Day 10 Dipole Moments and Flux

Day 9: Dipole moments and Electric Flux
Motion of a Charged Particle in a Uniform Electric Field
When a particle of charge q and mass m is placed in an electric field E, the electric
force exerted on the charge is qE. If that is the only force exerted on the particle, it
must be the net force, and it causes the particle to accelerate according to the particle
under a net force model. Therefore,
Fe= qE = ma
and the acceleration of the particle is
Given a pair of plates one of which is + charged and the other is
negatively charged. A electron is fired horizontally between the two
plates as shown.
Draw the electric field and the trajectory of the electron. What 2A topic does this remind you of?
Electric Dipole in an Electric Field
We defined an electric dipole as consisting of two charges of equal
magnitude and opposite sign separated by a distance 2a as shown at right.
We want to define a quantity that represents the electrical polarity of a of a
system of charges which we will call the Electric Moment.
The electric dipole moment is defined as the product of the magnitude of
charge and the distance of separation between the charges.
In this case it is the vector p directed from -q toward +q along the line
joining the charges and having magnitude
p = 2aq
Now suppose an electric dipole is placed in a uniform electric field E and makes an angle  with the field
as shown at right. We identify E as the field external to the dipole,
established by some other charge distribution, to distinguish it from the
field due to the dipole.
Each of the charges is modeled as a particle in an electric field.
Make a sketch showing the force on each of these particles and include the
net force on the dipole.
Therefore, the net force on the dipole is zero.
Will there be torque acting on this system? Remember that there is some angle between the F and the
moment vector as shown at right. Write three equations:
1. The torque on the positive charge due to the electric field
2. The torque on the negative charge due to the electric field
3. The net torque on the dipole.
The two forces produce a net torque on the dipole, however; the dipole is therefore described by the rigid
object under a net torque model. As a result, the dipole rotates in the direction that brings the dipole
moment vector into greater alignment with the field. The torque due to the force on the positive charge
about an axis through O in has magnitude Fa sin , where a sin  is the moment arm of F about O. This
force tends to produce a clockwise rotation. The torque about O on the negative charge is also of
magnitude Fa sin ; here again, the force tends to produce a clockwise rotation. Therefore, the magnitude
of the net torque about O is
= 2Fasin 
Now write this torque as a function of q, E,  and a.
Note several of these things are difficult to measure. However we can use our definition of the diplole
moment as p = 2aq to simplify this further.
 = pEsin 
Based on this expression, it is convenient to express the torque in vector form as the cross product of the
vectors p and E:
We can also model the system of the dipole and the external electric field as an isolated system for energy.
Let’s determine the potential energy of the system as a function of the dipole’s orientation with respect to
the field. To do so, recognize that work must be done by an external agent to rotate the dipole through an
angle so as to cause the dipole moment vector to become less aligned with the field. The work done is then
stored as electric potential energy in the system. Notice that this potential energy is associated with a
rotational configuration of the system. Previously, we have seen potential energies associated with
translational configurations: an object with mass was moved in a gravitational field, a charge was moved
in an electric field, or a spring was extended. Now we want the work associated with a rotation. Write an
expression for the work associated with a torque:
dW =  d
Our expression for torque previously was  = pEsin Plug this into the definition above and integrate it
between an angle of i and f to find an expression for the work done in rotating the dipole.
Note we have written this as a change in energy due to the work energy theorem.
Lets look at some convenient limits: Assume that you start with Ui = 0 at an angle of 90o. Now write an
expression for Uf.
Uf = -pE cos 
If you were particularly clever, you noticed that this is just the same as the dot product between the
moment and the electric field.
Lets compare this to the expression for the potential energy: mgh.
Rewriting mhg, we notice that the definition of the moment matches that from potential energy, with the
Electric field replacing the gravitational acceleration.
In the case of the gravitational field, it is a translational potential, such that a mass will tend to return to
the bottom position. For the electric field it is a rotational potential, such that the charge will tend to
return to an orientation where the charges are aligned with the field.
Molecules are said to be polarized when a separation exists between
the average position of the negative charges and the average position
of the positive charges in the molecule. In some molecules such as
water, this condition is always present; such molecules are called
polar molecules. Molecules that do not possess a permanent
polarization are called nonpolar molecules.
We can understand the permanent polarization of water by
inspecting the geometry of the water molecule. The oxygen atom in
the water molecule is bonded to the hydrogen atoms such that an
angle of 105o is formed between the two bonds. The center of the
negative charge distribution is near the oxygen atom, and the center
of the positive charge distribution lies at a
point midway along the line joining the
hydrogen atoms.
