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184_notes:examples:week5_flux_dipole [2017/09/22 15:54] – dmcpadden | 184_notes:examples:week5_flux_dipole [2018/07/24 15:02] (current) – curdemma | ||
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=====Example: | =====Example: | ||
Suppose you have a two charges, one with value $5 \mu\text{C}$, | Suppose you have a two charges, one with value $5 \mu\text{C}$, | ||
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===Representations=== | ===Representations=== | ||
* We represent the situation with the following diagram. | * We represent the situation with the following diagram. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole' | + | First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole' |
- | {{ 184_notes: | + | [{{ 184_notes: |
+ | |||
+ | Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are entering. Not only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize. | ||
- | FIXME Missing | + | We could write this as a comparison between |
+ | $$\Phi_{left}=-\Phi_{right}$$ | ||
- | You can imagine that if we were able to draw this in three dimensions, we would have just as many field lines entering the cylinder as exiting. Since flux is a measure of the " | + | Putting it together, we tentatively write: |
- | $$\Phi_{\text{cylinder}}=0$$ | + | $$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ |
We gain more confidence when we read the [[184_notes: | We gain more confidence when we read the [[184_notes: | ||
$$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | ||
Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above. | Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above. |