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--- 184_notes:gauss_ex [2017/09/08 21:33] – [Example] tallpaul
+++ 184_notes:gauss_ex [2021/06/04 00:36] (current) – schram45
@@ Line 1: / Line 1: @@
+Section 21.3 from Matter and Interactions (4th edition)
+/*[[184_notes:q_enc|Previous Page: Enclosed Charge]]*/
 ===== Putting Gauss's Law Together =====
-At this point, we have talked about how to find the electric flux through [[184_notes:e_flux|flat surfaces]] and through [[184_notes:eflux_curved|curved surfaces]] as well how to find the [[184_notes:q_enc|enclosed charge using charge density]]. These notes will go through two examples of how we find the electric field at a single point using electric flux, enclosed charge and symmetry arguments (Gauss's Law). Finally, we will discuss the advantages/disadvantages to using Gauss's Law.
+At this point, we have talked about how to find the electric flux through [[184_notes:e_flux|flat surfaces]] and through [[184_notes:eflux_curved|curved surfaces]] as well how to find the [[184_notes:q_enc|enclosed charge using charge density]]. These notes will go through two examples of how we find the electric field at a single point using electric flux, enclosed charge and symmetry arguments (Gauss's Law). You should notice many similarities between the steps that we use for Gauss's Law and those that we used for [[184_notes:amp_law|Ampere's Law]]. Finally, we will discuss the advantages/disadvantages to using Gauss's Law.
+{{youtube>ESRDPplcFho?large}}
 ==== Line of charge example ====
+[{{184_notes:electricflux11.jpg?350|Line of charge. What is the electric field at point P (teal dot)?  }}]
-{{184_notes:electricflux11.jpg?350}}
+Suppose we want to find the electric field at Point P (shown by the teal dot), which is a distance of $d=.05 m$ away from a very long line of charge (shown by the red line) with a charge density of $\lambda=4\:nC/m$. Instead of building up the electric field by splitting the line into chunks, we will use the symmetry of the line charge to solve Gauss's Law. Remember Gauss's Law says that the total electric flux through a closed Gaussian surface must be equal to the amount of charge (divided by a constant) inside the Gaussian surface.
+$$\Phi_{tot}=\int \vec{E} \bullet \vec{dA}=\frac{Q_{enclosed}}{\epsilon_0}$$
-Suppose we want to find the electric field at Point P, which is a distance of $d=.05 m$ away from a very long line of charge with a charge density of $\lambda=4\:nC/m$. Instead of building up the electric field by splitting the line into chunks, we will use the symmetry of the line charge to solve Gauss's Law. Remember Gauss's Law says that the total electric flux through a closed Gaussian surface must be equal to the amount of charge (divided by a constant) inside the Gaussian surface.
+=== Step 1 - Draw the electric field lines ===
-$$\Phi_{tot}=\int \vec{E} \cdot \vec{dA}=\frac{Q_{enclosed}}{\epsilon_0}$$
-=== Step 1 - Draw the electric field lines and determine a good Gaussian surface ===
+[{{  184_notes:electricflux12.jpg?350|Gaussian surface around a long line of charge}}]
-{{184_notes:electricflux12.jpg?350}}
+//__Assuming Point P is close to the center of a very long line of positive charge__//, the electric field close to the center  will point radially away from the line of charge, meaning the electric field vectors (shown by the red arrows) point horizontally away from the line and are the same size for a certain distance away from the line.
-//__Assuming Point P is close to the center of a very long line of positive charge__//, the electric field close to the center  will point radially away from the line of charge, meaning the electric field vectors point horizontally away from the line and are the same size for a certain distance away from the line. When we are picking our Gaussian surface, we want to pick a shape with sides that are either parallel to or perpendicular to the electric field vectors. In this case, a cylinder will work nicely. Remember that the choice of Gaussian surface is completely arbitrary, so we are picking a shape that will provide the simplest math. We will pick our cylinder to have a radius equal to $d=.05 m$, so that Point P is on the edge of the cylinder. The height of the cylinder doesn't particularly matter as long as it is small enough to not include the end points of line, so let's just call the height of the cylinder $h$. We will want the center of the cylinder to be centered along the line of charge so that the electric field along the surface of the cylinder is constant in direction and magnitude.
+=== Step 2 - Determine a good Gaussian surface and find the electric flux through the Gaussian surface ===
+When we are picking our Gaussian surface, we want to pick a shape with sides that are either parallel to or perpendicular to the electric field vectors. In this case, a cylinder will work nicely. Remember that the choice of Gaussian surface is completely arbitrary, so we are picking a shape that will provide the simplest math. We will pick our cylinder to have a radius equal to $d=.05 m$, so that Point $P$ is on the edge of the cylinder. The height of the cylinder doesn't particularly matter as long as it is small enough to not include the end points of the line, so let's just call the height of the cylinder $h$. We will want the center of the cylinder to be centered along the line of charge so that the electric field along the surface of the cylinder is constant in direction and magnitude.
