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184_notes:examples:week5_flux_two_radii [2017/09/17 20:09] – [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, let's consider how we can simplify the problem. Think about the nature of the electric field due to a point charge, and of the $\text{d}\vec{A}$ vector for a spherical shell. The magnitude of the electric field will be constant along the surface of a given sphere, since the surface is a constant distance away from the point charge. Further, $\vec{E}$ will always be parallel to $\text{d}\vec{A}$ on these spherical shells, since both point directly away from the point charge. A more in depth discussion of these symmetries can be found in the notes of [[184_notes: | + | Before we dive into calculations, let's consider how we can simplify the problem |
- | 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: |
- | $$\Phi_e=\int\vec{E}\bullet | + | Since the shell is a fixed distance from the point charge, the electric field has constant magnitude on the shell. Since $\vec{E}$ is parallel to $\text{d}\vec{A}$ and has constant magnitude (on the shell), the dot product simplifies substantially: |
- | In order to find electric flux, we must first find $\vec{A}$. Remember in the [[184_notes: | + | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} |
- | $$\vec{w}=3\text{ m }\hat{z}$$ | + | |
- | $$\vec{l}=5\text{ m }\cdot\cos 30^\circ (\hat{x})+5\text{ | + | Note that $E$ is a scalar value representing $\vec{E}$ |
- | Now, we can find the area vector: | + | |
+ | $$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}$$ | ||
+ | |||
+ | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we find $E=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we find $E=2.5\cdot | ||
+ | |||
+ | To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by $\text{d}A$. Since we are integrating this little piece over the entire shell, we end up with the area of the shell' | ||
+ | $$\int\text{d}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*} | \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. |