7-8-99

Potential energy

In discussing gravitational potential energy in PY105, we usually
associated it with a single object. An object near the surface of the
Earth has a potential energy because of its gravitational interaction
with the Earth; potential energy is really not associated with a single
object, it comes from an interaction between objects.

Similarly, there is an electric potential energy associated with
interacting charges. For each pair of interacting charges, the
potential energy is given by:

electric potential energy: PE = k q Q / r

Energy is a scalar, not a vector. To find the total electric potential
energy associated with a set of charges, simply add up the energy
(which may be positive or negative) associated with each pair of
charges.

An object near the surface of the Earth experiences a nearly uniform
gravitational field with a magnitude of g; its gravitational potential
energy is mgh. A charge in a uniform electric field E has an electric
potential energy which is given by qEd, where d is the distance moved
along (or opposite to) the direction of the field. If the charge moves
in the same direction as the force it experiences, it is losing
potential energy; if it moves opposite to the direction of the force,
it is gaining potential energy.

The relationship between work, kinetic energy, and potential energy, which was discussed in PY105, still applies:

An example

Two positively-charged balls are tied together by a string. One ball
has a mass of 30 g and a charge of 1 ; the other has a mass of 40 g and
a charge of 2 . The distance between them is 5 cm. The ball with the smaller charge
has a mass of 30 g; the other ball has a mass of 40 g. Initially they
are at rest, but when the string is cut they move apart. When they are
a long way away from each other, how fast are they going?

Let's start by looking at energy. No external forces act on this
system of two charges, so the energy must be conserved. To start with
all the energy is potential energy; this will be converted into kinetic
energy.

Energy at the start : KE = 0

PE = k q Q / r = (8.99 x 109) (1 x 10-6) (2 x 10-6) / 0.05 = 0.3596 J

When the balls are very far apart, the r in the equation for
potential energy will be large, making the potential energy negligibly
small.

Energy is conserved, so the kinetic energy at the end is equal to the potential energy at the start:

The masses are known, but the two velocities are not. To solve for the
velocities, we need another relationship between them. Because no
external forces act on the system, momentum will also be conserved.
Before the string is cut, the momentum is zero, so the momentum has to
be zero all the way along. The momentum of one ball must be equal and
opposite to the momentum of the other, so:

Plugging this into the energy equation gives:

Electric potential

Electric potential is more commonly known as voltage. The potential at a point a distance r from a charge Q is given by:

V = k Q / r

Potential plays the same role for charge that pressure does for fluids.
If there is a pressure difference between two ends of a pipe filled
with fluid, the fluid will flow from the high pressure end towards the
lower pressure end. Charges respond to differences in potential in a
similar way.

Electric potential is a measure of the potential energy per unit
charge. If you know the potential at a point, and you then place a
charge at that point, the potential energy associated with that charge
in that potential is simply the charge multiplied by the potential.
Electric potential, like potential energy, is a scalar, not a vector.

connection between potential and potential energy: V = PE / q

Equipotential lines are connected lines of the same potential. These
often appear on field line diagrams. Equipotential lines are always
perpendicular to field lines, and therefore perpendicular to the force
experienced by a charge in the field. If a charge moves along an
equipotential line, no work is done; if a charge moves between
equipotential lines, work is done.

Field lines and equipotential lines for a point charge, and for a constant field between two charged plates, are shown below:

An example : Ionization energy of the electron in a hydrogen atom

In the Bohr model of a hydrogen atom, the electron, if it is in the
ground state, orbits the proton at a distance of r = 5.29 x 10-11 m.
Note that the Bohr model, the idea of electrons as tiny balls orbiting
the nucleus, is not a very good model of the atom. A better picture is
one in which the electron is spread out around the nucleus in a cloud
of varying density; however, the Bohr model does give the right answer
for the ionization energy, the energy required to remove the electron
from the atom.

The total energy is the sum of the electron's kinetic energy and the
potential energy coming from the electron-proton interaction.

The kinetic energy is given by KE = 1/2 mv2.

This can be found by analyzing the force on the electron. This force is
the Coulomb force; because the electron travels in a circular orbit,
the acceleration will be the centripetal acceleration:

Note that the negative sign coming from the charge on the electron
has been incorporated into the direction of the force in the equation
above.

This gives m v2 = k e2 / r, so the kinetic energy is KE = 1/2 k e2 / r.

