Worksheet for Exploration 24.1: Flux and Gauss's Law

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1 Worksheet for Exploration 24.1: Flux and Gauss's Law In this Exploration, we will calculate the flux, Φ, through three Gaussian surfaces: green, red and blue (position is given in meters and electric field strength is given in N/C). Note that this animation shows only two dimensions of a threedimensional world. You will need to imagine that the circles you see are spheres. Flux is a measure of the electric field through a surface. It is given by the following equation: Φ = surface E da = surface E cosθ da, where E is the electric field, da is the unit area normal to the surface and θ 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). a. What is the magnitude of the electric field along the surface? i. When you measure the electric field be careful and precise. Spend sufficient time to take good measurements. E green = E red = E blue = b. In what direction does it point? c. What direction is normal to the Gaussian surface? i. that is relative to the electric field direction in each case. If the electric field, E, and the normal to the Gaussian surface, A, always point in the same direction relative to each other, and the electric field is constant, then the equation for flux becomes: Φ = Ecosθ da = EAcosθ d. 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 cosθ = 1. Therefore, for this case, Φ = EA. e. Calculate the flux for the surface you've chosen (remember that the surface area of a sphere is 4πR 2 ).

2 i. Place all flux answers below in part f. f. Calculate the flux for the other two surfaces. Φ Φ Φ green red blue = = = 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 Φ = q enclosed /ε 0. g. What is the magnitude and sign of the point charge? q=

3 Worksheet for Exploration 24.2: Symmetry and Using Gauss's Law Gauss's Law is always true: Φ = surface E da = q enclosed /ε 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 E, using an equation like surface E da = q enclosed /ε 0, 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 N/C). In practice, this means that you pick a Gaussian surface with the same symmetry as the charge distribution. *Note some surfaces may be selected so that E is constant on part of the surface. If E changes on other parts of the surface that may be OK as long as the field direction is parallel to the surface. In other words, it is OK to include surfaces with non-constant E as long as the flux through those surfaces is zero at each point. 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. a. 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? i. Note that you will need to be careful measuring the electric field strengths. E 1 = E 2 = E 3 = Same or not. Discuss. 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). b. 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? i. When you find E da cos(θ) and set da to 1, you have found the flux through a unit region of the surface. This is equivalent to finding the flux per unit area, or a flux density. The flux density may or may not be constant on given surfaces. Flux Density 1 = Flux Density 2 =

4 Flux Density 3 = c. In the context of your answers above, why is the sphere (as shown in the simulation) a better choice for using Gauss's law than the box? i. Also note whether the sphere can be a poor choice? Think about what you can do to make the flux density of the sphere non-uniform. Discuss or draw. Let's try another charge configuration. Put a sphere around part of 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). **For parts d and e note that your book may discuss a large or infinite flat plate that extends to fill the entire plane containing the rods. Such a plate produces a uniform electric field on either side of the plate directed in or out depending on the sign of the charge. Here the simulation limits the charge locations to the simulation window and is NOT IDEAL like the situations in your book. Hence you will observe slight non-uniformities in the electric fields. d. Would the value of E da = E da cosθ be the same at any three points on the Gaussian surface? i. Again with da =1 this is really a measure of flux density as above. Flux Density 1 = Flux Density 2 = Flux Density 3 = e. Explain, then, why you would not want to use a sphere for this configuration.

5 Now, put a box around part of 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). f. Find the value of E da = E da cosθ at three points on the top. Are they essentially the same? i. Essentially the same in light of comments ** above. Flux Density 1 = Flux Density 2 = Flux Density 3 = g. What about E da = E da cosθ on the sides? i. Note earlier comments * which should apply to the sides of the box (4 sides). Discuss why the flux is essentially zero along each of these four sides.

6 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). h. 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/2ε 0 [SINGLE PLATE] 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/ε 0 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). i. This second case where the result is E = σ/ε 0 is really two separate plates of charge. If the charges are the same then the region between the plates has a net field of zero, and outside as given which is due to contributions from both plates. A similar situation gives the field between two oppositely charged capacitor plates with the same formula. Sketch the two capacitor plates, the field for each plate, and then another sketch showing the total field in the region to the left, then between, then to the right of both plates.

7 Worksheet for Exploration 24.3: 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 N/C, and flux is given in N cm 2 /C). a. 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)? i. In each case sketch the plot of the Electric field magnitude vs. R. Conductor plot Insulator plot b. 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. Flux Insulator= Flux Conductor=

8 c. Why is the flux the same? i. Also discuss what the electric field is at the surface of the big Gaussian surface for each case. d. How much charge is on each sphere? How do you know? q insulator = q conductor = e. What do you expect the flux to be through a Gaussian surface inside the conductor? Why? Try it and explain the results. Flux Inside Conductor= Now try putting the same size small Gaussian surface inside the insulator. f. What flux value do you get? Flux inside insulator= g. How much charge is enclosed in this smaller surface? q enclosed smaller surface insulator = h. What is the ratio of the charge enclosed in the small surface to the total charge on the insulating sphere?

9 q q small = total i. 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. V V small = total j. 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. E Gauss Law = E measured = As a reminder, Gauss's law relates the flux to the charge enclosed (q enclosed ) in a Gaussian surface through the following equation: Φ = q enclosed /ε 0 (and Flux = Φ = E da= E cosθ da) where ε 0 is the permittivity of free space (8.85 x C 2 /Nm 2 ), E is the electric field, da is the unit normal to the surface, and θ is the angle between the electric field vector and the surface normal. The surface area of a sphere is 4πr 2.

10 Worksheet for Exploration 24.4: 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 N/C). 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). a. As a function of the distance away from the charge (as a function of r), what is the electric field of a point charge? i. That is, give the theoretical formula for magnitude of the electric field for the special case of a point charge. E point = b. 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? i. Prediction. ii. Go ahead and make measurements for each configuration. r= 2r= Configuration 1 E(r)= E(2r)= Configuration 2 E(r)= E(2r)= c. Which configuration, then, is a point charge?

11 d. 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: i. To use Gauss s law for the line charge you must consider the electric field direction for a line charge (consider it positive). Describe the direction of the E field. ii. Next you should write out both sides of Gauss s law (net flux on one side, charge enclosed on the other). The charge density is a linear charge density usually called λ (which means charge per unit length). Consider the length of the Gaussian can L and radius r. The total charge in the can is then λl. iii. The flux side of Gauss s law has three parts in this particular case. The end face of the cylindrical can near you, the end face away from you, (both of those are the flat surfaces) and the curved surface of the can. Write out the total flux as the sum of those three contributions. Then determine how to write out the flux through each surface (two should be really easy due to the orientations of E and the normals). The last surface should also be easy. You should now be able to solve and obtain an expression for E line. E Line = e. 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? f. Does the electric field of the other configuration agree with this? i. Remember you have measurements above to check this.

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