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184_notes:examples:week5_flux_two_radii [2017/09/18 13:18] – [Solution] tallpaul | 184_notes:examples:week5_flux_two_radii [2017/09/25 13:11] – [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=== | ||
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$$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | ||
* 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: |
- | $$\Phi_e=\int\vec{E}\bullet | + | FIXME (Quick something about why the dot product simplifies) Since $\vec{E}$ is constant with respect to $\text{d}\vec{A}$ (in this case, it is sufficient that $\vec{E}$ is parallel to $\text{d}\vec{A}$ |
- | We can rewrite | + | $$\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}$$ | + | FIXME (You could use this to justify why E is constant on area) We can rewrite |
- | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $\left|\vec{E}\right|=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we have $\left|\vec{E}\right|=2.5\cdot 10^6 \text{ N/C}$. | + | $$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}$$ |
+ | |||
+ | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we find $E=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we find $E=2.5\cdot 10^6 \text{ N/C}$. | ||
To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by $\text{d}A$. Since we are integrating this little piece over the entire shell, we end up with the area of the shell' | To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by $\text{d}A$. Since we are integrating this little piece over the entire shell, we end up with the area of the shell' | ||
- | $$\int\text{d}\left|\vec{A}\right|=\int\text{d}A=A=4\pi r^2$$ | + | $$\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 $A=4.52\cdot 10^{-2} \text{ m}^2$. | 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 $\Phi_e=\left|\vec{E}\right|A$. For our two shells: | + | 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*} | ||
- | \Phi_{\text{small}} &= 1.0\cdot 10^7 \text{ N/C } \cdot 1.13\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^{-9} \text{ Nm}^2\text{/ | + | \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^{-9} \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*} | \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. | 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. |