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184_notes:examples:week5_flux_dipole [2017/09/18 16:58] – [Example: Flux from a Dipole] tallpaul | 184_notes:examples:week5_flux_dipole [2018/07/24 15:02] (current) – curdemma | ||
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+ | [[184_notes: | ||
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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 $5 \mu\text{C}$, |
===Facts=== | ===Facts=== | ||
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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==== | ||
- | Before | + | First, notice that we probably do not want to do any calculations |
- | + | [{{ 184_notes:5_dipole_field_lines.png? | |
- | Since $\vec{E}$ is constant with respect to $\text{d}\vec{A}$ (in this case, it is sufficient that $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude), we can rewrite our flux representation: | + | |
- | + | ||
- | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ | + | |
- | + | ||
- | We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge' | + | |
- | + | ||
- | $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ | + | |
- | + | ||
- | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $E=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we have $E=2.5\cdot 10^6 \text{ N/C}$. | + | |
- | To figure out the area integral, notice | + | Notice |
- | $$\int\text{d}A=A=4\pi r^2$$ | + | |
- | The last expression, $4\pi r^2$, is just the surface area of a sphere. We can plug in values for $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $A=1.13\cdot 10^{-2} \text{ m}^2$. For the larger shell, we have $A=4.52\cdot 10^{-2} \text{ m}^2$. | + | |
- | Now, we bring it together to find electric flux, which after all our simplifications can be written | + | 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, |
- | \begin{align*} | + | $$\Phi_{left}=-\Phi_{right}$$ |
- | \Phi_{\text{small}} &= 1.0\cdot 10^7 \text{ N/C } \cdot 1.13\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^5 \text{ Nm}^2\text{/C} \\ | + | |
- | \Phi_{\text{large}} &= 2.5\cdot 10^6 \text{ N/C } \cdot 4.52\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^5 \text{ Nm}^2\text{/ | + | |
- | \end{align*} | + | |
- | We get the same answer for both shells! It turns out that the radius | + | 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 |