AP Physics C: Electricity & Magnetism Electric Fields

Hi, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about electric fields.0004

Our objectives are going to be to calculate the electric field due to 1 or more point charges.0007

To analyze electric field diagrams.0014

To calculate the electric field by various continuous charge distributions by integration and this is going to get a little bit involved for our second lesson.0017

This is one of the trickier parts of the course.0026

Take some time here, do not always expect to get that on the first try.0028

And then apply Gauss’s law to find the electric field for planar, spherical, and cylindrical symmetric charge distributions.0032

We are going to set some of the base line up in this lesson and then on the following lesson on Gauss’s law specifically.0039

We are really going to dive to that with a bunch of examples and finish up that once.0045

We will not finish this fourth objective here in this lesson itself.0049

As we get started, I would like to mention that a lot of this work in AP Physics C is fairly complicated especially the first time you see it.0054

The notation is a little tricky, there is a lot to think about, a lot to do.0061

You are not going to get all of these the first time through.0065

Doing this has to be an active pursuit.0068

As we come to the example problems, I recommend hitting the pause button and seeing if you can do the problems.0070

If you get stuck, play it a bit more, and if you can get through it, do so and then check your answer with what you see in the video.0075

There is a lot more active learning that needs to be done to really ingrain some of these concepts.0081

Please take the time to do that as you work through our example problems.0087

Moving on, what are electric fields? if the electrostatic force is a non contact or field force, it operates across some distance.0092

The property of space that allows a charged object to feel a force is called the electric field.0101

It is a made up concept to help us understand how that force works.0107

You can detect the presence of an electric field by placing a positive test charge,0111

add various points in space and then measuring the resulting force on that test charge.0116

The electric field strength factor which we symbolize with a E and you will see it0121

in textbooks sometimes with bold or with an arrow over top of it, it is known as a vector.0126

It is the amount of electrostatic force observed by a charge per unit of charge.0131

The electric field vector is the force it feels ÷ the amount of charge.0135

The direction of the electric field vector is a direction of a positive test charge would feel a force.0141

As we talk about electric fields, it is important to remember that these have units and that they are going to have some meaning.0147

The standard units of the electric field are units of force N/ charge N/ C.0153

We are going to find out later that is also equivalent to volt per meter but standard units you are going to see most of the time are N/ C.0163

Let us start off with a simple example and then we will work our way into more and more complex examples.0175

2 oppositely charged parallel metal plates 1 cm apart exert a force of 3.6 × 10 ⁻15 N on electron placed between plates.0182

Calculate the magnitude of the electric field strength, we want to know how strong the electric field is between the plates.0192

We can do this in a pretty straightforward manner saying that the electric field is = electrostatic force ÷ the amount of charge.0199

That is going to be 3.6 × 10 ⁻15 N ÷ our charge, it is the charge that we have on an electron0210

because it is an electron between those plates, the magnitude of the charge on electron is an elementary charge or 1.6 × 10⁻¹⁹ C.0221

When I go through and do this with my calculator, I come up with about 2.25 × 10⁴ N/ C.0232

Hopefully, a pretty straightforward example to get us started.0247

Let us talk about how we could visualize this electric field.0250

Since, you can actually see the electric field you can visualize it by drawing0254

what we call field lines to show the direction of the electric force on the positive test charts that we would put somewhere in space.0257

Electric field lines always point away from positive charges and toward negative charges,0264

away from the positives and toward the negatives.0270

Electric field lines never cross each other and they always intersect conductors at right angles to the surface.0274

Stronger fields have closer and denser lines, weaker field the lines are further apart.0281

The field strength and line density decreases as you move away from the charges.0286

To do just a simple one to begin with.0291

If we were to take a look at a positive point charge, we have electric field lines moving away from it, moving radially away.0293

Electric field is the same at all points that are in equal distance from the center of that point charge so you have the same density of lines.0303

You have the symmetry.0309

You can almost think of a positive charge when we draw electric field lines as kind of a magic blow dryer.0313

It is blowing air out from that point in all directions and what you are drawing is the vectors of the air coming out of it.0319

On the other hand, a negative charge works the same way but all the lines are coming into it.0325

A positive charge would be sucked in this direction.0331

You can almost think of these as the magic vacuum cleaner in space, it is sucking all the air in and you can draw just the wind vectors there.0334

Away from positive charges in the negative charges.0343

Also note that, as you get further away the distance between these increases quickly showing that the electric field strength drops off quickly.0348

We have another example of an inverse square law here, just like we had for the electrostatic force.0356

If we put these charges together to make what is called the dipole to charges, positives go toward negatives,0366

the electric field lines begins at the positive and ends at the negative.0372

They are symmetric and if we were to have a particle hearsay, positive test charge somewhere sitting right there for example,0376

it is going to feel a force in the direction of the field lines which we do not show all of them there but it would feel a force in roughly that direction.0384

Or if we put a positive charge over here, it would feel a force in that direction.0391