The potential energy, on the other hand, is PE = - k e2
/ r. Note that the potential energy is twice as big as the kinetic
energy, but negative. This relationship between the kinetic and
potential energies is valid not just for electrons orbiting protons,
but also in gravitational situations, such as a satellite orbiting the
Earth.

The total energy is:

KE + PE = -1/2 ke2 / r = - 1/2 (8.99 x 109)(1.60 x 10-19) / 5.29 x 10-11

This works out to -2.18 x 10-18 J. This is usually stated in energy units of electron volts (eV). An eV is 1.60 x 10-19
J, so dividing by this gives an energy of -13.6 eV. To remove the
electron from the atom, 13.6 eV must be put in; 13.6 eV is thus the
ionization energy of a ground-state electron in hydrogen.

 

 

7-6-99

Charge

    * there are two kinds of charge, positive and negative
    * like charges repel, unlike charges attract
    * positive charge comes from having more protons than electrons; negative charge comes from having more electrons than protons
    * charge is quantized, meaning that charge comes in integer multiples of the elementary charge e
    * charge is conserved

Probably everyone is familiar with the first three concepts, but what does it mean for charge to be quantized? Charge comes in multiples of an indivisible unit of charge, represented by the letter e. In other words, charge comes in multiples of the charge on the electron or the proton. These things have the same size charge, but the sign is different. A proton has a charge of +e, while an electron has a charge of -e.

Electrons and protons are not the only things that carry charge. Other particles (positrons, for example) also carry charge in multiples of the electronic charge. Those are not going to be discussed, for the most part, in this course, however.

Putting "charge is quantized" in terms of an equation, we say:

q = n e

q is the symbol used to represent charge, while n is a positive or negative integer, and e is the electronic charge, 1.60 x 10-19 Coulombs.

The Law of Conservation of Charge

The Law of conservation of charge states that the net charge of an isolated system remains constant.

If a system starts out with an equal number of positive and negative charges, there¹s nothing we can do to create an excess of one kind of charge in that system unless we bring in charge from outside the system (or remove some charge from the system). Likewise, if something starts out with a certain net charge, say +100 e, it will always have +100 e unless it is allowed to interact with something external to it.

Charge can be created and destroyed, but only in positive-negative pairs.

Table of elementary particle masses and charges: 

Electrostatic charging

Forces between two electrically-charged objects can be extremely large. Most things are electrically neutral; they have equal amounts of positive and negative charge. If this wasn¹t the case, the world we live in would be a much stranger place. We also have a lot of control over how things get charged. This is because we can choose the appropriate material to use in a given situation.

Metals are good conductors of electric charge, while plastics, wood, and rubber are not. They¹re called insulators. Charge does not flow nearly as easily through insulators as it does through conductors, which is why wires you plug into a wall socket are covered with a protective rubber coating. Charge flows along the wire, but not through the coating to you.

Materials are divided into three categories, depending on how easily they will allow charge (i.e., electrons) to flow along them. These are:

    * conductors - metals, for example
    * semi-conductors - silicon is a good example
    * insulators - rubber, wood, plastic for example

Most materials are either conductors or insulators. The difference between them is that in conductors, the outermost electrons in the atoms are so loosely bound to their atoms that they¹re free to travel around. In insulators, on the other hand, the electrons are much more tightly bound to the atoms, and are not free to flow. Semi-conductors are a very useful intermediate class, not as conductive as metals but considerably more conductive than insulators. By adding certain impurities to semi-conductors in the appropriate concentrations the conductivity can be well-controlled.

There are three ways that objects can be given a net charge. These are:

   1. Charging by friction - this is useful for charging insulators. If you rub one material with another (say, a plastic ruler with a piece of paper towel), electrons have a tendency to be transferred from one material to the other. For example, rubbing glass with silk or saran wrap generally leaves the glass with a positive charge; rubbing PVC rod with fur generally gives the rod a negative charge.
   2. Charging by conduction - useful for charging metals and other conductors. If a charged object touches a conductor, some charge will be transferred between the object and the conductor, charging the conductor with the same sign as the charge on the object.
   3. Charging by induction - also useful for charging metals and other conductors. Again, a charged object is used, but this time it is only brought close to the conductor, and does not touch it. If the conductor is connected to ground (ground is basically anything neutral that can give up electrons to, or take electrons from, an object), electrons will either flow on to it or away from it. When the ground connection is removed , the conductor will have a charge opposite in sign to that of the charged object.