Washing with soap and water is a household
scenario in which the dipole structure of
water is exploited. Grease and oil are made
up of nonpolar molecules, which are
generally not attracted to water. Plain water
is not very useful for removing this type of
grime. Soap contains long molecules called
surfactants. In a long molecule, the polarity
characteristics of one end of the molecule
can be different from those at the other end.
In a surfactant molecule, one end acts like a
nonpolar molecule and the other acts like a
polar molecule. The nonpolar end can
attach to a grease or oil molecule, and the
polar end can attach to a water molecule.
Therefore, the soap serves as a chain,
linking the dirt and water molecules
together. When the water is rinsed away,
the grease and oil go with it.
Shown below in the photograph at the left and in the detail at the right is a device called a Faraday cage.
In the detail photograph three suspensions of aluminum foil chunks on pieces of string can be seen. Each
of the suspensions include a string passing over the top of the cylindrical screen with small rectangular
chunks of aluminum foil, a few centimeters in size, attached to the ends of the string. Note that one end of
each string is outside the cylinder and one end of the string is inside the cylinder.
The question this week involves what the foils do when the van de Graff generator is turned on, putting
large negative charge onto the cylindrical screen and perhaps by conduction onto the foils. In particular,
when the cylinder is charged, which, if any, of the aluminum foils will move away from the cylinder,
either inward or outward.
When the cylinder is charged:
(a) only the inner foils will move away from the cylinder.
(b) only the outer foils will move away from the cylinder.
(c) both the inner and the outer foils will move away from the cylinder.
(d) neither the inner nor the outer foils will move away from the cylinder.
The answer is (b): only the outer foils will move away from the cylinder, as can be seen in the photograph
at the right and in an mpeg video by clicking your mouse on the photograph at the right.
In a device like this, the electric charge moves to the outside of the cage, leaving no net charge on the
inside. Therefore no effect is felt by the foils on the inside of the cage.
This works the same way as a solid metallic cylinder would work, except that using a screen allows us to
see inside and observe the effect of charging the cylinder on foils both outside and inside the cage.
You have been representing the electric field due to a configuration of electric charges by an arrow that indicates
magnitude and direction; using the principles of superposition and linearity, you can determine the length and
direction of the arrow for each point in space. This is the conventional representation of a “vector field.” An
alternative representation of the vector field involves defining electric field lines. Unlike an electric field vector,
which is an arrow with magnitude and direction, electric field lines are continuous. You can use a simulation
written by Cal Tech to explore some of the properties of electric field lines for some simple situations.
For this activity you will need:
• 1 computer
• 1 Coulomb computer simulation
Activity: Simulation of Electric Field Lines from Point Charges
Load the Coulomb program into the computer. Figure out how to turn off everything except electric field.
a. Set a single charge of ±1, 2, 3, or 4 Coulombs somewhere on the computer screen and run the program.
Sketch the lines in the space below. Show the direction of the electric field on each line by placing arrows
on them. Indicate how much charge you used on the diagram.
How many lines are there in the drawing? Are the lines more dense or less dense near the charge?
Explain. How does the direction of the lines depend on the sign of the charge?
b. Try another magnitude of charge. No need to sketch the result, but how many lines are there? Can you
describe the rule for telling how many lines will come out of, or start or end on, a charge in this simulation
if you know the magnitude of the charge in Coulombs?
c. Repeat the exercise using two charges with the same magnitude having unlike signs. Place them at two
different locations on the screen. After a few minutes sketch the lines in the space below. Indicate how
much charge you used on the diagram. Comment on why the lines are more or less dense near the charges.
How does the direction of the lines depend on the sign of the charge?
d. Summarize the properties of electric field lines. What does the number of lines signify? What does the
direction of a line at each point in space represent? What does the density of the lines reveal?
As you have just seen, we can think of an electrical charge as having a number of electric field lines, either
converging on it or diverging from it, that is proportional to the magnitude of its charge. We can now explore the
mathematics of enclosing charges with surfaces and seeing how many electric field lines pass through a given
surface. Electric flux is defined as a measure of the number of electric field lines passing through a surface. In
defining “flux” we are constructing a mental model of lines streaming out from the surface area surrounding each
unit of charge like streams of water or rays of light. Modern physicists do not really think of charges as having
anything real streaming out from them, but the mathematics that best describes the forces between charges is the
same as the mathematics that describes streams of water or rays of light. So, for now, let’s explore the behavior of
this model.