What if you have a negative charge?0397

Let us take and do an example with a negative charge.0400

If we put a negative charge, let us say we put it right over here that purple spot,0403

it is going to go in the opposite direction of the field line that would feel a force that way which only makes sense,0408

it would be attracted by that positive test charge and move to the right.0413

If we happen to have 2 positive charges or 2 negative charges, you get this other pattern as well,0417

where there is a point in the middle of those, where there is no force or everything cancels out.0423

Some examples of how you can draw electric field lines.0429

We did a couple point charge diagrams, let us talk about the electric field due to point charges in a little bit more detail.0434

Electric fields are caused by electrical charges, of course.0441

The electric field due to a point charge can be derived from the definition of the electric field and we just learned about in our previous lesson, Coulomb’s law.0445

If the electric field is the force ÷ the charge and the magnitude of the force is given by Coulomb’s law here.0453

Or if you want it to, we can go to our other notation and say that the force is 1/4 π E₀, remember that is equivalent to the K × Q1 Q2/ R².0462

We can start to put these together and say for example the electric field here is the force K Q1 Q2/ R² ÷ Q.0476

We end up with one of the Q cancels out makes a ratio of 1, we end up with KQ/ R².0489

Or if you are looking at it in this notation here, the electric field would be 1/ 4π E₀ × Q/ R².0498

We have our formulas for the electric field due to a point charge.0510

To find the electric field due to a multiple point charges, what we do is we take the vector sum of0517

all of the electric fields due to the individual point charges.0522

We use the law of super position to just add all of those up.0526

You will notice a bunch of comparisons electricity to gravity.0531

The formula electric forces K Q1 Q2/ R² compared to gravity G M1 M2/ R².0535

The only major difference mass vs. charge and the constant, the fudge factor.0542

The electric field strength is the force ÷ the charge.0548

With gravity, the gravitational field strength was the gravitational force ÷ the mass.0552

We can also write this, our formula is KQ/ R² for electricity, for gravity GM/ R², see they are parallels here.0559

The constants are different but that is just because we are trying to make the math workouts so our units come out to something that makes sense.0568

The charge units are Coulomb’s, the mass units are kilograms.0575

The math has tons of similarities.0578

One that we have not pointed out yet though that is a very important, gravity only attracts.0581

The electrostatics can attract and repel.0589

This is very important.0598

Let us do a problem where we have got the electric field from a couple of point chargers.0600

Find the electric field at the origin due to the three charges shown in the diagram.0607

We got a charge here at 0, 8 m which is 2 C, a huge amount of charge.0611

Over here at 2, 2 we have got a + 1 C charge and here at 8, 0, 8 m on the X axis we have -2 C charge.0618

How can we do that?0628

What I'm going to do is I'm going to find the electric field of the origin due to each of these 3 charges individually and then take their vector sum.0630

First thing is let us find the electric field at the origin due to do this green charge up here.0639

Since, this is a positive charge we should be able to see that down here at the origin,0645

the field lines go away from positive charges our electric field should be in that direction.0649

The electric field due to that + 2 C point charge at the origin is going to be KQ/ R² where K is 9 × 10⁹.0657

Our charge is 2 C ÷ our distance from the origin 8 m that we have to square or I get about 2.81 × 10⁸ N/ C down which is in vector form 0, -2.81 × 10⁸ N/ C.0670

There is the electric field at the origin due to this green charge up here.0700

Let us do it now for the red charge.0704

We can see if this is a -2 C charge, the electric field is going to go toward the -2 C charge so they should feel that the origin is going to go toward the right.0709

We already know the direction.0719

The magnitude then is given by KQ/ R² which is 9 × 10⁹ Nm²/ C² × 2 C charge.0722

I’m not going to worry about the magnitude since I already know the direction, ÷ the square of the distance 8 m between those.0733

Or again, we end up with 2.81 × 10⁸ N/ C to the right or in vector notation that is going to be 2.81 × 10⁸ N/ C in the x and nothing in the y, 0.0742

We have got to figure out our portion due to this blue charge is 1 C charge at 2, 2.0762

By inspection, we should be able to see that the field lines are going to get away from the positive charge so0770

the origin that is going to be going in that direction.0776

To find its magnitude however, our same formula E = KQ / R² where K is 9 × 10⁹.0780

Our charge now is 1 C and the square of this distance.0791

To do that, we got to do a little bit of math.0795

I'm going to make a right triangle realizing that this is 2, this is 2, so I can use that the Pythagorean Theorem to find out that distance that is 2 and 2.0799

This is going to be the √2² + 2² so R² is √2² + 2² is just √2² + 2²² or 8.0809

That is 9 × 10⁹/ 8 which will be 1.13 × 10⁹ N/ C down into the left in a 45° angle.0825

What we have been doing this in vector notation so what we do with that?0838

We do not have a straight x and y component right from our answer yet.0843

We got to do a little bit of math here and realize that that is going to be equal to.0847

