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184_notes:examples:week3_superposition_three_points [2018/01/24 18:03] – tallpaul | 184_notes:examples:week3_superposition_three_points [2018/05/29 14:27] – [Solution] curdemma | ||
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===== Example: Superposition with Three Point Charges ===== | ===== 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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* We can use superposition to add electric field contributions from the point charges (vector superposition): | * We can use superposition to add electric field contributions from the point charges (vector superposition): | ||
* We can use superposition to add electric potential contributions from the point charges (scalar superposition): | * We can use superposition to add electric potential contributions from the point charges (scalar superposition): | ||
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+ | ===Representations=== | ||
+ | [{{ 184_notes: | ||
===Goal=== | ===Goal=== | ||
* Find the electric field and electric potential at P. | * Find the electric field and electric potential at P. | ||
- | |||
- | ===Approximations & Assumptions=== | ||
- | * The electric potential infinitely far away from the point charge is $0 \text{ V}$. | ||
- | |||
- | ===Representations=== | ||
- | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | 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 the diagram. | + | 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: | + | [{{ 184_notes: |
- | First, let's find the contribution from Charge 1. The vector $\vec{r}_1$ points from the source to 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 |
Visually, this is what we know about $\hat{r_1}$, | Visually, this is what we know about $\hat{r_1}$, | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
Now, we can find $\vec{E}_1$ and $V_1$. Before we show the calculation, | Now, we can find $\vec{E}_1$ and $V_1$. Before we show the calculation, | ||
<WRAP TIP> | <WRAP TIP> | ||
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$$\vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{4R^2}\hat{x} = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ | $$\vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{4R^2}\hat{x} = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ | ||
$$V_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{2R} = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ | $$V_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{-Q}{2R} = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ | ||
- | These answers | + | These answers |
- | For Charge 2, we expect the following | + | For Charge 2, we expect the following |
- | {{ 184_notes: | + | [{{ 184_notes: |
- | We can determine | + | We can see that $\hat{r_2}=-\hat{y}$, |
$$\vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{y}) = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}$$ | $$\vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(-\hat{y}) = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}$$ | ||
$$V_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$ | $$V_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$ | ||
- | These answers should also make sense for a few reasons: 1) we now have a positive electric potential (coming from the positive | + | These answers should also make sense for a few reasons: 1) we now have a positive electric potential (coming from the positive |
- | For Charge 3, we expect the following | + | For Charge 3, we expect the following |
- | {{ 184_notes: | + | [{{ 184_notes: |
- | We can determine | + | We can see that $\hat{r_3}=-\hat{x}$ (Pointing from Charge 3 to $P$), and |
$$\vec{E}_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{4R^2}(-\hat{x}) = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ | $$\vec{E}_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{4R^2}(-\hat{x}) = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}$$ | ||
$$V_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{2R} = \frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ | $$V_3 = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{Q}{2R} = \frac{1}{8\pi\epsilon_0}\frac{Q}{R}$$ | ||
- | 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 this also makes sense. | + | 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: | + | [{{ 184_notes: |
- | Now that we have the individual contributions from the point charges, we can add together the potentials and use the principle of superposition to add together the electric field vectors | + | 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 |
$$\vec{E}_{tot}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_3 = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}-\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}-\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x} = \frac{1}{8\pi\epsilon_0}\frac{Q}{R^2}(-\hat{x}-2\hat{y})$$ | $$\vec{E}_{tot}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_3 = -\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x}-\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{y}-\frac{1}{16\pi\epsilon_0}\frac{Q}{R^2}\hat{x} = \frac{1}{8\pi\epsilon_0}\frac{Q}{R^2}(-\hat{x}-2\hat{y})$$ | ||
$$V_{tot}=V_1+V_2+V_3 = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}+\frac{1}{4\pi\epsilon_0}\frac{Q}{R}+\frac{1}{8\pi\epsilon_0}\frac{Q}{R} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$ | $$V_{tot}=V_1+V_2+V_3 = -\frac{1}{8\pi\epsilon_0}\frac{Q}{R}+\frac{1}{4\pi\epsilon_0}\frac{Q}{R}+\frac{1}{8\pi\epsilon_0}\frac{Q}{R} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$ | ||
- | So our total electric potential is positive - this should make sense as we have two positive charges and only one negative charge, and one of the positive charges is much closer to P than the other charges so it should have a stronger effect. We found that our electric points down and to the left, which makes sense as it points | + | So our total electric potential is positive - this should make sense as we have two positive charges and only one negative charge, and even better, |