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184_notes:examples:week5_flux_cylinder_line [2017/09/18 14:12] – tallpaul | 184_notes:examples:week5_flux_cylinder_line [2017/09/25 15:39] – [Solution] tallpaul | ||
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=====Example: | =====Example: | ||
- | Suppose you have a point charge with value $1 \mu\text{C}$. What are the fluxes | + | 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 |
===Facts=== | ===Facts=== | ||
- | * The point charge | + | * The cylinder |
- | * The two spheres have radii $3 \text{ | + | * The axis of the cylinder is aligned with the line charge. |
+ | * The line charge has linear charge density | ||
===Lacking=== | ===Lacking=== | ||
- | * $\Phi_e$ for each sphere | + | * $\Phi_e$ for the cylinder. |
- | * $\text{d}\vec{A}$ or $\vec{A}$, if necessary | + | |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
* There are no other charges that contribute appreciably to the flux calculation. | * 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 |
- | | + | * Line of charge is very very long |
===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 |
- | $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | + | $$\vec{E} = \frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ |
- | * 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 | + | {{ 184_notes:5_cylinder_line.png?400 |Charged Line and Cylindrical Surface}} |
====Solution==== | ====Solution==== | ||
- | Before we dive into calculations, let's consider how we can simplify | + | First, we evaluate |
+ | {{ 184_notes: | ||
+ | It' | ||
- | Since $\vec{E}$ is constant with respect to $\text{d}\vec{A}$ (in this case, it is sufficient that $\vec{E}$ | + | The electric field is parallel to the bases of the cylinder, so $\vec{E}\bullet\text{d}\vec{A}=0 on the bases$. So the flux through |
- | + | ||
- | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ | + | |
- | + | ||
- | We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge' | + | |
- | + | ||
- | $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ | + | |
- | + | ||
- | 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 $E=2.5\cdot 10^6 \text{ N/C}$. | + | |
- | + | ||
- | 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 | + | |
- | + | ||
- | Now, we bring it together to find electric | + | |
\begin{align*} | \begin{align*} | ||
- | \Phi_{\text{small}} & | + | \Phi_{\text{total}} &= \Phi_{\text{bases}}+\Phi_{\text{wall}} \\ |
- | \Phi_{\text{large}} & | + | &= 0 + \int_{\text{wall}}\vec{E}\bullet |
+ | & | ||
+ | | ||
+ | &= 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 the same answer for both shells! It turns out that the radius | + | Notice |
+ | $$\Phi_{\text{cylinder}}=\frac{Q_{\text{enclosed}}}{\epsilon_0}=\frac{\lambda l}{\epsilon_0}$$ | ||
+ | This is the same result! An alternative question for this example could have been: What is the electric |