Summary: Applications of Gauss Law
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1 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 1 Summary: Applications of Gauss Law 1. Field outside of a uniformly charged sphere of radius a: 2. An infinite, uniformly charged plane with charge density σ: 3. Field inside a univormly charged sphere of charge density ϱ: 4. A long coaxial wire: Suggested Reading: Griffiths: Chapter 2, Sections , pages 58-64; Sections , pages Weber and Arfken, Chapter 1, Section 1.13, pages Wangsness: Chapters 3 and 4, pages
2 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 2 Applications of Gauss Law: A uniformly charged sphere To find the electric field outside of some uniformly charged sphere, volume charge density ϱ, radius r 0, centred at the origin, simply construct a spherical Gaussian surface concentric with the centre of the sphere, with a radius R > r 0. Then, by Gauss Law: E da = Q S ɛ 0 where Q is the total charge enclosed, i.e., Q = ϱdv V
3 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 3 By symmetry the electric field at the surface of this sphere must have the same magnitude, E R and must point normal to the surface (any contributions to components tangential to the surface will have equal and opposite compensating contributions from elsewhere in the sphere). Thus 2π π E d A = E R dφ (ˆr ˆr)R 2 sin θdθ = 4πE R R 2 = Q ɛ 0 S 0 So that the electric field everywhere on this spherical surface of radius R is just E(R) = Q ˆr 4πɛ 0 R 2 which is precisely the electric field that would arise from a point charge Q located at the origin. Infinite uniformly charged plane: 0 A surface S extends to infinity in both the x and y directions at z = 0. It is covered with a uniform
4 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 4 charge density σ. Use Gauss Law to determine the electric field on either side of the plane. We construct a small pill-box shaped Gaussian surface S which penetrates the surface S (as shown in the figure above). By symmetry we know that the electric field, whatever its magnitude, must point perpendicular to the surface at any point (think about it!). Thus, in evaluating the total flux through the Gaussian surface S we need only worry about the top and bottom surfaces which are parallel to S. Since the charge density on the surface is uniform, the total charge enclosed by the Gaussian pill-box is just σ cross-sectional area of the pill-box (Note: if we like we can shrink the sides of the pill-box which are perpendicular to S down to an infinitesimal height). It is also obvious from symmetry that the value of E(x, y, z) cannot depend on either x or y. Finally, as we increase the height of the box above the plane, the flux through the top (or bottom) surface remains the same. Thus, E(x, y, z) is independent of z as well (except that below the plane it points in the z direction while above the plane it points in the +z direction. Gauss
5 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 5 Law states that Φ = E d Q encl A = S ɛ 0 Since E = E 0 ẑ (by symmetry) we can take it outside the integral and Φ = E da = E0 2A = σ A S ɛ 0 where A is the area of the top surface of the pillbox. The electric field above the plane then is just E = E 0 ẑ = σ ẑ 2ɛ 0 while below the plane it has the same magnitude but points in the opposite direction. We can now make the following observations: For a point charge at the origin, E 1/r 2 For an infinite line of charge, E 1/r For an infinite charged plane, E constant
6 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 6 Problem 2.12 Use Gauss law to find E inside a uniformly charged sphere having charge density ϱ Φ = E d Q encl A = S ɛ 0 By symmetry no one point on the surface S can be any different than any other point. Thus, E(r < a) = E r ˆr, where a is the radius of the charged sphere.
7 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 7 The charge enclosed by S is simply 2π π r Q encl = ϱdv = ϱ r 2 sin θdrdθdφ S Thus, = 4πr3 3 ϱ E(r < a) = ϱ 4πr 3 /3 = ϱ r ɛ 0 4πr 2 ɛ 0 3 Note: The charge outside of the Gaussian surface S does not contribute to the electric field on the surface S. What if the Gaussian surface S was not centred at the origin? Gauss Law still gives the net flux through the Gaussian surface as just Q encl /ɛ 0. However, we cannot easily evaluate the surface integral because
8 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 8 the electric field on this surface depends on position (i.e., we can t use symmetry to argue that it is constant). Problem 2.16 A long co-axial cable consists of an inner cylindrical rod (radius a), uniformly positively charged with volume charge density ϱ, and a concentric outer cylindrical tube, of radius b > a, with a negative surface charge density σ which ensures that charge is balanced (i.e., total charge is precisely 0). Use Gauss law to find E everywhere To find the electric field outside of the coaxial cable, construct a cylindrical Gaussian surface of length L concentric to the axis of the cable but with a radius r > b
9 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 9 By symmetry, E must point radially outward from the axis of the coaxial cable and can only be a function of r. For r b: Q encl = 0 and by Gauss Law S E da = E S da = Q encl ɛ 0 = 0 Here, S is the outer curved surface of the Gaussian cylinder but does not include the ends, for which there would be no contribution. Therefore, since S da = 2πr L 0 we conclude that outside
10 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture the coaxial cable E = 0, for r b For a r b: Construct a cylindrical Gaussian surface with radius r which now lies in between the inner conductor and the outer conductor Now, Q encl ɛ 0 = so that S = 1 ɛ 0 E da = E da = E 2πr L S V ϱdv = ϱ ɛ 0 πa 2 L E = ϱa2, radially outward if ϱ > 0 2ɛ 0 r For r a: Construct a cylindrical Gaussian surface inside the inner electrode (recall that it is a solid conductor having volume charge density ϱ).
11 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture Using the same arguments, we find that Q encl = E da ɛ = E da = E 2πr L 0 S S = 1 ϱdv = ϱ πr 2 L ɛ 0 ɛ 0 Giving V E = ϱr radially outward if r a 2ɛ 0 Summarizing, the electric field for all values of r is:
12 Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture pointing radially outward (in the direction ˆρ if cylindrical coordinates are used with the coaxial axis along the z axis.
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