The x component is going to the left so that is going to be -1.13 × 10⁹ N/ C in the x component cos 45°.0852

Our y component is going to be going down as well so that will be negative.0866

I end up with -1.13 × 10⁹ N/ C sin 45° or when I do the math there that is = -7.95 × 10⁸ N/ C in the x and -7.95 × 10⁸ N/ C in the y.0871

Just taking a minute to step back.0899

We now have the magnitude and the broken down vector notation of the electric fields due to each of these 3 charges.0900

To find the total electric field, we need to add these keeping in mind of their vectors.0908

We cannot just add their straight magnitudes, we have to add vector components.0912

To find the total electric field, we are going to add up all of the x components and all of the y components.0917

Our x components we have a 0 from the green one.0926

We have 2.81 × 10⁸ from our red charge and we have a -7.95 × 10⁸ N/ C from our blue charge.0930

We have these three to add up for our x and for the y component we are going to do the same basic thing.0949

We have -2.81 × 10⁸ N/ C due to our green charge.0958

We have no contribution in the y direction due to our red charge and we have -7.95 × 10⁸ contribution from our blue charge.0965

We add up the x, we add up the y to find out that our total electric field is going to be,0981

when we add all those up we are going to get about -5.14 × 10⁸ N/ C in the x direction and about -1.08 × 10⁹ N/ C in the y.0990

Our total electric field is going down into the left, you can figure out the angle if we wanted to.1013

But this gives us the vector sum of these 3 electric field contributions to give us our total.1019

Let us make sure we highlight that after all that work.1027

There is our total electric field at 0, 0 due to those 3 point charges.1031

It is kind of complicated but if you take it step by step, it is really not bad.1039

Let us take a look at an example where we are looking to find out where the electric field is 0.1046

Here on this line, we have 2 point charges, we have a + 1 C charge at x = -6 m and the + 2 C charge at x = 5 m.1051

The electric fields from both of these go away from positive charges so we got an electric field to the right here and to the left on that side of them.1063

Due to the red charge, we have an electric field to the right, electric field to the left.1072

There should be some point on this number line where the red and the blue exactly cancel out and there is no electric field.1078

By inspection, as I look at this, we have a smaller charge here so I would expect that1085

we would have the point at 0, maybe a little bit to the left of the middle of those 2 charges.1091

For 5 and 6, about -.5 would probably be halfway between them so I'm expecting we would probably get an answer that is somewhere over in this region.1098

Before we even go into the math, let us qualitatively take a look, make something of a guess and1108

then we will go see if everything matches up when we go through our math here.1113

What we are going to do to find this out is, let us say that this is our point and they do not know exactly where it is1118

but we will define the distance between our 1 C charge and our point where the electric field is 0, let us call that distance R.1123

If we call that R, then the distance from here to our red charge must be 11 -R because the total distance between them is 11 so R + 11 – R is 11 m.1133

We can go in and we can start doing our math to find out exactly where that point resides where the electric field is 0.1148

Looking at the blue charge first, the electric field due to that 1 C charge is K Q1/ R².1155

If we look at our red charge over here, the electric field due to that is going to be K Q2/ our distance is going to be 11 - R².1165

The total electric field we know has to be 0 at this point.1178

The total electric field which is going to be our electric field due to the blue charge1183

which is K × 1 C ÷ R² - this one was going to the left our red electric field contribution1188

K × 2 ÷ 11 - R² and we know all of that has to = 0.1203

We have got an algebra problem.1217

We have got K on the left, K on the right, 0 over here, so what we can do is we can factor out the K's1220

which implies 1/ R² from the left, after we cancel out our K’s must be equal to,1226

I’m going to move this portion to the right hand side.1240

We have 2 ÷ this 11- R it is kind of awkward to use that way.1242

I’m going to multiply that out 11 - R² is 11 -R × 11 - R or 121 11² -11 R -11 R is -22 R + R².1250

I’m trying to solve this for R so what I might do here is maybe I will cross multiply this and just make it look a little simpler,1267

to say then that we will have 2 R² = we still have our R² -22 R + 1211275

which implies then if I subtract R² from both sides, I get R² = and then we have R² =.1293

Let us do it a little bit differently.1310

I think we can save a step here.1311

We are going to put this into a quadratic equation so I could bring all of this over to the left, subtract all of this and say that R² + 22 R -121 = 0.1313

And that is a little bit slicker.1325

I'm taking 2 R² - R² to get R² -22 R or adding 22 R to both sides and then subtracting the 121 to rearrange.1326

We got a system that we can put in a quadratic formula because I do not see a very easy way to factor that in any other direction.1338

If we use the quadratic formula, we can then say that our R must be equal to, quadratic formula -B ± √B² - 4 AC/ 2 A.1345

In this case, -B is going to be our -22 ± √22² -4 A × C 1 × 121 is just going to be that -121/ 2 A and A is 1 so that is ÷ 2.1364

