Electrostatic NotesLevel 1: The Basics, Conduction and Insulation

from Princeton Review

IonizationMost 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 Chargefrom Princeton Review

Practice Problem: Finding the Number of Charges

Detecting ChargesWhen 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 likeOne 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.

InductionInduction in a Conductorfrom AP Advantage

Induction in an Insulator

Electroscopes

GroundingGrounding 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 Fieldfrom 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 Easyfrom Brooke’s Cole

Solution

SuperpositionSuperposition 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 ForcesIf the forces are parallel to each other, this is easy enough. Just add and subtract accordingly.

Practice Problem: Parallel Superposition

Solution

Forces at anglesThat 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 Fieldsfrom 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 Fieldsfrom Princeton Review

from Cliff Notes

Calculating Electric Field StrengthElectric fields can be calculated using this equation:

E= kQr2

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 ChargesIf 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:

E=|Fq |

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, DThe electric field vector from the +Q charge points down and from the –Q charge points to theright so the resultant field points down and rightThe electric field vector from the +Q charge points down and from the –Q charge points to theright so the resultant field points down and right

Summary of This SectionCalculates a relationship between two charges

To convert Describes something about one charge

F= kQqr2 E=F

qE= kQ

r2

Level 3: Electric Potential EnergyElectric 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 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= kQqr

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∞= kQ∞

=0

The problem below is a very typical electric potential energy problem.

Memorize This!

Practice Problem: Work to Assemble Charges

SolutionW =∆ U

W =U f −U i

We are moving these particles from infinity so…

W =U f −U∞

W =U f −0

W =U f

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 = kQqr

W =k (12× 10−6)(−22×10−6)

(0.4)

W =−5.9J

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 =W outside force=−W field

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:

K i+U i+W =K f +U f

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 1989B2For 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 EnergyThe 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 EnergySometimes, 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:

K i+U i+W outside force=K f +U f

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+W outside force=0+U f

W outside force=U f

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=mv2

2

AP-Like Practice Problem

SolutionIn this situation, no one is describing an outside force sweeping in and moving anything, so W ouide force=0 J . In addition, the charges are released from rest, so we can say K i=0 J .

K i+U i+W outside force=K f +U f

0+ kQqri

+0=K f +kQq

r f

kQqr i

− kQqr f

=K f

kQq( 1r i

−1r f )=K f

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=mv2

2

to find the velocity of the charge at that point.

Electric PotentialNotice 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=Uq 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 PotentialThe equation for calculating electric potential is:

V= kQr

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 PotentialAgain, electric potential is a scalar value. You can just add it up. No need to work about vectors.

Practice ProblemThe 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

Calculates a relationship between two charges

To convert Describes something about one charge

Vectors F= kQqr2 E=F

qE= kQ

r2

Scalars U = kQqr

V=Uq

V= kQr

AP Practice: 1987B2This 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