GAUSS LAW
ELECRIC FLUX
Dr Manjunatha S
Plasma ball, the colorful lines emanating from the
sphere give evidence of strong electric field.
Dr MANJUNATHA S, CS&IS
2
Gauss’s Law
Karl Friedrich Gauss
(1777-1855) – German mathematician
Dr MANJUNATHA S, CS&IS
3
Gauss’s Law
Already can calculate the E-field of an
arbitrary charge distribution using
Coulomb’s Law.
Gauss’s Law allows the same thing,
but much more easily…
… so long as the charge distribution
is highly symmetrical.
Karl Friedrich Gauss
(1777-1855) – German mathematician
Dr MANJUNATHA S, CS&IS
4
Electric Flux
Measures how much an
electric field wants to “push
through” or “flow through”
some arbitrary surface area
Definition : The total no of
lines of force emanation
from a charge.
Denoted by a
Plane Area
Greek letter Φ Dr MANJUNATHA S, CS&IS
Flux lines
Electric Flux:
General Equation for Electric Flux is
E-Flux through a surface depends on three things:
1. How strong the E-field is at each infinitesimal area.
2. How big the overall area A is after integration.
3. The orientation between the E-field and each
infinitesimal area.
Dr MANJUNATHA S, CS&IS
6
Electric Flux – Case 1
Easiest case:
• The E-field is uniform
• The plane is
perpendicular to the
field
Electric Flux
Dr MANJUNATHA S, CS&IS
7
Electric Flux – Case 1
Easiest case:
• The E-field is uniform
• The plane is
perpendicular to the
field
Electric Flux
Flux depends on how
strong the E-field is and
Dr MANJUNATHA S, CS&IS
8
how big the area is.
Electric Flux
We have used electric field lines to visualize
electric fields and indicate their strength.
We are now going to
count* the number of
electric field lines
passing through a
surface, and use this
count to determine the
electric field.
Dr MANJUNATHA S, CS&IS
*There are 3 kinds of people in this world: those who can count, and those who can’t.
E
9
The electric flux passing through a surface is the
number of electric field lines that pass through it.
Because electric field
lines are drawn
arbitrarily, we quantify
electric flux like this:
E=E . A, except that…
If the surface is tilted,
fewer lines cut the
surface.
Later we’ll learn about magnetic flux, which is
MANJUNATHA S, CS&IS
why I will use the subscript E onDrelectric
flux.
A
E
E

10
The green lines miss!
We define A to be a vector
having a magnitude equal to the
area of the surface, in a
direction normal to the surface.
A

E
The “amount of surface”
perpendicular to the electric
field is A cos .

Therefore, the amount of surface area effectively
“cut through” by the electric field is A cos .
AEffective = A cos  so E = EAEffective = EA
cos .
Remember the dot product ,E  E  A
Dr MANJUNATHA S, CS&IS
11
If the electric field is not uniform, or the surface
is not flat…
divide the surface
into infinitesimal
surface elements
and add the flux
through each…
E
A
 E  lim
Ai  0
 E  A
i
i
i
 E   E  dA
Remember, the
direction of dA is
normal to the surface.
Dr MANJUNATHA S, CS&IS
12
If the surface is closed (completely
encloses a volume)…
…we count* lines going
out as positive and lines
going in as negative…
E
dA
 E   E  dA
a surface integral,
therefore a double integral
For a closed surface, dA is normal to the surface and always
points away from the inside.
Dr MANJUNATHA S, CS&IS
13
*There are 10 kinds of people in this world: those who can count in binary, and those who can’t.
Electric Flux – Flux through a Closed Surface
The vectors dAi point
in different directions
At each point, they are
perpendicular to the
surface
By convention, they
point outward
Dr MANJUNATHA S, CS&IS
14
Flux through a Cube
Assume a uniform E-field
pointing only in +x
direction
Find the net electric flux
through the surface of a
cube of edge-length l, as
shown in the diagram.
Dr MANJUNATHA S, CS&IS
15
What the *!@* is this
 thing?
Nothing to panic about!
The circle just
reminds you to
integrate over a
closed surface.
Dr MANJUNATHA S, CS&IS
16
Question:
Give five different equations for electric flux. Which one do I
need to use?
Answer: use the simplest (easiest!) one that works.
 E  EA
Flat surface, E  A, E constant over
surface. Easy!
 E  EA cos  Flat surface, E not  A, E constant over
surface.
E  E  A
Flat surface, E not  A, E constant over
surface.
Surface not flat, E not uniform. Avoid, if
 E  E  dA
possible.

