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Physics 2113
Jonathan Dowling
Flux Capacitor (Schematic)
Physics 2113
Lecture: 09 WED 04 FEB
Gauss’ Law I
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
Angle
Matters Too
Weak
E-Field
θ
dA
Number of E-Lines
Through Differential
Area “dA” is a Measure
of Strength
Electric Field & Force Law Depends on Geometry
Point of Charge: Field Spreads in 3D
Like Inverse Area of Sphere = 1/(4πr2)
kQ 1
E= 2 µ 2
r
r
C = 2p r
Line of Charge: Field Spreads in 2D Like
Inverse Circumference of Circle = 1/(2πr)
2kl 1
E=
µ
r
r
Sheet of Charge: Field Spreads in 1D Like
A Constant — Does Not Spread!
E = 2p ks = constant
Electric Flux: Planar Surface
• Given:
– planar surface, area A
– uniform field E
– E makes angle θ with NORMAL to
plane
• Electric Flux:
 = E•A = E A cos 
• Units: Nm2/C
• Visualize: “Flow of Wind”
Through “Window”
E

normal
AREA = A=An
Electric Flux: The General Case
Air Flow Analogy
Electric 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) (2RL)E
(b) 2(R2)E
(c) Zero
(R2)E–(R2)E=0
What goes in —
MUST come out!
Hint!
Surface area of sides of cylinder: 2RL
Surface area of top and bottom caps (each): R2
dA
E
L
dA
R
(a)
+EA?
–EA?
0?
(b)
+EA?
–EA?
0?
(c)
+EA?
–EA?
0?
(c)
+EA?
–EA?
0?
Electric 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)R3 E = -13.33 Nm3/C
(b) 2R E = -20 Nm/C
(c) 4R2 E= -40 Nm2/C
What could produce such a field?
What is the flux if the sphere is not
centered on the charge?
Electric Flux: Example
r
(Inward!)
q
(Outward!)
Since r is Constant on the Sphere — Remove
E Outside the Integral!
Surface Area Sphere
Gauss’ Law:
Special Case!
Gauss’s Law: Gravitational Field vs Electric Field
r
M
r
q
ICPP: Compute the Surface Integral
For each of the four Surfaces where +
is a proton and – an electron