5.3.2: Explorations

Exploration 1: Flux and Gauss's Law

In this Exploration we will calculate the flux, \(\phi\), through three Gaussian surfaces: green, red, and blue (position is given in meters and electric field strength is given in newtons/coulonb). Note that this animation shows only two dimensions of a three-dimensional world. You will need to imagine that the circles you see are spheres. Restart.

Flux is a measure of the electric field through a surface.   It is given by the following equation:

\[\phi=\int_{\text{surface}}\mathbf{E}\cdot d\mathbf{A}=\int_{\text{surface}}E\cos\theta dA\nonumber\]

where \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is the unit area normal to the surface, and \(\theta\) is the angle between the electric field vector and the surface normal.

Move the test charge along one of the Gaussian surfaces (you must imagine that it is a sphere even though you can only see a cross section of it).

1. What is the magnitude of the electric field along the surface?
2. In what direction does it point?
3. What direction is normal to the Gaussian surface?

If the electric field, \(\mathbf{E}\), and the normal to the Gaussian surface, \(\mathbf{A}\), always point in the same direction relative to each other, and the electric field is constant, then the equation for flux becomes: \(\phi = E\cos\theta\int dA = EA\cos\theta\).

4. In the case of the point charge in (a)-(c), what is the angle between the electric field and the normal to the surface?

This means that \(\cos\theta = 1\). Therefore, for this case, \(\phi = EA\).

5. Calculate the flux for the surface you've chosen (remember that the surface area of a sphere is \(4\pi R^{2}\)).
6. Calculate the flux for the other two surfaces.

Because the electric field decreases as \(1/r^{2}\), but the area increases as \(r^{2}\), the flux is the same for all three cases. This is the basis of Gauss's law: The flux through a Gaussian surface is proportional to the charge within the surface. With twice as much charge, there is twice as much flux. Gauss's law says that \(\phi = q_{\text{enclosed}}/\varepsilon_{0}\).

7. What is the magnitude and sign of the point charge?

Exploration authored by Anne J. Cox.
Script authored by Mario Belloni.

Exploration 2: Symmetry and Using Gauss's Law

Gauss's Law is always true: \(\phi =\int_{\text{surface}}\mathbf{E}\cdot d\mathbf{A} = q_{\text{enclosed}}/\varepsilon_{0}\), 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 \(\mathbf{E}\), using an equation like \(\int_{\text{surface}}\mathbf{E}\cdot d\mathbf{A} = q_{\text{enclosed}}/\varepsilon_{0}\), \(\mathbf{E}\) has to be able to come out of the integral, and for that to happen, \(\mathbf{E}\) 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  \(\mathbf{E}\cdot d\mathbf{A} = E dA \cos\theta\) (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  \(\mathbf{E}\cdot d\mathbf{A} = E dA \cos\theta\) (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 \(\mathbf{E}\cdot d\mathbf{A} = E dA \cos\theta\) 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 \(\mathbf{E}\cdot d\mathbf{A} = E dA \cos\theta\) at three points on the top. Are they essentially the same?
7. What about \(\mathbf{E}\cdot d\mathbf{A} = E dA \cos\theta\) on the sides?

For the plate, using a box as a Gaussian surface means that \(\mathbf{E}\cdot d\mathbf{A} = E dA \cos\theta\) 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:

\[\int_{\text{surface}}\mathbf{E}\cdot d\mathbf{A}=E\int_{\text{surface}}dA=EA\text{ (for the surfaces where the flux is nonzero).}\nonumber\]

8. Knowing that the charge per unit area on the big plate is \(\sigma\), use Gauss's law to show that the expression for the electric field above or below a charged plate is \(E = \sigma /2\varepsilon_{0}\) 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.

Exploration 3: Conducting and Insulating Sphere

What is the difference between the electric fields inside and outside of a solid insulating sphere (with charge distributed throughout the volume of the sphere) and those inside and outside of a conducting sphere? Move the test charge to map out the magnitude of the electric field as a function of distance from the center (position is given in centimeters, electric field strength is given in newtons/coulonb, and flux is given in \(\text{N}\cdot\text{cm}^{2}\text{/C}\)). Restart.

1. Compare the electric fields inside and outside of the two spheres. What is the same and what is different (same total charge on both spheres)?
2. If a Gaussian surface larger than the two spheres is put around each, how will the flux through each compare? Why?

Try putting a big Gaussian surface around the insulator. The bar measures the flux. Now try around the conductor.

3. Why is the flux the same?
4. How much charge is on each sphere? How do you know?
5. What do you expect the flux to be through a Gaussian surface inside the conductor? Why? Try it and explain the results.

Now try putting the same size small Gaussian surface inside the insulator.

6. What flux value do you get?
7. How much charge is enclosed in this smaller surface?
8. What is the ratio of the charge enclosed in the small surface to the total charge on the insulating sphere?
9. What is the ratio of the volume of the small surface compared to the volume of the insulating sphere? Explain why the two ratios in (h) and (i) are the same.
10. Use Gauss's law for the smaller surface to calculate the field at that point inside the sphere. Verify that it agrees with the value on the graph.

As a reminder, Gauss's law relates the flux to the charge enclosed (\(q_{\text{enclosed}}\)) in a Gaussian surface through the following equation:

\[\phi = q_{\text{enclosed}}/\varepsilon_{0}\text{ (and Flux }=\phi =\int\mathbf{E}\cdot d\mathbf{A}=\int E\cos\theta dA)\nonumber\]

where \(\varepsilon_{0}\) is the permittivity of free space (\(8.85\times 10^{-12}\text{ C}^{2}\text{/N}\cdot\text{m}^{2}\)), \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is the unit normal to the surface, and \(\theta\) is the angle between the electric field vector and the surface normal. The surface area of a sphere is \(4\pi r^{2}\).

Exploration authored by Anne J. Cox.
Script authored by Mario Belloni and Anne J. Cox.

Exploration 4: Application of Gauss's Law

A point charge has radial (spherical) symmetry about the center of the charge, while a line charge has cylindrical symmetry about the center of the wire (position is given in meters and electric field strength is given in newtons/coulonb). However, a two-dimensional view of both can look the same. Restart.

Consider the two configurations. One is a point charge and one is a line of charge (pointing into and out of the screen). Which is which? The electric field is different for the two cases (and you use two different Gaussian surfaces).

1. As a function of the distance away from the charge (as a function of \(r\)), what is the electric field of a point charge?
2. Therefore, if you measure the electric field at some point and then measure it twice as far away, how much should the electric field be decreased?
3. Which configuration, then, is a point charge?
4. Use Gauss's law to find an analytic expression for the electric field around a line of charge. You may find the following diagram useful:

Figure \(\PageIndex{1}\)

5. If you measure the electric field at some point and then move twice as far away, how should the field drop off from a line of charge?
6. Does the electric field of the other configuration agree with this?

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

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