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Lecture #3.3
Gauss’ Law
We have seen how the principle of superposition can be used in order to calculate
electric field produced by different configurations of electric charges. It turns out,
however, that there is an easier way of resolving problems, especially when a problem
involves some symmetry.
Let us consider a uniform electric field which crosses some plane surface in
space. For simplicity, let this electric field be perpendicular to the surface, so we can look
at this as if the field flows through the surface and introduce the concept of electric flux.
Remember that the density of electric field lines per unit of area gets larger in the areas
where the field is stronger, so the electric flux, showing the number of electric field lines
crossing the given surface, is   EA . However, this is only true if the lines are
perpendicular to the surface. If they are parallel then the flux is going to be zero, so in
general we have
  EA cos  ,
(3.3.1)
where  is the angle between the direction of electric field and the direction of outward
normal to the surface. Equation 3.3.1 can be rewritten in vector notation as
  E  A,
(3.3.2)
where vector A is the external normal to the surface of the plane and has the absolute
value of this surface. If one has a complicated surface, which consist of several planes,
where each plane has area Ai then the total flux will be
   E  Ai ,
(3.3.3)
i
If we consider a closed surface then the flux is positive for electric field lines
leaving the enclosed volume of the surface and it is negative for the field lines entering
this enclosed volume.
In a simple situation when there is only one positive point-like charge q at the
center of the spherical surface of radius R, it produces the electric field which goes in all
radial directions from this charge perpendicular to the surface of the sphere, so that the
electric flux through the surface of the sphere is going to be
  EA 
kq
q
q
4R 2 
4 
2
4 0
0
R
,
This situation is a special case of the general law, which is known as Gauss’ Law. The
Gauss’ Law states that the electric flux through any closed surface is equal to the total
charge enclosed by this surface divided by the permittivity of free space  0 .

q
(3.3.4)
0
As a matter of fact the statement of the Gauss’ law is even more general than the
Coulomb’s law. From theoretical standpoint, one can state the Gauss’ law as the
fundamental principle of electrostatics and derive Coulomb’s law from it.
One can use Gauss’ law in order to study the behavior of electric field near the
surface of the conductor. We have already discussed that electric field is always
perpendicular to the surface of the conductor, since if it had a tangential component then
the charges would have to move along the surface which is not the case in equilibrium.
We also know that electric field inside of the conductor is zero. So, if we choose the
closed Gaussian surface around some small portion of the conductor’s surface of area A,
then we have
EA 
E
q
0
,
q


A 0  0
where  is the charge’s density per unit or area.
Now having Gauss’s law in hand, we can find the electric field inside of the
parallel plate capacitor. Let us neglect any effects that may occur close to the ends of the
capacitor and consider it as two infinitely big parallel planes. The total electric field
inside of this capacitor is a vector sum of the fields produced by each of the planes. Let
each of those plates carry the charge density (charge per unit of area)  . To apply
Gauss’s law we have to surround this plane by the close surface. It could be the
parallelepiped with two sides of the area A parallel to the surface and all other sides
perpendicular to the surface of the capacitor. The electric field is perpendicular to the
plane and so the total electric flux through the surface of this parallelepiped is going to be
  2 AE 
q
0
,
so the electric filed produce by the charged plate is
E
q


2 A 0 2 0
(3.3.5)
You can see that it is uniform field. It does not depend on the distance from the plate. The
total field inside of the capacitor due to both its plates (one positively and another one is
negatively charged but with the same charge density) is twice as large
E

0
(3.3.6)
The electric field outside of the capacitor is zero because electric fields caused by two
different plates have the same magnitude but opposite directions and cancel each other.
We shall apply our knowledge of the field inside of the capacitor to consider behavior of
electric charges in uniform fields.