1. Physics 2102 Lecture: 04 THU 28 JAN
Jonathan Dowling
Flux Capacitor (Schematic)
Physics Lecture: 04 THU 28 JAN
Gauss’ Law I
Carl Friedrich Gauss
1777 – 1855

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: Example
Note that E is NORMAL to both bottom and top cap
E is PARALLEL to curved surface everywhere
So: F = F1+ F2 + F = pR2E + 0 – pR2E = 0!
Physical interpretation: total “inflow” = total “outflow”! 1 dA 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! S
(One of Maxwell’s 4 equations!)

11. Properties of Conductors
Inside a Conductor in Electrostatic Equilibrium, the Electric Field Is ZERO. Why?
Because If the Field Is Not Zero, Then Charges Inside the Conductor Would Be Moving.
SO: Charges in a Conductor Redistribute Themselves Wherever They Are Needed to Make the Field Inside the Conductor ZERO.
Excess Charges Are Always on the Surface of the Conductors.

12. Gauss’ Law: Conducting Sphere
A spherical conducting shell has an excess charge of +10 C.
A point charge of -15 C is located at center of the sphere.
Use Gauss’ Law to calculate the charge on inner and outer surface of sphere
(a) Inner: +15 C; outer: 0
(b) Inner: 0; outer: +10 C
(c) Inner: +15 C; outer: –5 C R2 R1
–15C

13. Gauss’ Law: Conducting Sphere
Inside a conductor, E = 0 under static equilibrium! Otherwise electrons would keep moving!
Construct a Gaussian surface inside the metal as shown. (Does not have to be spherical!)
–5 C
+15C
Since E = 0 inside the metal, flux through this surface = 0
Gauss’ Law says total charge enclosed = 0
Charge on inner surface = +15 C
–15C
Since TOTAL charge on shell is +10 C,
Charge on outer surface = +10 C - 15 C = -5 C!

14. Faraday’s Cage Safe in the Plane!?
Given a hollow conductor of arbitrary shape. Suppose an excess charge Q is placed on this conductor. Suppose the conductor is placed in an external electric field. How does the charge distribute itself on outer and inner surfaces?
(a) Inner: Q/2; outer: Q/2
(b) Inner: 0; outer: Q
(c) Inner: Q; outer: 0
• Choose any arbitrary surface inside the metal
• Since E = 0, flux = 0
• Hence total charge enclosed = 0
All charge goes on outer surface!
Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect”

16. Field on Conductor Perpendicular to Surface
We know the field inside the conductor is zero, and the excess charges are all on the surface. The charges produce an electric field outside the conductor.
On the surface of conductors in electrostatic equilibrium,
the electric field is always perpendicular to the surface.
Why?
Because if not, charges on the surface of the conductors would move with the electric field.

17. Charges in Conductors
Consider a conducting shell, and a negative charge inside the shell.
Charges will be “induced” in the conductor to make the field inside the conductor zero.
Outside the shell, the field is the same as the field produced by a charge at the center!

18. Gauss’ Law: Conducting Plane
Infinite CONDUCTING plane with uniform areal charge density s
E is NORMAL to plane
Construct Gaussian box as shown.
Note that E = 0 inside conductor

19. Gauss’ Law: Conducting Example
Charged conductor of arbitrary shape: no symmetry; non-uniform charge density
What is the electric field near the surface where the local charge density is s?
(a) s/e0
(b) Zero
(c) s/2e0 +
E = 0
THIS IS A
GENERAL
RESULT FOR
CONDUCTORS!

20. Summary:
Gauss’ law provides a very direct way to compute the electric flux.
In situations with symmetry, knowing the flux allows to compute the fields reasonably easily.
Field of an insulating plate: 0 ;of a conducting plate: 0 .
Properties of conductors: field inside is zero; excess charges are always on the surface; field on the surface is perpendicular and E=0.