184_notes:examples:week3_superposition_three_points

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184_notes:examples:week3_superposition_three_points [2018/01/24 18:15] – tallpaul184_notes:examples:week3_superposition_three_points [2021/05/19 14:46] (current) – schram45
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-===== Superposition with Three Point Charges =====+[[184_notes:superposition|Return to superposition]] 
 + 
 +===== Example: Superposition with Three Point Charges =====
 Suppose we have a distribution of point charges in a plane near a point $P$. There are three point charges: Charge 1 with charge $-Q$, a distance $2R$ to the left of $P$; Charge 2 with charge $Q$, a distance $R$ above $P$; and Charge 3 with charge $Q$, a distance $2R$ to the right of $P$. Find the electric potential and the electric field at the point $P$. Suppose we have a distribution of point charges in a plane near a point $P$. There are three point charges: Charge 1 with charge $-Q$, a distance $2R$ to the left of $P$; Charge 2 with charge $Q$, a distance $R$ above $P$; and Charge 3 with charge $Q$, a distance $2R$ to the right of $P$. Find the electric potential and the electric field at the point $P$.
  
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 ===Representations=== ===Representations===
-{{ 184_notes:3_superposition.png?300 |Point Charge Distribution}}+[{{ 184_notes:3_superposition.png?300 |Point Charge Distribution}}] 
 + 
 +<WRAP TIP> 
 +===Assumptions=== 
 +  * Charge is constant: Simplifies the values of each charge meaning they are not charging or discharging over time. 
 +  * Charges are not moving: Simplifies the separation vectors of each charge as these would be changing if the charges were moving through space. 
 +</WRAP>
  
 ===Goal=== ===Goal===
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 In this example, it makes sense to set up coordinate axes so that the $x$-axis stretches from left to right, and the $y$-axis stretches from down to up. To make it easier to discuss the example, we'll also label the point charges as Charge 1, Charge 2, and Charge 3, as shown in an updated representation. In this example, it makes sense to set up coordinate axes so that the $x$-axis stretches from left to right, and the $y$-axis stretches from down to up. To make it easier to discuss the example, we'll also label the point charges as Charge 1, Charge 2, and Charge 3, as shown in an updated representation.
  
-{{ 184_notes:3_superposition_labels.png?400 |Point Charge Distribution with Labels}}+[{{ 184_notes:3_superposition_labels.png?400 |Point Charge Distribution with Labels}}]
  
 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>
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 For Charge 2, we expect the following representation to be accurate, again since the separation vector $\vec{r}_2$ points from the source to the observation ($2\rightarrow P$), and Charge 2 is positive: For Charge 2, we expect the following representation to be accurate, again since the separation vector $\vec{r}_2$ points from the source to the observation ($2\rightarrow P$), and Charge 2 is positive:
-{{ 184_notes:3_superposition_2.png?400 |E-vector and r-hat for Charge 1}}+[{{ 184_notes:3_superposition_2.png?400 |E-vector and r-hat for Charge 1}}]
 We can see that $\hat{r_2}=-\hat{y}$, and We can see that $\hat{r_2}=-\hat{y}$, and
 $$\vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{y}) = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}$$ $$\vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{y}) = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}$$
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 For Charge 3, we expect the following representation to be accurate, since Charge 3 is positive: For Charge 3, we expect the following representation to be accurate, since Charge 3 is positive:
-{{ 184_notes:3_superposition_3.png?400 |E-vector and r-hat for Charge 1}}+[{{ 184_notes:3_superposition_3.png?400 |E-vector and r-hat for Charge 1}}]
 We can see that $\hat{r_3}=-\hat{x}$ (Pointing from Charge 3 to $P$), and We can see that $\hat{r_3}=-\hat{x}$ (Pointing from Charge 3 to $P$), and
 $$\vec{E}_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{4R^2}(-\hat{x}) = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ $$\vec{E}_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{4R^2}(-\hat{x}) = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$
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 Again, we get a positive electric potential from the positive charge (so this is good), and we get an electric field that points away from the positive charge, so we are happy. Again, we get a positive electric potential from the positive charge (so this is good), and we get an electric field that points away from the positive charge, so we are happy.
  
-{{ 184_notes:3_superposition_total.png?400 |E-vector superposition}}+[{{ 184_notes:3_superposition_total.png?400 |E-vector superposition}}]
  
 Now that we have the individual contributions from the point charges, we can use the principle of superposition to add together the electric field vectors and the electric potentials. See above the representation for superposition of electric fields. Now that we have the individual contributions from the point charges, we can use the principle of superposition to add together the electric field vectors and the electric potentials. See above the representation for superposition of electric fields.
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  • by tallpaul