An example of induction using a negatively charged object and an initially-uncharged conductor (for example, a metal ball on a plastic handle).

(1) bring the negatively-charged object close to, but not touching, the conductor. Electrons on the conductor will be repelled from the area nearest the charged object.

(2) connect the conductor to ground. The electrons on the conductor want to get as far away from the negatively-charged object as possible, so some of them flow to ground.

(3) remove the ground connection. This leaves the conductor with a deficit of electrons.

(4) remove the charged object. The conductor is now positively charged.

A practical application involving the transfer of charge is in how laser printers and photocopiers work.

Why is static electricity more apparent in winter?

You notice static electricity much more in winter (with clothes in a dryer, or taking a sweater off, or getting a shock when you touch something after walking on carpet) than in summer because the air is much drier in winter than summer. Dry air is a relatively good electrical insulator, so if something is charged the charge tends to stay. In more humid conditions, such as you find on a typical summer day, water molecules, which are polarized, can quickly remove charge from a charged object.

Try this at home

See if you can charge something at home using friction. I got good results by rubbing a Bic pen with a piece of paper towel. To test the charge, you can use a narrow stream of water from a faucet; if the object attracts the stream when it's brought close, you know it's charged. All you need to do is to find something to rub - try anything made out of hard plastic or rubber. You also need to find something to rub the object with - potential candidates are things like paper towel, wool, silk, and saran wrap or other plastic.

Coulomb's law

The force exerted by one charge q on another charge Q is given by Coulomb's law: 

r is the distance between the charges.

Remember that force is a vector, so when more than one charge exerts a force on another charge, the net force on that charge is the vector sum of the individual forces. Remember, too, that charges of the same sign exert repulsive forces on one another, while charges of opposite sign attract.

An example

Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 x 10-6 C. The charges in the other two corners are -3.0 x 10-6 C. What is the net force exerted on the charge in the top right corner by the other three charges? 

To solve any problem like this, the simplest thing to do is to draw
a good diagram showing the forces acting on the charge. You should also
let your diagram handle your signs for you. Force is a vector, and any
time you have a minus sign associated with a vector all it does is tell
you about the direction of the vector. If you have the arrows giving
you the direction on your diagram, you can just drop any signs that
come out of the equation for Coulomb's law.

Consider the forces exerted on the charge in the top right by the other three:

You have to be very careful to add these forces as vectors to get the
net force. In this problem we can take advantage of the symmetry, and
combine the forces from charges 2 and 4 into a force along the diagonal
(opposite to the force from charge 3) of magnitude 183.1 N. When this
is combined with the 64.7 N force in the opposite direction, the result
is a net force of 118 N pointing along the diagonal of the square.

The symmetry here makes things a little easier. If it wasn't so symmetric, all you'd have to do is split the vectors up in to x and y components, add them to find the x and y components of the net force, and then calculate the magnitude and direction of the net force from the components. Example 16-4 in the textbook shows this process.

The parallel between gravity and electrostatics

An electric field describes how an electric charge affects the region around it. It's a powerful concept, because it allows you to determine ahead of time how a charge will be affected if it is brought into the region. Many people have trouble with the concept of a field, though, because it's something that's hard to get a real feel for. The fact is, though, that you're already familiar with a field. We've talked about gravity, and we've even used a gravitational field; we just didn't call it a field.

When talking about gravity, we got into the (probably bad) habit of calling g "the acceleration due to gravity". It's more accurate to call g the gravitational field produced by the Earth at the surface of the Earth. If you understand gravity you can understand electric forces and fields because the equations that govern both have the same form.

The gravitational force between two masses (m and M) separated by a distance r is given by Newton's law of universal gravitation: 

A similar equation applies to the force between two charges (q and Q) separated by a distance r:

The force equations are similar, so the behavior of interacting masses is similar to that of interacting charges, and similar analysis methods can be used. The main difference is that gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive. The charge (q or Q) plays the same role in the electrostatic case that the mass (m or M) plays in the case of the gravity.

A good example of a question involving two interacting masses is a projectile motion problem, where there is one mass m, the projectile, interacting with a much larger mass M, the Earth. If we throw the projectile (at some random launch angle) off a 40-meter-high cliff, the force on the projectile is given by:

F = mg

This is the same equation as the more complicated equation above, with G, M, and the radius of the Earth, squared, incorporated into g, the gravitational field.

