Electric Flux and Gauss’s Law
Gauss’s law states that the electric flux through a
closed surface is proportional to the charge enclosed
by the surface:

s
E.da 
1
0
Qenc
Note: A charge outside the surface will contribute
nothing to the total flux. This is the essence of
Gauss’s Law.
Gauss’s Law in differential form
Gauss’s law in it’s integral form:
1
 E.da  Qenc
0
s
Using divergence theorem:

s
E.da   (.E )d
v
Qenc in terms of charge density :
Qenc    d
v
So Gauss’s law becomes:
 
v (.E )d  v   0

d

Since this holds for any volume, the integrands must be
equal:
1
.E 

0
Gauss’ Law and Coulomb law
The angle  between E and dA is
zero at any point on the surface, we
can re-write Gauss’ Law as
A
spherical
Gaussian
surface centered on a point E has the same value at all points on
charge q
the surface
E can be moved out,
Integral is the sum
of surface area
Coulomb’s Law
Applications of Gauss’s Law
Gauss’s law can be used to find the electric field in
systems with simple configurations.
Applications of Gauss’ Law
Gauss’s law is always true, but it is not always useful.
In what conditions Gauss’s Law can not be used ?
If charge density is not uniform, or our chosen Gaussian
surface (closed) is such that E is not in the same
direction as da at every point through it would still have
been true that the flux of E is (1/ϵ0)q
Why
We could not pull E out of the integral.
When we can use Guass Law?
Symmetry is crucial
Criteria for calculation of E using Guass’s Law:
Construct a closed surface called Gaussian surface
near the charge distribution, over which
i. magnitude of E have constant value for
uniform charge distributions.
ii. angle () between E and normal to this surface
have constant value for symmetrical charge
distributions and symmetrical surfaces so that
ECos can be taken outside the integral sign.
There are only three kinds of symmetry which we can
use for calculation of E:
1. Spherical symmetry: Make your Gaussian surface
a concentric sphere.
2. Cylindrical symmetry: Make your Gaussian
surface a coaxial cylinder.
3. Plane symmetry: Use a Gaussian “pillbox” which
straddles the surface.
Using Cylindrical Symmetry
Question: Evaluate the electric field that arises from an
infinite line of charge.
Pick a point P for evaluation of E.
Understand the symmetry
The only variable on which E may depend is R, the
distance from the line of charge.
There is no angular dependence because the line is
cylindrically symmetric, i.e. it does not matter at which
angle about the line you view it.
There is no axial dependence because
the line of charge is infinitely long,
i.e. there is no preferred position
along the line.
The direction of E is directly away
from the line, perpendicular to it at all
locations.
There is no azimuthally or axial
components because the line is
infinite and has no extent in the
transverse directions.
Construct the Gaussian surface
A cylindrical tube having a radius
equal to the distance between the point
of evaluation and the line of charge.
This surface can be divided in 3 parts.
Surface 2: The electric field anywhere on this surface has
the same direction as the infinitesimal area vector and has
a constant value everywhere on the surface.
Surface 1 and 3: Since a Gaussian surface must be
closed, the tube is then capped with flat endcaps, 1 and 3.
The field vector is perpendicular to the area vector so there
is no flux through the endcaps.
Examine the Gaussian surface
The flux through the endcap 1 is zero since A1 is
perpendicular to E everywhere on this surface. Similarly,
the flux through surface 3 is zero.
Since the electric field is everywhere constant on surface
2 and points in the same direction as the surface vector at
any point, the flux through surface 2 will be the field
magnitude E times the area of the tube wall.
Evaluate the electric flux
through the Gaussian surface
Evaluate the charge enclosed by the Gaussian surface
Apply Gauss's law for the result
What is the electric field from an infinitely long wire
with linear charge density of +100 nC/m at a point
10cm away from it ?
+++++++++++++++++++++++++++++++++++++++++
.
R =10 cm
Ey

Ey 
2 0 R
2k
Ey 
R
2  9 109 Nm2 100 109 C / m
Ey 
 18 103 N / C
0.1m
Line of Charge by Direct Integration 1
Problem: Evaluate the electric
field that arises from an
infinite line of charge.
Evaluate the contribution from the infinitesimal
charge element
Considering a positive test charge at point P,
the direction of the electric field is shown.
The strength of the electric field
contribution from the infinitesimal charge
shown is proportional to dq and inversely
proportional to the square of the distance
separating point P and the charge element.
Exploit symmetry as appropriate
The horizontal components of the two contributions
will balance.
We need only consider the vertical contribution to the
electric field by each infinitesimal charge.
Substituting these terms, dEy becomes
Setting up the integrating over the charge distributing
Expressing the integrand in terms
of theta and switching the limits of
integration:
Problem 2: (a)Electric field inside a conductor.
Ans: E=0 inside the a conductor.
Why
If we put a conductor in an external field (E0). Due to this
electric field electrons will start to move in the opposite
direction of the electric field. After some time they will
reach to the end of the material, the charges pile up: plus
on the right hand side and minus on the left hand side.
These induced charges produce a field of it’s own in the
opposite direction to E0.
The field of the induced charges tends to cancel off the
original field.
(b) Charge inside a conductor:?
Gauss’s Law
.E   /  0
If E=0 so  is also zero.
(c) Any net charge resides on the surface.
If we place a Gaussian surface just inside the surface of the
charged conductor,
E = 0; at all points on the Gaussian surface.
According to Gauss’ Law the net charge enclosed the Gaussian
surface must also be zero.
Therefore the excess charged must be outside the
Gaussian surface and must lie on the actual surface of
the conductor.
(d) A conductor is an equipotential:
Let a and b are two points inside or on the surface of a
given conductor
b
V (a)  V (b)    E.dl  0
a
hence
V (a)  V (b)
(e) E is perpendicular to the surface, just outside a
conductor:
• Think an electric field at
some arbitrary angle at the
surface of a conductor.
• There is a component

