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--- 184_notes:examples:week5_flux_two_radii [2017/09/18 13:28] – [Solution] tallpaul
+++ 184_notes:examples:week5_flux_two_radii [2021/06/04 00:47] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:eflux_curved|Return to Electric Flux through Curved Surfaces notes]]
 =====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 $6 \text{ cm}$?
@@ Line 9: / Line 11: @@
   * $\Phi_e$ for each sphere
   * $\text{d}\vec{A}$ or $\vec{A}$, if necessary
-===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.
 ===Representations===
@@ Line 21: / Line 18: @@
 $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$
   * We represent the situation with the following diagram. Note that the circles are indeed spherical shells, not rings as they appear.
-{{ 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}}]
+<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.
+  * There is no background electric field.
+  * The electric fluxes through the spherical shells are due only to the point charge.
+The first three assumptions ensure that there is nothing else contributing or affecting the flux through our spheres in the model.
+  * Perfect spheres: This will simplify our area vectors and allows us to use geometric equations for spheres in our calculations.
+  * Constant charge for the point charge: Ensures that the point charge is not charging or discharging with time.
+</WRAP>
 ====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.
+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.
+[{{ 184_notes:electricflux4.jpg?400 |Area-vectors and E-field-vectors point in same direction}}]
-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:
 $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$
-We can rewrite $E$ (scalar value representing $\vec{E}$ as a magnitude, with a sign indicating direction along point charge'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.
+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.
 $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$
-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 $E=2.5\cdot 10^6 \text{ N/C}$.
+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's surface: