Chapter 27 Summary

This chapter has one topic, Gauss’ Law. In words Gauss’s states that the net electric flux passing through a closed surface is proportional to the net charge inside the surface. For this statement to be useful, you need to understand what “net flux through a closed surface” means! It is given below as a more precise mathematical statement,

∫ E ^. ds = q/ε₀ .

The integral on the left side of the equation is the electric flux, the net amount of Electric Field “piercing” through a closed surface. (Think about the quills of the hedgehog!) I could not find an integral symbol with a circle on the integral sign signifying that the surface is closed, but keep in mind that Gauss’s law always applies to a flux integral over a closed surface. The surface in the integral is cleverly called a “Gaussian Surface.” The ds vector is equal in magnitude to a little area element on the surface and the direction of ds is along an outward pointing normal vector, that is a vector perpendicular to the surface.

The other important piece of information is that in electrostatics, the study of stationary charges, there cannot be an electric field inside a conductor. If there were, the field would cause the charges to move contradicting our assumption that the charges were stationary! In electrostatics, the only place charges can reside is on the surface of conductors. And the electric field at the surface of a conductor has to be perpendicular to the surface because if there was some piece of the electric field parallel to the surface, the charges there would move along the surface!

1. Gauss’s Law is a concise way of representing Coulomb’s law while emphasizing that the electric field is more fundamental than the electric force because the field concept did away with the “spooky” force acting at a distance difficulty talked about in Chapter 25.

2. To be able to actually evaluate the integral in Gauss’s Law to find an electric field, the charge distribution has to be very symmetric. The three kinds of charge distributions that allow for “simple” calculations of electric fields using Gauss’s law are: spherically symmetric, cylindrically symmetric, and flat infinite planes with a uniform charge distribution.

3. To use Gauss’s Law for a charge distribution that is spherical, we choose a shape for the “Gaussian” surface that matches the symmetry of the electric field. In this case we choose spherical Gaussian surfaces concentric with the charge distribution. Because the charge distribution is symmetric, the electric field strength will only depend on r, the distance from the center of the charge distribution, and will point either radially toward or away from the center depending on the sign of the net charge inside the radius r.

4. For a point charge q, the integral on the left becomes simple because E and ds are parallel if q>0 and antiparallel if q<0. The magnitude of E is constant on a spherical surface so it can be pulled outside the integral sign. Now the integral is just the sum of a bunch of little areas over the surface of a sphere which is 4 π r². So the integral on the left equals E 4 π r² which is equal to q/ε₀. Solving for E we just get the familiar result,

E = q/4 π ε₀ r² .

5. This works for any spherically symmetric charge distribution. The area of the Gaussian surface is always 4 π r² and the q on the right side of the equation is the net charge inside the surface.

6. Exactly the same argument works for cylindrically symmetric charge distributions accept that we need to pick a Gaussian surface with the appropriate symmetry, cylindrical in this case. The electric field around a cylindrically symmetric charge distribution is going to point radially away from the axis if the net charge inside is positive and radially toward the axis if the net charge is negative. The electric field is perpendicular to the curved surface of the cylinder and parallel to the top and bottom circular ends of the cylinder. Hence the electric flux is zero through the ends because the normal vector is perpendicular to E and the net electric flux through a Gaussian cylindrical surface is just E 2 π r L, where L is the length of the cylinder and r is its radius. As in the spherical case, the net flux is caused by the net charge, q, inside the Gaussian cylindrical surface.

7. For infinite planes, it is easy to see that the electric field has to be perpendicular to the charged surface. The appropriate shape for the Gaussian surface here is what is commonly called a “pillbox,” a squat cylinder. Now the electric field is perpendicular to the circular ends of the pillbox and parallel to the short height of the pillbox. If the area of the ends is A, then the net electric flux is either E A or 2 E A. The first case gives the answer for a conducting surface where the Electric Field is zero inside the conductor. The second case works for a plane of charge where the Electric Field pierces both ends of the pillbox. Again the net electric flux is caused by the net charge, q, inside the Gaussian pillbox.

8. Although Gauss’ Law is true for any shape closed surface, the integral for the net flux can be devilishly hard to calculate. Hence we will use Gauss’s law to find the electric field only when the charge distribution has one of the symmetries listed above.