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--- 184_notes:examples:week5_flux_two_radii [2017/09/18 12:35] – [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 flux 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?400 |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 point directly away 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.
-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:electricflux4.jpg?400 |Area-vectors and E-field-vectors point in same direction}}]
-$$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = \left|\vec{E}\right|\int\text{d}\left|\vec{A}\right|$$
+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:
-We can rewrite $\left|\vec{E}\right|$ since it is constant on the surface of a given spherical shell. We use the formula for the electric field from a point charge.
+$$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$
-$$\left|\vec{E}\right| = \frac{1}{4\pi\epsilon_0}\frac{|q|}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{|q|}{r^2}$$
+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.
-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 $\left|\vec{E}\right|=1.0\cdot 10^7 \text{N/C}$. For the larger shell, we have $\left|\vec{E}\right|=2.5\cdot 10^6 \text{N/C}$.
+$$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$
+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 $E=2.5\cdot 10^6 \text{ N/C}$.
+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:
+$$\int\text{d}A=A=4\pi r^2$$
+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 $A=4.52\cdot 10^{-2} \text{ m}^2$.
-In order to find electric flux, we must first find $\vec{A}$. Remember in the [[184_notes:e_flux#Area_as_a_Vector|notes on flux]] that area can be a vector when we define it as a cross product of width and length vectors. Here, we can use the following for width and length, with width being the top of the rectangle, and pointing out of the page, and length being the longer side, and pointing at an upwards angle:
+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:
-$$\vec{w}=3\text{ m }\hat{z}$$
-$$\vec{l}=5\text{ m }\cdot\cos 30^\circ (\hat{x})+5\text{ m }\sin 30^\circ\hat{y} = 2.5\sqrt{3}\text{ m } \hat{x} + 2.5\text{ m }\hat{y}$$
-Now, we can find the area vector:
 \begin{align*}
-\vec{A} &= \vec{l}\times\vec{w} \\
+\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} \\
-        &= (2.5\sqrt{3}\text{ m } \hat{x} + 2.5\text{ m }\hat{y}) \times (3\text{ m }\hat{z}) \\
+\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}
-        &= 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{ m}^2 \hat{y}
-\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} \\
-        &= (8\text{ V/m } \hat{x}) \bullet (7.5\text{ m}^2\hat{x} - 7.5\sqrt{3}\text{ m}^2 \hat{y}) \\
-        &= 60\text{ Vm}
 \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.