Transcript Document

Ch 24: Gauss’s Law
24.1 Electric Flux
Think of flowing water as an analogy where the flow lines would be mass flow
(mass/time) per area. Here E is field lines per area. We define a quantity called
electric flux, E which is EA in this simple case. Electric flux is total number of
lines through the area. In the water analogy we would have mass flow.
24.1 Electric Flux
The total electric flux E which goes through the vertical plane A also goes
through the diagonal plane. Noting that EA = EAcos, we see that
E = E•A, where A is a vector normal to the area of value equal to the
area, is a consistent definition for E.
For small area patches dE = E•dA
E = E•A = EAcosi
E is positive.
Outflow is positive.
E is negative.
Inflow is negative.
E is zero.
Inflow equals outflow.
Ch 24: Gauss’s Law
24.2 Gauss’s Law
CT1: S1
equals
CT2: S2
equals
CT3: S3
equals
-3Q
CT4: S4
equals
A. -3Q/0
B. -2Q /0
C. -Q /0
D. 0
E. Q /0
F. 2Q /0
G. 3Q /0
CT5: A icosahedron has 20 equal triangular faces as
pictured above. Assume a charge q is placed at the
center of the icosahedron (an equal distance from each
face). Using the symmetry of the situation, determine
how much electric flux goes through each face.
A. q/0
B. q/40
C. q/60
D. q/200
E. none of the above
Gauss's law: using superposition qin is
the sum of the charges enclosed by the
Gaussian surface.

E
=
 E  dA
=
qin / 0
A. Symmetries:
1) spherical 2) cylindrical (linear) 3) planar
B. Method:
1) note symmetry
2) draw appropriate Gaussian surface
3) calculate electric flux E
4) set  E = qin / 0
5) solve for E
24.4 Perfect Conductors in
Electrostatic Equilibrium
1. E = 0 inside perfect conductors
2. The charge must reside on the surface of
a perfect conductor
3. E =  / 0 n where n is a unit vector
normal to the surface and pointing
outward.
4.  is greatest at points of least radii of
curvature (i.e. pointy)
Before Class Assignment 2/6/08
13
1
1
correct
I don’t know
no explanation
P24.44 (p. 689)
24.3 Application of Gauss’s Law to Various Charge Distributions
CT6: At a distance r >c
from the common
center, the electric field
is
2Q
-3Q
A. 2Q/40r2
B. -3Q/40r2
C. -Q/40r2
D. -Q/40r
E. constant
P24.39 (p. 689)
CT7: The electric field in the
conducting cylinder is
A. 0
B. /20
C. /40
D. /20r
E. /40r
F. r/40
G. r/40
CT8: The electric field between the
wire and the cylinder is
A. 0
B. /20
C. /40
D. /20r
E. /40r
F. r/40
G. r/40
P24.35 (p. 688)
Left
2
2
1
Gaussian
Surface 1:
A squat
cylinder
3
3
Right
1
Gaussian
Surface 2:
A squat
cylinder
E
conducting
plates so 
on each side
A.
B.
C.
D.
E.
Table 24-1, p.754
Summary
Gauss’s Law: E =  E  dA = qin/0
• Spherical Symmetry:
use concentric, spherical Gaussian surface
• Cylindrical Symmetry:
use concentric, cylindrical Gaussian surface,
assume far from ends of long thin cylinder
• Planar Symmetry:
use a disk shaped Gaussian surface,
assume far from edges of planar surface
E will be constant over the surface and either parallel or
normal to the area so
 E  dA
will either be EA or 0. You
may have to integrate to get qin.
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