Download Phy 211: General Physics I

Phy 213: General Physics III
Chapter 23: Gauss’ Law
Lecture Notes
Electric Flux
1. Consider a electric field passing through a region in
space. The electric flux , FE, is the scalar product
of the average electric field vector and the normal
area vector for the region, or:
FE= Eavg  A = Eavg A cos
2. Conceptually, the electric flux represents the flow of
electric field lines (normal) through a region of
space
E
A

Eavg
Gauss’ Law
1. Gauss’ Law is a fundamental law of nature relating
electric charge to electric flux
2. According to Gauss’ Law, the total electric flux through
any closed (“Gaussian”) surface is equal to the enclosed
charge (Qenclosed) divided by the permittivity of free space
(eo):
qenclosed
FE=  Ei  Ai=
or,
eo
i
FE=
 E  dA =
qenclosed
eo
3. Gauss’ Law can be used to determine the electric field (E)
for many physical orientations (distributions) of charge
4. Many consequences of Gauss’ Law provide insights that
are not necessarily obvious when applying Coulomb’s
Law
Gauss’ Law for a Point Charge
Consider a point charge, q
q
r
Gaussian sphere
1. An appropriate “gaussian” surface for a point charge is a
sphere, radius r, centered on the charge, since all E fields
will be ∟ to the surface
2. The electric flux is: FE= E  A = E 4 r2


3. Applying Gauss’ Law yields E:

F E = E 4 r
2

=
qenclosed
eo
qenclosed
 E =
4e or2
Spherical Symmetry: Charged Sphere
1. Spherical symmetry can be exploited for any spherical shape
2. Consider a hollow conducting sphere, radius R & charge q:
q
R
Gaussian
sphere
r
3. The flux through a “gaussian” sphere is: FE= E  A = E  4 r2 
4. Applying Gauss’ Law: FE= E  4 r =
2
qenclosed
eo
qenclosed
5. For r > R: E =
4e or2
qenclosed
6. For r < R: E =
= 0 qenclosed= 0C
2
4e or
qenclosed
 E=
4e or2
Inside a hollow
conductor, E = 0
Spherical Symmetry: Charged Sphere (2)
Consider a solid non-conducting sphere with radius R & uniform
charge density, r:
r
R
r
Gaussian sphere
1. Applying Gauss’ Law: FE= E  4 r =
2
ρ

4
3
 R3 
2. For r > R:
qenclosed
E =
=
2
4e or
3. For r < R:
qenclosed
E =
=
2
4e or
4e or 2
4e or2
ρ  34  r3 
qenclosed
eo
qenclosed
 E=
4e or2
ρR 3
=
3e or2
ρr
=
3e o
Inside a uniformly charged insulator, E ≠ 0
Spherical Symmetry: Charged Spheres (3)
Consider 2 concentric hollow conducting spheres with radii R1 &
Gaussian sphere
R1 and charges q1 & q2:
q1
R1
r
R2
q2
1. Applying Gauss’ Law: FE= E  4 r =
2
qenclosed
eo
qenclosed
 E=
4e or2
qenclosed
=0
2. For r < R1: E =
2
4e or
qenclosed
q1
3. For R2 > r > R1: E =
=
2
4e or
4e or2
q1+q2 

qenclosed
=
4. For r > R2: E =
2
2 {opposite charge subtracts}
4e or
4e or
Application: Line of Charge
1. Cylindrical symmetry is useful for evaluating a “line of
charge” as well as charged cylindrical shapes
2. Consider an infinitely long line of charge, charge density l:
l
+++++++++++++++++++++++++++++++++++++
+
0
r
x
3. The flux through each of the 3 surfaces is:
FE=F1+F2+F3=F3 since F1= F2 = 0
qenclosed l x
FE= E3 2 r  x=
=
eo
eo
l
4. Applying Gauss’ Law: E =
2e or
Planar Symmetry: A Charged Sheet
1. A Gaussian “pill-box” is an appropriate Gaussian
surface for evaluating a charged flat sheet
2. Consider a uniformly charged flat sheet with a
surface charge density, s:
Gaussian
“pill-box”
3. The flux through the pill-box surface is:
FE=Fside+Fface1+Fface2=2Fface
Fface =Fface1=Fface2 
4. The electric field:
FE=2 E  r2 =
qenclosed
eo
qenclosed s r2
s
 E=
=
=
2
2
2e or
2e or 2e o
Application: 2 Parallel Charged Plates
1. Two oppositely charged conducting
plates, with charge density s:
2. Surface charges draw toward each
other on the inner face of each plate
3. To determine the electric field within
the plates, apply a Gaussian “pill-box”
to one of the plates
+q
Gaussian
“pill-box”
4. The flux through the pill-box surface
is: FE=Fside+Finner face+Fouter face=Fouter face
where F
inner face
=F side =0
5. The electric field is:
2
q
q
s
r
s
2
enclosed
enclosed
FE= E  r =
 E=
=
=
2
2
eo
e or
e or e o


-q
Consequences of Gauss’ Law
1. Electric flux is a conserved quantity for an enclosed
electric charge
2. The electric field inside a charged solid conductor
is zero
3. The electric field inside a charged hollow conductor
or non-conductor is zero
4. Geometrical symmetries can make the calculation
of an electric field less cumbersome even for
complicated charge distributions