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Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces.

Units of Chapter 22 Electric Flux Gauss’s Law Applications of Gauss’s Law Experimental Basis of Gauss’s and Coulomb’s Laws

22-1 Electric Flux Electric flux: Electric flux through an area is proportional to the total number of field lines crossing the area. Figure 22-1. (a) A uniform electric field E passing through a flat area A. (b) E┴ = E cos θ is the component of E perpendicular to the plane of area A. (c) A┴ = A cos θ is the projection (dashed) of the area A perpendicular to the field E.

22-1 Electric Flux Example 22-1: Electric flux. Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°. Solution. The flux is EA cos θ = 3.5 N·m2/C.

22-1 Electric Flux Flux through a closed surface:

22-2 Gauss’s Law The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law: This can be used to find the electric field in situations with a high degree of symmetry.

22-2 Gauss’s Law For a point charge, Therefore, Solving for E gives the result we expect from Coulomb’s law: Figure 22-7. A single point charge Q at the center of an imaginary sphere of radius r (our “gaussian surface”—that is, the closed surface we choose to use for applying Gauss’s law in this case).

22-2 Gauss’s Law Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives: Figure 22-8. A single point charge surrounded by a spherical surface, A1, and an irregular surface,A2. Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface.

22-2 Gauss’s Law Finally, if a gaussian surface encloses several point charges, the superposition principle shows that: Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.

22-2 Gauss’s Law Conceptual Example 22-2: Flux from Gauss’s law. Consider the two gaussian surfaces, A1 and A2, as shown. The only charge present is the charge Q at the center of surface A1. What is the net flux through each surface, A1 and A2? Solution: The net flux through A1 is Q/ε0, as it encloses charge Q. The net flux through surface A2 is zero, even though the field is not zero on the surface.

22-3 Applications of Gauss’s Law Example 22-3: Spherical conductor. A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? Figure 22-11. Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2). b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero. c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

22-3 Applications of Gauss’s Law Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0). Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03).

22-3 Applications of Gauss’s Law Example 22-5: Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere. Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05. b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05.

22-3 Applications of Gauss’s Law Example 22-6: Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends. Solution: If the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere. Therefore, a cylindrical gaussian surface will allow the easiest calculation of the field. The field is parallel to the ends and constant over the curved surface; integrating over the curved surface gives E = λ/2πε0R.

22-3 Applications of Gauss’s Law Example 22-7: Infinite plane of charge. Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane. Solution: We expect E to be perpendicular to the plane, and choose a cylindrical gaussian surface with its flat sides parallel to the plane. The field is parallel to the curved side; integrating over the flat sides gives E = σ/2ε0.

22-3 Applications of Gauss’s Law Example 22-8: Electric field near any conducting surface. Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by E = σ/ε0 where σ is the surface charge density on the conductor’s surface at that point. Solution: Again we choose a cylindrical gaussian surface. Now, however, the field inside the conductor is zero, so we only have a nonzero integral over one surface of the cylinder. Integrating gives E = σ/ε0.

22-3 Applications of Gauss’s Law The difference between the electric field outside a conducting plane of charge and outside a nonconducting plane of charge can be thought of in two ways: The field inside the conductor is zero, so the flux is all through one end of the cylinder. The nonconducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.

22-3 Applications of Gauss’s Law Conceptual Example 22-9: Conductor with charge inside a cavity. Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor? Solution: The field must be zero within the conductor, so the inner surface of the cavity must have an induced charge totaling –q (so that a gaussian surface just around the cavity encloses no charge). The charge +Q resides on the outer surface of the conductor.

22-3 Applications of Gauss’s Law Procedure for Gauss’s law problems: Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). Draw the surface. Use the symmetry to find the direction of E. Evaluate the flux by integrating. Calculate the enclosed charge. Solve for the field.

22-4 Experimental Basis of Gauss’s and Coulomb’s Laws In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero. Figure 22-22. (a) A charged conductor (metal ball) is lowered into an insulated metal can (a good conductor) carrying zero net charge. (b) The charged ball is touched to the can and all of its charge quickly flows to the outer surface of the can. (c) When the ball is then removed, it is found to carry zero net charge.

Summary of Chapter 22 Electric flux: Gauss’s law: Gauss’s law can be used to calculate the field in situations with a high degree of symmetry. Gauss’s law applies in all situations, and therefore is more general than Coulomb’s law.