A little bit of calculator work and you should come up with 2 possible answers, either R = 4.56 or R = -26.6.1390

In here, we got to use a little bit of common sense and say which one of these make sense?1402

If R is 4.56 that means this distance over here, pretty close to our guess a bit, that make sense.1406

If R was -26.6 we are way to the left of the screen and that is not going to make any sense.1413

We have got a left vector from this electric field, left there and they are going to add together.1419

You are not going to get 0.1423

We can cross that one out as not making sense for our problem.1425

Therefore, we know that R must equal 4.56.1428

Where on this diagram is that 0 point?1434

If we start at -6 and we go 4.56 to the right that tells us that our x position1438

must be -6 + 4.56 or about -1.44 m on our number line.1445

The exact answer should be right around there.1454

Finding the electric field due to a couple of point charges we can use the law of super position.1461

Let us take a look at a little bit more straight forward example for a moment, stepping back.1468

A distance of 1 m separates the centers of 2 small charge sphere, they exert a gravitational force FG and electrostatic force FE on each other.1473

If the distance between the center of the spheres is increased to 3 m, we are tripling that distance,1485

what happens to the gravitational force and electrostatic force?1491

If we triple the distance, they get further apart, of course the force has to get smaller so we can get rid of those choices.1495

Again, this is an inverse square law relationship for both the gravity and the electrostatic force.1503

Inverse square law, triple the distance 1/9 of the force.1512

The correct answer here must be A.1517

The direction of the electric field due to a point charge.1523

In the diagram here below, P is a point that is near a sphere with charge -2 C.1527

What is the direction of the electric field at point P?1532

If you remember our rules for electric field lines, electric field lines go in to negative charges.1535

Over here at P, we must have a force to the left.1545

The electric field there must be pointing to the left, assuming that would be a positive test charge.1550

To the left would be the correct answer.1557

Moving on, let us see if we can get into a little bit deeper derivations.1565

To do that, we are going to have to talk for a moment about charge density.1570

Let us assume we have something like a line that is uniformly charged.1574

The total charge ÷ the length of that line would give you the linear charge density which1578

we are going to give the abbreviation the symbol λ Λ, it is amount of charge per unit length.1583

In other cases, we may be looking at a distribution of charge that is on some sort of surface.1590

A linear charge density does not make sense but instead we are going to look at total charge per unit area1595

that we are going to give the symbol sigma Σ, a surface charge density.1602

We may also be looking at objects that are three dimensional that have some volume.1608

When we have charged uniformly distributed through a volume, it can have some volume charge density taking the total charge ÷ the unit volume.1616

And that we are going to give the symbol ρ Ρ.1625

We are going to use these definitions fairly regularly throughout the rest of the course.1627

A linear charge density, charge ÷ length.1631

An area charge density, charge ÷ area.1635

The volume charge density, charge ÷ volume.1638

Let us put this in to play as we try and find the electric fields due to a semicircle of charge.1644

We have a thin insulating semicircle of charge total charge Q at some radius R centered around at point C.1652

We want to find the electric field at point C due to the semicircle of charge.1660

Before I even begin the math, the first thing I’m going to do is observe for a minute.1665

If this is symmetrical around C and it is uniformly charged, there should not be any horizontal component of electric field.1669

That should all cancel out.1677

We should only have a vertical component that we need to deal with.1679

By symmetry, we can simplify the problem and work in just the horizontal direction.1683

As we do this, the first thing I’m going to do is I'm going to define some linear charge density where our linear charge density,1688

the charge per unit length on this, is the amount of charge on that length ÷ its length,1695

which in this case is going to be its total charge Q ÷ the length of that semicircle.1703

If the distance around an entire circle is its circumference 2 π R, half of that for a semicircle must be π R.1710

Our linear charge, hence the Λ must be charged ÷ π R.1719

As we do this, we are also going to break this up and our strategy is going to be to take and look at little tiny pieces of the semicircle.1725

And as we go through it, we are going to find the electric field due to each of those pieces and then add them all up.1733

I'm going to define this little piece and draw a couple of lines here just to make this a little bit clear.1740

It is not quite as clear as we wanted there.1751

It looks something like that.1755

What we are going to do is we are going to call this R Θ so that this angle right there is a little differential of Θ, a little tiny piece of Θ.1757

The amount of charge that we are enclosing up here in this little piece, that length is going to be RD Θ.1769

The amount of charge that is enclosed there which we are going to call DQ, a little tiny bit of the charge1778

is going to be that length RD Θ × the linear charge density in that area which we just defined as Λ.1783

The differential of charge, the tiniest little bit of charge and that little bit is RD Θ × Λ.1794

To find then the electric field due to that just that little bit of charge, we will find the differential.1802

The little tiny piece of electric field which we are only looking in the y direction is going to be, we could use KQ/ R².1808

