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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 [2017/09/25 14:08] – [Solution] tallpaul | ||
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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=== | ||
Line 20: | Line 20: | ||
{{ 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?300 |Dipole Electric Field Lines}} | ||
- | Since $\vec{E}$ is constant with respect to $\text{d}\vec{A}$ (in this case, it is sufficient | + | 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, |
- | $$\Phi_e=\int\vec{E}\bullet | + | We could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality, |
+ | $$\Phi_{left}=-\Phi_{right}$$ | ||
- | We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge' | + | Putting it together, we tentatively write: |
- | + | $$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ | |
- | $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| | + | We gain more confidence when we read the [[184_notes: |
- | + | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | |
- | 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 $E=2.5\cdot 10^6 \text{ N/C}$. | + | Since the total charge |
- | + | ||
- | To figure out the area integral, 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 as $\Phi_e=EA$. For our two shells: | + | |
- | \begin{align*} | + | |
- | \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{/ | + | |
- | \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 | + |