CHAPTER 22: GAUSS’S LAW
• It turns out that there are four fundamental equations that govern electricity
& magnetism, whether you are dealing with electro/magneto-statics OR
electro/magneto-dynamics
• These four equations are known as Maxwell’s equations, although many
brilliant people besides just James Clerk Maxwell contributed to their
development
• This chapter focuses on the first of these equations, known as Gauss’s law,
which is actually equivalent to Coulomb’s law but is more general
• This law can be a powerful tool for calculating the electric field due to some
charge distribution. And once you know the electric field, you can use it to
calculate the force on any other charge distribution placed in this field.
• So let’s first qualitatively discuss Gauss’s law before diving into the math.
As was the case in the last chapter, there is some charge distribution for
which you need to calculate the electric field that it creates. You could use
the “clunky” method of last chapter and calculate it directly via some tedious
integral. And sometimes that is your only choice. However, given enough
symmetry, Gauss’s law provides a simpler and much more elegant route to
obtaining the field.
• The idea is to surround the charge distribution, for which you want to
calculate the field, by some imaginary, closed surface. You will want to
choose the shape of this surface based on whatever symmetry you have.
• Recall the discussion of field lines at the end of the last chapter. The density
of field lines is proportional to the magnitude of the electric field in that
region. Field lines which go from inside the surface to the outside will be
counted as “positive” and field lines which go from outside the surface to the
inside will be counted as “negative”.
• The algebraic sum of the number of field lines crossing any closed
surface will be proportional to the net charge enclosed by that surface
• This statement is very general
• The shape as well as the size of the “gaussian surface” is arbitrary as long as
it completely encloses the charges in question
• The presence of charges outside the surface does not invalidate the above
statement, although the shapes of the field lines crossing the surface may
change
• NOTE: I will use the term gaussian surface throughout this chapter. This is
NOT a real, physical surface. It is an imaginary, mathematical surface that
one chooses in order to apply Gauss’s law.
ELECTRIC FLUX
• Now we must quantify this notion of “number of field lines crossing a
surface,” and we will do so using the concept of electric flux
• Imagine we have a fan blowing air through a loop. The amount of air
flowing through the loop depends on its orientation with respect to the
direction of the airflow.
• The dots in the figure to the right represent the tips of the arrows in the
figure on the left. These figures illustrate the fact that when the loop is tilted
at some angle with respect to the airflow, the amount of air flowing through
it is reduced.
• So not only do the number of field lines crossing the surface depends on the
magnitude of the electric field, but also on the orientation of the field with
respect to the surface itself
• Now let’s consider a small “patch” of area of our gaussian surface. Its
orientation in space can be uniquely defined by a vector normal to
(perpendicular to) the surface. When calculating the flux, we only want to
consider the component of the electric field parallel to this normal vector.
(This is analogous to how we define work – only the component of the force
parallel to the displacement contributes to the work). So, if 𝜙𝜙 is the angle
��⃗ we want is 𝐸𝐸 cos 𝜙𝜙.
between these two vectors, then the component of 𝑬𝑬
Finally, we need to multiply this by the magnitude of the area of the “patch”
to obtain the flux through it.
• Just like with work, we can express this more compactly using a dot product.
We will use the symbol Φ𝐸𝐸 to denote electric flux.
• For a uniform electric field crossing a flat surface, the electric flux is defined
��⃗ = 𝐸𝐸𝐸𝐸 cos 𝜙𝜙
𝑬𝑬 ⋅ 𝑨𝑨
as: Φ𝐸𝐸 = ��⃗
• ��⃗
𝑨𝑨 is called the “area vector” – it has magnitude equal to the surface area and
its direction is perpendicular to the surface. Because of this it is often
��⃗ = 𝐴𝐴𝒏𝒏
�, where 𝒏𝒏
� is a unit vector normal (perpendicular) to the
denoted 𝑨𝑨
surface. For a closed surface, it is directed perpendicularly outward (as
��⃗.
opposed to inward). 𝜙𝜙 is the angle between ��⃗
𝑬𝑬 and 𝑨𝑨
��⃗ are parallel (i.e., ��⃗
• Electric flux is maximum when ��⃗
𝑨𝑨 and 𝑬𝑬
𝑬𝑬 is perpendicular
��⃗ are perpendicular (i.e., ��⃗
𝑬𝑬 is
to the surface) and it is zero when ��⃗
𝑨𝑨 and 𝑬𝑬
parallel to the surface)
• Our definition of flux so far only applies to a situation where both the
electric field is uniform and the surface is flat. What if the surface is curved
and/or the electric field varies with position?
