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

2. 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)

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

4. 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” q E
AREA = A=An
normal

5. Electric Flux: General Surface
For any general surface: break up into infinitesimal planar patches
Electric Flux F = 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 F
Outward E gives positive flux F E dA dA E
Area = dA

6. Electric Flux: Example
(pR2)E–(pR2)E=0
What goes in —
MUST come out! dA 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

7. Electric Flux: Example1 dA
Note that E is NORMAL to both bottom and top cap
E is PARALLEL to curved surface everywhere
So: F = F1+ F2 + F = pR2E pR2E = 0!
Physical interpretation: total “inflow” = total “outflow”! 2 dA 3 dA

8. 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 = p 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?

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

10. 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!)

11. Examples

12. 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

13. 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.
R = 1 mm
E = ?
1 m
Approximate as infinitely long line -- E radiates outwards.
Choose cylindrical surface of radius R, length L co-axial with line of charge.

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

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

16. 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.