I’m going to use 1/ 4π E₀ version, 1/ 4π Ε 0 is the same as K × our charge DQ ÷ the distance between them R²1817

and we are looking at just the y components, so sin Θ.1832

Once we have that we can start to get set up and do all of or math.1838

The hardest part of these problems is just getting into the habit learning how to set these up.1841

There is no better way than to just do a problem after problem after problem and check as you go on.1846

Let us take this a little bit further.1854

We have our RD Θ here, our differential of the angle if that is Θ and we just said that the differential of the electric field1858

in the y direction due to this little bit of charge was 1/ 4π Ε 0 DQ/ R² sin Θ which implies then since we know that DQ = Λ RD Θ.1866

We just said that on our previous screen that the differential EY = 1/ 4π E₀ and replace DQ with Λ RD Θ.1881

Λ RD Θ we will make that stand out ÷ R² × sin Θ.1903

To add up all these little bits of the electric field what we are going to do is an operation called integration.1918

It is adding all of these tiny little pieces up to get a whole.1925

We will integrate on both sides, do the same thing to both sides you get the same thing.1929

We are going to integrate, to get the entire electric field we need to integrate from Θ = 0 all the way across until we get to Θ = π.1937

Because once around an entire circle is 2 π, halfway around the semicircle would be π.1946

We are going to integrate from Θ = 0 to π.1952

What we can do, what is nice over here is you can pull constants out of the integration.1959

1/ 4π E₀ that is all a constant anywhere in our problem, that is not going to change.1964

Let us rewrite this again and say that this then is, here I see a simplification.1972

R and R² we can simplify there.1979

We can write this as integral of DEY = the integral from Θ= 0 to π.1983

We have 1/ 4π E₀.1991

We got R in the denominator, we have got Λ sin Θ D Θ.1995

We write it that way before we really get into the math, it is like simplifying as much as possible.2005

What we can do is we can pull all of the constants out of the integral.2011

The left hand side is the integral of all these little pieces of tiny pieces of the y give us the total EY.2015

The electric field in the y direction or the y component of that is going to be equal to, pulling other constants,2023

Λ is a constant 4π E₀ and R are all constants for the purposes of this problem.2029

That will be Λ / 4π E₀ R integral from Θ = 0 to π of sin Θ D Θ = Λ/ 4π E₀ R, the integral of the sin is the opposite of the cos.2034

We have the opposite of that cos of Θ evaluated from 0 to π which implies then the y component of the electric field2061

that is going to be Λ/ 4π E₀ R and then we have – cos π - cos of 0.2071

Cos of π is -1 so --1 is going to give us a 1 here, 1 – cos 0 is 1.2089

Two negatives we will get 1 there.2098

What I end up with is the y component of my electric field is going to be Λ/ 4π E₀ R.2100

All of this is just 2.2112

Therefore, the y component of our electric field becomes 2 Λ / 4π E₀ R.2119

2/4 I can reduce to 1/2 or Λ/ 2 π E₀ R.2128

There we have our y component of our electric field but we can even take that a bit further.2138

That is a good answer but we know the total charges so we can substitute that and get rid of that Λ knowing that2144

we already defined previously Λ = Q/ π R to say then that our y component of the electric field is Q / π R × whatever we had before 2 π E₀ R,2151

which implies then that the electric field, the y component is going to be, we got a π² and R², so that is going to be our total charge Q ÷ 2 E₀.2177

We have got π² and then R².2193

Quite a bit of math involved but our strategy is pretty straightforward.2203

Find and break up our little bit of charge.2208

Find the electric field due to that little bit of charge and then we just add that up for all of those little bits of charge to make the whole.2212

That is what we are really did with all of this math.2219

Probably worth doing another couple of these.2225

Thin, straight, uniform line of charge, find the electric field at distance D from a long straight and slating rod of length L at a point P2229

which is perpendicular to the wire and equidistance from the end of the wire, assuming the wire is uniformly charged.2239

We have got this point P lined up right with the center of this length L of an insulating rod that is uniformly charged.2245

Our strategy again is going to be very similar.2253

Let us even write that down to make sure we have got it nice and clear.2257

First thing we are going to do is divide the total charge Q into smaller charges Δ Q.2261

We are going to find the electric field due to each Δ Q.2275

Once we have done that, we are going to add up the electric field due to each Δ Q to get our total electric field.2288

And finally, as I look at this problem it looks like we can make another symmetry argument and should be pretty obvious that2307

if this is centered on our wire then we are not going to have to worry about any vertical components of the electric field.2313

It is all going to be horizontal.2320

We are going to use symmetry to show that the y component of the electric field is 0 or therefore we are only going to worry about EX,2321

the X component due to symmetry.2332

It is always nice to simplify these wherever you can because they are complicated on their own.2337

We have some uniform charge on this insulating rod, if they are uniformly charged we can talk again about a linear charge density.2342

Λ was going to be our total charge Q divided by our total length of the wire L.2352

Before we did the math, we are going to set this up as carefully as possible.2357

