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184_notes:examples:week5_flux_two_radii [2017/09/18 13:19] – [Solution] tallpaul | 184_notes:examples:week5_flux_two_radii [2017/09/25 13:26] – [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: |
- | $$\Phi_e=\int\vec{E}\bullet | + | Since the shell is a fixed distance from the point charge, the electric field has constant magnitude on the shell. Since $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude (on the shell), the dot product simplifies substantially: |
- | We can rewrite | + | $$\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}$$ | + | Note that $E$ is a scalar value representing |
- | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $\left|\vec{E}\right|=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we have $\left|\vec{E}\right|=2.5\cdot 10^6 \text{ N/C}$. | + | $$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 10^5 \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{/ |