Physics 102 Conference 3 Gauss s Law Conference 3 Physics 102 General Physics II Monday, February 10th, 2014 3.1 Quiz Problem 3.1 A spherical shell of radius R has charge Q spread uniformly over its surface. Find the electric field inside and outside the shell. We ll use Gauss s law for r < R, the charge enclosed by a (fictitious) Gaussian sphere is zero, so that E(r) 4 π r 2 = 0 and E = 0. For r > R, the charge enclosed by a Gaussian sphere is Q, and E(r) 4 π r 2 = Q/ so that E(r) = Q 4 π r 2. 1 of 6
3.1. QUIZ Conference 3 Problem 3.2 An infinite line of charge has constant λ (charge-per-unit-length). electric field a distance s from the line of charge. Find the Once again, Gauss s law for a Gaussian cylinder that encloses the line of charge, and has radius s, length l, the charge enclosed is: λ l, and then E(s) 2 π s l = λ l/ gives E = λ 2 π s. Problem 3.3 An infinite sheet of charge has constant charge-per-unit-area σ, find the electric field a height z above the sheet. For a Gaussian pillbox with top surface area A and height z above the plane (and extending to z below), the charge enclose is σ A, and Gauss s law read: E(z) A 2 = σa/ then E(z) = σ 2. 2 of 6
3.2 Flux Flux is defined in terms of a surface and an electric field. Given a surface with vector area element da (normal to the surface) and an electric field E, the flux through the surface is defined to be: Φ = E da. (3.1) Problem 3.4 A point charge sits at the back corner of a cube. Find the electric flux through the face of the cube shown below. Find flux of electric field through this face Charge q sits at the back left corner Think of the larger cube that encloses the point charge, shown below: The flux through this large cube, which encloses the point charge, is q by Gauss s law. The shaded face is 1/24 of the total flux, so the flux through that face is just Φ = q 24. 3 of 6
Problem 3.5 A sphere of radius R has charge-per-unit-volume ρ = α r for constant α. What are the units of α? Find the electric field inside and outside the sphere. In both cases, the electric field depends only on the distance to the origin, and points away from the center of the sphere. What changes in the application of Gauss s law is the amount of charge enclosed by the Gaussian sphere: For a Gaussian sphere of radius r < R, we have Q enc = r 0 4 π r 2 (α r) d r = 4 π r4 4 α. (3.2) The integral of E over the surface is: E(r) da = E(r) da = E(r) 4 π r 2. Putting the two together in Gauss s law gives: E da = Q enc E(r) 4 π r 2 = α π r4 (3.3) so that we end up with E(r) = α r2 4. α must be a charge/ l 4 so that α r is a charge-per-unit-volume. Outside the sphere, our Gaussian surface encloses all of the charge, so we can take the integral in (3.2) out to R, giving: Q enc = 4 π R4 4 α, (3.4) and then E(r) 4 π r 2 = Q enc / gives E(r) = α R4 4 r 2. These match at r = R as expected. 4 of 6
Problem 3.6 There is an infinite slab with charge-per-unit-volume ρ 0 extending from z = d to z = d. At the center of the slab, we carve out a hollow sphere of radius d find the electric field a height h above the center of the slab, directly over the hollow sphere. ẑ E =? h d ˆx ŷ z =0 d 0 empty The electric field above the infinite slab, with no hole cut out, is, from Gauss s Law (applied to a cylinder of height 2 z centered at the x y plane, with top area A): E da = Q enc E(z) 2 A = ρ 0 2 d A (3.5) E(z) = ρ 0 d The electric field above a uniformly charged sphere with constant ρ 0 is E s (h) = 4 ρ 0 3 π d3 4 π. The superposition of these two configurations gives the target configuration, so the superposition of these two fields is the electric h 2 field: E = ρ 0 d ẑ ρ 0 d 3 ẑ. (3.6) 3 h2 5 of 6
Problem 3.7 An infinite line charge has constant charge-per-unit-length λ. Surrounding the line charge is a cylindrical shell of radius R, and carrying a constant chargeper-unit-area σ. Given λ, what must σ be in order to get zero electric field for all points outside the cylindrical shell? For that σ, what is the electric field in between the line of charge and the shell? R =? The electric field outside of a wire with constant λ is E = 2 π s this is also the electric field in between the line charge and the cylindrical shell (it does not depend on σ at all). The electric field outside a cylinder with constant σ is, via Gauss s law: E(s) 2 π s l = σ 2 π R l/ giving E(s) = σ R s. The total electric field outside the configuration is, then, E(s) = 1 [ σ R + λ ]. (3.7) s 2 π Then σ = λ 2 π R is the charge density on the cylindrical surface that will ensure that the electric field outside the entire configuration is zero. Note that it has the correct dimensions to be a surface charge. λ 6 of 6