AP Physics C: Electricity & Magnetism Electric Charge & Coulomb's Law

Hello, everyone, and welcome to www.educator.com.0000

I am Dan Fullerton, and today we are going to start a series of lessons for AP Physics C,0003

specifically the Electricity and Magnetism course.0008

Let us tag in by talking about electric charge and Coulomb’s law.0011

Some of our objectives are going to be to calculate the charge on an object and explain the law of conservation of charge.0016

That is going to be a big idea throughout the course.0022

Describe differences between conductors and insulators.0025

Explain the difference between conduction and induction.0029

Talk about polarization and inducing charges or induction, and use Coulomb’s law in the principle of super position,0033

which is going to come up a lot in this course to solve for the force of the charge particle due to other point chargers.0040

With that, why do not we dive in and start talking about electric charges.0047

First off, matter is made up of atoms, and atoms contain protons which are positive,0052

electrons which are negative, and neutrons in the nucleus which are neutral.0057

Most atoms are neutral, they have the same number of protons and the same number of electrons.0062

Their positive charges that are exactly balanced by the negative charges.0067

Atoms that have lost or gained electron have a net charge or no longer neutral, so they are called ions.0071

Also, the fundamental unit of charge is known as a Coulomb with the abbreviation C.0079

The Coulomb is a very large amount of electric charge.0084

Typically, we are going to be talking about units much smaller than the Coulomb.0088

Not always, but the vast majority of the time.0092

The smallest isolated unit of an electric charge is called the elementary charge.0095

It is the charge on a single electron or a single proton.0100

1 elementary charge is equal the 1.6 × 10⁻¹⁹ C.0103

You are going to see that come up again and again in the course, this fundamental unit.0110

It is also in your formula sheet for the AP Physics C test.0115

Like charges repel while opposites attract, you probably knew that already.0119

Electric charges conserve this law of conservation of charge is a huge idea again in AP Physics C.0123

What that really means is if we have any close system, whatever charge we start with,0132

is the same charge we our going to have when we our done.0136

If we start with a neutral atom and we pull off 2 electrons, we have -2 elementary charges over here,0138

we our left with must be 2 positive, 2 elementary charges, so that the sum total of the charge is still 0.0144

It is still neutral, that law of conservation of charge.0149

Let us take a look at an example or 2 and see how this works out.0156

First off, Mitt is a cat that possesses an axis of 6 × 10⁶ electrons, what is the net charge on N?0160

The way that I will go about doing that is let us find the charge abbreviated with a Q =6 × 10⁶ elementary charges and0168

the charge on each one of those electrons is -1.6 × 10⁻¹⁹ C because electrons0181

are negatively charged for every elementary charge or electron.0191

Therefore, when we multiply through we are going to have our e cancel out0196

6 × 10⁶ × -1.6 × 10⁻¹⁹ give us an answer of right around -9.6 × 10⁻¹³ C.0201

Pretty straightforward, let us do another one.0218

Let us find a charge on Α particle.0221

An Α particle also known as a helium nucleus consists of 2 protons and 2 neutrons, what is the charge of an Α particle?0224

The first thing we have to recognize here is that the neutrons are neutral, they have no charge.0234

They do not contribute any charge to the answer to our question.0239

Instead, we need to focus on these 2 protons.0242

The charge Q is +2 elementary charges which is going to be 2 × 1.6 × 10⁻¹⁹ C for a total of 3.2 × 10⁻¹⁹ C.0245

Again, notice our answer how much smaller than it is than the Coulomb, 10⁻¹⁹ C.0266

Typically, we are going to be dealing with quantities of charge that are smaller than Coulomb’s.0270

Not always, but quite a bit of time.0276

Let us move on for a moment and talk about conductors and insulators.0279

Conductors allow electric charges to move freely.0283

They have a very low resistivity, another term it is going to be good to know.0286

Insulators, on the other hand, do not allow electric charge to move freely.0291

Therefore, they have a very high resistivity.0295

Resistivity is a material property.0298

It tells you how freely that material allows electrons to move and it gets the symbol ρ ρ.0301

The Greek letter ρ, it is like a squiggly P.0306

Materials that have a high resistivity things like a glass, plastic, my flannel blanket at home, they are all good insulators.0310