It should be obvious that the number of field lines passing through a surface depends on how that surface is
oriented relative to the lines. The orientation of a small surface of area A is usually defined as a vector that is
perpendicular to the surface and has a magnitude equal to the surface area. By convention, the normal vector points
away from the outside of the surface. The normal vector is pictured below for two small surfaces of area A and A'
respectively. In the picture it is assumed that the outside of each surface is white and the inside grey.
A diagram showing how the normal vector representing an area points from inside to outside. The inside of a surface is
shaded grey.
Activity: Drawing Normal Vectors
Use the definition of “normal to an area” given above to draw normals to the surfaces shown below. Let the
length of the normal vector in cm be equal in magnitude to each area in cm . Don’t just draw arrows of
arbitrary length.
By convention, if an electric field line passes from the inside to the outside of a surface, we say the flux is positive.
If the field line passes from the outside to the inside of a surface, the flux is negative.
How does the flux through a surface depend on the angle between the normal vector to the surface and the
electric field lines? In order to answer this question in a concrete way, you can make a mechanical model of some
electric field lines and of a surface. What happens to the electric flux as you rotate the surface at various angles
between 0 degrees (or 0 radians) and 180 degrees (or  radians) with respect to the electric field vectors? To make
your model you will need to arrange nails in a 10  10 array poking up at 1/4" intervals through a piece of
Styrofoam. The surface can be a copper loop painted white on the “outside.” You will need:
• Styrofoam or 3/8" plywood (5"  5" square)
• 100 nails, approximately 4" in length (mounted on the Styrofoam or plywood)
• 1 wire loop (4"  4" square)
• 1 paper, 5"  5" with 1/4" graph rulings (to affix to the mounting square to help with spacing the nails)
• 1 protractor
Once the model is made you can perform the measurements with a protractor and enter the angle in radians
and the flux into a computer data table for graphical analysis.
Apparatus designed to determine how many uniformly spaced flux lines will pass
through an imaginary surface area as a function of the angle between the direction of
the flux lines and the normal vector representing the surface area.
Activity: Flux as a Function of Surface Angle
a. Use your mechanical model, a protractor, and some calculations to fill in the following data table, or you
may want to feed data directly into your graphing software for later display. Hint: Interested in being lazy
but clever? Use symmetry to determine the angles corresponding to negative flux without making more
Number of lines
b. Plot the flux  as a function of radians. Look very carefully at the data. Is it a line or a curve? What
mathematical function might describe the relationship?
c. Try to confirm your guess by constructing a spreadsheet model and overlay graph of the data and the
mathematical function you think matches the data. Affix or sketch the plots in the space to the right of the
data table. Summarize your procedures and conclusions in the following space.
d. What is the definition of the vector dot product of two vectors in terms of vector magnitudes andthe angle
 
between them? Can you relate the scalar value of the flux, , to the dot product of the vectors E and A ?
In the simulation, the yellow ring is the surface across which we determine the electric flux. The flux is
caused by the electric field, the green lines, caused by one or two electric charges. You can grab the edge
of the yellow ring and change its shape. If you move the pointer inside the ring and press down on the
mouse, you can drag the center of the ring to other positions on the screen.
Questions 1 – 4.
Question 1: Flux Into or Out of an Oval
Design your own experiments to see how electric charge affects the flux into or out of the oval. In your
experiments, you can:
• change the shape of the oval;
• move the center of the oval so that it surrounds the charge or does not surround the charge; and
• change the magnitude and sign of the electric charge.
When finished, develop in words a qualitative rule to determine the electric flux flowing into or out of the
oval. Give examples to support your statements.
Question 2: Electric Flux with Two Charges
In the simulation, click the "two charges" configuration. You can now adjust the sign, magnitude and
separation of two electric charges. Repeat the experiments such as done in Question 1 to see if your rule
applies for this two charge system. You can:
• change the shape of the ring;
• change the position of the center of the ring (move it all over); and
• change the magnitudes and signs of the electric charges.
Question 3: First way to determine electric flux
Change the simulation back to "one charge." The meter indicates the electric flux Φ into or out of the oval
and the net electric charge Q inside the oval.
• What happens to the flux if you double or triple the positive electric charge inside the oval?
• What happens to the flux if you double or triple the negative electric charge inside the oval?
• Find an equation with a proportionality constant that relates the electric flux into or out of the oval and
the electric charge inside the oval.
