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Physics 2102 Jonathan Dowling. Physics 2102 Lecture 3. Flux Capacitor (Schematic). Gauss’ Law I. Michael Faraday 1791-1867. Version: 1/22/07. What are we going to learn? A road map.
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Physics 2102 Jonathan Dowling Physics 2102 Lecture 3 Flux Capacitor (Schematic) 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 currentsMagnetic field Magnetic force on moving charges • Time-varying magnetic field Electric Field • More circuit components: inductors. • Electromagneticwaveslight waves • Geometrical Optics (light rays). • Physical optics (light waves)
STRONG E-Field Angle Matters Too Weak E-Field dA Number of E-Lines Through Differential Area “dA” is a Measure of Strength What? — The Flux!
E q normal AREA = A=An Electric Flux: Planar Surface • Given: • planar surface, area A • uniform field E • E makes angle q with NORMAL to plane • Electric Flux: F = E•A = E A cosq • Units: Nm2/C • Visualize: “Flow of Wind” Through “Window”
E dA E Area = dA dA Electric Flux: General Surface • For any general surface: break up into infinitesimal planar patches • Electric Flux F = EdA • 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 F • Outward E gives positive flux F
dA (pR2)E–(pR2)E=0 What goes in — MUST come out! dA Electric Flux: Example E • Closed cylinder of length L, radius R • Uniform E parallel to cylinder axis • What is the total electric flux through surface of cylinder? (a) (2pRL)E (b) 2(pR2)E (c) Zero L R Hint! Surface area of sides of cylinder: 2pRL Surface area of top and bottom caps (each): pR2
dA 1 2 dA 3 dA Electric Flux: Example • Note that E is NORMAL to both bottom and top cap • E is PARALLEL to curved surface everywhere • So: F = F1+ F2 + F3 =pR2E + 0 - pR2E = 0! • Physical interpretation: total “inflow” = total “outflow”!
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? • (4/3)pR2 E = -13.33p Nm2/C (b) 2pR2 E = -20p Nm2/C (c) 4pR2 E= -40p Nm2/C What could produce such a field? What is the flux if the sphere is not centered on the charge?
r q Electric Flux: Example (Inward!) (Outward!) Since r is Constant on the Sphere — Remove E Outside the Integral! Surface Area Sphere Gauss’ Law: Special Case!
S 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! (One of Maxwell’s 4 equations!)
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
R = 1 mm E = ? 1 m 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.
R = 1 mm E = ? 1 m Gauss’ Law: cylindrical symmetry (cont) • Approximate as infinitely long line -- E radiates outwards. • Choose cylindrical surface of radius R, length L co-axial with line of charge.
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.