The modern world as we know it would not be possible without electricity and magnetism. Television, computers, circuits, microwaves, etc. all rely on the principles of electricity and magnetism. I doubt if the scientists working to understand these principles in the 1700's and 1800's ever dreamed of the technology we have today.
A simple way to ``charge" an object is to rub it. For example if you rub a glass or plastic rod with silk, some of the electrons get rubbed off onto the silk. As a result the rod becomes positively charged and the silk becomes negatively charged. Another example is combing your hair. Now there are more protons than electrons in your hair, so it's positively charged. The comb, on the other hand, has more electrons and is negatively charged. You can then use the charged comb to pick up bits of paper which are slightly positively charged. (Actually the molecules in the paper are polarized by the charged comb. They are polarized in such a way as to be attracted to the comb. More on this later.)
Things don't like to be charged. They like to be neutral. The negatively charged comb attracts positively charged objects. The electrons from the comb want to jump to the positively charged hair. Once the hair and comb are neutral, there is no longer any desire to transfer electrons and there is no attraction. Jumping charges create sparks and lightning.
Charge is conserved: That means that charge doesn't spontaneously appear or disappear. Electrons can go from your hair to the comb. But if you lose 4 electrons from your hair, your comb gains 4 electrons.
Units: The MKS unit of charge is the Coulomb. One Coulomb (1C) is a lot of charge. A lightning bolt consists of 20C of charge flowing from sky to earth. An electron has a tiny amount of charge: . All electrons are the same; each has 1e of negative charge. Protons have 1e of positive charge. We say that charge is ``quantized" because it comes in these discrete packets. You can find objects with q = 1e, 2, e, or 10e, but never non-integer amounts of charge like 3.5e.
The opposite of conductors are insulators. Charge does not flow in insulators; it's stuck. Rubber and glass are examples of insulators. In these materials the electrons are stuck in covalent bands. You can charge insulators by rubbing them because they are good at ``holding charge." For example, you can charge a glass rod or a plastic comb by rubbing it.
Now suppose that we bring the positively charged rod near an insulator. Charge is not free to roam in an insulator but we can distort the charge distribution and polarize the insulator. Microscopically the electron cloud in an insulator molecule near the rod will be attracted toward the rod, leaving a deficit of negative charge in the part of the molecule that is further away from the rod. This deficit of negative charge will be positively charged. So the molecule will have more positive charge at one end than the other. We say that the rod has polarized the molecule. The induced charge distribution makes the molecule attracted toward the rod. This is how van der Waals interactions work. A polarized molecule induces other molecules to be polarized in such a way as to be attracted to the initially polarized molecule. This polarization explains why a comb that has been rubbed against your hair can attract bits of neutral paper.
Coulomb's Law
Let q denote the amount of charge, e.g., 3e.
The electrostatic force between charge q1
and q2 a distance r12 apart
=2.0 true in
Coulomb.eps
is given by Coulomb's law
The constant
, where
and is dimensionless.
is called the
permittivity constant and it has dimensions.
| = | |||
| = |
Principle of Superposition
Coulomb's law tells us the force
that charge q2 exerts on q1. What if we have more than 2
charges? Suppose we have 5 charges.
Direction of the force in Coulomb's Law
We have said that charges q1 and q2 feel a force whose
magnitude is given by Coulomb's law.
=3.0 true in
axes.eps
Notice that the force
on q1 due to
q2 is equal and opposite to the force
on q2
due to q1.
=2.5 true in
twoChargeExmple.eps
Example: Suppose we have 2 point charges q1 and +e
located at
and q2= + 2e located at
. What is the
force felt by q1?
Is the force attractive or repulsive? It's repulsive.
Magnitude: First calculate the magnitude of the force
| = |
| = | |||
| = | |||
| = | d2 + d2 | ||
| = | |||
| = |
Method II: Vectors:
| = | |||
| = |
Now add a 3rd charge q3 = q2 = +2e at (d,0). What is the force on q1?
The principle of superposition tells us
| = | |||
| = | |||
| = |
| = | |||
| = |
| = | |||
| = | |||
| = |
Electric Field
Coulomb's law tells us that a charge q1 exerts a force on q2 a
distance r away. How does it exert a force without even touching q2?
We ``explain" this action-at-a-distance by saying that q1 sets up
an electric field around itself.
What do we mean by a field?