Question 4: Second way to determine electric flux
The green electric field lines represent the electric field surrounding the source charges. Develop a rule for
the electric flux passing out of or into the oval by counting the electric field lines passing out of or into the
One convenient way to
is to
 
use the dot product dA , so that   E  A . Flux is a scalar. If the electric field is not uniform or if the
surface subtends different angles with respect to the electric field
 must
 calculate the flux by
 lines, then we
breaking the surface into infinitely many infinitesimal areas dA so that d  E  A , and then taking the
integral sum of all of the flux elements. This gives
   d   E  dA (flux through a surface)
Some surfaces, like that of a sphere or that representing a rectangular box, are closed surfaces—
that is, they have no holes or breaks in them. Because we want to study the amount
of flux passing
through closed surfaces, there is a special notation to represent the integral of E  dA through a closed
surface. It is represented as follows:
 
   dA   E • dA (flux through a closed surface)
Question 5: Find the electric flux leaving a sphere that surrounds a +3.0 ?µC point charge. We use the
oval shown in Simulation 2 to help in our thinking. The definition for electric flux that we will use is:
where the integral is over a closed surface--a sphere in this example.
5a: Choosing a symmetric Gaussian surface. This integration method is almost always used with surfaces
that have considerable symmetry--a sphere in this example. Grab the side of the oval ring and change it
into a circle of any radius--as best you can. Center the circle about the point charge. Press down on the
pointer when it is outside the circle and move the pointer onto the circle and observe the electric field.
Move the pointer to all sides of the ring and observe the electric field. If it is in fact a circle centered on
the point charge, the electric field reading should be fairly constant (don't worry about minor variations).
If the field varies dramatically on different parts of the "circle", you will either have to reshape the circle
or center it better. When done, record the electric field on the circle and leave it as is.
5b: In the flux integral, what is the direction of
--a small part of the sphere's surface area?
5c: For the situation shown in the simulation, can you see why
for every element on the
5d: Why can you take E outside the integral--the last step below?
5e: Why can we now simply multiply the magnitude of the electric field at the surface times the total area
of the sphere's surface--the last step below?
= E (4πr2).
Summary of the Integral Method for Calculating Electric Flux:
• The direction of dA is perpendicular to the surface and outward from its inside.
• Often for the situations in which we calculate electric flux by integration, EdA = EdA because E and dA
are parallel.
• Often, we choose surfaces on which the electric field has the same magnitude at each point on the
surface, which allows us to move the electric field outside the integral sign:
• This leaves only an integral over the area which equals the area of the surface. We get:
= E (Area)
Question 6: Change the charge to Q1 = 4 µC and pull the yellow ring in the simulation off to the side so
that there is no charge inside it. Determine the electric flux leaving or entering the ring in the simulation
using the charge inside the ring and the number of flux lines crossing the ring surface methods (Methods 1
and 2). Compare your predictions with the value shown in the simulation.
Question 7: Click the "two charges" button and set Q1 = +4 µC and Q2 = -2 µC. Make the yellow oval
into a big circle that surrounds Q2 but not Q1 (see Question 7 in the worksheet). Determine the electric flux
leaving or entering the oval in the simulation using the charge inside the oval and the number of flux lines
crossing the oval surface methods (Methods 1 and 2). Check your predictions against the flux value shown
in the meter.
Question 8: Change the radius of the Gaussian sphere to R = 15 mm. Use the charge inside and integral
methods (Methods 1 and 3) to determine the electric flux leaving the 15 mm-radius sphere surrounding the
+10 x 10-10-C charge that is uniformly distributed in the 10 mm-radius solid sphere. The electric field at
the surface of the 15-mm radius sphere is 40,000 N/C (we learn later that 1 N/C = 1 V/m).
Question 9: Change the charge distributed in the solid sphere to -10.0 x 10-10 C. Repeat the calculations in
Question 8 only this time with the 15mm-radius sphere surrounding the -10 x 10-10 C charge. The electric
field at the surface of the 15-mm radius sphere is 40,000 N/C and points into the spherical surface.
Electric Flux:
The Electric Flux E is the product of Component of the Electric Field Perpendicular to a Surface times the
Surface Area.
(SI: V.m = N.m2/C)
Unit Vectors
The electric flux becomes negative when the direction of the E-field and the normal to the surface are in
opposite directions relative to each other, i.e. when  > 90o.
Flux over a Closed Surface:
A Closed Surface is the surface of some Spacial Volume like a sphere or a cylinder. The volume could have
any shape.
