184_notes:examples:week3_superposition_three_points

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184_notes:examples:week3_superposition_three_points [2018/05/29 14:25] – curdemma184_notes:examples:week3_superposition_three_points [2018/05/29 14:27] – [Solution] curdemma
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 First, let's find the contribution from Charge 1. The separation vector $\vec{r}_1$ points from the source to the observation ($1\rightarrow P$), so $\vec{r}_1 = 2R\hat{x}$, and $$\hat{r_1}=\frac{\vec{r}_1}{|r_1|}=\frac{2R\hat{x}}{2R}=\hat{x}$$ First, let's find the contribution from Charge 1. The separation vector $\vec{r}_1$ points from the source to the observation ($1\rightarrow P$), so $\vec{r}_1 = 2R\hat{x}$, and $$\hat{r_1}=\frac{\vec{r}_1}{|r_1|}=\frac{2R\hat{x}}{2R}=\hat{x}$$
 Visually, this is what we know about $\hat{r_1}$, and what we expect for $\vec{E}_1$, since Charge 1 is negative: Visually, this is what we know about $\hat{r_1}$, and what we expect for $\vec{E}_1$, since Charge 1 is negative:
-{{ 184_notes:3_superposition_1.png?400 |E-vector and r-hat for Charge 1}}+[{{ 184_notes:3_superposition_1.png?400 |E-vector and r-hat for Charge 1}}]
 Now, we can find $\vec{E}_1$ and $V_1$. Before we show the calculation, though, we need to make an assumption about the electric potential. Now, we can find $\vec{E}_1$ and $V_1$. Before we show the calculation, though, we need to make an assumption about the electric potential.
 <WRAP TIP> <WRAP TIP>
  • 184_notes/examples/week3_superposition_three_points.txt
  • Last modified: 2021/05/19 14:46
  • by schram45