Post on 17-Dec-2015
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Physics 2113 Physics 2113 Lecture: 09 WED 04 FEBLecture: 09 WED 04 FEB
GaussGauss’’ Law I Law I
Flux Capacitor (Schematic)
Physics 2113
Jonathan Dowling
Carl Friedrich Gauss1777 – 1855
What Are We Going to Learn?What Are We Going to Learn?A Road MapA 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!What? — The Flux!STRONGE-Field
WeakE-Field
Number of E-Lines Through Differential
Area “dA” is a Measure of Strength
dAθ
Angle Matters Too
Electric Field & Force Law Electric Field & Force Law Depends on GeometryDepends on Geometry
Point of Charge: Field Spreads in 3D Like Inverse Area of Sphere = 1/(4πr2)
Line of Charge: Field Spreads in 2D Like Inverse Circumference of Circle = 1/(2πr)
Sheet of Charge: Field Spreads in 1D Like A Constant — Does Not Spread!
Electric Flux: Planar SurfaceElectric Flux: Planar Surface
• Given: – planar surface, area A– uniform field E– E makes angle θ with NORMAL to
plane
• Electric Flux:
Φ = E•A = E A cos θ• Units: Nm2/C• Visualize: “Flow of Wind”
Through “Window”
θ
E
AREA = A=An
normal
Electric Flux: The General CaseElectric Flux: The General Case
Air Flow Analogy
Electric Flux: ICPPElectric Flux: ICPP
• Closed cylinder of length L, radius R
• Uniform E parallel to cylinder axis
• What is the total electric flux through surface of cylinder?
(a) (2πRL)E
(b) 2(πR2)E
(c) Zero
Hint!Surface area of sides of cylinder: 2πRLSurface area of top and bottom caps (each): πR2
L
R
E
(πR2)E–(πR2)E=0What goes in — MUST come out!
dA
dA
(a) +EA? –EA? 0?
(b) +EA? –EA? 0?
(c) +EA? –EA? 0?
(c) +EA? –EA? 0?
Electric Flux: ICPPElectric Flux: ICPP• 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)πR3 E = 13.33π Nm3/C
(b) 2πR E = 20π Nm/C
(c) 4πR2 E= 40π Nm2/C
What could produce such a field?
What is the flux if the sphere is not centered on the charge?
q
r
Electric Flux: ExampleElectric Flux: Example
Since r is Constant on the Sphere — RemoveE Outside the Integral!
Gauss’ Law:Special Case!
(Outward!)
Surface Area Sphere
(Inward!)
q
r
Gauss’s Law: Gravitational Field vs Electric FieldGauss’s Law: Gravitational Field vs Electric Field
r
M
ICPP: Compute the Surface IntegralFor each of the four Surfaces where + is a proton and – an electron