= Differential Area associated with the surface. Its direction is perpendicular (or normal) to the
If you can find a surface over which the E-field is constant then E can be taken out of the surface integral
and the electric flux reduces to a simple multiplication of E and the area of the surface.
Flux through a Cube
Consider a uniform electric field E oriented in the x direction in
empty space. A cube of edge length l, is placed in the field, oriented
as shown. Find the net electric flux through the surface of the cube.
a) What faces have a zero flux?
b) Write the integrals for the remaining faces.
c) Sum the flux over all six faces.
How is the flux passing through a closed surface related to the enclosed charge? Let’s pretend we live in a twodimensional world in which all charges and electric field lines are constrained to lie in a flat two-dimensional space—
of course, mathematicians call such a space a plane.1
For this project you will need:
• 1 computer
• 1 Coulomb software simulation†
Open the Coulomb program on the computer again and set it to sketch lines for some nutty, creative mix of
charges. Don’t be too creative or the lines will take forever to sketch out, limit to 3 charges of low magnitude. You
should do the following:
1. Open the Coulomb simulation and place some positive and negative charges at different places on the screen.
Then start the program to calculate and display the electric field lines in two dimensions.
2. Either sketch or print out the screen configuration showing the charges and the associated “E-field” lines. If
you fail at this, use the image below, noting that the bottom left charge is negative and the other two are
If you haven’t already read it, we recommend that you read E.A. Abbott’s book entitled Flatland; A Romance of Many
Dimensions (Dover, New York, 1952). It’s a delightful piece of late nineteenth-century political satire in the guide of a
mathematical spoof.
In a two-dimensional map of flux lines it is not possible to assign a fixed number of lines to a charge and to assign spacings
between coming from infinity unambiguously. Thus, the line densities may not look proper in some of the Coulomb software
plots. This should not matter to students completing this exercise.
3. Draw arrows on each of the lines indicating in what direction a small positive test charge would move. Note:
“Small” means that the test charge does not exert enough forces on the charge distributions that create the Efield to cause the field to change noticeably.
4. Figure out what the two-dimensional equivalent of a “closed surface” ought to look like and draw several
“closed surfaces” on your diagram.
Having done all of this preparation, you should be ready to discover how the net number of lines passing
through a surface is related to the net charge enclosed by the surface.
Activity: Gauss’ Law in Flatland
a. Place a replica of the charge configuration you designed and the associated field lines in the space
b. Draw some two-dimensional closed “surfaces” in pencil in the space above. Some of them should
enclose charge, and some should avoid enclosing charge. Count net flux lines coming out of each
“surface.” Note: Consider lines coming out of a surface as positive and lines going into a surface as
negative. The net number of lines is defined as the number of positive lines minus the number of
negative lines.
Charge enclosed by the
arbitrary surface
Lines of flux in and out
of the surface
c. What is the apparent relationship between the net flux and the net charge enclosed by a twodimensional “surface”?
Gauss’ Law in Three Dimensions
If you were to repeat the simulated exploration you just performed in a three-dimensional space, what do you
think would be the appropriate expression for Gauss’ law?
Activity: Statement of Gauss’ Law
a. Express the three-dimensional form of Gauss’ law in words.
b. Express the law using an equation.
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.
An electrical conductor is a material that has electrical charges in it that are free to move. If a charge in a
conductor experiences an electric field, it will move under the influence of that field since it is not bound (as it
would be in an insulator). Thus, we can conclude that if there are no moving charges inside a conductor, the
electric field in the conductor must be zero.
Let’s consider a conductor that has been touched by a charged plastic rod so that it has an excess of
negative charge on it. Where does this charge go if it is free to move? Is it distributed uniformly throughout the
conductor? If we know that E = 0 inside a conductor, we can use Gauss’ law to figure out where the excess
charge on the conductor is located.
Diagram showing how an imaginary Gaussian surface can be constructed just inside
the surface of the conductor.
Activity: Where Is the Excess Charge in a Metal?
a. Consider a conductor with an excess charge of Q. If there is no electric field inside the conductor (by
definition), then what is the amount of excess charge enclosed by the Gaussian surface just inside the
surface of the conductor? Hint: Use the three-dimensional Gauss’ law here.
b. If the conductor has excess charge and it can’t be inside the Gaussian surface according to Gauss’ law,
then what’s the only place the charge can be?
c. Given the fact that as like charges, the excess charges will repel each other, is the conclusion you
reached in part b. above physically reasonable? Explain. Hint: How can each unit of excess charge
that is repelling every other unit of excess charge get as far away as possible from the other excess
charges on the conductor?
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