At this point, we have talked about how to find the electric flux through flat surfaces and through curved surfaces as well how to find the 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 charg and symmetry arguments (Gauss's Law). Finally, we will discuss the advantages/disadvantages to using Gauss's Law.

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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's Law.

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Assuming Point P is close to the center of a very long line of negative charge, the electric field close to the center will point radially towards the line of charge, meaning the electric field vectors point horizontally toward 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.

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. $$\Phi_{tot}=\int \vec{E}_{top} \cdot \vec{dA}_{top}+ \int \vec{E}_{bottom} \cdot \vec{dA}_{bottom}+\int \vec{E}_{side} \cdot \vec{dA}_{side}$$ 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: $$\Phi_{top}=\int \vec{E}_{top} \cdot \vec{dA}_{top} = 0$$ 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. $$\Phi_{bottom}=\int \vec{E}_{bottom} \cdot \vec{dA}_{bottom} = 0$$ These answers should make sense since there are no electric field lines that poke through the top or bottom surfaces of the Gaussian cylinder.

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. $$\Phi_{side}=\int \vec{E}_{side} \cdot \vec{dA}_{side} = \int |\vec{E}_{side}| |\vec{dA}_{side}| cos(0)$$ $$\Phi_{side}=\int E_{side} dA_{side}$$

• Different way to say Q makes E-field
  • Symmetry required
  • Always true, rarely useful
  • Not computational

Inside a cylinder of charge

Sphere of charge