They are all of a very high resistivity.0319

Materials with a very low resistivity are good conductors, things like gold and silver.0322

Most of the metals typically are very good conductors.0327

Let us talk about how we charge something and we will start by talking about charging by conduction.0333

When things are charged by contact we call that conduction.0337

Here is a trick you can try, rub a balloon against your hair for a little bit,0341

some electrons from the atoms in your hair are transferred to the balloon, an example of conduction.0345

The balloon therefore, picks up some extra electrons it becomes negatively charged,0350

while your hair, by the law of conservation of charge must become positively charged.0355

We can test by now by taking the balloon and going and trying to stick it onto a wall or something like that.0359

Oftentimes, you can get it to a stick which is showing that it is charged in the lowest charge and by common sense0366

you know it is probably the balloon.0370

Conductors may be charged by contact.0373

The charge conductor brought into contact with an identical neutral conductor will share the charge across the conductors.0375

We will show example of that here right quick.0385

Here we have a conductor carrying a net charge of 8 elementary charges and0389

it spread into contact with an identical conductor that has no net charge.0393

What is the charge on each conductor after we bring them together and then separate them?0397

Once we touch them together, if they are identical spheres they are going to share the charge across them.0402

Each one is going to have 4 elementary charges and they split them apart to each have 4.0407

What is that in Coulomb’s?0413

We have 4 elementary charges which is 4 × 1.6 × 10⁻¹⁹ C or 6.4 × 10⁻¹⁹ C on each of those conducting spheres.0415

A useful tool for detecting charges known as the electroscope, basically what it is, is an insulating bottle of some sort0436

that has a metal rod in it and down at the bottom you have a couple of conducting foil leaves that are very easy to move.0444

If you bring something like a positively charged rod near the metal head,0451

the electrons are free to move in this metal rod trying to get as close as they can, opposites attract.0457

When that happens, as the electrons move to the top, you are left with a net positive charge here on the bottom0464

and then the leaves at the bottom, because they have the same charge, tend to spread apart.0470

It is kind of a charge detector.0474

We could also charge something by induction,0479

charging a conductor with out coming into contact with another charged object.0481

Here is an example of how we might do that with an electroscope.0485

If we start out and bring a positive rod near the head of the electroscope, the negative charges try and congregate near that positive rod.0489

It attracts the negative charges leaving a net positive charge down here at the bottom of the electroscope and the leaves spread apart.0497

What we do though, is we connect that metal rod to ground.0506

You can think of connecting the ground as connecting to the earth.0511

The earth is like this giants sink or source of charges.0514

It is kind of like a magic purse for any electrical charge.0519

If you need more negative charges connect it to ground, it will be pulled up.0522

If you have extra charges to get rid off, put it in the ground.0525

It is a great place to go store them for a while.0528

What we do is while we have a positive rod here but not touching, we attach the metal rod to ground and0530

because we have a positive charge, we want to have negative charges here.0537

The ground wire sucks up all these negative charges so we have a bunch of negative charges now in the head of the metal.0541

Then we disconnect the ground wire and the negative charges are stuck on the negative rod, on the metal rod.0547

Therefore, the leaves spread apart again but it is because they are charged negatively as we bring that rod away.0556

We ended up having a rod that was neutral to begin with, over here on the right when it is done it is negatively charged.0561

But we never touched our positively charged rod to it all we did was connect it to ground.0569

That would be an example of charging by induction.0575

Let us take a look at some examples.0580

A positively charged glass rod attracts object x.0582

The net charge of object x maybe, our choices are A it may be 0 or negative, B it may be 0 or positive,0586

C it must be negative, or D it must be positive.0594

Let us think about that.0599

If we take a positively charged glass rod and we bring it near some object x,0600

if it is a negatively charged they will definitely be attracted, opposites attract, so that could be true.0609

But also, if we have a positively charged rod and we bring it near a neutral object0617

what can happen is if we look closely, the entire thing is negative, or it is neutral.0624