• In that case, we have to divide the surface into many small patches (small
enough to be approximately flat and the field to be approximately uniform
everywhere on it). Treating each patch as an infinitesimal piece with area 𝑑𝑑𝑑𝑑
��⃗ = 𝒏𝒏
�, we have 𝑑𝑑𝑨𝑨
�𝑑𝑑𝑑𝑑. Each patch then contributes
and unit vector 𝒏𝒏
•
•
•
•
•
•
��⃗ ⋅ 𝑑𝑑𝑨𝑨
��⃗ to the flux and integrating over the entire surface, we obtain
𝑑𝑑Φ𝐸𝐸 = 𝑬𝑬
��⃗ = ∫ 𝐸𝐸 cos 𝜙𝜙 𝑑𝑑𝑑𝑑. This is called a
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
the total flux through it: Φ𝐸𝐸 = ∫ ��⃗
surface integral.
Now that we have quantified the notion “number of field lines crossing a
surface” by developing the concept of flux, we can restate our previous
assertion as follows:
The electric flux through any closed surface is proportional to the net
charge enclosed by that surface
��⃗ refers to the integral of ��⃗
The quantity ∫ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
𝑬𝑬 over any surface. When we
��⃗, this refers specifically to a closed surface.
write ∮ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
So, mathematically, the above statement in bold can be written as ∮ ��⃗
𝑬𝑬 ⋅
��⃗ ∝ 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒
𝑑𝑑𝑨𝑨
𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 represents the net charge enclosed by the surface over which you are
integrating the electric field
To evaluate the constant of proportionality between flux and enclosed
charge, let’s consider a positive point charge q and a spherical surface of
radius R centered on the charge
• As we saw in the last chapter, the electric field of a point charge is given by
𝑞𝑞
��⃗
𝑬𝑬 = 𝑘𝑘 2 𝒓𝒓�. The unit vector 𝒓𝒓� means that the field points radially away from
𝑟𝑟
q (or toward, if q is negative) which, in turn, means that everywhere on the
surface of a sphere it is perpendicular to the surface. In other words, on a
��⃗ and 𝑑𝑑𝑨𝑨
��⃗ are everywhere parallel (cos 𝜙𝜙 = 1).
� = 𝒓𝒓� and so 𝑬𝑬
sphere, 𝒏𝒏
��⃗ = ∮ 𝐸𝐸 cos 𝜙𝜙 𝑑𝑑𝑑𝑑 = ∮ 𝐸𝐸𝐸𝐸𝐸𝐸
• Φ𝐸𝐸 = ∮ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
• Since the magnitude of the electric field varies as 1/r2 for a point charge, it is
constant when it is evaluated at a fixed radius from q as it would be on a
spherical surface. Thus, we can pull E outside of the integral.
• Now we have Φ𝐸𝐸 = 𝐸𝐸 ∮ 𝑑𝑑𝑑𝑑 = 𝐸𝐸𝐸𝐸 = �𝑘𝑘
𝑞𝑞
𝑅𝑅 2
� (4𝜋𝜋𝑅𝑅2 ), so Φ𝐸𝐸 = 4𝜋𝜋𝜋𝜋𝜋𝜋
• The point charge q is the only charge inside of our spherical surface, so the
constant of proportionality between flux and enclosed charge is 4𝜋𝜋𝜋𝜋. 𝑘𝑘 is
related to another constant known as the vacuum permittivity: 𝜀𝜀0 =
• And finally we have
1
4𝜋𝜋𝜋𝜋
��⃗ = 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒
Gauss’s law: ∮ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
𝜀𝜀
0
• Don’t worry about the physical significance of the constants 𝜀𝜀0 and 𝑘𝑘 (they
have none) – the fact that there even is a constant of proportionality is
merely a consequence of the system of units chosen (this is the same reason
there is a constant of proportionality G in the force law for gravitation). And
the fact that there are two constants is for purely historical reasons. It
probably would have been better to define the unit of charge in such a way
that k = 1, and there are systems of units that do so but SI is the
internationally accepted system.
• Gauss’s law is one of the four fundamental equations that describe the
behavior of electromagnetic fields. It applies to all electric fields, no matter
how complex the charge distributions producing them.
• The integral can be computed over any closed surface you want and 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 is
always the net charge enclosed by that particular surface
• NOTE: A subtle point to recognize regarding Gauss’s law – ��⃗
𝑬𝑬 is due to all
the charges that are present, whether they are inside or outside the closed
surface. Nonetheless, the result of the integral will always be proportional to
only the net charge inside the closed surface.
• Note that for a spherical surface with a point charge located at its center, the
flux does not depend on the radius of the sphere. This makes sense, though,
considering that as r increases, the electric field decreases at a rate ∝ 1/r2
while the surface area increases at a rate ∝ r2. These two effects cancel each
other and the flux is the same regardless of the size of the spherical surface
you choose.