If this total length is L, I'm going to call this Y = - L / 2 and we will call this end L/ 2.2362

I have got our distance D toward point P and let us take just a little bit of charge, a little bit of length of that wire and2372

we will find out what that is going to look like as far as the electric field contribution.2379

I'm going to draw a line.2385

There is our electric field due to that little bit I, we are only going to be worrying about the x component of that.2392

If we do this, we will define this angle as Θ to that little piece I.2399

We will have some y component, some y value, y distance from our center line which will call yi to that little piece of line i with charge Qi or DQ.2404

As we do this, the length of this is probably going to be important.2416

The distance from our charge due to a point where we want to know the electric field.2419

We will call that Ri and by the Pythagorean theorem I can see that that is going to be equal to the √D² + yi².2423

All right, that looks pretty good.2438

Finally, let us write our equation for the x component of the electric field due to this little bit I before I get heavy into the math.2440

The electric field due to this little portion I in the x direction is going to be 1/ 4π E₀, our K value again × the charge DQ ÷ RI²2449

the distance from our charge to our point × the cos of Θ I because we are after the x component.2466

I think we have got everything we need for our setup.2474

It is time to start getting into the math and plugging through this thing.2478

Let us dive in.2484

The electric field in the x direction = 1/ 4π E₀ × Δ Q in that little piece ÷ RI² × cos of Θ I which implies then2487

we know that RI = √y I² + D² and the cos of Θ I, as we look here cos SOHCAHTOA.2509

Adjacent / hypotenuse is going to be D/ RI.2522

cos Θ = D/ RI and we can write this as EI in the x direction is 1/ 4π E₀ Δ Q/ y I² + D² is RI².2527

We can get rid of the square root sign × cos Θ I which is D/ RI which is y I² + D² √ ½.2546

Let us get rid of that Δ Q now.2560

Our Δ Q we know is Q/ L × our little bit of differential of y up and down that length.2565

Then the x component due to that little piece I is 1/ 4π E₀.2575

We have our Q, we have our D, we have our Dy, be careful because D and Dy are different.2585

D does not mean D multiplied Dy.2595

Dy here is differential of y, little tiny piece of Dy.2597

Q D Dy/ L, capital L is still down there.2601

We still have our YI² + D² but because we have y² + d² × y² + d² ^½.2606

I can write that as 2³/2, which implies then that the EI in the x direction.2616

Let us do the whole thing, let us integrate both sides.2628

Let us say that just by looking at that little piece, the x component is going to be the integral from y = - L / 2 to L / 2.2632

We are going to add up from here all the way to the top, all of those little pieces of 1/ 4π E₀ × we have got Q × D/ L.2641

We also have multiplied by our DY/ y I² + D²³/2,2657

which implies then that the x component of our electric field is, we can pull other constants.2671

Q and D are not going to change.2680

Our L is not going to change and 4π E₀ is not going to change.2682

Those are all constants for the purposes of our problems so they can come out of the integral sign.2686

We have QD / 4π E₀ L × the integral from - L/ 2 to L/ 2 of DY/ yI² + D²³/2.2691

This is a tricky integral, not one that I expect you to be able to do with that.2718

When we get to an integral like this where I like to go is the back of a calculus book, the front.2722

It is a great place to go and look up some of those integration formulas.2727

If you can find a form that this fits where it solved for you, let somebody else do the work.2732

Use the formula that gives you the integration and that is exactly what I'm going to do here.2736

I have looked this up already.2741

This implies then using the formula that I looked up, the integral of some Dx/ A² + x²³/2 = x ÷ a² × √a² + x² + constant of integration.2744

This fits that same format.2771

Instead of Dx, we have got Dy, our x² we got yi², a² we got D².2773

We can use this formula to do our integration.2780

Taking the next step using this formula, then our electric field x component is Q D/ 4π E₀ L.2785

I'm very careful to make sure that my solution is using the same piece as the same correlation with this formula.2798

I use green to help highlight what we have done here.2807

Instead of x we have y/, instead of a² we have our constant D², then we have got √a² + x² which for us will be √y² + D² ^½ .2810

We are evaluating this whole integration as for - L/ 2 to L/ 2.2830

Evaluated from - L/ 2 to L/ 2 which is equal to,2836

Our left hand side over here still stays the same Q × D/ 4π E₀ L.2844

We have got D² here and the D up here, we can do some simplification and leave us with just a D in the denominator.2855

We will put a D down here and that one can go way.2865

We also now have to deal with all of this mess.2872

Let us just go right in, I will do this over here to give us some room.2878

L/ 2/ L / 2² + D² ^½ - -L/ 2 ÷ -L / 2² + D² ^½.2883

A little bit more simplification, that is = Q/ 4π E₀ DL this piece, × L/ 2 - -L/ 2 that is going to be the same denominator when you square that.2909

