Download Electric Potential Energy

Electrostatic Notes
Level 1: The Basics, Conduction and Insulation
from Princeton Review
Ionization
Most of the time, matter does not have any charge. This is because the protons and electrons perfectly cancel out. In
some situations, however, an atom might gain or lose electrons. This process is called ionization. An ion is just a term for
any atom that has a charge- this could mean it has a slightly positive charge or a slightly negative charge.
The most important thing you need to remember throughout this entire section is this, the Law of Conservation of
Charge-charges cannot be created or destroyed. They can be exchanged, passed around, added and subtracted to
objects, but overall, you can’t destroy charge any more than you can destroy matter (which is to say that you, know,
sometimes when we get out in the furthest reaches of physics this whole things breaks down a little, but hey, let’s not
talk about that now…)
Elementary Charge
from Princeton Review
Practice Problem: Finding the Number of Charges
Detecting Charges
When we are talking about objects that are ionized (have an unbalanced charge) we frequently use the term insulator
and conductor. When you put charges on an insulator, the charges will pretty much stay put- they will stay exactly
where you want them to. We think of insulators as not having any charge, but this isn’t actually true. We think this
because if you touch an insulator, it isn’t likely to give you a shock, because it is holding on to the extra charge it has. A
conductor, on the other hand, will allow the charge to flow. This is why you don’t want to stick a fork into a toaster. The
conductors inside will easily transfer any charge it has straight into your hand.
Like repels like
One key idea behind electrostatics is the concept that opposites attract and likes repel. This means that electrons (and
positive charges) will want to get as far away from each other as conceivably, mathematically possible.
from REA:
Because the charges what to get as far away from each other as
possible, they will always spread out along the outside of a conductor.
This means that you will never have a charge inside on a conductor. It
you put one there, it would immediately transfer to the outside. A
Faraday Cage is a fancy name for a conductor that is hollow inside. If
you place an object inside, it would essentially be the safe from any
charge you would zap the outside with.
Induction
Induction in a Conductor
from AP Advantage
Induction in an Insulator
Electroscopes
Grounding
Grounding occurs when we provide a path of conductors from the object to the actual ground. The ground is basically a
giant ocean of electrons. It can accept tons of extra electrons without any problem. It can also give tons of electrons
without any problem. Below, I show you an example of how grounding can alter the charge on an object.
Level 2: Electric Force and Electric Field
from Brookes Cole
Notice that the notes above say Coulomb’s Law helps us calculate the magnitude of the electric force. It doesn’t say
anything about the direction. This is actually the easiest way to deal with electric force. We use Coulomb’s Law to find
the electric force first, and then we use a good mix of logic and superposition to find the rest. We will get into this next.
For now, just keep this in mind.
Practice Problem: Coulomb’s Law Easy
from Brooke’s Cole
Solution
Superposition
Superposition is just a fancy way of saying “adding vectors”. Remember that when we add vectors, we need to take into
account the direction. And force is a vector. Hold on to your hats, kids, because this is going to take a little trig.
For Parallel Forces
If the forces are parallel to each other, this is easy enough. Just add and subtract accordingly.
Practice Problem: Parallel Superposition
Solution
Forces at angles
That wasn’t too bad. So, of course, it won’t be heavily featured on the AP test. What you are much more likely to see are
problems where the forces need to be added up at angles.
Practice Problem: Superposition
Electric Fields
from AP Advantage
The earth has a gravitational field around it, caused by its mass. This means that objects with mass are attracted to it
gravitationally. Objects farther away from the earth, will still feel attracted to it, but they won’t feel that attraction with
the same intensity. Objects closer to the earth with feel the attraction more. Notice that in this description, I didn’t
mention how big the mass actually is. I don’t need to tell you the object is an elephant or a golf ball for you to
understand that the further we get away, the less force that object would feel.
Electric field strength is very similar. It is a concept that gives us a picture of how strong the force of attraction/repulsion
could be near a charge. Electric force tell about the relationship between two charges- how attracted or repulsed they
are by one another. Electric field strength only tells about one charge. It tells us about how attractive or repelling that
object is in general.