When we bring the positively charged rod near it, the atoms that are nearest to the rod,0633

the electrons on those atoms are attracted to the positive rod.0638

What they will do is actually spend a little bit more time orbiting the nucleus over towards the positive rod0642

so with just a little tiny bit of a shift in the material.0649

But the electrons spending a little bit more time on this side of the nucleus than this side,0654

that gives us a negative charge over on this side of the object and we can still get attraction.0658

A charged object can attract a neutral object so we know that the answer must be 0 or negative.0665

The only way to prove 2 objects are charged is repulsion.0674

You can charge a neutral object with a charged object.0678

If you want to prove that they are both charged they have to repel.0682

This, by the way, is known as polarization of those atoms.0687

The whole thing is electrically neutral but because electrons spend a little bit more time over here0693

you get a little bit stronger force of attraction than you do repulsion in the net attraction occurs.0698

Let us talk about that a little bit more.0706

When a charged object is brought in your conductor, the electrons and the conductor are free to move.0709

But when a charged object is brought in your insulator, the electrons are not free to move.0713

They spend a little more time on one side of their orbit trading a net separation of charge known as polarization.0717

The distance between the shifted positive and negative charges multiplied by the charge is known as the electric dipole moment.0726

We just have a couple of diagrams showing that with the conductor at charged rod in a conductor.0734

The electrons spend time here, it is easy to see the attraction with the charged rod and insulator,0740

where the charges are not free to move inside the atoms and set themselves.0746

The electron spends a little bit more time toward the positive side and you still get that net force of attraction.0750

We have been talking about this force of attraction, repulsion between charges,0758

let us go formalize that a little bit and see if we can calculate it using what is known as Coulomb’s law.0762

Charged objects apply a force upon each other known as the electrostatic force or sometimes called the Coulombic force.0768

Similar to gravity, the force of attraction or repulsion is determined by the amount of charge and the distance between the charges.0775

It is actually another inverse square law.0781

It is the inverse square of the distance between the centers of those charges.0784

If we have charge 1 Q1 over here, Q2 over here, the magnitude of the force between them can be found0788

by taking some constant, some fudge factor K × the first charge in Coulomb’s ×0795

the second charge in Coulomb’s ÷ the square of the distance between their centers.0801

Be careful here, we are using R is the distance between the centers of the object.0805

In this case, we our not talking about our radius itself, R is the distance between their centers.0810

When we do this, we end up with K constant of roughly 9 × 10⁹ Nm²/ C².0816

Notice, how similar this is to the force of gravity.0828

Universal gravitational constant × the first mass × second mass × the square of the distance between them.0832

The K and G are different fudge factor to make the units work out right.0840

You have the same form and they work in a very similar manner.0844

The other big difference between gravity and the Coulombic force is that the Coulombic force can repel whereas gravity only ever attracts.0847

Let us take a look at Coulomb’s law now in vector form.0859

Here we have charge 1 and charge 2, R 1 2 is the vector from 1 to 2 and0862

we our also going to define a unit vector in the direction from 1 to 2 which we our going to call R ̂ 1 to 2.0871

These are both positive what we expect is a force of 1 on 2, recalling it over to the right.0878

Writing that mathematically, the force of 1 on 2 is equal to our fudge factor K 9 × 10⁹ Nm²/ C² × the magnitude of the first charge,0885

the magnitude of the second charge ÷ the square of the distance between them and the R ̂ 1 2 was just saying the direction.0897

It is from 1 to 2 on the direction of that line.0904

Another way to write this is we can take this K and write it as 1/ 4π Ε0, the permittivity of free space.0908

9 × 10⁹ or 1/ 4π E₀ are the same thing.0917

We are going to use this form of that fudge factor quite regularly in the E and M course0924

because it is going to help us with some derivations later.0929

But really they are the same thing.0932

This is a constant and this is a constant, they have the same value.0934

Where, if we write it this way Ε 0 is 8.85 × 10⁻¹² C²/ Nm².0937

K= 1/ 4π E₀.0945

If you are doing these calculations and you see this, easy enough to just swap in K for it if you want.0948

9 × 10⁹ or it looks like you can maybe cancel a π or E₀.0954

Go from K that way, they are the exact same thing.0959

Determining the electrostatic force, we have 3 protons separated from a single electron by a distance of 1 micron.0966

