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+ | Section 21.3 from Matter and Interactions (4th edition) | ||
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+ | [[184_notes: | ||
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===== Putting Gauss' | ===== Putting Gauss' | ||
- | At this point, we have talked about how to find the electric flux through [[184_notes: | + | At this point, we have talked about how to find the electric flux through [[184_notes: |
+ | {{youtube> | ||
==== Line of charge example ==== | ==== Line of charge example ==== | ||
- | FIXME Add Figure | + | [{{184_notes: |
- | 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 $\lambda=-4C/m$. 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' | + | 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 |
+ | $$\Phi_{tot}=\int \vec{E} \bullet \vec{dA}=\frac{Q_{enclosed}}{\epsilon_0}$$ | ||
- | === Step 1 - Draw the electric field lines and determine a good Gaussian surface | + | === Step 1 - Draw the electric field lines === |
- | FIXME Add Figure | + | |
- | // | + | [{{ 184_notes: |
+ | |||
+ | // | ||
+ | |||
+ | === 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' | ||
- | === 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. | 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. | ||
- | $$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$ | + | $$\Phi_{tot}=\int \vec{E}_{top} \bullet |
- | Starting with the top surface, the electric field vectors would still be pointing radially away from the line of charge (since we are still in the middle of the line of charge) and the area vector for the top surface would point perpendicularly away from the top surface (up in this case). This means at every location on the top surface, the electric field vectors would be perpendicular to the dA vectors so the dot product for the top surface is zero: | + | Starting with the top surface, the electric field vectors would still be pointing radially away from the line of charge (since we are still in the middle of the line of charge) and the area vector |
- | $$\Phi_{top}=\int \vec{E}_{top} \cdot \vec{dA}_{top} = 0$$ | + | $$\Phi_{top}=\int \vec{E}_{top} \bullet |
- | 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 |
- | $$\Phi_{bottom}=\int \vec{E}_{bottom} \cdot \vec{dA}_{bottom} = 0$$ | + | $$\Phi_{bottom}=\int \vec{E}_{bottom} \bullet |
These answers should make sense since there are no electric field lines that poke through the top or bottom surfaces of the Gaussian cylinder. | These answers should make sense since there are no electric field lines that poke through the top or bottom surfaces of the Gaussian cylinder. | ||
- | This leave the electric flux through the side of the cylinder. We know that the electric field points radially away from the center of the line. Since we have circular cylinder, any dA along the curved side would also point radially away from the surface. Thus, for any point along the side of the cylinder, the electric field vector and the dA vector would be parallel. This means the dot product reduces to a simple multiplication of the magnitudes. | + | This leaves |
- | $$\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 |
$$\Phi_{side}=\int E_{side} dA_{side}$$ | $$\Phi_{side}=\int E_{side} dA_{side}$$ | ||
+ | |||
+ | Since the side surface of the cylinder is the same distance away from the line of charge, the magnitude of the electric field is constant at every point along the side. This means that E is constant over the whole surface area, so we can pull it out of the integral. | ||
+ | $$\Phi_{side}=E_{side} \int dA_{side}$$ | ||
+ | Then the integral of dA is just A (or rather adding up all the little pieces of area gives you the total area). | ||
+ | $$\Phi_{side}=E_{side}A_{side}$$ | ||
+ | We can then rewrite the surface area of the side of the cylinder in terms of the radius of the cylinder (d) and the height of the cylinder (h). | ||
+ | $$\Phi_{side}= E_{side} 2\pi d h$$ | ||
+ | So including the fact that the electric flux through the top and bottom of the cylinder, we get the total electric flux to be: | ||
+ | $$\Phi_{tot}= E_{side} 2\pi d h$$ | ||
+ | |||
=== Step 3 - Find the amount of charge enclosed === | === Step 3 - Find the amount of charge enclosed === | ||
+ | Now, we need to figure out how much charge is enclosed by the Gaussian cylinder that we chose. Since we said that the height of the cylinder was h, then we can use the given charge density to determine the enclosed charge: | ||
+ | $$Q_{encl} = \lambda l = \lambda h$$ | ||
+ | Checking the units shows that we have $C = \frac{C}{m}m$, | ||
=== Step 4 - Solve for electric field and determine the direction === | === Step 4 - Solve for electric field and determine the direction === | ||
- | + | Now we can plug the total electric flux and the enclosed charge into Gauss' | |
+ | $$\Phi_{tot}= \frac{Q_{enclosed}}{\epsilon_0}$$ | ||
+ | $$E_{side} 2\pi d h=\frac{\lambda h}{\epsilon_0}$$ | ||
+ | $$E_{side} =\frac{\lambda h}{\epsilon_0 2\pi d h}$$ | ||
+ | $$E_{side} =\frac{\lambda}{\epsilon_0 2\pi d}$$ | ||
+ | $$E_{side} =\frac{4*10^{-9}}{8.85*10^{-12}* 2\pi*0.05}=1439\: | ||
+ | This is then the magnitude of the electric field anywhere along the side of the cylinder, including Point P. If we want to write the electric field vector at Point P, then we need to add the direction to the field magnitude we just found. Before (in Step 1), we already determined that the electric field would point radially away from the line of charge. So the electric field at Point P will point in the $+\hat{x}$ direction (in cartesian coordinates) or in the $\hat{r}$ direction (in cylindrical coordinates). Thus, we found the electric field at Point P to be: | ||
+ | $$\vec{E}_P =\frac{\lambda}{\epsilon_0 2\pi d} \hat{x}$$ | ||
+ | $$\vec{E}_P = 1439\: | ||
+ | |||
+ | ==== Gauss' | ||
+ | Thus, to summarize, the steps for using Gauss' | ||
+ | - 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' | ==== Advantages and Disadvantages of Using Gauss' | ||
- | *Different way to say Q makes E-field | + | While there are many advantages |
- | *Symmetry required | + | |
- | *Always | + | |
- | *Not computational | + | |
==== Example ==== | ==== Example ==== | ||
- | Inside a cylinder | + | [[: |
- | + | ||
- | Sphere of charge | + |