So, you've seen a field before, in the form of g. Electric fields operate in a similar way. An equivalent electrostatics problem is to launch a charge q (again, at some random angle) into a uniform electric field E, as we did for m in the Earth's gravitational field g. The force on the charge is given by F = qE, the same way the force on the mass m is given by F = mg.

We can extend the parallel between gravity and electrostatics to energy, but we'll deal with that later. The bottom line is that if you can do projectile motion questions using gravity, you should be able to do them using electrostatics. In some cases, you¹ll need to apply both; in other cases one force will be so much larger than the other that you can ignore one (generally if you can ignore one, it'll be the gravitational force). 

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Electric field

To help visualize how a charge, or a collection of charges, influences the region around it, the concept of an electric field is used. The electric field E is analogous to g, which we called the acceleration due to gravity but which is really the gravitational field. Everything we learned about gravity, and how masses respond to gravitational forces, can help us understand how electric charges respond to electric forces.

The electric field a distance r away from a point charge Q is given by:

Electric field from a point charge : E = k Q / r2

The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by :

F = qE

If q is positive, the force is in the same direction as the field; if q is negative, the force is in the opposite direction as the field.

Learning from gravity

Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/s2 and points straight down. If you threw a mass through the air, you know it would follow a parabolic path because of gravity. You could determine when and where the object would land by doing a projectile motion analysis, separating everything into x and y components. The horizontal acceleration is zero, and the vertical acceleration is g. We know this because a free-body diagram shows only mg, acting vertically, and applying Newton's second law tells us that mg = ma, so a = g.

You can do the same thing with charges in a uniform electric field. If you throw a charge into a uniform electric field (same magnitude and direction everywhere), it would also follow a parabolic path. We're going to neglect gravity; the parabola comes from the constant force experienced by the charge in the electric field. Again, you could determine when and where the charge would land by doing a projectile motion analysis. The acceleration is again zero in one direction and constant in the other. The value of the acceleration can be found by drawing a free-body diagram (one force, F = qE) and applying Newton's second law. This says:
qE = ma, so the acceleration is a = qE / m.

Is it valid to neglect gravity? What matters is the size of qE / m relative to g. As long as qE / m is much larger than g, gravity can be ignored. Gravity is very easy to account for, of course : simply add mg to the free-body diagram and go from there.

The one big difference between gravity and electricity is that m, the mass, is always positive, while q, the charge, can be positive, zero, or negative.

What does an electric field look like?

An electric field can be visualized on paper by drawing lines of force, which give an indication of both the size and the strength of the field. Lines of force are also called field lines. Field lines start on positive charges and end on negative charges, and the direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed at that point. If the charge is positive, it will experience a force in the same direction as the field; if it is negative the force will be opposite to the field.

The fields from isolated, individual charges look like this: 

When there is more than one charge in a region, the electric field lines will not be straight lines; they will curve in response to the different charges. In every case, though, the field is highest where the field lines are close together, and decreases as the lines get further apart.

An example

Two charges are placed on the x axis. The first, with a charge of +Q, is at the origin. The second, with a charge of -2Q, is at x = 1.00 m. Where on the x axis is the electric field equal to zero? 

This question involves an important concept that we haven't
discussed yet: the field from a collection of charges is simply the
vector sum of the fields from the individual charges. To find the
places where the field is zero, simply add the field from the first
charge to that of the second charge and see where they cancel each
other out.

In any problem like this it's helpful to come up with a rough
estimate of where the point, or points, where the field is zero is/are.
There is no such point between the two charges, because between them
the field from the +Q charge points to the right and so does the field
from the -2Q charge. To the right of the -2Q charge, the field from the
+Q charge points right and the one from the -2Q charge points left. The
field from the -2Q charge is always larger, though, because the charge
is bigger and closer, so the fields can't cancel. To the left of the +Q
charge, though, the fields can cancel. Let's say the point where they
cancel is a distance x to the left of the +Q charge.

Cross-multiplying and expanding the bracket gives:

Solving for x using the quadratic equation gives: x = 2.41 m or x = -0.414 m

The answer to go with is x = 2.41 m. This corresponds to 2.41 m to the left of the +Q charge. The other point is between the charges. It corresponds to the point where the fields from the two charges have the same magnitude, but they both point in the same direction there so they don't cancel out.

