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--- 184_notes:examples:week4_two_segments [2017/09/13 00:42] – [Solution] tallpaul
+++ 184_notes:examples:week4_two_segments [2021/05/25 14:28] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:dq|Return to  $dQ$ ]]
 ===== 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.
-{{ 184_notes:4_two_segments.png?350 |Axes with Two Segments}}
 ===Facts===
@@ Line 8: / Line 8: @@
   * 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 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===
-  * 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====
-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  $\text{d}Q$  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>
+=== 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,  $\lambda$ .
+  * 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  $P$ .
+</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  $\text{d}Q$  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}}
@@ Line 42: / Line 63: @@
  $\vec{E}_{-Q}=\int_0^L\frac{1}{4\pi\epsilon_0}\frac{-Q\text{d}x}{L\cdot|(r_x-x)\hat{x}+r_y\hat{y}|^3}((r_x-x)\hat{x}+r_y\hat{y})$
-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*}
 \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*}
 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.