It is just going to give us one total L so that is × L ÷ L/ 2² + D² ^½.2925

Putting all of this together, we got an L down here and L up here.2939

We can simplify that, I end up with the x component of the electric field at point P due to this line of charge as Q/ 4π E₀ D × L/ 2² + D²¹/2.2944

We will make it nice and pretty and put in the square root sign there.2969

There is our answer.2973

Almost feel I need a nap after that.2978

There is a lot involved, step by step.2981

Each little step is not that difficult but there is a lot of bookkeeping, a lot of keeping track of the details.2984

When you see these problems, take your time, do not let them intimidate you but it is all about practice.2990

Just doing it again and again and again, nobody gets it on the first time.2995

It is scary the first couple of times you look at it and it is supposed to be, stick with it.3000

Let us take the solution and do a little bit more with a lower here though.3008

Let us see what would happen if this line was infinitely long?3012

In that case, the electric field in the x direction we will find that out by taking the limit as our length of that wire3018

approaches infinity of our answer from before, Q/ 4π E₀ D × L/ 2² + √D².3028

That can give us the answer.3048

You got to be careful as we did the limit because as L gets very big you would think that this is just going to dominate.3051

You also have to remember that Q increases and goes toward infinity as L does.3057

We cannot do it quite that simply.3061

Instead, we can look at this and say you know as L approaches infinity, L/ 2² is going to get so much bigger than D² that this D² really is not going to matter.3063

The √L/ 2² is going to simplify to L/ 2 in the limit.3073

We would get Q/ 4π E₀ D L/ 2 but we also know if you recall Λ is Q/ L or Q = Λ L.3079

As L gets very big having a set charge is going from infinity does not make sense.3098

We got to put it in terms of a linear charge density.3103

We would write this as Λ L/, we have here L / 2 × 4 is just going to give us L/ 2 L.3106

We will have 2 π E₀ DL and our L make a ratio of 1.3120

We can state then that the electric field in the x direction due to that infinitely long line of charge would be Λ/ 2 π E₀ D.3129

That looks a whole lot nicer.3142

We are going to come back to a problem like this and solve it in a different way,3145

a much smoother way in our next lesson when we talk about Gauss’s law.3149

How about the electric field if the distance D is infinite?3154

Instead of an infinite line, we are an infinite distance away.3157

We can follow the same basic strategy, the horizontal component of an electric field3161

is going to be the limit, as D approaches infinity of Q/ 4π E₀ D × √L/ 2² + D².3168

As D approaches infinity now, this L/ 2 piece is not going to matter.3189

The √D² is going to be D so we are going to get Q/ 4π E₀.3195

We had 1 D, we got a D from this piece, D².3204

That look pretty familiar, that looks like it is a point charge.3209

Instead of D, think of R, how far you are.3213

As you get infinitely far away from this wire, it is going to start looking like a point charge.3216

Imagine you are zillions of miles away, you cannot tell it is a line.3221

Electrically you cannot tell either, it starts to act like a point charge.3224

That answer should make sense as well.3227

You are very long distance away that piece of wire starts to act like a point charge.3229

Let us do another derivation here.3239

I think we need all the practice we can get.3241

We are going to try and find the electric field at a point on the axis here P that is perpendicular to a ring,3243

a thin insulating ring that is uniformly charge at some radius R.3249

We are going to use the same strategies again, all of the real physics work is in the setup.3255

By symmetry as I look at this, I can see that the only electric field is going to be in z direction.3261

Everything else along the x and the y cancel out so that will help a little bit.3267

We can also define for this problem a linear charge density Λ which is Q/ L.3272

It is going to be Q/ the length of our ring is a circumference 2 π R.3280

Or Δ Q as we look at some portion of the ring is going to = our linear charge density × our radius × the arc length that we go through here.3289

I'm going to call this as we go up to here, let us call this direction φ Φ.3306

Times some little bit of Φ, differential of Φ, D Φ.3314

Once we have done that, we can start looking at some of the geometry of the problem here.3320

We need to know the electric field at point P due to this little bit of charge.3325

I will draw my vector distance here being nice happy green color.3330

This will be our electric field due to that little bit of charge.3342

This distance we will call Ri which in this case we can use the Pythagorean Theorem, it is going to be we have got 1 piece of the triangles Z, the other piece R.3353

This will be the √z² here is the component + R² the constant, and the radius of our hoop here, our ring of charge.3368

We will call our angle here Θ i.3379

What else do we need to define?3385

Usually we can get that little piece of electric field defined before we really dive in.3387

Let us say that the electric field due to that little piece we are looking at just as z component 1/ 4π E₀.3394

We have got our charge Δ Q ÷ our distance Ri².3403

It looks like we are dealing with the cos of Θ I, the horizontal component as we have set this up currently.3409

We can do a little bit more work here.3420

We know that RI is z² + R² we already defined that.3421

RI = √z² + R² and cos of Θ i.3426

We also look at the cos that is the adjacent / the hypotenuse that is going to be, our adjacent side is our z component and our hypotenuse is Ri again.3434