E 
 E  dA
Closed surface. Most general. Most
complex.
If the surface is closed, you may be able to “break it up” into simple
segments and still use E=E·A for each segment.
Dr MANJUNATHA S, CS&IS
17
Quick Quiz:
What would happen to the E-flux if we
change the orientation of the plane?
ANSWER: The
electric flux decreases
Dr MANJUNATHA S, CS&IS
18
Quick Quiz:
Which plane area experiences the
most flux?
Answer:
Figure 1, small angle between plane and E.
Dr MANJUNATHA S, CS&IS
19
Quick Quiz:
What are the three things on which
Eelctric flux depends?
Answer: Angle between flux and the
plane, Electric field and plane area
Dr MANJUNATHA S, CS&IS
20
Electric Flux – Calculating E-Flux
 The surface integral means the integral must be
evaluated over the surface in question… more in a
moment.
 The value of the flux will depend both on the field
pattern and on the surface
 The units of electric flux are N.m2/C
 The net electric flux through a surface is directly
proportional to the number of electric field lines
passing through the surface.
Dr MANJUNATHA S, CS&IS
21
Example: 1
A cylinder of radius R is immersed in a uniform
electric field E with its axis (i) parallel to the field
(ii) perpendicular to the field. Prove that the E
through the cylinder in each case is zero.
dA
c
a
b
dA
dA
Dr MANJUNATHA S, CS&IS
22
Electric Flux Density
The number of lines passing through a normal
surface is called flux density and it is denoted
by D.
Dr MANJUNATHA S, CS&IS
23
Electric Flux Density
A point charge is surrounded by an isolated
electric field.
Gaussian Surface
Line density D
N
r
A
Radius r
Dr MANJUNATHA S, CS&IS
24
Consider the field near a positive point charge q has
a radial field indicated by flux lines
Then, imagine a surface (radius r) surrounding q.
These lines indicate the field direction, ie.,the
direction of force.
Radius r
r
The electric field intensity E at
radius r due to charge q will be
at any point
We know that
Gaussian
Surface
Dr MANJUNATHA S, CS&IS
25
The electric field becomes
We can write as
The area, A =1/4πr2 and
the Electric density, D= ε0 E
Hence the Electric flux D, is found and its
SI unit is C/ mDr2MANJUNATHA S, CS&IS
26
Example 2.
Write an equation for finding the total number of
lines N leaving a single positive charge q.
Radius r
r
E is proportional to N/A
and is equal to kq/r2 at any
point.
Gaussian
Surface
Dr MANJUNATHA S, CS&IS
27
Substitute E and A value in the above equation
Dr MANJUNATHA S, CS&IS
28
Quick Quiz
Which surface – S1, S2
or S3 – experiences the
most electric flux?
Dr MANJUNATHA S, CS&IS
29
Gauss’s Law or Gauss’s Theorem
Gauss’s Law is just a flux calculation
We’re going to build imaginary surfaces –
called Gaussian surfaces – and calculate the
E-flux.
Gauss’s Law only applies to closed surfaces.
Gauss’s Law directly relates electric flux to
the charge distribution that creates it.
Dr MANJUNATHA S, CS&IS
30
Gauss’s Law or Gauss’s Theorem
The surface integral of the normal component of the
electric flux density D over any closed surface equal
charge enclosed
Or
The surface integral of the normal component of
electric intensity E over a closed surface equal to
1/εo time of the total charge.
 NET
  qenclosed
  E  dA 
surface
Dr MANJUNATHA S, CS&IS
0
31
Gauss’s Law
The net number of electric field lines is crossing
any closed surface in an outward direction
numerically equal to the net total charge within
that surface.
 NET
 qenclosed

