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184_notes:examples:week5_flux_two_radii [2017/09/18 12:30] – [Solution] tallpaul | 184_notes:examples:week5_flux_two_radii [2017/09/25 13:27] – [Solution] tallpaul | ||
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* There are no other charges that contribute appreciably to the flux calculation. | * There are no other charges that contribute appreciably to the flux calculation. | ||
* There is no background electric field. | * There is no background electric field. | ||
- | * The electric | + | * The electric |
===Representations=== | ===Representations=== | ||
Line 21: | Line 21: | ||
$$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | ||
* We represent the situation with the following diagram. Note that the circles are indeed spherical shells, not rings as they appear. | * We represent the situation with the following diagram. Note that the circles are indeed spherical shells, not rings as they appear. | ||
- | {{ 184_notes: | + | {{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | Before we dive into calculations, | + | Before we dive into calculations, |
- | 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: | + | {{ 184_notes: |
- | $$\Phi_e=\int\vec{E}\bullet | + | Since the shell is a fixed distance from the point charge, the electric field has constant magnitude on the shell. Since $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude (on the shell), the dot product simplifies substantially: |
- | We can rewrite | + | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ |
- | $$\left|\vec{E}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ | + | Note that $E$ is a scalar value representing |
- | In order to find electric flux, we must first find $\vec{A}$. Remember in the [[184_notes: | + | $$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}$$ |
- | $$\vec{w}=3\text{ | + | |
- | $$\vec{l}=5\text{ m }\cdot\cos 30^\circ (\hat{x})+5\text{ m }\sin 30^\circ\hat{y} = 2.5\sqrt{3}\text{ m } \hat{x} + 2.5\text{ m }\hat{y}$$ | + | 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 find $E=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we find $E=2.5\cdot 10^6 \text{ N/C}$. |
- | Now, we can find the area vector: | + | |
+ | To figure out the area integral, notice | ||
+ | $$\int\text{d}A=A=4\pi r^2$$ | ||
+ | The last expression, | ||
+ | |||
+ | 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*} | \begin{align*} | ||
- | \vec{A} & | + | \Phi_{\text{small}} & |
- | | + | \Phi_{\text{large}} & |
- | &= 7.5\sqrt{3}\text{ m}^2 (-\hat{y}) + 7.5\text{ m}^2\hat{x} \\ | + | |
- | &= 7.5\text{ m}^2\hat{x} - 7.5\sqrt{3}\text{ | + | |
- | \end{align*} | + | |
- | Again, since the rectangle is not a closed surface, the choice for direction of $\vec{A}$ was arbitrary, and it would have been just fine to define it oppositely. Anyways, we can proceed to determine the electric flux: | + | |
- | \begin{align*} | + | |
- | \Phi_e &= \vec{E}\bullet\vec{A} \\ | + | |
- | | + | |
- | &= 60\text{ | + | |
\end{align*} | \end{align*} | ||
+ | |||
+ | We get the same answer for both shells! It turns out that the radius of the shell does not affect the electric flux for this example. You'll see later that electric flux through a closed surface depends //only// on the charge enclosed. The surface can be weirdly shaped, and we can always figure out the flux as long as we know the charge enclosed. |