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184_notes:examples:week3_superposition_three_points [2018/01/24 18:10] – [Solution] tallpaul | 184_notes:examples:week3_superposition_three_points [2018/02/03 21:15] – tallpaul | ||
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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): | ||
+ | |||
+ | ===Representations=== | ||
+ | {{ 184_notes: | ||
===Goal=== | ===Goal=== | ||
* Find the electric field and electric potential at P. | * Find the electric field and electric potential at P. | ||
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- | ===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: | ||
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These answers seem to make sense: We have a negative electric potential, which we would expect from a negative charge. We also found that our electric field points toward Charge 1, which we would again expect because Charge 1 is negative. | These answers seem to make sense: We have a negative electric potential, which we would expect from a negative charge. We also found that our electric field points toward Charge 1, which we would again expect because Charge 1 is negative. | ||
- | For Charge 2, we expect the following | + | For Charge 2, we expect the following |
{{ 184_notes: | {{ 184_notes: | ||
We can see that $\hat{r_2}=-\hat{y}$, | We can see that $\hat{r_2}=-\hat{y}$, | ||
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These answers should also make sense for a few reasons: 1) we now have a positive electric potential (coming from the positive Charge 2), 2) the electric field points away from Charge 2 (which we could expect since Charge 2 is positive), and 3) both the magnitude of the electric field and the electric potential are larger than those from Charge 1, which makes sense because Charge 2 is closer to Point $P$ than Charge 1 was. | These answers should also make sense for a few reasons: 1) we now have a positive electric potential (coming from the positive Charge 2), 2) the electric field points away from Charge 2 (which we could expect since Charge 2 is positive), and 3) both the magnitude of the electric field and the electric potential are larger than those from Charge 1, which makes sense because Charge 2 is closer to Point $P$ than Charge 1 was. | ||
- | For Charge 3, we expect the following | + | For Charge 3, we expect the following |
{{ 184_notes: | {{ 184_notes: | ||
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 | ||
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{{ 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, |