GAUSS'S LAW
Electric Flux over a Closed Surface = Charge enclosed by the Surface divided by o.
(o = the permittivity of free space 8.854x10-12 C2/(N m2) = 1/4k)


A more intuitive statement: the total number of electric field line entering or leaving a closed
volume of space is directly proportional to the charge enclosed by the volume.
If there is no net charge inside some volume of space then the electric flux over the surface of that
volume is always equal to zero.
Ideal Conductors:

An Ideal Conductor is typically a metal object in which any excess charges are free to move
around inside or on the surface of the conductor.
Gauss's Law shows that at Equilibrium (when the charges are at rest):
a. any excess charge resides only on the surface of the conductor.
b. the Electric Field at every point inside a conductor is always Zero.
c. the Electric Field always emerges from (or enters into) a conductor perpendicular to the surface.
d. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius
of curvature of the surface is smallest.
e. the magnitude of the Electric Field near the surface at a point where the Surface Charge Distribution at
that point is , is equal to
E = /o
Ideal Conductor
A metal Conductor is simulated with an excess of positive charge. An external charge magnitude
can be varied to observe the behavior of the charge on the surface of the conductor.
Verbal discussion of the above points:
a.
Any excess charges have the same sign and thus repel each other. The charges arrange themselves on the
surface so that they have the maximum separation between them.
b.
If the electric field inside a conductor were not zero then any free charges would move. Alternately, if the
excess charges are on the surface of the conductor, then any Gaussian surface inside the conductor would
have no charge in it. Thus Gauss's law shows that E must be zero everywhere inside the conductor.
It is useful to envision that the excess charges arrange themselves in such ways that the vector
sum of their combined electric fields inside the conductor always adds up to zero. This means
that the surface charge density becomes larger in regions where there is a larger surface
curvature, i.e. the greater the curvature of the surface, the larger the concentration of charges.
c.
If the electric field were not perpendicular to the surface of the conductor, then there would be some
component of the E-field parallel to the surface. The excess charges would move along the surface until
the arrange themselves so the their parallel E-fields cancel out and the come to rest.
d.
If you get infinitesimally close to the surface of a conductor, then it will look like a very large flat plate
with a surface charge density of  at that location. Applying Gauss's Law to a cylinder near the surface
we get,
MICROWAVE DEMOS
Microwaves are a type of wave that are sandwiched between radio waves and infrared radiation
on the electromagnetic spectrum. In the case of microwave ovens, the commonly used wave
frequency is roughly 2,450 megahertz (2.45 gigahertz).
Waves in this frequency range have an interesting property: They're absorbed by water, fats and
sugars. Once absorbed, they're converted directly into atomic motion -- heat. These waves boast
another interesting, related property, too: They're not absorbed by most plastics, glass or
ceramics. Metal reflects microwaves, which is why metal pans don't work well in a microwave
oven. It's also why the devices have metal walls -- for reflection.
Fork? Predict what happens:
CD? Predict what happens:
Lit toothpick: Predict what happens:
GAUSS’S LAW CALCULATIONS
USING GAUSS’ LAW TO CALCULATE ELECTRIC FIELDS
Gauss’ law can be stated mathematically by means of the expression
  q
 e   E • dA  in
0
-12
2
2
where  0 is the permittivity of free space (  0 = 7.85  10 C /N•m ). Gauss’ law is typically used to
compute the electric field at some distance from a uniform charge distribution that is symmetric. We will
be interested in two types of symmetry—cylindrical symmetry and spherical symmetry. Learning to
apply Gauss’ law will take some practice. The key is to pick a closed surface with the same symmetry as
the electric field associated with a charge distribution. Thus, a spherical surface works for point charges
and spherical distributions and a cylindrical surface works for line charges and cylindrical distributions.
Geometry Review: Circles, Spheres, and Cylinders
Before we wade into the use of Gauss’ law, let’s review some geometry for some fairly simple symmetric
shapes.
Activity: Some Geometry of Circles and Spheres
a. What is the equation for the circumference of a circle of radius r? If the radius doubles, what
happens to the circumference?
b. What is the equation for the area of a circle of radius r? If the radius doubles, what happens to
the area?
c. What is the equation for the volume of a sphere of radius r? If the radius doubles, what happens
to the volume?
d. Find the derivative of the volume V as a function of r. Show how this derivative can be used to
determine how much the volume of a sphere would increase (that is, the
factor dV) if the radius of the sphere were increased from r to r + dr.
Hint: Consider this increase as being the volume of a shell of thickness dr
surrounding the sphere.
e. If the letter S is used to represent the surface area of a sphere, what is the volume dV of a thin
shell of thickness dr that surrounds a sphere of radius r in terms of S and dr?
f. Use the derivative dV/dr from part d. and the idea that a spherical shell represents a volume
2
increase to show that the surface area of a sphere can be represented by the equation S = 4r .
g. If the radius of a sphere doubles, how much does its surface area increase?
h. If a charge Q is spread uniformly throughout the volume of an insulating sphere (that is, charge
cannot move around inside it) of radius r, what fraction of the charge lies within a radius of r/2?
Warning: The answer is not 1/2.
Spherical Symmetry, Charge Density, and Gauss’ Law
After the geometry review you just completed, you should be ready to use Gauss’ law to find the equation
describing the electric field at a distance r from a point charge. You should also be able to derive an
equation for the electric field at a distance r from an insulator with a uniform, spherically symmetric
charge distribution. You can find both these derivations in almost any calculus-based introductory
physics text.
The concept of charge density is very useful in figuring out how much charge is contained within
a given radius r in a charged sphere. Before you get started with your derivations of E-field equations,
take a moment to read about the concept of charge density.
If a volume element dV contains a small amount of charge dq, then the charge density is given by
the equation:

dq
dV
Thus, if the charge is distributed in a spherically symmetric manner, the amount of charge
contained within a radius r is given by:
r
r
r
0
0
0
q   dV   4r 2 dr  4  r 2 dr
In the case where the charge is uniformly distributed,  is a constant given by:

3Q
4R 3
where Q is the total charge in the sphere and R is the sphere’s radius. In this case, then, the charge
contained within a radius r is given by:
q

r
0
r
dV  4  r 2dr
0
Gauss’ Law and the Point Charge
Let’s begin by using Gauss’ law in the form
  q
 e   E • dA  in
0
to find the electric field magnitude at any distance, r, from a point charge +q.
Activity: Gauss’ Law and a Point Charge
a. Write down the general expression for Gauss’ law.


b. Explain why the dot product E  dA can be replaced with the expression EdA inside the integral
where E is the magnitude of the electric field and dA is the magnitude of an element
 of area on
the surface of the spherical shell. Hint: What is the angle between the vector E and the vector
dA at any point on the spherical surface?
c. Explain why the magnitude of the electric field is the same at all points on the spherical shell.
Hint: Think of Coulomb’s law and the definition of electric field.
d. If the magnitude of the electric field, E, is the same at all points on the spherical surface, explain
why E can be factored out of the integral


.
E

dA

e. Since 4r is the equation for the area of the surface of a sphere, what is the equation for dA?
2
f. Finally, what is the magnitude of the electric field, E, as a function of the central charge, q, and
the distance from it, r?
Gauss’ Law and the Spherically Symmetric Charge Distribution
Use some of the ideas from the last two activities about charge density and about using Gauss’ law to find
the magnitude of the electric field near a point charge to find the values of the magnitude of electric field
in the vicinity of a spherically symmetric distribution of charges.
In the next activity you are to compute the magnitude of the electric field at a distance r from the
center of a uniformly charged sphere of radius R with a total charge of Q throughout its volume, where r
< R (that is, the point in question is inside the sphere!).
Activity: Gauss’ Law and Spherically Symmetric Charge Distributions
a. To use Gauss’ Law to find the electric field, you’ll need to know the charge enclosed inside a
sphere of radius r in which charges are distributed uniformly through the larger sphere of radius
3
R. Start by using the equation for the volume of a sphere of radius r (V = 4/3 r ) to show that the
charge q that lies within a radius r is given by
r3
qQ 3
R
b. Next, use Gauss’s law along with the equation you just derived to show that the magnitude of the
electric field inside a uniformly charged sphere of radius R having a total charge of Q is

