Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
184_notes:examples:week5_flux_dipole [2017/09/18 18:05] – [Solution] tallpaul | 184_notes:examples:week5_flux_dipole [2018/07/24 15:02] (current) – curdemma | ||
---|---|---|---|
Line 1: | Line 1: | ||
+ | [[184_notes: | ||
+ | |||
=====Example: | =====Example: | ||
- | Suppose you have a two charges, one with value $5 \mu\text{C}$, | + | Suppose you have a two charges, one with value $5 \mu\text{C}$, |
===Facts=== | ===Facts=== | ||
Line 18: | Line 20: | ||
===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 | + | 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: | ||
+ | |||
+ | 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. | ||
+ | |||
+ | We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality, | ||
+ | $$\Phi_{left}=-\Phi_{right}$$ | ||
+ | |||
+ | Putting it together, we tentatively write: | ||
+ | $$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ | ||
+ | We gain more confidence when we read the [[184_notes: | ||
+ | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | ||
+ | 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. |