Physics 2102 Jonathan Dowling Flux Capacitor Schematic Physics

Physics 2102 Jonathan Dowling Flux Capacitor (Schematic) Physics 2102 Lecture 3 �Gauss’ Law I Michael Faraday 1791 -1867 Version: 1/22/07

What are we going to learn? A road map • Electric charge Electric force on other electric charges Electric field, and electric potential • Moving electric charges : current • Electronic circuit components: batteries, resistors, capacitors • Electric currents Magnetic field Magnetic force on moving charges • Time-varying magnetic field Electric Field • More circuit components: inductors. • Electromagnetic waves light waves • Geometrical Optics (light rays). • Physical optics (light waves)

What? — The Flux! STRONG E-Field Angle Matters Too Weak E-Field d. A Number of E-Lines Through Differential Area “d. A” is a Measure of Strength

Electric Flux: Planar Surface • Given: – planar surface, area A – uniform field E – E makes angle with NORMAL to plane • Electric Flux: F = E • A = E A cos • Units: Nm 2/C • Visualize: “Flow of Wind” Through “Window” E q normal AREA = A=An

Electric Flux: General Surface • For any general surface: break up into infinitesimal planar patches • Electric Flux F = E d. A • Surface integral • d. A is a vector normal to each patch and has a magnitude = |d. A|=d. A • CLOSED surfaces: – define the vector d. A as pointing OUTWARDS – Inward E gives negative flux F – Outward E gives positive flux F E d. A Area = d. A E d. A

Electric Flux: Example • Closed cylinder of length L, radius R • Uniform E parallel to cylinder axis • What is the total electric flux through surface of cylinder? (a) (2 p. RL)E (b) 2(p. R 2)E (c) Zero (p. R 2)E–(p. R 2)E=0 What goes in — MUST come out! Hint! Surface area of sides of cylinder: 2 p. RL Surface area of top and bottom caps (each): p. R 2 d. A E L d. A R

Electric Flux: Example d. A • Note that E is NORMAL to both bottom and top cap • E is PARALLEL to curved surface everywhere • So: F = F 1+ F 2 + F 3 = p. R 2 E + 0 - p. R 2 E = 0! • Physical interpretation: total “inflow” = total “outflow”! 1 2 3 d. A

Electric Flux: Example • • Spherical surface of radius R=1 m; E is RADIALLY INWARDS and has EQUAL magnitude of 10 N/C everywhere on surface What is the flux through the spherical surface? (a) (4/3)p. R 2 E = -13. 33 p Nm 2/C (b) 2 p. R 2 E = -20 p Nm 2/C (c) 4 p. R 2 E= -40 p Nm 2/C What could produce such a field? What is the flux if the sphere is not centered on the charge?

Electric Flux: Example r (Inward!) q (Outward!) Since r is Constant on the Sphere — Remove E Outside the Integral! Surface Area Sphere Gauss’ Law: Special Case!

Gauss’ Law: General Case • Consider any ARBITRARY CLOSED surface S -- NOTE: this does NOT have to be a “real” physical object! • The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED! • The results of a complicated integral is a very simple formula: it avoids long calculations! S (One of Maxwell’s 4 equations!)

Examples

Gauss’ Law: Example Spherical symmetry • Consider a POINT charge q & pretend that you don’t know Coulomb’s Law • Use Gauss’ Law to compute the electric field at a distance r from the charge • Use symmetry: – draw a spherical surface of radius R centered around the charge q – E has same magnitude anywhere on surface – E normal to surface r q E

Gauss’ Law: Example Cylindrical symmetry • Charge of 10 C is uniformly spread over a line of length L = 1 m. • Use Gauss’ Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line. • Approximate as infinitely long line -- E radiates outwards. • Choose cylindrical surface of radius R, length L co-axial with line of charge. E=? 1 m R = 1 mm

Gauss’ Law: cylindrical symmetry (cont) R = 1 mm • Approximate as infinitely long line -- E radiates outwards. • Choose cylindrical surface of radius R, length L co-axial with line of charge. E=? 1 m

Compare with Example! if the line is infinitely long (L >> a)…

Summary • Electric flux: a surface integral (vector calculus!); useful visualization: electric flux lines caught by the net on the surface. • Gauss’ law provides a very direct way to compute the electric flux.