  E  dA 
surface
The net E-flux
through a closed
surface
0
Charge inside the
surface
Dr MANJUNATHA S, CS&IS
32
Gauss’s Law – confirming Gauss’s Law
Assume a single positive point charge of
magnitude q sits at the center of our
imaginary Gaussian surface, which we
choose to be a sphere of radius r.
 NET
 
  E  dA
surface
At every point on the sphere’s
surface, the electric field from the
charge points normal to the sphere.
Dr MANJUNATHA S, CS&IS
33
Gauss’s Law – confirming Gauss’s Law
 NET
 
  E  dA 
surface

 EdA cos
surface
EdA

surface
Dr MANJUNATHA S, CS&IS
34
Gauss’s Law – confirming Gauss’s Law
Now we have:
 NET 
 EdA
surface
The electric field due to the point
charge is constant all over the
sphere’s surface. So…
 NET  E
 dA
surface
Dr MANJUNATHA S, CS&IS
35
Gauss’s Law – confirming Gauss’s Law
This, we can work with.
 dA
 NET  E
surface
We know the magnitude of the electric
field E at the sphere’s surface due to a
point charge at a distance r away from
the charge.
E po int ch arg e
ke q
 2
r
Dr MANJUNATHA S, CS&IS
36
Gauss’s Law – confirming Gauss’s Law
Thus:
 NET
ke q
 2  dA
r surface
And, this surface integral is easy.
2
dA

4


r

sphere
Dr MANJUNATHA S, CS&IS
37
Gauss’s Law – confirming Gauss’s Law
Therefore,
 NET
ke q
2
 2  (4  r )
r
But, we can rewrite
Coulomb’s constant.
ke 
1
40
Dr MANJUNATHA S, CS&IS
38
Gauss’s Law – confirming Gauss’s Law
Therefore:
 NET
ke q
 2  (4  r 2 )
r
But, we can rewrite Coulomb’s
constant.
ke 
1
40
Dr MANJUNATHA S, CS&IS
39
  NET
q