This is terrible, but the only way I know how to explain this is to relate it to people. Electric force is similar to the amount
of attraction two people feel toward one another. It’s about the relationship between them. Electric field strength is
similar to how attractive a person is- how much that person is, in general, able to attract other people.
The key idea here is that the electric field lines show the direction a positive charge would move. If we put a negative
electric charge in the field, it would go the exact opposite direction.
from Princeton Review
Adding Electric Fields
from Princeton Review
from Cliff Notes
Calculating Electric Field Strength
Electric fields can be calculated using this equation:
𝐸=
𝑘𝑄
𝑟2
I put an absolute value on that because, again, we can use this equation to find the magnitude of the electric field, but
we are going to need some logic and trig to figure out the direction of the electric field strength. Let’s start by first
practicing finding the electric field around one charge.
Practice Question
Solution
AP Practice Question: Comparing Points
The diagram above shows an isolated, positive charge Q. Point (B) is twice as far away from Q as point A. The
ratio of the electric field strength at point A to the electric field strength at point B is
(A) 8 to 1 (B) 4 to 1 (C) 2 to 1 (D) 1 to 1 (E) 1 to 2
E ∝ 1/r2 so if r × 2, E ÷ 4
Calculating Electric Field Strength Near Multiple Charges
If we want to find out the electric field strength caused by multiple charges, we are going to again have to rely on the
concept of superposition. Remember, you are going to need to care about their direction.
I think it is very useful, when solving a problem like this, to use a little logic first, before I get out my calculator and start
pounding away. Let’s look at the charge on the left. How is this going to contribute to the electric field strength at Point
P. Well, if I put a test charge at Point P, it would be repelled by the +Q charge and want to move to the right. What about
the influence of the –Q? If we had a test charge at Point P, it would be attracted toward the –Q, urging it to the right. So
both of the electric fields will combine to push the charge to the right.
Very often, the AP test will follow up a question like the one above by asking you to calculate the force a new charge
would feel if it was put at this point. To do this, you should recall the relationship between electric field strength and
electric force:
𝐹
𝐸=| |
𝑞
These are very common, very easy questions. If you know the electric field strength and you know the charge you are
curious about, it is a simple plug n’ chug.
I’ve seen a ton of questions just like this:
Practice: Find the position where E=0
Solution
AP Practice Problem: Electric Field Strength and Magnitude
Charges –Q and +Q are located on the x– and y–axes, respectively, each at a distance d from the origin O, as
shown above.
Answers: D, D
The electric field vector from the +Q charge points down and from the –Q charge points to the
right so the resultant field points down and right
The electric field vector from the +Q charge points down and from the –Q charge points to the
right so the resultant field points down and right
Summary of This Section
Calculates a relationship between two
charges
𝑘𝑄𝑞
𝐹= 2
𝑟
To convert
Describes something about one charge
𝐸=
𝐹
𝑞
𝐸=
𝑘𝑄
𝑟2
Level 3: Electric Potential Energy
Electric Potential Energy really freaks people out, but it shouldn’t. It’s not all that different from the gravitational
potential energy that we are used to. Let’s say we were interested in a the potential
energy of a book held a few feet off the ground. When we calculate this energy, we
Memorize This!
are essentially calculating the amount of work it took to put the energy at this
position. We define the ground as zero, and we use the conservation of energy
theorem to find how much potential energy at that position. In a sense, we are
describing an energy relationship between the book and the earth.
Electric potential energy is very similar. When we do these calculations, we are
finding the energy it took to bring two (or three or four…) charges together. Here is equation for electric the electric
potential energy between two charges:
𝑈=
𝑘𝑄𝑞
𝑟
This equation is, by itself, actually quite useless. What we really care about is the change in electric potential energy
because that will tell us how much work it took to assemble the particles.
∆𝑈 = 𝑊
This is much more useful, because we can use it to calculate how much energy (work) it takes to move particles around.