3 protons + 3 elementary charges separated from a single electron -1 elementary charge by a distance of 1 micron or 1 × 10⁻⁶ m.0973

Find the force between them and is it attractive or repulsive.0986

Let us do the easy part first, attractive or repulsive, they are opposites they must attract.0990

Let us find the magnitude of that force.0999

Our first charge, Q1 is 3 elementary charges which is 3 × 1.6 × 10⁻¹⁹ C or 4.8 × 10⁻¹⁹ C.1001

Our second charge, Q2 is the charge of our single electron which is -1 elementary charge1020

or -1.6 × 10⁻¹⁹ C and the distance between them is 10⁻⁶ m.1027

Therefore, when we go to find the electrostatic force that is going to be K Q1 Q2/ R².1039

K is 9 × 10⁹ Nm²/ C² × Q1 4.8 × 10⁻¹⁹ C × Q2 -1.6 × 10⁻¹⁹ C ÷ the square of the distance between them which is 10⁻⁶ m².1049

When I go through and do my math, plug it in my calculator,1082

I come up with something like the electrostatic force is about equal to 6.9 × 10⁻¹⁶ N.1085

Of course that must be attractive.1095

Moving on, a beam of electrons is directed into the electric field between 2 oppositely charged parallel plates.1105

What is the direction of the electrostatic force exerted on the electrons by the electric field?1113

Let us see, our electron beam is actually going to be just a bunch of electrons traveling to the right.1119

As the electrons travel to the right, their negative charges they see the positive charge up here,1127

they are going to feel a force towards the positive due to this plane.1132

And down here they see a negative, their negative they are going to be repelled1139

that is also going to push them in the upper directions. Our net force must be up.1143

Let us see if we can put this together into a slightly more involved problem.1146

We have 3 point charges located at the corners of a right triangle as shown where Q1 and1149

Q2 are both 3 micro Coulomb’s, 3 × 10⁻⁶ C and Q3 is – 4 micro Coulomb’s.1160

If Q1 and Q2 are each 1 cm from Q3, find the net force on Q3.1167

Let us start by looking at the force of 1 on 3, we will call that F 1, 3 and that is going to be equal to1175

K Q1 Q3 / R² which is 9 × 10⁹ Nm²/ C² × our charge on Q1 3 micro Coulomb’s1187

or 3 × 10⁻⁶ C × Q3 which we said was -4 × 10⁻⁶ C.1200

I’m just going to work with magnitudes for now and figure out the directions when we are done ÷ the square of the distance between them.1210

1 cm is 0.01 m and we have to square that.1218

I end up with about 1080 N when I plug that into my calculator.1223

We are looking at the force of 1 on 3.1229

If this is negative and this is positive, we are going to have a force upward of on 3 due to 1.1231

That is 1080 N up or I could write that in unit vector notation 1080 N in the J ̂ direction.1240

Let us take a look at the force on Q3 due to Q2.1253

The force of 2 on 3 is going to be K Q3 Q2/ R² which is going to be 9 × 10⁹ × our first charge 3 micro C,1260

3 × 10⁻⁶ × our second charge 4 × 10⁻⁶ / the square of the distance between them again 0.01 m² or 1080 N.1277

As we look at this one, we have a positive charge here1293

and negative charge here so the force on 3 due to 2 is going to be to the right.1296

1080 N to the right or 1080 N in the I ̂ direction, in the x direction unit vector notation.1302

If we want the total force, we will use the law of super position and just add these together that is going to be,1309

we will have this component 1080 N to the right in the i ̂ direction + we have our 1080 N in the J ̂ direction.1319

Or if you wanted to write this in component form that is just 1080 N in the x, 1080 N in the y.1336

If we wanted to know the magnitude of that force, we have 1080 N up, 1080 N to the right which makes a right triangle.1348

We can use the Pythagorean Theorem to figure that out.1354

A total force then, the magnitude of that total force is going just be √1080² + 1080²1358

or about 1527 N and the direction it is going to be an angle of about 45° NE something like that.1370

1527 N, one to the right, one up the combination.1387

Let us take a look at one last example here in this lesson and see how it goes.1395

Putting all of this stuff together, we have 2 identically charged balls of mass 5 mg hanging from the ceiling by a light string of length 20 cm.1399