E E 
1
4 0 

Qr
1
where k e 
3
4 0
R
More microwave fun:
Steel Wool: Predict what happens:
Soap: Predict what happens:
Christmas Tree Ornament: Predict what happens:
http://media.pearsoncmg.com/bc/aw_young_physics_11/pt2a/Media/Electricity/1108GaussLaw/Main.htm
l
Question 1: Electric Field Outside a Charged Sphere
A solid sphere of 10-mm radius has +10 x 10-10 C of electric charge distributed uniformly
throughout the sphere. Use Gauss's law to determine the electric field caused by this charge at a
distance of 15 mm from the center of the sphere. The answer is shown in the meter.
Question 2: Electric Field Outside a Charged Spherical Shell:
A 10-mm radius spherical shell, like a basketball, has +10 x 10-10 C of electric charge distributed
uniformly on its surface. Use Gauss' law to determine the electric field caused by this charge at a
distance of 15 mm from the center of the sphere. After your prediction, you can click the "Shell"
Object type and set R = 15 mm and compare the answer with your prediction.
Question 3: Electric field inside the charged spherical shell
Leave the Object type on "Shell." Use Gauss' law to predict the value of the electric field inside
the spherical shell. If you want, you can calculate the magnitude of the electric field at R = 5 mm
from the center of the 10-mm radius spherical shell. After your prediction, move the R slider to 5
mm to check your prediction.
Question 4: Electric Field Inside a Charged Solid Sphere
Predict the value of the electric field 5.0 mm from the center of a 10-mm radius solid sphere. The
sphere is uniformly charged with +10 x 10-10 C of electric charge. (a) First, determine the electric
charge inside a 5.0 mm radius Gaussian surface. (b) Then, use Gauss' law to determine the
electric field 5.0 mm from the center of the charged sphere. When finished, you can check your
result with the simulation.
Question 5: Electric Field Inside a Charged Solid Sphere
Show that the electric field INSIDE the solid uniformly charged sphere varies as:
E = [k Q/(Rcharged sphere)2](R/Rcharged sphere)
where Rcharged sphere is the radius of the charged sphere (10 mm in the simulation), Q is the total
charge on the sphere (adjustable with the Q slider), and R is the distance from the center of the
sphere at the position where the electric field is being calculated.
Activity: Some Geometry of Cylinders
a. Consider a cylinder of radius r and length L. What is its volume in
terms of , r, and L?
.
b. If a charge Q is spread uniformly throughout the volume of a cylinder of radius r and length L,
what fraction of the charge lies within a radius of r/2? Warning: The answer is not 1/2.
c. What is the surface area of the cylinder in terms of , r, and L? Hint: Don’t neglect the ends.
What happens to the electric field when there is cylindrical symmetry? Can Gauss’ law be used to find
the electric field at a distance r from an infinite line of electrical charges? How about the electric field at
a distance r from an infinitely long insulator with a uniform charge distribution that has cylindrical
symmetry? Consult an introductory text for hints if necessary.
Activity: Gauss’ Law and Cylindrical Symmetry
a. Use Gauss’ law to calculate the electric field at a distance r from a very long, straight, uniformly
charged cylinder that has a charge per unit length of . (Mathematically, we can treat a wire
which is physically very long as if it were infinitely long.) Hint: Using a symmetry argument,
explain why you can neglect the electric field perpendicular to the two ends of the cylinder.
–9
b. Assume that the charge per unit length  = 5.00  10 C/m. Plug some numbers into the
equation you derived to find the magnitude of the electric field at a distance of 5.00 cm from the
line of charge. As usual, you should express your result using the appropriate number of
significant figures.
-12
Hint: 0 = 7.85  10
2
2
C /N • m
9
2
2
ke = 1/(40) = 7.99  10 N • m /C
c. What is the direction of the electric field? Explain the reasons for your answer.
d. How do we know that the E-field lines are perpendicular to the wire (that they have no
component that is parallel to the wire)?
GAUSS’ LAW FOR ELECTRICAL AND GRAVITATIONAL FORCES
Gauss’ law states that the net electric flux through any closed surface is equal to the net charge inside the
surface divided by 0 where 0 = 1/(4ke). Mathematically this is represented by an integral of the dot
product of the electric field and the normal to each element of area over the closed surface:
  q
 c   E • dA  enclosed
0
The fact that electric field lines spread out so that their density (and hence the strength of the
electric field) decreases at the same rate that the area of an enclosing surface increases can ultimately be
2
derived from the 1/r dependence of electrical force on distance. Thus, Gauss’ law should also apply to
gravitational forces.
Activity: The Gravitational Gauss’ Law
State a gravitational version of Gauss’ law inwords and then represent it with an equation. Hint:


If the electric field vector, E , is defined as Fe q , let’s define the gravitational field vector Y as

