Physics 2102 Lecture: 05 FRI 23 JAN Gauss’ Law I Version: 1/22/07 Flux Capacitor (Schematic) Physics 2102 Jonathan Dowling Carl Friedrich Gauss 1777 – 1855

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 Weak E-Field Number of E-Lines Through Differential Area “dA” is a Measure of Strength dA  Angle Matters Too

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

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

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  RL)E (b) 2(  R 2 )E (c) Zero Hint! Surface area of sides of cylinder: 2  RL Surface area of top and bottom caps (each):  R 2 L R E (  R 2 )E–(  R 2 )E=0 What goes in — MUST come out! dA

Electric Flux: Example Note that E is NORMAL to both bottom and top cap E is PARALLEL to curved surface everywhere So:  =   +   +     R 2 E + 0 –  R 2 E = 0! Physical interpretation: total “inflow” = total “outflow”! 3 dA 1 2

Electric Flux: Example Spherical surface of radius R=1m; 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)  R 2 E =   Nm 2 /C (b) 2  R 2 E =  20  Nm 2 /C (c) 4  R 2 E=  40  Nm 2 /C What could produce such a field? What is the flux if the sphere is not centered on the charge?

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