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| 184_notes:examples:week5_flux_cylinder_line [2017/09/18 14:12] – tallpaul | 184_notes:examples:week5_flux_cylinder_line [2021/06/04 00:54] (current) – schram45 |
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| | [[184_notes:q_enc|Return to Enclosed Charge notes]] |
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| =====Example: Flux through a Cylinder on a Line of Charge===== | =====Example: Flux through a Cylinder on a Line of Charge===== |
| Suppose you have a point charge with value $1 \mu\text{C}$. What are the fluxes through two spherical shells centered at the point charge, one with radius $3 \text{ cm}$ and the other with radius $6 \text{ cm}$? | Suppose you have a line of charge with a uniform linear charge density of $\lambda=15\mu\text{C/m}$. What is the electric flux through a cylinder with radius $R=0.5 \text{ m}$, and length $l=3 \text{ m}$ that is placed so that its axis is aligned with the line of charge? Feel free to use the electric field due to an infinite uniform line of charge: $\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$ (where $\hat{r}$ points away from line, and $r$ is the distance from the line). |
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| ===Facts=== | ===Facts=== |
| * The point charge has charge $q=1 \mu\text{C}$. | * The cylinder has radius $R=0.5 \text{ m}$, and length $l=3 \text{ m}$. |
| * The two spheres have radii $3 \text{ cm}$ and $6 \text{ cm}$. | * The axis of the cylinder is aligned with the line charge. |
| | * The line charge has linear charge density $\lambda=15\mu\text{C/m}$. |
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| ===Lacking=== | ===Lacking=== |
| * $\Phi_e$ for each sphere | * $\Phi_e$ for the cylinder. |
| * $\text{d}\vec{A}$ or $\vec{A}$, if necessary | |
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| ===Approximations & Assumptions=== | |
| * There are no other charges that contribute appreciably to the flux calculation. | |
| * There is no background electric field. | |
| * The electric fluxes through the spherical shells are due only to the point charge. | |
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| ===Representations=== | ===Representations=== |
| * We represent the electric flux through a surface with: | * We represent the electric flux through a surface with: |
| $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A}$$ | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A}$$ |
| * We represent the electric field due to a point charge with: | * We represent the electric field due to an infinite line of uniform charge density with: |
| $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | $$\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ |
| * 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. |
| {{ 184_notes:5_flux_two_radii.png?300 |Point charge and two spherical shells}} | [{{ 184_notes:5_cylinder_line.png?400 |Charged Line and Cylindrical Surface}}] |
| ====Solution==== | |
| Before we dive into calculations, 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 are directed along the radial direction from the point charge. A more in-depth discussion of these symmetries can be found in the notes of [[184_notes:eflux_curved#Making_Use_of_Symmetry|using symmetry]] to simplify our flux calculation. | |
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| 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: | <WRAP TIP> |
| | ===Approximations & Assumptions=== |
| | There are a few approximations and assumptions we should make in order to simplify our model. |
| | * There are no other charges that contribute appreciably to the flux calculation. |
| | * The cylinder is aligned with respect to the line so that its bases are perpendicular to the line, and its wall is parallel (as described). This is just a geometric simplification for the model, and ensures the electric field through the cylinder wall is constant as the wall will be at a uniform distance away from the line of charge at any point. |
| | * Line of charge is very very long: This allows us to use the electric field equation provided in the problem statement, and ensures the electric field is constant through the wall of our cylinder. |
| | </WRAP> |
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| $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ | ====Solution==== |
| | First, we evaluate the situation qualitatively. Consider the electric field vectors from the charged line near the surface of the cylinder: |
| We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge's radial axis) since it is constant on the surface of a given spherical shell. We use the formula for the electric field from a point charge. | [{{ 184_notes:5_line_field_lines.png?400 ||Positively Charged Line -- Electric Field Lines}}] |
| | It's a little tough to demonstrate the electric field vectors with only two dimensions to draw on, but you can imagine that the thicker arrows point out of the page more, and the thinner arrows point into the page more (but the magnitude of the arrows are all the same). In essence, each vector points directly away from and perpendicular to the line of charge, as indicated in the formula for electric field from a line charge. |
| $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ | |
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| 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 $E=2.5\cdot 10^6 \text{ N/C}$. | |
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| 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's surface: | |
| $$\int\text{d}A=A=4\pi r^2$$ | |
| 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 $A=4.52\cdot 10^{-2} \text{ m}^2$. | |
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| 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: | The electric field vectors are parallel to the bases of the cylinder, so $\vec{E}\bullet\text{d}\vec{A}=0$ on the bases. So the flux through the bases should be $0$. For the wall of the cylinder, the electric field vectors are perpendicular to the surface, which means they are parallel to the area-vectors. These facts will greatly simplify our integral calculation of the flux. |
| \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{/C} \\ | \Phi_{\text{total}} &= \Phi_{\text{bases}}+\Phi_{\text{wall}} \\ |
| \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{/C} | &= 0 + \int_{\text{wall}}\vec{E}\bullet \text{d}\vec{A} \\ |
| | &= \int_{\text{wall}}E\hat{r}\bullet \text{d}A\hat{r} \\ |
| | &= E\int_{\text{wall}}\text{d}A \\ |
| | &= EA_{\text{wall}} \\ |
| | &= \frac{\lambda}{2\pi R \epsilon_0}\cdot 2\pi R l \\ |
| | &= \frac{\lambda l}{\epsilon_0} |
| \end{align*} | \end{align*} |
| | When we plug in values for $\lambda$, $l$, and $\epsilon_0$, we get $\Phi_{\text{cylinder}}=5.08\cdot 10^6\text{ Vm}$. |
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| 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. | Notice that in the [[184_notes:gauss_ex|next section of notes]], we define "Gauss' Law", which states that the total flux through a close surface is the amount of charge enclosed divided by $\epsilon_0$. A quick check for this example shows us that the charge enclosed by the cylinder covers a length of $l$, which means the charge is $\lambda l$. If we had used Gauss' Law, we would have quickly found that |
| | $$\Phi_{\text{cylinder}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\lambda l}{\epsilon_0}$$ |
| | This is the same result! An alternative question for this example could have been: What is the electric field due to a uniformly charged line? If we were not given the electric field at the beginning, we could have used symmetry arguments and Gauss' Law to work backwards, starting with the charge enclosed, and then using the integral formula for electric flux to solve for the electric field. There are sometimes electric fields that we do not know off-hand, and Gauss' Law is often the best tool to find them. |