Fg m by analogy.
We can use the new version of Gauss’ law to calculate the gravitational field at some distance from the
surface of the earth just as we can use the electrical Gauss’ law to determine the electric field at some
distance from a uniformly charged sphere. This is useful in figuring out the familiar force “due to
gravity” near the surface of the earth and at other locations.
Activity: The Gravitational Force of the Earth

a. Use Gauss’ law to show that the magnitude of the gravitational field, Y , at a height h above the
2
surface of the earth is given by GMe/(Re + h) . Hints: How much mass is enclosed by a

spherical shell of radius Re + h? Does Y have a constant magnitude everywhere on the surface
of the spherical shell? Why? If so, can you pull it out of the Gauss’ law integral?

b. Calculate the gravitational field, Y , at the surface of the earth
3
24
(h = 0). The radius of the earth is Re  6.38  10 km and its mass Me  5.98  10 kg. Does the

result look familiar? How is Y related to the gravitational acceleration ag?
E-field of Non-Conducting Spherical Charge Distribution Using Gauss's Law
Problem
A non-conducting sphere of radius R = 2.90 m has a total charge Q = 1.40 mC distributed through out the
interior of the sphere in such a way that the charge density increases according to
.
(A) Determine the value of the constant  and its units.
(B) Determine the magnitude and direction of the electric field both inside and outside the sphere.
Sketch and Process:
A spherical volume is filled with fixed charges that radially increase from the center of the
sphere in a uniform manor.
Relevant Physics:
The key to using Gauss's law is to find a surface over which the electric field is constant in both
magnitude and direction. This allows us to take the E-field out of the integral for the electric flux so that
the electric flux is just the product of E and the area of the surface A, E = E A.
Since the charge distribution  in this problem is only a function of r, the electric field due to this
charge distribution at any point (inside or outside the charged sphere) will also be radially
outward for a positive charge distribution. Thus, a sphere (with any radius r) is a Gaussian
surface over which the E-field is not only constant but is also parallel to the surface normal.
Separate from above is calculating the charge inside the Gaussian surface chosen. Inside a
spherical volume of radius r, where r <= R , the charge can be found using the charge
distribution formula,
The volume of a sphere is given by V = 4/3  r3. Taking the differential of both sides we can find
the differential volume, dV = 4  r2 dr. Note that is just the surface area of a sphere (A = 4  r2)
times the thickness of a differential shell dr.
For a radius r less than that of the non-conducting sphere of radius R, the charge will be less than
the total charge in the non-conducting sphere. If r > R, then the charge inside the Gaussian
sphere will be the same as that in the non-conducting sphere of radius R
Gauss's Law applied to a spherical volume would become
Observe that if the charge distribution had been only a point charge at the origin, then this last
formula would go over to that we already found for the E-field of point charge using Coulomb's
Law. This is why we say that Gauss's Law is a more general form of Coulomb's Law in that
Coulomb's Law is a specal case of Gauss's Law.
(A) Find .
Since we know the total charge in the whole sphere QTotal and the charge distribution (r)
Solving for 
(B) Find E
Since the charge distribution is spherical, the E-field will be radial outwards or inwards. In addation, the
charge distribution is positive so that the E-field is radial outwards.
It is easy to prove this outside the sphere. Imagine the sphere to be made up an infinite number of
circular rings perpendicular to line between the origin and some point. For every charge on one
side of a ring there is another charge on the opposite side of that ring. The perpendicular
components of the E-field cancel leaving only the components parallel to the line, i.e. the radial
components.
Inside: Applying Gauss's Law at a distance r < R inside the sphere, the electric field at that
distance E = E(r) can be found since we can also find the charge Q(r) inside this sphere of
radius r < R since we are also given the charge distribution (r).
Using the results of part A we can express this in terms of the total charge in the sphere Q =
QTotal,
This is not a general result for spherical charge distributions. For example, if the charge density
had been constant throughout the sphere so that (r) = constant, then the E-field would have
been a linear function of the radius instead of a quadratic.
Outside: Here we pick a sphere with r > R as our Gaussian surface. All the charge Q in the
sphere is inside this Gaussian surface where r > R,
Not surprisingly the E-field will look like the E-field of a point charge Q at the origin.
Graphically this looks like,
Final Thoughts: You are trapped in a lightning storm in your car.
What is your best course of action. Give a detailed description
to support your answer.
1) Get out of the car and run to the nearest tree
2) Get out of the car and lie down flat on the
ground.
3) Seek the highest point nearby and put up your
umbrella
4) Seek the lowest point nearby preferably a ditch
or ravine.
5) Stay in your car