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Physics 2102
Jonathan Dowling
Flux Capacitor (Schematic)
Physics 2102
Lecture: 04 THU 28 JAN
Gauss’ Law I
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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
dA
Number of E-Lines
Through Differential
Area “dA” is a Measure
of Strength
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”
E
q
normal
AREA = A=An
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
E
Area = dA
dA
Electric Flux: Example
• 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
(pR2)E–(pR2)E=0
What goes in —
MUST come out!
Hint!
Surface area of sides of cylinder: 2pRL
Surface area of top and bottom caps (each): pR2
dA
E
L
dA
R
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 + F3
= pR2E + 0 – pR2E
= 0!
• Physical interpretation:
total “inflow” = total
“outflow”!
dA
1
2
3
dA
dA
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?
(a) (4/3)pR2 E = -13.33p 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?
Electric Flux: Example
r
q
E = - 2 rˆ
r
(Inward!)
dA = dArˆ
q



(Outward!)
E  dA = EdAcos(180) = -EdA
Since r is Constant on the Sphere — Remove
E Outside the Integral!
 kq
F =  E  dA = -E  dA = - 2 4 pr 2  Surface Area Sphere
 r 
q
=4 p  = -q /0 Gauss’ Law:
4p0
Special Case!
Gauss’ Law: General Case
• Consider any ARBITRARY
CLOSED surface S -- NOTE: this
does NOT have to be a “real”
physical object!
S
• 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!
F

Surface
E  dA =
q
0
(One of Maxwell’s 4 equations!)
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.
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
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
• Since E = 0 inside the metal, flux
through this surface = 0
• Gauss’ Law says total charge
enclosed = 0
• Charge on inner surface = +15 C
Since TOTAL charge on shell is +10 C,
Charge on outer surface = +10 C - 15 C = -5 C!
+15C
–15C
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
• Choose any arbitrary surface
inside the metal
• Since E = 0, flux = 0
• Hence total charge enclosed = 0
All charge goes on outer surface!
Safe in the
Plane!?
Inside cavity is “shielded”
from all external electric
fields! “Faraday Cage
effect”
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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.
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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!
•
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 ?
(a) /0
(b) Zero
+
+
+
+ ++
+
(c) /20
Applying Gauss' law, we have,
+ +
+
+
+
E=0
A
0
= AE

Solving for the electric field, we get E =
0
THIS IS A
GENERAL
RESULT FOR
CONDUCTORS!
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: /2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.