The easiest way to do this is to have a point where U=0, in order to make calculations easier. When we talk about
gravitational potential energy, the ground is usually are reference point where potential energy is zero. In the case of
electric charges, we say that infinity is zero. This makes a lot of sense:𝑈∞ =
The problem below is a very typical electric potential energy problem.
𝑘𝑄
∞
=0
Practice Problem: Work to Assemble Charges
Solution
𝑊 = ∆𝑈
𝑊 = 𝑈𝑓 − 𝑈𝑖
We are moving these particles from infinity so…
𝑊 = 𝑈𝑓 − 𝑈∞
𝑊 = 𝑈𝑓 − 0
𝑊 = 𝑈𝑓
Many textbooks will actually jump to just using the last equation (you’ll see an example of this below). The AP test,
however, often rewards points for making it obvious that the electric potential energy is zero at infinity, so I would stick
to at least writing the line above that, if not the whole thing.
𝑊=
𝑊=
𝑘𝑄𝑞
𝑟
𝑘(12 × 10−6 )(−22 × 10−6 )
(0.4)
𝑊 = −5.9 𝐽
Okay, what does this number actually tell us? It tells us how much work an outside force would need to due to bring
these charges together. If the value had been positive, it means that we would have to break a sweat pushing these two
particles together- we would expect that for particles that have the same sign and are repulsed by one another. In order
for that to happen, those particles would steal 5.9 J of energy from us in order to come together. In this situation, the
work is negative, which means the universe naturally wants these particles to be together and they will rush toward one
another without our help. In fact, the negative implies that we can actually get energy out of this interaction, if we find a
way to harness it.
What is doing the work to bring these particles together, in this situation, where the particles are naturally drawn
together and the work is negative? The universe- the same way the universe does the work (through gravity) of pulling a
book to the floor. Every now and then, you will be asked to find the work the field would do in bringing the charges
together. That’s easy- just throw a negative in front of your answer.
∆𝑈 = 𝑊𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑜𝑟𝑐𝑒 = −𝑊𝑓𝑖𝑒𝑙𝑑
This is a good equation to memorize- maybe one of the equations to memorize for the AP test, because it is applicable
for pretty much any situation in physics where we don’t take kinetic energy into account. It is an outgrowth of
conservation of energy theorem, which states:
𝐾𝑖 + 𝑈𝑖 + 𝑊 = 𝐾𝑓 + 𝑈𝑓
Keep this equation in the back of your mind. We are going to return to it in a minute. For now, let’s do some practice.
AP Practice 1989B2
For this problem, you can solve everything but d. Skip it for now and come back later. I’ll teach you that in a minute.
1989B2 Solution
The Scalar Nature of Electric Potential Energy
The best thing about dealing with electric potential energy is that it is scalar. I know that scalars have a bad rap with you,
but really, they are there to make your life easier. When we add vectors like electric field strength and force, we have to
do tons of hideous trig, and while we might feel smart afterward, they are time consuming. Electric potential energy is a
scalar, which means that if we want to add a couple electric potentials together- we can just add them together. It can
feel like sweet, sweet victory when you are rushing through a test.
from the Princeton Review
Solution
Electric Potential Energy to Kinetic Energy
Sometimes, the problem isn’t just asking us to figure out how much work it would take to assemble a system. We
occasionally have to calculate how fast an object would be moving after it was released from a particular position. Again,
we return to the Mother of All Equations, the conservation of energy:
𝐾𝑖 + 𝑈𝑖 + 𝑊𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝐾𝑓 + 𝑈𝑓
Up until now, we’ve treated the initial and final kinetic energies to be zero, and we’ve dealt with the reference point of
having no potential energy at infinity, so this equation was reduced to:
0 + 0 + 𝑊𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑜𝑟𝑐𝑒 = 0 + 𝑈𝑓
𝑊𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝑈𝑓
There are problems where we don’t keep these values at zero. Of course, potential energy at infinity is always zero, but
sometimes we are just curious about moving an object from one position to another. In that case, we need to make
heavy use of the constant acceleration equation, and keep in mind that:
𝐾=
𝑚𝑣 2
2
AP-Like Practice Problem
Solution
In this situation, no one is describing an outside force sweeping in and moving anything, so 𝑊𝑜𝑢𝑖𝑑𝑒 𝑓𝑜𝑟𝑐𝑒 = 0 𝐽. In
addition, the charges are released from rest, so we can say 𝐾𝑖 = 0𝐽.
