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

What Are We Going to Learn?
A Road Map
• Electric charge
- Electric
- Electric force on other electric charges 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)

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

• Given:
– planar surface, area A
– uniform field E
– E makes angle q with NORMAL to plane
• Electric Flux:
Φ = E•A = E A cos θ
• Units: Nm 2 /C
• Visualize: “Flow of Wind”
Through “Window”
E
θ normal
AREA = A= A n

• For any general surface: break up into infinitesimal planar patches
• Electric Flux
Φ = ∫ 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 Φ
– Outward E gives positive flux Φ
Area = dA dA
E 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) (2 π R L )E
(b) 2( π R 2 )E
(c) Zero ( π R 2 )E–( π R 2 )E=0
What goes in —
Hint!
MUST come out!
Surface area of sides of cylinder: 2 π R L
Surface area of top and bottom caps (each): π R 2 dA dA
E
R
L

• Note that E is
NORMAL to both bottom and top cap
• E is PARALLEL to curved surface everywhere
• So: Φ =
= 0!
Φ
1
+ Φ
2
+ Φ
3
= π R 2 E + 0 – π R 2 E
• Physical interpretation: total “inflow” = total
“outflow”!
1 dA
2 dA
3 dA

• 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) π R 2 E = − 13.33
π Nm 2 /C
(b) 2 π R 2 E = − 20 π Nm 2 /C
(c) 4 π R 2 E= − 40 π Nm 2 /C
What could produce such a field?
What is the flux if the sphere is not centered on the charge?

q r
!
E
= !
d
!
A r q
2
= + ( )
(Inward!)
ˆ (Outward!)
!
E
• d
!
A
=
EdA cos(180
°
)
= !
EdA
!
=
Since r is Constant on the Sphere — Remove
E Outside the Integral!
#
!
E
" d
!
A
= $
E
# dA
=
&
%
' $ kq r 2 )
(
*
( )
Surface Area Sphere
= $ q
4
+,
0
( ) = $ q /
,
0
Gauss’ Law:
Special Case!

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
! "
%
Surface
!
E
# d
!
A
$
0 q
(One of Maxwell’s 4 equations!)

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.

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
R
2
R
1
–15C
( a) Inner: +15 C; outer: 0
(b) Inner: 0; outer: +10 C
(c) Inner: +15 C; outer: –5 C

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!

Faraday ’ ’ s Cage
• 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
Safe in the
Plane!?
• 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”

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.

• 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!

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
Applying Gauss' law, we have,
A "
!
0
= AE
Solving for the electric field, we get E =
!
"
0

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
σ
?
(a)
σ / ε
0
(b) Zero
(c)
σ /2 ε
0
+
+
+
+
+
+
+
+
+
+
+
+
E   =   0
Applying Gauss' law, we have,
A "
!
0
= AE
Solving for the electric field, we get E =
!
"
0
THIS IS A
GENERAL
RESULT FOR
CONDUCTORS!

• 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: σ /2 ε
0 plate: σ / ε
0
.
;of a conducting
• Properties of conductors : field inside is zero; excess charges are always on the surface; field on the surface is perpendicular and E= σ / ε
0.