A field is any physical quantity that takes on different values at different
points in space (and maybe even time). Think of a topographic map that shows
a terrain. Each point (x,y) is associated with a height h(x,y) above
sea level. This is called a scalar field since only one number is
associated with (x,y). A vector field has a vector associated with
each point in space. For example, consider a river. At each point you can
assign a velocity
telling how fast the water is flowing and
in which direction.
is a velocity field; it is a vector field.
An electric field is also a vector field. At each point the electric
field
is the force that a positive unit magnitude test
charge q0 feels. A test charge is a spy charge. It feels the force
of the other charges but they don't feel it. (Test charges are always
positive.) If
, then we just divide by q0 to
get , thus
Field of a point charge
Since Coulomb's law says that the force
exerted on q0 by q is
, the electric field produced by the point
charge q is
Rules for drawing field lines
How can we visualize the fields. How can we draw them? We could draw vectors to represent what's happening:
A somewhat better representation is to connect the arrows to form field lines:
We can't tell the strength from the length of the arrows anymore, but we can from the density of lines. Close in where the field is strong the density is high. In 3 dimensions, we would have a pin cushion with the density of lines decreasing as 1/r2. These are some rules you should follow in drawing field line:
Superposition
We've seen the electric field produced by a point charge. What is the
field produced by more than one charge? According to the principle of
superposition, the force that a test charge q0 feels is the sum of
the forces produced by each of the real changes:
Electric Dipole
As a simple example let's consider 2 point charges: +q and -q(q>0)separated by a distance d. Both charges lie on the z axis. This
charge configuration is called an electric dipole. It has an electric dipole moment which is a vector whose magnitude is
qd. The direction points from the negative charge to the positive
charge. So
in this case.
=1.5 true in
dipole.eps
In general the dipole moment's magnitude is the charge times the distance
between the charges. (Notice that points opposite to .)
What is the electric field at a point along the z-axis? First let's determine the direction of . If we put a positive test charge q0 above the dipole, it is closer to +q than to -q. So it feels the repulsion of +q more than the attraction of -q. So . If we place q0 below the dipole on the z-axis, it feels the attraction of -q more than the repulsion of +qbecause it's closer to -q. So it is attracted to the dipole. The force and hence the electric field is in the direction. The field of a dipole looks like:
Notice that on the z-axis . But away from it the field lines are curved. How do we know that a point on z-axis the field has ? Why wouldn't tilt one way or the other? The answer is by symmetry. If it did tilt, which way would it tilt? Right? Left? Backwards? Forwards? Nothing in the problem favors any of these directions. So straight along the z-axis is the direction of . But if our point of observation is to the right , then could bend to the right . In other words our point of observation breaks the symmetry.
=2.5 true in
dipoleCalc.eps
Now let's calculate the magnitude of
for a point
on the z-axis. By superposition we add the fields
due to each charge:
=
=
=
=
=
Let's assume that the point of observation is far away from the dipole so that . Then we can expand the terms in [ ] by the binomial theorem (or equivalently, the Taylor expansion). Recall that the binomial theorem states
So let . Then we get
=
=
=
If we plug in
, we get
Notice that if we had only a point charge +q located at the origin, the electric field along the z-axis would be
Notice that the dipole's electric field falls off faster with distance than a single point charge . This is because far away from the dipole, the electric field of the minus charge kind of cancels the electric field of the positive charge: -q +q = 0.
Continuous Charge Distributions
Sometimes we have continuous
charge distributions rather than discrete charges. One way to think of a
continuous charge distribution is to imagine charged paint, i.e., paint
that has lots of positive (or negative) point charges dissolved in it.
If we paint a line or a ring then the amount of charge per unit length
Recipe to find from continuous charge distribution
Typical Problem: Given
, Find .
Recipe for solution:
| Etot,x | = | ||
| Etot,y | = | ||
| Etot,z | = | (1) |
Point Charge in a Uniform Field
So far we have been considering the electric field set up by charges or
charge distributions. What happens if a charge finds itself in an
electric field created by other charges? Answer: the charge feels a
force
| (2) |
Notice that the direction of the force depends on the sign of q. always points in the direction a positive charge wants to go.
=2.5 true in
PtchargeInE.eps
We refer to as the external field since it is not produced by qbut rather, acts on q. If the charge is free to move, then it will
accelerate according to
Millikan Oil Drop Experiment
Millikan used this to prove that charge was quantified. He shot charged oil drops into an field to counteract the gravitational field. In other words, he used the field to stop the drop from falling.
He knew m and g. (He knew the oil's density and could measure its size in a microscope.) So by measuring how big a field he needed, he could deduce the charge. More precisely, he found that charge was quantized.