-=== Step 2 - Find the electric flux through the Gaussian surface ===
 Now that we have a Gaussian surface, we can find the electric flux at the surface of the cylinder. The cylinder has three surface - the flat top, the flat bottom, and the curved side of the cylinder - so we need to account for the flux through all three surfaces to find the total electric flux.
-$$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$
+$$\Phi_{tot}=\int \vec{E}_{top} \bullet \vec{dA}_{top}+ \int \vec{E}_{bottom} \bullet \vec{dA}_{bottom}+\int \vec{E}_{side} \bullet \vec{dA}_{side}$$
-Starting with the top surface, the electric field vectors would still be pointing radially away from the line of charge (since we are still in the middle of the line of charge) and the area vector for the top surface would point perpendicularly away from the top surface (up in this case). This means at every location on the top surface, the electric field vectors would be perpendicular to the dA vectors so the dot product for the top surface is zero:
+Starting with the top surface, the electric field vectors would still be pointing radially away from the line of charge (since we are still in the middle of the line of charge) and the area vector (shown by the green arrow) for the top surface would point perpendicularly away from the top surface (up in this case). This means at every location on the top surface, the electric field vectors would be perpendicular to the dA vectors so the dot product for the top surface is zero:
-$$\Phi_{top}=\int \vec{E}_{top} \cdot \vec{dA}_{top} = 0$$
+$$\Phi_{top}=\int \vec{E}_{top} \bullet \vec{dA}_{top} = 0$$
-Similarly for the bottom surface, the electric field vectors point radially away from the line of charge, but the dA vectors would point down (perpendicular to the surface). So the electric flux through the bottom surface is also zero.
+Similarly for the bottom surface, the electric field vectors point radially away from the line of charge, but the dA vectors (shown by the green arrow) would point down (perpendicular to the surface). So the electric flux through the bottom surface is also zero.
-$$\Phi_{bottom}=\int \vec{E}_{bottom} \cdot \vec{dA}_{bottom} = 0$$
+$$\Phi_{bottom}=\int \vec{E}_{bottom} \bullet \vec{dA}_{bottom} = 0$$
 These answers should make sense since there are no electric field lines that poke through the top or bottom surfaces of the Gaussian cylinder.
-This leaves the electric flux through the side of the cylinder. We know that the electric field points radially away from the center of the line. Since we have circular cylinder, any dA along the curved side would also point radially away from the surface. Thus, for any point along the side of the cylinder, the electric field vector and the dA vector would be parallel. This means the dot product reduces to a simple multiplication of the magnitudes.
+This leaves the electric flux through the side of the cylinder. We know that the electric field points radially away from the center of the line. Since we have circular cylinder, any dA along the curved side (shown by the green arrows) would also point radially away from the surface. Thus, for any point along the side of the cylinder, the electric field vector and the dA vector would be parallel. This means the dot product reduces to a simple multiplication of the magnitudes.
-$$\Phi_{side}=\int \vec{E}_{side} \cdot \vec{dA}_{side} = \int |\vec{E}_{side}| |\vec{dA}_{side}| cos(0)$$
+$$\Phi_{side}=\int \vec{E}_{side} \bullet \vec{dA}_{side} = \int |\vec{E}_{side}| |\vec{dA}_{side}| cos(0)$$
 $$\Phi_{side}=\int E_{side} dA_{side}$$
@@ Line 52: / Line 58: @@
 $$\vec{E}_P =\frac{\lambda}{\epsilon_0 2\pi d} \hat{x}$$
 $$\vec{E}_P = 1439\:N/C\:\hat{x}$$
+==== Gauss's Law Steps Summary ====
+Thus, to summarize, the steps for using Gauss's Law are:
+  - Figure out and draw the electric field around the charge distribution.
+  - Choose a Gaussian surface that a) goes through your observation point, b) has area vectors that are either parallel or perpendicular to the electric field vectors, and c) has a constant electric field along the Gaussian surface. This allows you simplify the electric flux integral.
+  - Find the amount of charge enclosed by the Gaussian surface (maybe using charge density if you need a fraction of the total charge).
+  - Solve for the electric field and determine the direction.
 ==== Advantages and Disadvantages of Using Gauss's Law ====
@@ Line 57: / Line 71: @@
 ==== Example ====
-Inside a (insulated) sphere of charge (part b could be for a conductor)
+  * [[:184_notes:examples:Week5_gauss_ball|Gauss' Law Application -- A Ball of Charge]]
+    * Video Example: Gauss's Law Application -- A Ball of Charge
+{{youtube>8PEh2KA4Dxk?large}}