**CONCEPTUAL QUESTION: A spherical surface surrounds a point charge. If a
second point charge is brought very close to the sphere, will the flux thru the
surface change? Will the electric field on the surface change (either in direction or
magnitude)?
USING GAUSS’s LAW TO CALCULATE ELECTRIC FIELDS
• It must be emphasized again that Gauss’s law holds for any surface
enclosing any charge distribution
• However it is only practical for calculating electric fields for charge
distributions with sufficient symmetry
SPHERICAL SYMMETRY
• In order for a charge distribution to have spherical symmetry, the charge
density can only depend on the distance (not the direction) from some
central point. So, anywhere on a spherical surface of a given radius, the
charge density has a constant value.
• In this case, the appropriate gaussian surface to use is a sphere of some
radius r centered on the point of symmetry. On a surface such as this, two
important simplifications can be made to the surface integral.
1. Because of spherical symmetry, the electric field either points radially
outward or radially toward the point of symmetry. The area vector always
��⃗
points radially outward from a closed sphere. This means that ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
either equals EdA (outward) or –EdA (toward).
2. Also because of spherical symmetry, the electric field has constant
magnitude everywhere on the sphere so we can pull it outside of the
surface integral.
��⃗ =
• With these two simplifications, Gauss’s law becomes ∮ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
∮ ±𝐸𝐸𝐸𝐸𝐸𝐸 = ±𝐸𝐸 ∮ 𝑑𝑑𝑑𝑑 = ±4𝜋𝜋𝑟𝑟 2 𝐸𝐸
**SEE EXAMPLE #1 and #2**
• STRATEGIES FOR APPLYING GAUSS’s LAW
1. Examine the symmetry of the problem (spherical, line, or planar) and
construct a gaussian surface on which both the magnitude of the field and
its direction relative to the surface is constant.
2. Calculate the flux. Based on #1, 𝐸𝐸 cos 𝜙𝜙 should be constant on the
gaussian surface so you can pull it outside of the integral leaving only
𝑑𝑑𝑑𝑑, the integral of which equals the surface area.
3. Calculate the amount of charge enclosed by the gaussian surface. This is
not necessarily equal to the total charge if the gaussian surface lies inside
the charge distribution.
4. Set the flux equal to the enclosed charge divided by 𝜀𝜀0 .
5. Solve for the magnitude of the electric field. Its direction should be
obvious from the symmetry.
LINE SYMMETRY
• This type of symmetry is sometimes called cylindrical symmetry
• For a charge distribution to have this type of symmetry, the charge density
can only depend on the perpendicular distance from some line, called the
axis of symmetry. As a result, the magnitude of the electric field has the
same dependence and, because of line symmetry, must point radially toward
or away from the axis of symmetry.
• The appropriate gaussian surface in this case is a cylinder of radius r and
length l, oriented such that the axis of symmetry passes through the centers
of the circular ends.
��⃗ either equals EdA or –EdA
• As with the case of spherical symmetry, ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
and E is constant in magnitude on the curved part of the cylindrical surface.
��⃗ are perpendicular on the circular ends so cos 𝜙𝜙 = 0 everywhere
• ��⃗
𝑬𝑬 and 𝑑𝑑𝑨𝑨
and there is no flux through the ends
• So, once again the flux is the product of the magnitude of the electric field
and the surface area. Only the curved surface contributes to the flux so
𝐴𝐴 = 2𝜋𝜋𝜋𝜋𝜋𝜋
��⃗ = ±2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋. As with spherical symmetry, the
• Gauss’s law becomes ∮ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
positive sign is for an outwardly directed field and the negative sign is for an
inwardly directed field.
**SEE EXAMPLE #3**
PLANAR SYMMETRY
• For a charge distribution to have planar symmetry, the charge density can
only depend on the distance from a plane. Thus the magnitude of the electric
field has this same dependence and the only direction it can have that
respects this symmetry is perpendicular to the orientation of the plane of
symmetry.
• We need a gaussian surface that extends equal distances on either side of the
symmetry plane and whose ends are perpendicular to the electric field and
whose sides are parallel to the field. A cylinder will do the trick.
• No field lines cross the sides of the cylinder so the flux across this part of the
closed surface is zero. Because of the symmetry, the electric field cannot
depend on position parallel to the symmetry plane, so it is constant over the
ends of the gaussian cylinder and is also perpendicular to these two surfaces.
• The flux through each end is therefore 𝐸𝐸𝐸𝐸 and the total flux through the
��⃗ = 2𝐸𝐸𝐸𝐸.
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
entire closed surface is Φ𝐸𝐸 = ∮ ��⃗
• Now you find the charge enclosed by the gaussian surface, which is only the
portion of the planar charge distribution defined by the circular intersection
of it with the imaginary cylinder.
• Then solve for the electric field.
