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184_notes:examples:week5_flux_two_radii [2017/09/22 15:39] – dmcpadden | 184_notes:examples:week5_flux_two_radii [2021/05/29 21:09] – schram45 |
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| [[184_notes:eflux_curved|Return to Electric Flux through Curved Surfaces notes]] |
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=====Example: Flux through Two Spherical Shells===== | =====Example: Flux through Two Spherical Shells===== |
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 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 $6 \text{ cm}$? |
* There is no background electric field. | * There is no background electric field. |
* The electric fluxes through the spherical shells are due only to the point charge. | * The electric fluxes through the spherical shells are due only to the point charge. |
| * Perfect spheres. |
| * Constant charge for the point charge (no charging/discharging). |
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===Representations=== | ===Representations=== |
$$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ | $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$ |
* 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:5_flux_two_radii.png?300 |Point charge and two spherical shells}} | [{{ 184_notes:5_flux_two_radii.png?300 |Point charge and two spherical shells}}] |
====Solution==== | ====Solution==== |
Before we dive into calculations, let's consider how we can simplify the problem by thinking 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 FIXME (You could just pull in the picture that exists in the notes for this). 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. | Before we dive into calculations, let's consider how we can simplify the problem by thinking 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. See below for a visual. 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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| [{{ 184_notes:electricflux4.jpg?400 |Area-vectors and E-field-vectors point in same direction}}] |
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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 $\text{d}\vec{A}$ and has constant magnitude), we can rewrite our flux representation: | 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: $\vec{E}\bullet \text{d}\vec{A} = E\text{d}A$. We can now rewrite our flux representation: |
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$$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ |
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FIXME (You could use this to justify why E is constant on area) We can rewrite $E$ since it is constant on the surface of a given spherical shell. (Where $E$ is scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge's radial axis.) We use the formula for the electric field from a point charge. | Note that $E$ is a scalar value representing $\vec{E}$ as a magnitude, which includes a sign indicating direction along the point charge's radial axis. We can rewrite $E$ using the formula for the electric field from a point charge. |
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$$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ | $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ |