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Section 21.3 from Matter and Interactions (4th edition) | Section 21.3 from Matter and Interactions (4th edition) | ||
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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: | ||
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{{youtube> | {{youtube> | ||
==== Line of charge example ==== | ==== Line of charge example ==== | ||
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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\: | 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\: | ||
- | $$\Phi_{tot}=\int \vec{E} \cdot \vec{dA}=\frac{Q_{enclosed}}{\epsilon_0}$$ | + | $$\Phi_{tot}=\int \vec{E} \bullet |
=== Step 1 - Draw the electric field lines and determine a good Gaussian surface === | === Step 1 - Draw the electric field lines and determine a good Gaussian surface === | ||
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- | // | + | // |
=== Step 2 - Find the electric flux through the Gaussian surface === | === Step 2 - Find the electric flux through the Gaussian surface === | ||
Now that we have a Gaussian surface, we can find the electric flux at the surface of the cylinder. The cylinder has three surface - the flat top, the flat bottom, and the curved side of the cylinder - so we need to account for the flux through all three surfaces to find the total electric flux. | Now that we have a Gaussian surface, we can find the electric flux at the surface of the cylinder. The cylinder has three surface - the flat top, the flat bottom, and the curved side of the cylinder - so we need to account for the flux through all three surfaces to find the total electric flux. | ||
- | $$\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} \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 (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: | 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: | ||
- | $$\Phi_{top}=\int \vec{E}_{top} \cdot \vec{dA}_{top} = 0$$ | + | $$\Phi_{top}=\int \vec{E}_{top} \bullet |
Similarly for the bottom surface, the electric field vectors point radially away from the line of charge, but the dA vectors (shown by the green arrow) would point down (perpendicular to the surface). So the electric flux through the bottom surface is also zero. | Similarly for the bottom surface, the electric field vectors point radially away from the line of charge, but the dA vectors (shown by the green arrow) would point down (perpendicular to the surface). So the electric flux through the bottom surface is also zero. | ||
- | $$\Phi_{bottom}=\int \vec{E}_{bottom} \cdot \vec{dA}_{bottom} = 0$$ | + | $$\Phi_{bottom}=\int \vec{E}_{bottom} \bullet |
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 (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. | 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. | ||
- | $$\Phi_{side}=\int \vec{E}_{side} \cdot \vec{dA}_{side} = \int |\vec{E}_{side}| |\vec{dA}_{side}| cos(0)$$ | + | $$\Phi_{side}=\int \vec{E}_{side} \bullet |
$$\Phi_{side}=\int E_{side} dA_{side}$$ | $$\Phi_{side}=\int E_{side} dA_{side}$$ | ||
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==== Example ==== | ==== Example ==== | ||
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