Ink Jet Printing
A moving particle is affected in the same way by an field:
. Thus a charged particle passing
through an field is deflected.
=2.5 true in
inkjet.eps
Which particle is positively charged and which is negatively charged?
Gauss' Law
Gauss' law is a way of formulating Coulomb's law that makes it easy
to find given a charge distribution
if the
charge distribution has some symmetry, e.g. spherical, cylindrical,
planar, etc. To understand Gauss' Law, we need to understand what
flux is.
Flux
Recall that when we talked about field lines, we said that a high
density of field lines meant was strong and a low density
meant that was weak. The flux through a surface is
proportional to the number of field lines piercing a surface. Here
are some analogies:
For an field, we can think of field lines piercing a surface. We want the component of to the surface. The official definition of flux is
Perhaps it's easier to think of a sum over pieces of the surface. Think of dividing the surface into pieces . Each piece is so small that it can be considered flat. is a vector whose magnitude is the area of the piece and whose direction is perpendicular to the surface. (``Perpendicular'' to the surface is also called "normal" to the surface.)
If an electric field passes through this surface, then
| (3) |
| (4) |
The dot product picks out the component of perpendicular to the surface. If is perpendicular to the surface, then is parallel to because is perpendicular to the surface. So because . But if is parallel to the surface, then . So the maximum flux occurs when is perpendicular to the surface, just as in our examples.
Notice that the flux has different signs in the following
In the limit that the area elements become infinitesimal elements da, the sum becomes an integral.
.)
Calculate the net flux through a thin box. points outward.
| (5) |
what comes in one side goes out the other no net flux.
Gauss' Law deals with the flux through closed surfaces:
| (6) |
Gauss' Law says
Suppose an imaginary or ``Gaussian" surface encloses some blob of charge qenc. Gauss' law says that the total flux through the surface is proportional to the charge enclosed. qenc is the total amount of charge enclosed. Notice that if q < 0 and if q > 0.
Examples:
Notice that the more charge that is enclosed, the more flux there is. This is like the light bulb in the bag - brighter light means more flux.
In (b), notice that the flux even though .
Gauss' Law is a useful trick for finding if you are given a symmetrical charge distribution. If the distribution is unsymmetrical, it's too hard to do the integral . But for certain symmetrical distributions, you can choose a Gaussian surface so that you don't really have to do an integral. In some cases, the integral winds up being zero because the field E=0 or because . In other cases the E is constant on the Gaussian surface and the integral . The symmetries where this happens are spherical, cylindrical, and planar. What follows are the easy examples of using Gauss' law to find the field.
First let's go over the basic strategy for solving problems using Gauss' Law.
Recipe for Solving Problems with Gauss' Law
Typical Problem:
Given the charge distribution
, find
.
Recipe:
Let's apply this recipe to some examples.
Point Charge
As an example, suppose we are given a point charge q>0. What is
? The point charge and its field are spherically
symmetric. So let's surround it with a spherical Gaussian surface of
radius r.
=2.0 true in
ptChargeGauss.eps
Gauss' Law says
(7)
The point charge is in the center. The field points
radially outward
.
| (8) |
| (9) |
This is exactly what we got from Coulomb's Law. In fact Gauss' law is equivalent to Coulomb's Law.
Line of Charge
Consider a straight, infinitely long line of positive charge with a
charge per unit length . Find a distance r from
the line.
=2.5 true in
lineCharge.eps
Solution: Note the cylindrical symmetry and ( points radially outward). The Gaussian surface is a can of radius r and height h. on side. on the top and the bottom. at a distance r is a constant.
=
qenc
=
=
E
=
=
Plane of Charge
Consider an infinite plane of positive charge with uniform surface
charge density . Assume the sheet is insulating so that the
charge stays fixed. Find a distance r from the sheet.
Solution: Planar symmetry. points away from the sheet. to the sheet. Draw Gaussian pillbox with ends parallel to the sheet such that at ends. No flux through sides.
Note that E a distance r from the sheet is constant, i.e. it is
constant on the end face of the box.
The area of the end of the box is A, so the flux through
that face is EA. There are 2 faces, so the total flux is .
According to Gauss' law,
| (10) |
| (11) |
| (12) |
| (13) |
Spherical Shell of Charge:
Consider a spherical shell of uniform charge density. Let q > 0
be the total charge of the shell. The shell has radius R. Find
for r > R (outside) and r < R (inside).
Solution: Spherical Symmetry
Outside (r > R):
points radially outward in direction.
