Ch. 24
n̂
E  A E
Electric Flux
•Electric flux is the amount of electric field going across a surface
•It is defined in terms of a direction, or normal unit vector,
perpendicular to the surface
•For a constant electric field, and a flat surface, it is easy to calculate
•Denoted by E
ˆ
 E  AE cos 
 E  AE  n
•Units of Nm2/C
•When the surface is flat, and the fields are constant, you
can just use multiplication to get the flux
•When the surface is curved, or the fields are not constant,
n̂
you have to perform an integration
 E   E  nˆ dA
E  0
Warmup 03
Ex 1 - Let there be a constant electric field in the positive x direction of magnitude 2 N/C.
What is the flux through a cube that is 2m on a side sitting such that one of its sides is
parallel to the x-y plane? This is like 24.1 on page 728. Work out on board.
CT -1- A cylindrical piece of insulating material is placed in an external electric field, as
shown. The net electric flux passing through the surface of the cylinder is
A. positive.
B. negative.
C. zero.
Is the answer independent of the shape?
Warmup 03
Ex- Serway (24-3) A 40 cm diameter loop is rotated
in a uniform electric field until the position of maximum
electric flux (through the loop) is found.
The flux in this position is 5.20 x 105 Nm2/C.
What is the electric field strength?
Solve on Board
Total Flux Out of Various Shapes
A point charge q is at the “center” of a (a) sphere (b) joined hemispheres
(c) cylinder (d) cube. What is the total electric flux out of the shape?
b
a
ke q
    
E  nˆ 
n̂
E
2
a
 E   E  nˆ dA
ke q
 2 4 a 2
a
 E  4 ke q
E
E1
E2
E3
ke q
 E1  EA  2 2 a 2
a
q
 E1  2 ke q n̂ q
E2  0
a
n̂
E
 E 3  2 ke q
E n̂ E
 E  2 Ecap   Elat
 E  4 ke q


2 ke qb
 2  2 ke q 

2
2
b a 

q
4 ke qb
 E  4 ke q

b2  a 2
Electric Flux (Hard example)
A point charge q is at the center of a cylinder of radius a and height 2b.
What is the electric flux out of (a) each end and (b) the lateral surface?
ke q dA  2 sds
n̂
top
E 2
s
r
E
b
ke q
ke qb
ke qb
r
E  nˆ  2 cos   3 
r
2
2 3/ 2
r
r
 b  s  n̂ z
r  s b
2
2
•Consider a ring of radius s and thickness ds
 E   E  nˆ dA 
a

2 ke qbsds
2
b
 s
2 ke qb
0

2 3/ 2

2 ke qb
b2  s 2
a
a
0

E
b
q
b
a
2
2
r

z

a
lateral surface
 E  2 ke q 
b 2  a 2 E  nˆ  ke qa r 3
dA  2 adz
b
b
2
2 ke qa dz
4 ke qb
2

k
qz
e
ˆ
 E   E  ndA  


3/
2
2
2
a 2  b2
b  a  z 
a2  z 2
Gauss’s Law
•No matter what shape you use, the total electric flux out of a region
containing a point charge q is 4keq = q/0.
Why is this true?
•Electric flux is just measuring how many field
lines come out of a given region
q
•No matter how you distort the shape, the field lines
come out somewhere
 E  4 ke q
E  q 0
•If you have multiple charges inside the region their effects add
•However, charges outside the region do not contribute
E   q1  q2  q3   0
q4
q3
E 
q1
q2
qin
0
ϕE =
qin
E ∙ dA =
ϵ0
Warmup 03
Using Gauss’s Law to find total charge
A cube of side a has an electric field of
constant magnitude |E| = E pointing
directly out on two opposite faces and
directly in on the remaining four faces.
What is the total charge inside the cube?
A) 6Ea20
B) – 6Ea20 C) 2Ea20
D) – 2Ea20 E) None of the above
E
a
E
a
qin   0  E   0  E  nˆ A   0  2 Ea  4 Ea
2
qin  2 Ea 2 0
E
E
6
i 1
E
2

