.) |
\begin{align*} | \begin{align*} | ||
\vec{E} &= \vec{E}_{+Q}+\vec{E}_{-Q} \\ | \vec{E} &= \vec{E}_{+Q}+\vec{E}_{-Q} \\ | ||
- | &= \int_0^L\frac{1}{4\pi\epsilon_0}\frac{Q\text{d}y}{L\cdot|\vec{r}_P-y\hat{y}|^3}(\vec{r}_P-y\hat{y}) + \int_0^L\frac{1}{4\pi\epsilon_0}\frac{-Q\text{d}x}{L\cdot|\vec{r}_P-x\hat{x}|^3}(\vec{r}_P-x\hat{x}) \\ | + | &= \int_0^L\frac{1}{4\pi\epsilon_0}\frac{Q\text{d}y}{L\cdot|r_x\hat{x}+(r_y-y)\hat{y}|^3}(r_x\hat{x}+(r_y-y)\hat{y}) + \int_0^L\frac{1}{4\pi\epsilon_0}\frac{-Q\text{d}x}{L\cdot|(r_x-x)\hat{x}+r_y\hat{y}|^3}((r_x-x)\hat{x}+r_y\hat{y}) \\ |
- | &= \frac{Q}{4\pi\epsilon_0L}\left(\int_0^L\frac{\text{d}y}{|\vec{r}_P-y\hat{y}|^3}(\vec{r}_P-y\hat{y}) - \int_0^L\frac{\text{d}x}{|\vec{r}_P-x\hat{x}|^3}(\vec{r}_P-x\hat{x})\right) \\ | + | &= \frac{Q}{4\pi\epsilon_0L}\left(\int_0^L\frac{\text{d}y}{|r_x\hat{x}+(r_y-y)\hat{y}|^3}(r_x\hat{x}+(r_y-y)\hat{y}) - \int_0^L\frac{\text{d}x}{|(r_x-x)\hat{x}+r_y\hat{y}|^3}((r_x-x)\hat{x}+r_y\hat{y})\right) \\ |
\end{align*} | \end{align*} | ||
At this point we have the integrals set up, which you could solve by hand if you so desire or plug them into Wolfram Alpha, Mathematica, | At this point we have the integrals set up, which you could solve by hand if you so desire or plug them into Wolfram Alpha, Mathematica, | ||