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--- 184_notes:examples:week4_two_segments [2018/02/03 21:22] – [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.
@@ Line 37: / Line 39: @@
 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}}