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184_notes:gauss_ex [2017/06/29 17:57] – dmcpadden | 184_notes:gauss_ex [2018/05/15 17:01] – curdemma | ||
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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 $\lambda=-4\:nC/m$. 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' | + | 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 |
$$\Phi_{tot}=\int \vec{E} \cdot \vec{dA}=\frac{Q_{enclosed}}{\epsilon_0}$$ | $$\Phi_{tot}=\int \vec{E} \cdot \vec{dA}=\frac{Q_{enclosed}}{\epsilon_0}$$ | ||
=== Step 1 - Draw the electric field lines and determine a good Gaussian surface === | === Step 1 - Draw the electric field lines and determine a good Gaussian surface === | ||
- | FIXME Add Figure | ||
- | // | + | {{ 184_notes: |
+ | |||
+ | // | ||
=== Step 2 - Find the electric flux through the Gaussian surface === | === 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. | ||
$$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$ | $$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$ | ||
- | 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 |
$$\Phi_{top}=\int \vec{E}_{top} \cdot \vec{dA}_{top} = 0$$ | $$\Phi_{top}=\int \vec{E}_{top} \cdot \vec{dA}_{top} = 0$$ | ||
- | 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 |
$$\Phi_{bottom}=\int \vec{E}_{bottom} \cdot \vec{dA}_{bottom} = 0$$ | $$\Phi_{bottom}=\int \vec{E}_{bottom} \cdot \vec{dA}_{bottom} = 0$$ | ||
These answers should make sense since there are no electric field lines that poke through the top or bottom surfaces of the Gaussian cylinder. | 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) |
$$\Phi_{side}=\int \vec{E}_{side} \cdot \vec{dA}_{side} = \int |\vec{E}_{side}| |\vec{dA}_{side}| cos(0)$$ | $$\Phi_{side}=\int \vec{E}_{side} \cdot \vec{dA}_{side} = \int |\vec{E}_{side}| |\vec{dA}_{side}| cos(0)$$ | ||
$$\Phi_{side}=\int E_{side} dA_{side}$$ | $$\Phi_{side}=\int E_{side} dA_{side}$$ | ||
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$$E_{side} =\frac{\lambda h}{\epsilon_0 2\pi d h}$$ | $$E_{side} =\frac{\lambda h}{\epsilon_0 2\pi d h}$$ | ||
$$E_{side} =\frac{\lambda}{\epsilon_0 2\pi d}$$ | $$E_{side} =\frac{\lambda}{\epsilon_0 2\pi d}$$ | ||
- | $$E_{side} =\frac{-4*10^{-9}}{8.85*10^{-12}* 2\pi*0.05}=-1439\: | + | $$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. | + | 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\: | ||
==== 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 | + |