Gaussian surface is concentric spherical surface outside the shell of
charge. Gaussian sphere has radius r. E is constant on the sphere
so we can take it out of the integral.
| = | |||
| qenc | = | q | |
| = | |||
| E | = |
This is the same field as a point charge q at the center of the sphere. Thus ``a shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell's charge were concentrated at the center of the shell."
Inside (r<R):
By symmetry, if points in any direction, it will be
radial, i.e., along with . Draw a Gaussian sphere inside the
shell. Gaussian sphere has a radius r. E is a constant on the
Gaussian sphere so we can take it out of the integral.
| (14) |
| qenc=0 | (15) |
| (16) |
Solid Sphere of Charge
Suppose we have a solid sphere of uniform charge density. The radius
of the sphere is R. The total charge contained in the sphere is
Q > 0. Find
both outside (r > R) and inside (r < R) the sphere.
One Solution: Spherical symmetry. Divide the sphere into spherical shells and use superposition to add up the contributions to from the shells.
Outside
(r >R): is the same as for a point charge Q at the
center of the sphere.
What is
in terms of Q, the total charge? Q and
are proportional to the volume since the charge density
is uniform, so
| = | |||
| E(r) | = | ||
=2.5 true in
EvsrSolidSphere.eps
Another Solution
Calculate qenc using the uniform charge density
Outside (r>R)
Draw a spherical Gaussian surface. is radial and points in
the direction. So
| = | |||
| qenc | = | ||
| = | Q | ||
| = | |||
| E | = |
Inside (r<R): Spherical Gaussian surface inside the ball of charge.
=
qenc
=
=
=
E
=
E
=
This is the same answer we got before.
Charged Isolated Conductor
An isolated conductor can hold charge because there's nowhere for the
charge to go. If we put excess charge on a conductor, it resides on
the surface of the conductor, not in the interior. How do we know
this? Because
everywhere inside a conductor. If
somewhere inside, then the free charges (conduction
electrons) would feel a force
and they would move
in response to the force. But an isolated conductor doesn't have
flowing charges. So
inside. Gauss' Law tells us that
If the Gaussian surface lies inside the conductor, on the gaussian surface which implies no excess charge is enclosed. (There can be charge enclosed, but it must consist of equal amounts of positive and negative charge.) So all the excess charge is on the surface.
When you first dump charge on a conductor, it runs around until all the forces balance out. The charges, which all have the same sign, try to get as far apart as possible. When they get to the surface, they've gone as far as they can go. So they stop.
Notice that is perpendicular to the surface. If had any components tangent to the conductor's surface, the charge would run along the surface.
Let's suppose that a charged isolated conductor has a surface charge density of . For an irregularly shaped conductor, may vary along the surface. Let's find at the surface. Consider a small element of surface - small enough to be flat and to have . Draw a Gaussian pillbox. surface. E = 0 inside the conductor, so we only get flux through the ``front" side of the box. (The ``front" is parallel to the conductor's surface.)
| (17) |
qenc
Notice that this is different from the charged insulating sheet which had going out the front and back, giving .
Appendix
Charged Ring Problem:
Find the electric field a distance z above
the center of a circular ring of radius P which carries a uniform charge
density of .
=2.5 true in
chargedRing.eps
Solution:
First let's ask, ``what do we expect by symmetry?"
There is no preferred x or y direction. The system has azimuthal
symmetry, i.e., if we rotate the ring in the x-y plane about
the z-axis,
things are the same. So
has no x or y component:
Ex = Ey = 0. If did have a component in the x-yplane, which way would it point without showing favoritism? Another way
to see this is to note that the charge on opposite sides of the circle
produce fields whose x and y components cancel. So is parallel to the .
Calculate . To do this, we use the principle of superposition. We divide the ring into segments, each of length dsand charge . Then we calculate the field due to this segment. Finally we add up all the fields to get the total field produced by the ring.
So the magnitude of dE is given by Coulomb's law
=2.5 true in
chargedRing2.eps
Plug in and r2 = R2 + z2 to get
Since the total
, we just want the
z-component of :
We can express in terms of R and z:
| (18) |
| dEz | = | ||
| = | |||
| = |
| (19) |
| (20) |
| (21) |
| (22) |
This is the field of a point charge. So far from the ring, the ring looks like a point charge. (It's a good idea to take limits to see if we get sensible results).
If z=0, . This says that in the center of the ring, E=0. This is because the field produced by one bit of the ring is cancelled by a bit of charge on the opposite side of the ring.
Next: About this document ...
Clare Yu
2002-02-01