When we say light string in Physics that means you can neglect the mass of the string, makes life a whole lot simpler.1411

The total angle between them is 12° find the magnitude of the charge on each ball.1416

The first thing I'm going to do is let me draw in a little bit here,1424

I'm going to draw a center line between these just so that I can see what is going on here.1428

I'm going to call the angle write there θ.1434

If the whole angle between them is 12°, this θ is 6° and that θ must be 6°.1440

Let us call this ball 1 and that ball 2.1448

If we start from there, I always like to look at free body diagrams to help me identify forces and what is going on.1452

Starting with a free body diagram let us look at the ball on the left, ball number 1.1459

We have our object which I will draw as a dot, our forces on it we have its weight mg down,1464

we have the tension from the string T, and we have the force of electrostatic repulsion to the left FE.1474

I have a force in an angle and that always makes things more complicated.1486

What I want to do is break up this tension into components along the x and y axis.1490

The way I’m going to do that is I'm going to redraw my free body diagram to make a pseudo free body diagram1495

where I still have MG down, I have the electrostatic force to the left but now I'm going to break up this tension into a component1501

that is vertical and the component that is horizontal.1511

And looking at my angle here or if I were to look at it here as well, it is pretty easy to see1515

that the component to the right is going to be T sin θ that is opposite R θ and this component,1522

the vertical component will be T cos θ because the horizontal is adjacent to our angle as we have defined θ.1530

I can start writing Newton’s second law equation to solve the problem.1539

Let us start with a net force in the x direction, I can say net force in the x direction1543

and I just look at my pseudo free body diagram and write in all the forces in the x direction.1550

I have T sin θ in the positive x direction and I have the electrostatic force in the negative x direction1555

and because this is an equilibrium it is not accelerating, I know that the total must be 0.1567

Therefore, I could write then that T sin θ = electrostatic force.1575

There is one equation we can use, let us try the same thing with the y direction, net force in the y direction1586

and I look at my pseudo free body diagram again, I have T cos θ in the positive y direction – MG.1595

It is not accelerating in the y direction so that = 0.1607

Therefore, I can state that T cos θ must be equal to MG.1611

I got my 2 equations, looks like I can probably do some simplifications here.1621

I'm going to pull a little trick, as I look at this, if that is equal to that and that is equal to that,1626

if I divide the first equation by the second, that also still has to be equal.1632

T sin θ/ T cos θ = FE/ MG.1636

Let us write that, T sin θ/ T cos θ must be equal to our electrostatic force ÷ gravity.1643

T/ T is 1, sin θ/ cos θ is the tan θ so this implies then that tan θ =FE/ MG.1661

or I could rewrite this to say that the electrostatic force equals MG tan θ1677

but the electrostatic force we can get from Coulomb’s law that is equal to K Q1 Q2/ R².1684

Q1 and Q2 are the same, these are identical spheres so this is just Q and Q.1698

The trick then is this R is the distance between them, we could figure that out implies then1703

we know the length of the string and we know the angle, we can use a little trig1712

to realize that distance from here to the center point must be L sin θ.1715

This 20 cm sin θ.1721

The distance between the first and the second sphere must be twice that.1724

Our R must be 2 × the length of the string sin θ.1728

If I rewrite this again, we can say that K Q1 × Q2 if they are the same thing,1734

just Q that is K Q² / R² is going to be 4 L² sin² θ must be equal to MG 10 θ1741

which implies then that our charge itself, rearranging this and solving for just Q must be equal to the √4 MG L² sin² θ tan θ / K.1761

That is quite the equation.1782

Let us substitute in what we know to say that Q = 4 × 5 g, 0.005 kg × G 9.8 m /s² on the surface of the earth ×1784

our length 0.2 m² × the square of the sin of our angle which is 6° × the tan of angle 6°/ 9 × 10⁹ K N m²/ C².1800

Take the square root of that and I come up with about 3.16 × 10⁻⁸ C, the charge on each of those spheres.1821

Hopefully, that gets you a great start on electric fields and Coulomb’s law, electric charges and Coulomb’s law.1837

Come back for the next one and thank you very much for watching www.educator.com.1843

I will see you again real soon, make it a great day.1847