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Physics 1402.001 - Week Two • Electric Fields – Chapter 22 • Continuous Charge Distributions • Gauss’ Law Clicker Question No. One Three charges +q, +Q, and –Q are placed at the corners of an equilateral triangle as shown. The net force on charge +q due to the other two charges is A. up. B. down. C. along a diagonal. D. to the left. E. to the right. Clicker Question No. Two The force between two very small charged bodies is found to be F. If the distance between them is tripled without altering their charges, the force between them becomes A. 9F B. 3F C. F/3 D. F/9 E. 1/F 3 Clicker Question No. One Three charges +q, +Q, and –Q are placed at the corners of an equilateral triangle as shown. The net force on charge +q due to the other two charges is A. up. B. down. C. along a diagonal. D. to the left. E. to the right. Clicker Question No. Two The force between two very small charged bodies is found to be F. If the distance between them is tripled without altering their charges, the force between them becomes A. 9F B. 3F C. F/3 D. F/9 E. 1/F 3 Electric Field Lines for Two Positive Charges Enet Enet Ex Ey Electric Field Lines for Two Positive Charges Electric Fields from Continuous Charge Distributions E1 = kq/r1p2 E2 = kq/r2p2 R1p = sinθ1/yp R2p = sinθ2/yp E1 = kq sin2θ1/yp2 E2 = kq sin2θ2/yp2 Electric Fields from Continuous Charge Distributions Electric Fields from Continuous Charge Distributions Ex = kλ [1/r2 – 1/r1] Electric Fields from Continuous Charge Distributions • • • • Ez = kλ [1/r2 – 1/r1] r1 = z + L/2 r2 = z – L/2 Ez = kλ [1/(z – L/2) – 1/(z + L/2)] • Ez = kλL [1/(z2 – (L/2)2] and when z >>L/2 • Ez ~ kQ/z2 Electric Fields from Continuous Charge Distributions Electric Fields from Continuous Charge Distributions Clicker Question No. Three A conducting circular disk has a uniform positive surface charge density. Which of the following diagrams best represents the electric field lines from the disk? (The disk is drawn as a cross–section.) A. 1 B. 2 C. 3 D. 4 E. None of the diagrams. Clicker Question No. Three A conducting circular disk has a uniform positive surface charge density. Which of the following diagrams best represents the electric field lines from the disk? (The disk is drawn as a cross–section.) A. 1 B. 2 C. 3 D. 4 E. None of the diagrams. Electric Fields from Continuous Charge Distributions Electric Fields for Infinite Line of Charge Ez = (kλ/R) (sinθ2 – sinθ1) Ez = (kλ/R) (sinθ2 – sinθ1) ER = (kλ/R) (cosθ2 – cosθ1) ER = - (kλ/R) (cosθ2 – cosθ1) θ1 = 0 θ2 = π Ez = 0 ER = 2 (kλ/R) Flux of Electric Field Through a Surface Φ=E•A Flux of Electric Field Through a Surface Φ=E•A Φ2 = Φ1 Flux of Electric Field Through a Surface • The Electric Flux Φ is defined as the number of Electric Field E lines passing through an Area A • That is Φ = E • A • If the area A is perpendicular to the E field, then Φ = E x A • If the area A is at an angle θ to the E field, then • Φ = E • A = E A cosθ Flux Through a Closed Surface • For a continuously closed surface that defines an ‘inside’ and an ‘outside’ the total flux is defined by • Φ = ∫S E • dA = ∫S E • n dA • Any charge that resides outside surface S will therefore contribute Φ = 0 to calculation of total flux Flux of Electric Field Through a Surface Φ = ∫E • dA Only charges inside bounded surface contribute a net Electric flux passing through surface boundary Electric Field Lines – Electric Dipole Clicker Question No. Four Thefigureshowsasurfaceenclosingthechargesqand –q.Thenetfluxthroughthesurfacesurroundingthe twochargesis A. q/ε0 B. 2q/ε0 C. –q/ε0 D. zero E. –2q/ε0 Clicker Question No. Four Thefigureshowsasurfaceenclosingthechargesqand –q.Thenetfluxthroughthesurfacesurroundingthe twochargesis A. q/ε0 B. 2q/ε0 C. –q/ε0 D. zero E. –2q/ε0 Gauss’ Law – Point Charge Φ = E • A = Qnet/εo Φ = E(4πR2) = +Q/εo E = Q/[(4πεo) R2] E = kQ/R2 Gauss’ Law The net Electric Field passing through any surface fully enclosing electric charges is proportional to the net charge residing inside the surface. Φ = E • A = Qnet/εo Gauss’ Law Electric Field Lines for Two Positive Charges Clicker Question No. Five The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is Clicker Question No. Five Gauss’ Law The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is Φ = E • A = Qnet/εo Φ = (+2q – q)/εo = q/εo Clicker Question No. Six The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is Clicker Question No. Six The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is Electric Fields from Line of Charge Gauss’ Law – Infinite Line of Charge Ez = (kλ/R) (sinθ2 – sinθ1) ER = - (kλ/R) (cosθ2 – cosθ1) θ1 = 0 θ2 = π Ez = 0 ER = 2 (kλ/R) ER = 2/4πεo (λ/R) ER = 1/2πεo (λ/R) Gauss’ Law – Infinite Line of Charge Gauss’ Law – Distributed Charge ER (2π R L) = λ L / εo ER = λ L / (2π R L εo ) = λ / (2π εo R ) ER = 2 k λ / R Gauss’ Law – Uniformly Charged Slab Gauss’ Law – Uniformly Charged Slab Symmetry of Gaussian Surfaces Gauss’ Law – Thin Shell of Charge • Line symmetry – choose a Cylindrical Surface • Plane symmetry – choose a Cubic or Cylindrical surface • Point symmetry – choose a Spherical surface Gauss’ Law – Thin Shell of Charge Gauss’ Law – Uniformly Charged Sphere Gauss’ Law – Uniformly Charged Sphere Clicker Question No. Seven A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for ra < r < rb1 is Clicker Question No. Seven A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for ra < r < rb1 is Clicker Question No. Eight A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for rb1 < r < rb2 is Clicker Question No. Eight Clicker Question No. Nine A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for rb1 < r < rb2 is A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for r > rb2 is Clicker Question No. Nine A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for r > rb1 is