Differences

This shows you the differences between two versions of the page.

--- 184_notes:dipole_sup [2018/08/30 15:29] – dmcpadden
+++ 184_notes:dipole_sup [2020/08/17 17:29] (current) – dmcpadden
@@ Line 1: / Line 1: @@
-[[184_notes:comp_super|Next Page: Superposition and the Computer]]
+/*[[184_notes:comp_super|Next Page: Superposition and the Computer]]
-[[184_notes:superposition|Previous Page: Superposition]]
+[[184_notes:superposition|Previous Page: Superposition]]*/
 ===== Dipole Superposition Example =====
@@ Line 15: / Line 15: @@
 First, we will find the electric field from the positive charge, which is given by:
- $E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(r_{+ \rightarrow P})^3}\vec{r}_{+ \rightarrow P}$
+ $E_{+}=\frac{1}{4\pi\epsilon_0}\frac{q_{+}}{(r_{+ \rightarrow P})^3}\vec{r}_{+ \rightarrow P}$  where  $\vec{r}_{+ \rightarrow P}= \langle d/2, h,0 \rangle$  because it points from the positive charge to the location of Point P. Note that this equation for the r-vector is highly dependent on your choice of origin. In this case, we have placed the origin in between the two point charges and a distance h below Point P.
-FIXME where  $\vec{r}_{+ \rightarrow P}= \langle d/2, h,0 \rangle$  because it points from the positive charge to the location of Point P. In this equation,  $r_{+ \rightarrow P}$  is the magnitude of  $\vec{r}_{+ \rightarrow P}$  so
+In the electric field equation,  $r_{+ \rightarrow P}$  is the magnitude of  $\vec{r}_{+ \rightarrow P}$  so
  $r_{+ \rightarrow P}=\sqrt{(d/2)^2+h^2}$