𝐾𝑖 + 𝑈𝑖 + 𝑊𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝐾𝑓 + 𝑈𝑓
0+
𝑘𝑄𝑞
𝑘𝑄𝑞
+ 0 = 𝐾𝑓 +
𝑟𝑖
𝑟𝑓
𝑘𝑄𝑞 𝑘𝑄𝑞
−
= 𝐾𝑓
𝑟𝑖
𝑟𝑓
1 1
𝑘𝑄𝑞 ( − ) = 𝐾𝑓
𝑟𝑖 𝑟𝑓
If you plug n’ chug, you’ll find the final kinetic energy is +0.016 J.
If we were then given the mass of the charges (or were told they were elementary particles, where we could find the
mass) we could then use
𝐾=
𝑚𝑣 2
2
to find the velocity of the charge at that point.
Electric Potential
Notice that electric potential energy was all about a relationship between two objects. Specifically, it dealt with how
much energy it took to bring two (or more) charges together. In this way, electric potential energy is similar to force- you
are talking about a connection between two objects. Electric field strength, on the other hand, described something
about an individual object- it essentially gave us a mathematical hint about how much force another charge would feel if
it were placed nearby.
Electric force  about a relationship between two existing particles
Electric field strength  about one charge, describing the conditions near that object.
Electric potential is similar to electric field strength- it gives us a mathematical hint about the energy near one charged
particle. That’s right- you heard me. There is a difference in physics between electric potential energy and electric
potential. Please don’t throw the computer. I need them all.
The electric potential (V) tells us the electric potential energy per unit charge, just like the electric field strength told us
the electric force per unit charge.
𝑉=
𝑈
𝑞
just like 𝐸 =
𝐹
𝑞
Electric Potential energy  about a relationship between two existing particles
Electric Potential about one charge, describing the conditions near that object.
Calculating Electric Potential
The equation for calculating electric potential is:
𝑉=
𝑘𝑄
𝑟
we measure electric potential in volts. Of course, looking at this, we can also say the electric potential at infinity is zero.
Practice Problem
Solution
The Scalar Nature of Electric Potential
Again, electric potential is a scalar value. You can just add it up. No need to work about vectors.
Practice Problem
The following problem uses Angstroms, the one with the weird little A symbol, which is just 10-10 m.
Solution
Summary of All the Equations So Far
Vectors
Scalars
Calculates a relationship
between two charges
𝑘𝑄𝑞
𝐹= 2
𝑟
𝑈=
𝑘𝑄𝑞
𝑟
To convert
𝐸=
𝐹
𝑞
𝑉=
𝑈
𝑞
Describes something
about one charge
𝑘𝑄
𝐸= 2
𝑟
𝑉=
𝑘𝑄
𝑟
AP Practice: 1987B2
This great problem is extremely representative of the type of electrostatics questions they have put out year after year.
Really take the time to get familiar with it.
1987B2. Object I, shown above, has a charge of + 3 x 10 –6 coulomb and a mass of 0.0025 kilogram.
a. What is the electric potential at point P, 0.30 meter from object I?
Object II, of the same mass as object I, but having a charge of + 1 x 10 –6
b. How much work must be done to bring the object II from infinity to point P?
coulomb, is brought from infinity to point P, as shown above.
c. What is the magnitude of the electric force between the two objects when they are 0.30 meter apart?
d. What are the magnitude and direction of the electric field at the point midway between the two objects?
The two objects are then released simultaneously and move apart due to the electric force between them. No other forces act on the
objects.
e. What is the speed of object I when the objects are very far apart?
Solution
AP Practice: 2006B3
2006B3: Solution