**SEE EXAMPLE #4**
• Gauss’s law is really only a practical tool for calculating the electric field
when a gaussian surface can be constructed which encloses the charge
distribution such that 𝐸𝐸 cos 𝜙𝜙 has a constant value on all portions of the
closed surface (not necessarily the same value on each portion of the
surface). If this is possible, we are able to take 𝐸𝐸 cos 𝜙𝜙 outside of the surface
integral and solve for the electric field strength.
• The fields due to more complicated charge distributions can sometimes be
understood by considering the fields of simpler distributions that you have
calculated using either Coulomb’s law or Gauss’s law
CONDUCTORS
• In the previous chapter we defined a conductor as a material containing
charges which are free to move about. The figure below shows a conducting
sphere which is then placed in a region where there is a uniform electric
field
• The free charges are then accelerated by the field – positive charges are
pushed in the direction of the field and negative charges are pushed opposite
the direction of the field. This charge separation creates an internal field in a
direction opposite to the applied field.
• As more and more charge builds up, the internal field that the charge
separation creates gets larger and larger until eventually it cancels the
applied field inside the conductor (see figure), the free charges experience
zero net force from that point onward, and the conductor reaches
electrostatic equilibrium.
For a conductor in electrostatic equilibrium, the electric field is zero
everywhere inside
• If this were not true and there were a nonzero electric field inside, then
because there are free charges, there would be a net motion of charge and the
conductor would not be in electrostatic equilibrium.
• Normally a conductor, like most objects, is electrically neutral – it has the
same number of protons and electrons. But we can add excess charge to the
conductor and even if we inject, say, excess electrons into the interior,
because of the repulsive electric force, they will move as far apart from each
other as possible. So one might expect them all to reside on the surface of
the conductor.
• We can show this must be the case using Gauss’s law. Imagine we have a
piece of charged conducting material and we place our gaussian surface near
the actual surface of the conductor (see figure below).
• The electric field is zero everywhere inside the conductor if it is in
��⃗ is also
electrostatic equilibrium, so the flux through this surface ∮ ��⃗
𝑬𝑬 ⋅ 𝑑𝑑𝑨𝑨
zero. This, in turn, means that the gaussian surface cannot enclose any of the
excess charge.
• We can take our gaussian surface to be as close as we like to the actual
surface of the conductor, and, as long as it is inside the conductor, the flux
will always be zero. Thus we are forced to conclude that any excess charge
must reside on the surface.
If a conductor in electrostatic equilibrium carries a net charge, the excess
charge must lie entirely on the conductor’s surface
• There cannot be any electric field inside a conductor, but there can be an
electric field right on the conductor’s surface. However, this electric field
cannot have a component tangent to the surface. If it did, it would exert a
force on the free charges at the surface and cause them to move, violating
the assumption that the conductor is in electrostatic equilibrium.
The electric field right at the surface of a charged conductor in equilibrium
must be perpendicular to the conductor’s surface
• Although any excess charge on a conductor will be found only at the
surface, it may not be uniformly distributed
• At sharp points, the density of the charge is higher, and therefore the electric
field is stronger. Regions of the surface with a smaller curvature (more
smooth) have a lower charge density and the field is weaker here.
• This is depicted in the figure below – where the field strength is stronger, the
field lines are drawn closer together and where the field strength is weaker,
the field lines are more spaced out.
• Below, in figure (a), we have a conductor with excess charge qC. We have
already shown that the electric field inside must be zero and that any excess
charge must be located entirely on the surface.
• But what if there is a cavity inside the conductor, as in figure (b)? If we
place a gaussian surface in between the outer surface of the conductor and
the inner surface of the cavity, the electric field must be zero everywhere on
this imaginary surface. This means that the flux across this surface is also
zero and therefore there cannot be any net charge inside. The only place
there could be a net charge is on the inner surface of the cavity, but we have
shown there cannot be a net charge anywhere inside the gaussian surface.
• Finally, what if we place a point charge +q inside the cavity, as in figure (c)?
The flux across our gaussian surface is still zero, so there must be zero net
charge inside of it. The only place there could be a –q charge to cancel out
the +q point charge is on the inner surface of the cavity, as shown in the
figure. If we place a gaussian surface outside of the conductor, we enclose a
net charge of qC + q, which must reside entirely on the outer surface of the
conductor.
**SEE EXAMPLE #5**
• A conducting box (often called a Faraday cage – see figure below) can be
used to exclude electric fields from a region of space; this is called
electrostatic shielding
• Metal walls are ideal for screening, but wire screen or mesh can be used as
well
• This is also why it is safe inside a metal vehicle during a lightning storm…as
long as you are not touching anything metal inside the car that is in any way
connected to the body
**SEE EXAMPLE #6**
**SEE EXAMPLE #7**