E
a
CT –2 - Which of the following statements is (are) true?
A. When there are more electric field lines leaving a gaussian surface than entering it
then there is a net negative charge enclosed by the surface.
B. Gauss's law can be used to find the electric field if the total charge inside a closed
surface is known even if the distribution of that charge is not.
C. The electric flux through a closed surface is completely independent of the size and
shape of the surface.
D. Two of the above
E. None of the above
CT-3
S
Q2
Q3
Q1
Which of the following statements is (are) true?
A. The electric field E at any point on the surface S is determined only by the charges
inside S (Q1 and Q2).
B. The electric flux E through surface S is determined only by the charges inside S (Q1
and Q2).
C. The field E at any point on S is determined by all the charges (Q1, Q2 and Q3).
D. The electric flux E through S determined by all the charges (Q1, Q2 and Q3).
E. Two of the above.
F. None of the above statements is true!
Using Gauss’s Law to find flux
A very long box has the shape of a regular pentagonal prism.
Inscribed in the box is a sphere of radius R with surface charge
density . What is the electric flux out of one lateral side of the box?
Perspective
view
•The flux out of the end caps is negligible
•Because it is a regular pentagon, the flux from the
End view
other five sides must be the same
 E  qin  0
5 E ,side   A  0
5E ,side  4 R2  0
E ,side  4 R2 5 0
Using Gauss’s Law to find E-field
A sphere of radius a has uniform charge density  throughout. What
is the direction and magnitude of the electric field everywhere?
•Clearly, all directions are created equal in this problem
•Certainly the electric field will point away from the sphere at all points
•The electric field must depend only on the distance
E  rˆE
•Draw a sphere of radius r around this charge
•Now use Gauss’s Law with this sphere qin   0  E
 a3
E  rˆ
3 0 r 2
ˆ
V    0 AE  n
4
3
 a 3    0 4 r 2 E
r
Is this the electric field everywhere?
E
a
Using Gauss’s Law to find E-field (2)
A sphere of radius a has uniform charge density  throughout. What
is the direction and magnitude of the electric field everywhere? [Like
example 24.3]
•When computing the flux for a Gaussian surface, only include the
qin   0  E
electric charges inside the surface
E  rˆE
ˆ
V    0 AE  n
4
3
 r 3    0 4 r 2 E
E  rˆ  r 3 0
rˆ  a 3 0 r
E
 rˆ  r 3 0
3
r/a
2
r
for r  a,
for r  a.
E
a
Electric Field From a Line Charge
What is the electric field from an infinite
line with linear charge density ?
n̂
E
n̂
E  rˆE
L
r
 rˆ
E
2 0 r
•Electric field must point away from the line charge, and depends only
on distance
•Add a cylindrical Gaussian surface with radius r and length L
•Use Gauss’s Law
qin   0  E
•The ends of the cylinder don’t contribute
•On the side, the electric field and the normal are parallel

 L   0  E  nˆ da   0 EA   0 E  2 rL 
E
2 0 r
Electric Field From a Plane Charge
What is the electric field from an infinite
plane with surface charge density ?
E
E  kˆ E
n̂
n̂
n̂
 kˆ
E
2 0
•Electric field must point away from the surface, and depends only on
distance d from the surface
•Add a box shaped Gaussian surface of size 2d  L  W
•Use Gauss’s Law
qin   0  E
•The sides don’t contribute
•On the top and bottom, the electric field and the normal are parallel 
 LW   0  E  nˆ da   0 EA   0 E  2LW 
E
2 0
CT-4 - The electric charge per unit area is +for infinite plate 1 and –for
infinite plate 2.The magnitude of the electric field associated with plate 1 is 
and the electric field lines for this plate are as shown. When the two are placed
parallel to one another, the magnitude of the electric field is
A. / between, 0 outside.
B. / between, ± /(2) outside.
C. zero both between and outside.
D. ±/(2o) both between and outside.
E. none of the above.
Serway 24-35
Solve on
Board
Warmup 04
Conductors and Gauss’s Law
•Conductors are materials where charges are free to flow in response to
electric forces
•The charges flow until the electric field is neutralized in the conductor
Inside a conductor, E = 0
•Draw any Gaussian surface inside the
conductor
qin   0  E   0  E  nˆ dA  0
In the interior of a conductor,
there is no charge
The charge all flows to the surface
Warmup 04
Electric Field at Surface of a Conductor
•Because charge accumulates on the
surface of a conductor, there can be
electric field just outside the conductor
•Will be perpendicular to surface
n̂
•We can calculate it from Gauss’s Law
•Draw a small box that slightly penetrates
the surface
•The lateral sides are small and have no flux through them
•The bottom side is inside the conductor and has no electric field
•The top side has area A and has flux through it   E  n
ˆ A  EA
E
•The charge inside the box is due to the surface charge 
qin   A
•We can use Gauss’s Law to relate these
qin   0  E
 A   0 EA

E
nˆ
0
Where does the charge go?
A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm
has q = +80 nC of charge put on it. What is the surface charge
density on the inner surface? On the outer surface?
A) 20 nC/cm2 B) 5 nC/cm2 C) 4 nC/cm2 D) 0
E) None of the above
80 nC
•The Gaussian surface is entirely contained in the
conductor; therefore E = 0 and electric flux = 0
•Therefore, there can’t be any charge on the inner surface
1 cm
•From the symmetry of the problem, the charge will be
2 cm
uniformly spread over the outer surface
cutaway
20 nC
q
80 nC
2
ˆ
r
 

 5 nC/cm E  2
2
view
r 0
A 4  2 cm 
The electric field:
•The electric field in the cavity and in the conductor is zero
•The electric field outside the conductor can be found from Gauss’s Law
Warmup 04
CT - 5. Which of the following is true?
A. The electric field inside a charged insulating sphere must be zero.
B. . The electric field inside a charged conducting sphere must be zero.
C. The charge on a conducting spherical shell will always be equally distributed on the
inner and outer surface regardless of the presence of other charges in the vicinity of
the shell.
D. Two of the above
E. Three of the above
Serway 24-47
Solve on
Board
Ex. Serway 24-45. A long, straight wire is surrounded by a hollow metal cylinder, the
axis of which coincides with the wire. The wire has a charge per unit length of , and the
cylinder has a charge per unit length of 2. From this information, use Gauss's law to
find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b)
the electric field outside the cylinder a distance r from the axis.
Solve on
Board