0
Thus, we have confirmed Gauss’s law:
Dr MANJUNATHA S, CS&IS
40
Questions
+Q +
–
– 3Q
• If the electric field is zero for all points on the
surface, is the electric flux through the surface
zero?
• If the electric flux is zero for a closed surface, can
there will be charges inside the surface?
• What is the flux through the surface shown?
Why?
Dr MANJUNATHA S, CS&IS
41
Flux due to a Point Charge
A spherical surface surrounds a point charge.
What happens to the total flux through the
surface if:
(A)the charge is tripled,
(B)the radius of the sphere is doubled,
(C)the surface is changed to a cube, and
(D)the charge moves to another location inside
the surface?
Dr MANJUNATHA S, CS&IS
42
Applying Gauss’s Law
Gauss’s Law can be used to
(1) find the E-field at some position relative to a
known charge distribution, or
(2) to find the charge distribution caused by a known
E-field.
In either case, you must choose a Gaussian surface to use.
Dr MANJUNATHA S, CS&IS
43
Applying Gauss’s Law
Choose a surface such that…
1. Symmetry helps: the E-field is constant over the
surface (or some part of the surface)
2. The E-field is zero over the surface (or some
portion of the surface)
3. The dot product reduces to EdA (the E-field and
the dA vectors are parallel)
4. The dot product reduces to zero (the E-field and
the dA vectors are perpendicular)
Dr MANJUNATHA S, CS&IS
44
Spherical Charge Distribution
An insulating solid sphere of radius a has a uniform volume charge
density ρ and carries total charge Q.
(A) Find the magnitude of the E-field at a point outside the sphere
(B) Find the magnitude of the E-field at a point inside the sphere
Dr MANJUNATHA S, CS&IS
45
Spherical Charge Distribution
Dr MANJUNATHA S, CS&IS
46
Spherical Charge Distribution
Find the E-field a distance r from
a line of positive charge of
infinite length and constant
charge per unit length λ.
Dr MANJUNATHA S, CS&IS
47
Spherical Charge Distribution
Find the E-field due to an infinite
plane of positive charge with
uniform surface charge
density σ
Dr MANJUNATHA S, CS&IS
48
Conductors in Electrostatic Equilibrium
• In an insulator, excess charge stays put.
• Conductors have free electrons and, correspondingly, have different
electrostatic characteristics.
• You will learn four critical characteristics of a conductor in
electrostatic equilibrium.
• Electrostatic Equilibrium – no net motion of charge.
Dr MANJUNATHA S, CS&IS
49
Conductors in Electrostatic Equilibrium
• Most conductors, on their own, are in electrostatic equilibrium.
• That is, in a piece of metal sitting by itself, there is no “current.”
Dr MANJUNATHA S, CS&IS
50
Conductors in Electrostatic Equilibrium
Four key characteristics
1.
The E-field is zero at all points inside a conductor, whether hollow or
solid.
2.
If an isolated conductor carries excess charge, the excess charge
resides on its surface.
3.
The E-field just outside a charged conductor is perpendicular to the
surface and has magnitude σ/ε0, where σ is the surface charge
density at that point.
4.
Surface charge density is biggest where the conductor is most
pointy.
Dr MANJUNATHA S, CS&IS
51
Conductors (cont.) – Justifications
Einside = 0
• Place a conducting slab in an external
field, E.
• If the field inside the conductor were
not zero, free electrons in the
conductor would experience an
electrical force.
• These electrons would accelerate.
• These electrons would not be in
equilibrium.
• Therefore, there cannot be a field
inside the conductor.
Dr MANJUNATHA S, CS&IS
52
Conductors (cont.) – Justifications
Einside = 0
• Before the external field is applied,
free electrons are distributed evenly
throughout the conductor.
• When the external field is applied,
charges redistribute until the
magnitude of the internal field equals
the magnitude of the external field.
• There is a net field of zero inside the
conductor.
• Redistribution takes about 10-15s.
Dr MANJUNATHA S, CS&IS
53
Conductors (cont.) – Justifications
Charge Resides on the Surface
• Choose a Gaussian surface inside
but close to the actual surface
• The electric field inside is zero
(prop. 1)
• There is no net flux through the
gaussian surface
• Because the gaussian surface can
be as close to the actual surface as
desired, there can be no charge
inside the surface
Dr MANJUNATHA S, CS&IS
54
Conductors (cont.) – Justifications
Charge Resides on the Surface
• Since no net charge can be inside
the surface, any net charge must
reside on the surface
• Gauss’s law does not indicate the
distribution of these charges, only
that it must be on the surface of
the conductor
Dr MANJUNATHA S, CS&IS
55
Conductors (cont.) – Justifications
E-Field’s Magnitude and Direction
• Choose a cylinder as the Gaussian
surface
• The field must be perpendicular to
the surface
– If there were a parallel
component to E, charges
would experience a force and
accelerate along the surface
and it would not be in
equilibrium
Dr MANJUNATHA S, CS&IS
56
E-Field’s
Magnitude
and Direction
Conductors
(cont.)
– Justifications
E-Field’s Magnitude and Direction
• The net flux through the surface
is through only the flat face
outside the conductor
– The field here is
perpendicular to the surface
• Applying Gauss’s law
A
 E  EA 
0

E
0
Dr MANJUNATHA S, CS&IS
57
E-Field’s
Magnitude
and Direction
Conductors
(cont.)
– Justifications
E-Field’s Magnitude and Direction
• The field lines are
perpendicular to both
conductors
• There are no field lines
inside the cylinder
Dr MANJUNATHA S, CS&IS
58
Sphere inside a Spherical Shell
A solid insulating sphere of radius a carries a uniformly
distributed charge, Q. A conducting shell of inner radius b and
outer radius c is concentric and carries a net charge of -2Q.
Find the E-field in regions 1-4 using Gauss’s Law.
Dr MANJUNATHA S, CS&IS
59
Example:3
Consider a thin spherical shell of radius 14.0 cm
with a total charge of 32.0 μC distributed
uniformly on its surface. Find the electric field
(a)10.0 cm and
(b)20.0 cm from the center of the charge
distribution.
k Q  8.99  10  32.0  10 
E

 7.19 M N
6
9
e
2
r
 0.200
2
Dr MANJUNATHA S, CS&IS
C
60
Example 4
A uniformly charged, straight filament 7.00 m in length
has a total positive charge of 2.00 μC. An uncharged
cardboard cylinder 2.00 cm in length and 10.0 cm in
radius surrounds the filament at its center, with the
filament as the axis of the cylinder. Using reasonable
approximations, find (a) the electric field at the surface
of the cylinder and (b) the total electric flux through the
cylinder.