E
perpendicular to the surface,

E
so charges will move in this
direction until they reach the

E||
surface, and then, since they
cannot leave the surface,
they stop.
• There is also a component
parallel to the surface, so
there will be forces on
charges in this direction.
Since they are free to move, they will move to nullify any parallel
component of E.
In a very short time, only the perpendicular component is left.
(f) E inside the cavity of a conductor( if cavity has
no charge) :
Zero
If E=0 within an empty cavity, there is no charge on the
surface of the cavity.
Using this concept you can make a metal car during the
thunderstorm:
If lightning strikes, you will not be electrocuted.
(f) E just outside the conductor:
Consider a section of the surface that is
small enough to neglect any curvature
and assume the section is flat.
Embed a tiny cylindrical Gaussian
surface with one end cap full inside the
conductor and the other fully outside
and cylinder is perpendicular to the
surface.
The electric field just outside the
surface must also be perpendicular to
that surface otherwise surface
charges would be subject to motion.
We now sum the flux through the Gaussian surface.
There is no flux through the internal endcap. Why ??
There is no flux through the curved surface. Why ??
The only flux through the Gaussian surface is that
through the external endcap where E is perpendicular.
Assuming a cap area of A, the flux through the cap is
EA.
The charge qenc enclosed by the Gaussian surface lies on
the conductor’s surface area A. If  is the charge per
unit area , the qenc is equal to  A. Therefore Gauss’ Law
becomes
oEA =  A
E =  / o
(conducting surface)
Using Planar Symmetry
Find the electric field of an infinite plane
carries a uniform surface charge .
Draw a Gaussian pillbox extending equal
distances above and below the plane.
From symmetry , E must be perpendicular
to the surface and endcaps.
The surface charge is positive so E
emanates from the surface.
Since the field lines do not piece the
curved surface there is no flux.
Using Gauss’s law:
1
 E.da  
Qenc
0
 E.da  EA  EA  2EA
Qenc   A
2 EA 
1
0
 A;
Using Planar Symmetry
A cross-section of a thin, infinite conducting plate having
(a) positive charge (b) negative charge uniform charge
densities (1) on both the surfaces. Find the field away
from the sheets.
(a)
away from the plate.
For figure (b) plate is a identical plate with negative
charge. In this case E points inward.
Problem: Two infinite parallel planes carry equal but
opposite uniform charge densities 2. Find the field in
each of the three regions
i. To the left of the both
ii. Between them
iii. To the right of the both
(iii)
(i)
(ii)
Left plate produces a field (1/0), points away from it,
the left in region (i), and to the right in regions (ii) and (iii)
Right plate produces a field (1/0), points toward it, the
right in region (i) and (ii) to the left in region (iii)
(i)
(iii)
(ii)
In (i) and (iii) regions field due to both planes will cancel
each other. So E=0
In (ii) region points toward right and amount of
2
E   
 0 
Using Spherical Symmetry
Q. Find the field (a) inside and (b) outside of a
uniformly charged spherical shell of radius R.
x
Ans.(a)Since fields lines can not
cross, the spherical
symmetry
requires E field to be in the
radially direction and uniform on
the Gaussian surface, therefore
Gauss’ Law:
r
Ans(b) Outside the hollow sphere
We define a Gaussian surface
around the sphere for which r>R.
Again since fields lines can not cross, the spherical
symmetry requires E field to be in the radially
direction and uniform on the Gaussian surface,
therefore Gauss’ Law becomes
Same as point charge
Summarized result:
Using Gauss’ Law, we are able to prove two theorems
for a spherical shell.
A shell of uniform charge attracts and repels a charged
particle that is outside the shell as if all shell’s charge
were concentrated at the center.
If a charged particle is located inside a shell of
uniform charge, there is no net electric force on the
particle from the shell.
Q. Find the field outside of a uniformly charged solid
sphere of radius R and total charge q.
Ans: Draw a spherical (Gaussian) surface at radius
r>R.
1
Gauss’s law says for this surface:
 E.da  
S
Qenc  q
0
Qenc
Symmetry allows us to extract E from under the
integral sign: because E points radially outward, as
does da so drop the dot product.
 E.da  
S
E da
S
Also the magnitude of E is constant over the Gaussian
surface, so it comes outside the integral:

E da  E
S
 da  E 4 r
S
E 4 r 
2
E
1
0
1
4 0 r
2
q
qrˆ
2
Q. Find the field (a) outside and (b) inside of a
uniformly charged solid sphere of radius R with
uniform charge density (C/m3).
Ans (a): Outside (a>R)
 E.da  E 4 a
2

S
Qenc
1
0
4
3
 R 
3
 R3
E
3 0 a 2
Qenc
Ans (b): inside (R>a)
Still we have spherical symmetry centered on the
center of the charge.
Choose a Gaussian surface= sphere of radius a
 E.da  E 4 a
S
Qenc
2

1
0
4 3
 a 
3
 a3
a
E

2
3 0 a
3 0
Qenc
a
E
3 0
Inside the sphere
R
E
2
3 0 a
3
Outside the sphere