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Gauss’s Law
The electric flux through a
closed surface is proportional
to the charge enclosed
Definition of Flux
• The amount of field, material or
other physical entity passing
through a surface.
• Surface area can be represented as
vector defined normal to the
surface it is describing
• Electric Flux is defined by the
equation:
 
 E dA
 
surface
Electric Flux
• The amount of
electric field
passing through a
surface area
• The units of electric
flux are N-m2/C
E 

E dA
Carl Friedrich Gauss 1777-1855
Yes it’s the same guy
that gave you the
Gaussian
distribution and …
Ceres t 4.6 year , d=4.6 Au
To give you some
perspective he was born 50
years after Newton died
(1642-1727). Predicted the
time and place of the first
asteroid CERES (Dec. 31,
1801). Had the unit of
magnetic field named after
him and of course had
much to do with the
development of
mathematics
Gauss’ Law in E& M
• Uses symmetry to determine E-field due to
a charge distribution
• Method: Considers a hypothetical surface
enclosing some charge and calculates the
E-field
• The shape of that surface is
“EVERYTHING”
KEY TO USING Gauss’s Law
• The shape of the surrounding surface is one
that MIMICS the symmetry of the charge
distribution …..
E 

E dA
The electric flux passing through a spherical
surface surrounding a point charge
Use a spherical surface with area
A  4 r 2
Electric field for a point charge at r
E
q
4 0 r 2
Due to symmetry, E and dA are parallel and E is constant
over all of A. So…
E 



q
q
2
E d A  E  dA  EA  
4

r

2 
4

r
0
0




Gauss’s Law
The previous example is true in general:
The total flux passing through a closed surface is
proportional to the charge enclosed within that
surface.
E 

closed
surface
E  dA 
q
o
Note: The area vector points outward
The Gaussian Surface
An imaginary closed surface created to enable
the application of Gauss’s Law
What is the total flux through each surface?
Solving problems with Gauss’s Law
1. Charge densities –
It is convenient to define charge densities for
linear, surface and volume charge distributions
2. Symmetry and coordinate systems –
Choose that coordinate system that most nearly
matches the symmetry of the charge distribution.
For example, we chose spherical coordinates to
determine the flux due to a point charge because of
spherical symmetry.
Solving problems with Gauss’s Law
3. Determining the Electric Field from Gauss’s
Law – Symmetry must be present in the
charge distribution so that over the Gaussian
surface E and A vectors are parallel and E is
constant over the surface.
Consider three examples: (1) the long straight
line of charge, (2) the infinite plane sheet of
charge, and (3) a charged sphere.
Example - Long straight line of charge
Looking at the diagram (b), we can determine that the
problem has a cylindrical symmetry.
Therefore the Gaussian surface is a cylinder of radius r
and length l.
There are three surfaces to consider. The upper and
lower circular surfaces have normals parallel to the z
axis which are perpendicular to the electric field, thus
contribute zero to the flux.
The area to be evaluated for the integral is that of the
curved sides of the cylinder, 2rl. The charge
enclosed is ll.
Example - Long straight line of
charge
Since the field has radial symmetry, it is also
constant at a fixed distance of r.
qenclosed
   E dA  EA 
0
qenclosed
ll
l
E


o A
 0 2 rl  0 2 r
Gauss’ Law: Determining the E-field near the
surface of a nonconducting (insulating) sheet
  q
 E  dA 
o
EA 
q
o
A
EA 
o

E
o
and also from other side so,
E

2 o
Results of other geometries
Uniformly charged dielectric
infinite plane sheet
Uniformly charged dielectric
(insulating) sphere
Er  a


r
3 o
Er  a 
a
3
3 o r 2

E
2 o
Can you derive these results?
Conclusions Gauss’ LAW
1.
2.
Only the charge enclosed within a volume defined
by a closed surface contributes to the net electric
flux through the surface.
That net flux through the surface is proportional to
the charge enclosed within the volume.
Conclusions
3. Gaussian surface is an imaginary closed surface
necessary to solve a problem using Gauss’s Law
4. Gauss’s Law can be used to determine the electric field
of a charge distribution if there is a high degree of
symmetry
Conclusion: Gauss’ Law and
Conductors
5. Applying Gauss’s Law to the
interior of an electrostatically
charged conductor we
conclude that the electric field
within the conductor is zero
6. Any Net charge on a conductor
must reside on its surface
Shell Theorems:
Conductors
1) A shell of uniform charge attracts
or repels a charge particle that is
outside the shell as though all
charge is concentrated at the
center.
2) If a charged particle is located
inside such a shell, there is no
electrostatic force on the particle
from the shell

E 
1 q
r̂
4 o r 2

E  0
Electric Field at surface of conductor is perpendicular to
the surface and proportional to the charge density at the
surface
  q
Shown that the excess charge resides  E  dA  
o
on the outer surface of conductor
q
surface.
E  dA 
o
Unless the surface is spherical the
q
charge density  (chg. per unit area) EA 
o
varies
A
However the E-field just outside of a EA 
o
conducting surface is easy to

determine using Gauss’ Law
E 
o
Gauss’ Law can be used to determine
E-field in cases where we have:
• Spherical symmetry
• Cylindrical symmetry
• Planar symmetry