Both sides previous revision Previous revision Next revision | Previous revision |
184_notes:examples:week4_two_segments [2017/09/13 00:44] – [Solution] tallpaul | 184_notes:examples:week4_two_segments [2021/05/25 14:28] (current) – schram45 |
---|
| [[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 →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. |
| |
{{ 184_notes:4_two_segments.png?350 |Axes with Two Segments}} | |
| |
===Facts=== | ===Facts=== |
* 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 →rP=rxˆx+ryˆy | * The point P is at the arbitrary location →rP=rxˆx+ryˆy |
| * The electric field due to a point charge is →E=14πϵ0qr3→r
|
| * The electric field at P is the superposition of contributions from the two segments: →EP=→E+Q+→E−Q
|
| |
===Lacking=== | ===Goal=== |
* →EP | * Find →EP. |
| |
===Approximations & Assumptions=== | |
* The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments. | |
* →EP 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?350 |Axes with Two Segments}} |
* 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, we can treat 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
| <WRAP TIP> |
| === Approximation === |
| We begin with an approximation, which will make our calculations simpler, and makes sense based on our representation: |
| * The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments. |
| </WRAP> |
| |
| 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, λ. |
| * 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. |
| * Write an expression for d→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. |
| </WRAP> |
| |
| 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 |
| |
| <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}} |
\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, or some other computation program. | 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, or some other computation program. |
| |