The field around a charged conductor

A conductor is in electrostatic equilibrium when the charge distribution (the way the charge is distributed over the conductor) is fixed. Basically, when you charge a conductor the charge spreads itself out. At equilibrium, the charge and electric field follow these guidelines:

    * the excess charge lies only at the surface of the conductor
    * the electric field is zero within the solid part of the conductor
    * the electric field at the surface of the conductor is perpendicular to the surface
    * charge accumulates, and the field is strongest, on pointy parts of the conductor

Let's see if we can explain these things. Consider a negatively-charged conductor; in other words, a conductor with an excess of electrons. The excess electrons repel each other, so they want to get as far away from each other as possible. To do this they move to the surface of the conductor. They also distribute themselves so the electric field inside the conductor is zero. If the field wasn't zero, any electrons that are free to move would. There are plenty of free electrons inside the conductor (they're the ones that are canceling out the positive charge from all the protons) and they don't move, so the field must be zero.

A similar argument explains why the field at the surface of the conductor is perpendicular to the surface. If it wasn't, there would be a component of the field along the surface. A charge experiencing that field would move along the surface in response to that field, which is inconsistent with the conductor being in equilibrium.

Why does charge pile up at the pointy ends of a conductor? Consider two conductors, one in the shape of a circle and one in the shape of a line. Charges are distributed uniformly along both conductors. With the circular shape, each charge has no net force on it, because there is the same amount of charge on either side of it and it is uniformly distributed. The circular conductor is in equilibrium, as far as its charge distribution is concerned. 

With the line, on the other hand, a uniform distribution does not
correspond to equilbrium. If you look at the second charge from the
left on the line, for example, there is just one charge to its left and
several on the right. This charge would experience a force to the left,
pushing it down towards the end. For charge distributed along a line,
the equilibrium distribution would look more like this:

The charge accumulates at the pointy ends because that balances the forces on each charge.

Electric flux

A clever way to calculate the electric field from a charged conductor is to use Gauss' Law, which is explained in Appendix D in the textbook. Gauss' Law can be tricky to apply, though, so we won't get into that. What we will do is to look at some implications of Gauss' Law. It's also a good time to introduce the concept of flux. This is important for deriving electric fields with Gauss' Law, which you will NOT be responsible for; where it'll really help us out is when we get to magnetism, when we do magnetic flux.

Electric flux is a measure of the number of electric field lines passing through an area. To calculate the flux through a particular surface, multiply the surface area by the component of the electric field perpendicular to the surface. If the electric field is parallel to the surface, no field lines pass through the surface and the flux will be zero. The maximum flux occurs when the field is perpendicular to the surface. 

Permittivity

Even though we won't use this for anything, we should at least write down Gauss' law:

Gauss' Law - the sum of the electric flux through a surface is equal to the charge enclosed by a surface divided by a constant , the permittivity of free space.

What is the permittivity of free space? It's a constant related to the
constant k that appears in Coulomb's law. The relationship between the
two is this:

Implications of Gauss' Law

Gauss' Law is a powerful method of calculating electric fields. If you have a solid conducting sphere (e.g., a metal ball) that has a net charge Q on it, you know all the excess charge lies on the outside of the sphere. Gauss' law tells us that the electric field inside the sphere is zero, and the electric field outside the sphere is the same as the field from a point charge with a net charge of Q. That's a pretty neat result. 

The result for the sphere applies whether it's solid or hollow. Let's
look at the hollow sphere, and make it more interesting by adding a
point charge at the center.

What does the electric field look like around this charge inside the hollow sphere? How is the negative charge distributed on the hollow sphere? To find the answers, keep these things in mind:

    * The electric field must be zero inside the solid part of the sphere
    * Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone

We know that the electric field from the point charge is given by kq / r2. Because the charge is positive, the field points away from the charge.

If we took the point charge out of the sphere, the field from the negative charge on the sphere would be zero inside the sphere, and given by kQ / r2 outside the sphere.

The net electric field with the point charge and the charged sphere, then, is the sum of the fields from the point charge alone and from the sphere alone (except inside the solid part of the sphere, where the field must be zero). This is shown in the picture: 

How is the charge distributed on the sphere? The electrons must
distribute themselves so the field is zero in the solid part. This
means there must be -5 microcoulombs of charge on the inner surface, to
stop all the field lines from the +5 microcoulomb point charge. There
must then be +2 microcoulombs of charge on the outer surface of the
sphere, to give a net charge of -5+2 = -3 microcoulombs.