9
2ke 2 8.99  10 N  m
E

r
2


C 2  2.00  106 C 7.00 m 


0.100 m
E  EA cos  E  2 r  cos0


E  51.4 kN C ,radially outw ard
E  5.14  104 N C 2  0.100 m   0.020 0 m   1.00  646 N  m 2 C
Dr MANJUNATHA S, CS&IS
61
Example: 4
A square plate of copper with 50.0-cm sides has no net
charge and is placed in a region of uniform electric
field of 80.0 kN/C directed perpendicularly to the
plate. Find
(a) the charge density of each face of the plate and
(b) the total charge on each face.



  8.00  104 8.85  1012  7.08  107 C m 2

7
Q   A  7.08  10
 0.500
2
C
Q  1.77  107 C  177 nC
Dr MANJUNATHA S, CS&IS
62
Example:5.
How many electric field lines pass
through the Gaussian surface drawn
below.
First we find the NET
charge Sq enclosed by
the surface:
Sq = (+8 –4 – 1) = +3mC
Gaussian
surface
-4 mC
q1
q2
-
+8
mC
+
q4
-1 mC
q3
N = +3 mC = +3 x 10-6 lines
Dr MANJUNATHA S, CS&IS
-
+
+5
mC
63
Example 6.
A solid sphere (R = 6 cm) having net charge +8 mC is
inside a hollow shell (R = 8 cm) having a net charge of
–6 mC. What is the electric field at a distance of 12 cm
from the center of the solid sphere?
Draw Gaussian sphere at
radius of 12 cm to find E.
Gaussian
surface
-
8cm
-
Sq = (+8 – 6) = +2 mC
12
cm
Dr MANJUNATHA S, CS&IS
-
+8 mC
-6 mC
-
6 cm
- -
64
Example 6 (Cont.)
What is the electric field at a distance of 12 cm
from the center of the solid sphere?
Gaussian
surface
-
8cm
-
12
cm
-
-6 mC
-
+8 mC
-
6 cm
- -
Sq
2 x 10-6C
E

2
 0 (4 r ) (8.85 x 10-12 Nm2 C2 )(4 )(0.12 m) 2
2 m C
6 N
E

1.25
x
10
C
2
 0 (4 r )
Dr MANJUNATHA S, CS&IS
E = 1.25 N/C
65
Charge on Surface of Conductor
Since like charges repel,
you would expect that
all charge would move
until they come to rest.
Then from Gauss’s Law
...
Gaussian Surface just inside
conductor
Charged Conductor
Since charges are at rest, E = 0 inside conductor, thus:
All charge is on surface; None inside Conductor
Dr MANJUNATHA S, CS&IS
66
Example 7.
Use Gauss’s law to find the E-field just outside
the surface of a conductor. The surface
charge density, σ = q/A.
Consider q inside the pillbox. Elines through all areas outward.
E1
E3
A
+
+
+
E3
+ +
E-lines through sides cancel by
symmetry.
E3
+ + +
+
+
E+3
E2
+ +
Surface Charge Density 
The field is zero inside the conductor, so E2 = 0
q

E

 A 0
Dr MANJUNATHA S, CS&IS 0
0
oE1A + oE2A = q
67
Example 7 (Cont.)
Find the field just outside the surface if
 = q/A = +2 C/m2.
Recall that side fields
cancel and inside field
is zero, so that
E3
2 x 10-6 C/m 2
E
-12 Nm 2
8.85 x 10
C2
E3
A
+
E3
+
q

E1 

0 A 0
E1
+ +
+
+ + +
+
+
E+3
E2
+ +
Surface Charge Density 
E = 226,000 N/C
Dr MANJUNATHA S, CS&IS
68
Thank You
Thank You