point charge | charged plate

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Gauss's Law is always true: Φ = ∫_surface E · dA = q_enclosed/ε₀, but it isn't always useful for finding the electric field, which is what we are usually interested in. This should not be too surprising, because to find E, using an equation like ∫_surface  E · dA = q_enclosed/ε₀, E has to be able to come out of the integral, and for that to happen, E needs to be constant on a surface. This is where symmetry comes in. Gauss's law is only useful for calculating electric fields when the symmetry is such that you can construct a Gaussian surface so that the electric field is constant over the surface, and the angle between the electric field and the normal to the Gaussian surface does not vary over the surface (position is given in meters and electric field strength is given in newtons/coulonb). In practice, this means that you pick a Gaussian surface with the same symmetry as the charge distribution. Restart.

Consider a sphere around a point charge. The blue test charge shows the direction of the electric field. There is also a vector pointing in the direction of the surface normal to the sphere.

1. By moving the surface normal vector on the sphere and putting the test charge at three different points on the surface, find the value of  E · dA = E dA cosθ (set dA = 1) at these three points (read the electric field values in the yellow text box). Are they the same? Why or why not?

Now, put a box around the same point charge. The test charge now shows the direction of the electric field, and the smallest angle between the vector and a vertical axis is shown (in degrees). The red vector points in the direction of the surface normal to the box (two sides show).

2. By moving the surface normal vectors on the box and putting the test charge at three different points on the top surface, find the value of  E · dA = E dA cosθ (set dA = 1) at these three points. Are they the same? Why or why not?
3. In the context of your answers above, why is the sphere a better choice for using Gauss's law than the box?

Let's try another charge configuration. Put a sphere around a charged plate (assume the gray circles you see are long rods of charge that extend into and out of the screen to create a charged plate that you see in cross section).

4. Would the value of E · dA = E dA cosθ be the same at any three points on the Gaussian surface?
5. Explain, then, why you would not want to use a sphere for this configuration.

Now, put a box around a charged plate (assume the points you see are long rods of charge that extend into and out of the screen to create a charged plate that you see in cross section).

6. Find the value of E · dA = E dA cosθ at three points on the top. Are they essentially the same?
7. What about E · dA = E dA cosθ on the sides?

For the plate, using a box as a Gaussian surface means that E · dA = E dA cosθ is a constant for each section (top, bottom, and sides) and the electric field is a constant on the surface. This means you can write:

∫_surface E · dA = E ∫_surface  dA = EA (for the surfaces where the flux is nonzero).

8. Knowing that the charge per unit area on the big plate is σ, use Gauss's law to show that the expression for the electric field above or below a charged plate is E = σ/2ε₀ and the direction of the electric field is away from the plate for a positively charged plate. In your textbook you will probably also see an expression that says that the electric field is σ/ε₀ above or below the charged sheet. This holds true for conductors where σ is the charge/area on the top surface and there is the same amount of charge/area on the bottom surface (there is no net charge inside a conductor).

Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian and modified by Anne J. Cox.