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| Both sides previous revision Previous revision | Last revisionBoth sides next revision | ||
| 184_notes:dipole_sup [2018/08/30 15:29] – dmcpadden | 184_notes:dipole_sup [2018/08/30 15:30] – dmcpadden | ||
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| First, we will find the electric field from the positive charge, which is given by: | 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 | ||
| $$r_{+ \rightarrow P}=\sqrt{(d/ | $$r_{+ \rightarrow P}=\sqrt{(d/ | ||