184_notes:dipole_sup

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184_notes:dipole_sup [2018/08/30 15:29] dmcpadden184_notes:dipole_sup [2020/08/17 17:29] (current) dmcpadden
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-[[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 ===== ===== Dipole Superposition Example =====
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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 
 $$r_{+ \rightarrow P}=\sqrt{(d/2)^2+h^2}$$ $$r_{+ \rightarrow P}=\sqrt{(d/2)^2+h^2}$$
  
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  • Last modified: 2018/08/30 15:29
  • by dmcpadden