Rewriting this, Ei in the z direction is 1/ 4π E₀ × our charge Δ Q and pulls up there ÷ Ri²3448

is going to be z² + R² × cos Θ i which is going to be z / √z² + R².3463

If you want to simplify that before moving on which is probably a good idea, we have 1/ 4π E₀.3481

We will throw in our z, we have got z/ z² + R²³/2 × our Δ Q.3491

Before you get too nervous, this one is actually a little simpler than the last one.3510

Let us see if we can dive into the math now and solve this one.3514

Electric field due to a little bit i in the z direction is going to be 1/ 4π E₀ z/ z² + R²³/2 Δ Q.3522

But we know what Δ Q is.3541

Δ Q is just going to be our linear charge × Λ × the radius × D Φ.3546

Then we get that E in the z direction is going to be the integral from Φ = 0 all the way around the circle this time so from Φ= 0 to 2 π of 1/ 4π E₀.3557

We have our z/ z² + R²³/2 Λ R D Φ.3576

Next up, electric field in the z direction = let us pull all of our constants out of here.3590

4π E₀ a constant in our problems z constant, z² must be a constant, R a constant, Λ is a constant, R again is a constant.3597

We can pull most of these out of the integral sign so we have Λ R/ 4π E₀ z / z² + R²³/2 integral from 0 to 2π of D Φ.3609

That looks so much friendlier.3632

The reason we can do that is none of these depend on our variable of integration here Φ.3634

They are all constants with respect to any changes in Φ.3639

The integral of DΦ is just going to be Φ.3644

Therefore, we can say that Ez = the integral of DΦ is Φ from 0 to 2 π that is just going to be 2 π.3647

We have 2 π from this × Λ R ÷ 4π E₀ × z ÷ z² + R²³/2.3656

This is much better.3682

Our Λ we know what that is equal to, so this implies then knowing that Λ we already defined as Q/2 π R.3684

We can say that the z component of our electric field is = we have 2 π Λ was Q/ 2 π R.3695

We still have R up here and the 4π E₀ down here, a z/ z² + R²³/2.3714

It looks like we can make a couple of simplifications here.3729

2 π/ 2 π does make a ratio of 1.3732

We have got R and R, those make a ratio of 1 and that looks pretty good.3735

We can rewrite this then the electric field in the z direction is going to be 1/ 4π E₀.3742

We got a Q and a z ÷ we are left with z² + R²³/2.3753

Hopefully, you are starting to see a pattern in these.3772

The same sort of strategy applied again and again and again.3775

If we take a single ring, we just solved for the electric field there but what happens if we start solving for electric field due to a bunch of rings?3780

That is going to be our next problem.3789

A uniformly charged disk, define the electric field due to a uniformly charged3792

and slating disk of some radius R to point B perpendicular to the disk as shown on the diagram.3797

To do this when we are going to rely heavily on what we just did.3804

Realizing we already talked about how you find a ring of charge, our strategy is going to be to take our ring that we just solve for.3808

We know the electric field due to that ring and just say let us start with a little ring at the center,3823

find its electric field and make a bigger ring and add it, and a bigger ring.3828

We are going to integrate from the radius is 0 all the way to R, of all of these little rings to get the total electric field at point B.3834

Relying heavily on what we have done previously.3843

Again, we can make a symmetry argument that we only have to deal with the z component because we are centered on this disk.3846

This time we are going to define a surface charged area of Σ which is total charge Q ÷ A.3853

In this case, our area of the disk is π R².3861

If this is our ring here in purple, the charge of that ring is Δ Q on our ring there.3869

We will call the radius of our ring some Δ,3878

Let me make that a lowercase just to help differentiate what we are doing.3883

A Δ R, as we go out, we will call this little r, this variable that we are going to change.3888

We can look at the charge contained Δ Q.3895

Δ Q is going to be that surface charge density × the area of this ring.3900

To find the area of the ring, we have got to consider here is our ring, what happened if we cut it 1 point and then spread it out to make a line.3905

We get a little skinny rectangle.3913

The length of that rectangle is going to be Σ × the area, the length of that is going to be 2 π × Ri and the thickness of that rectangle is going to be Δ R.3915

With that in place, we can start to find our electric field due to some little bit of our disk, our little bit being that single ring.3934

The electric field in the z direction due to that little ring is 1/ 4π E₀ again × we have our z Δ Q due to that little piece of i/ z² + Ri²³/2.3944

Coming from the solution to our previous problem, we just did that work.3973

This implies then that Δ Q, using Δ Q = Σ × 2 π Ri Δ R that the electric field due to that ring i in the z direction3978

is 1/ 4π E₀ × z × Δ Q i which we just define over hear in green.3997

Σ 2π Ri Δ R ÷ z² + Ri²³/2.4007

There is our starting point then we are going to integrate this,4024

we are going to add these all up as we go from a little tiny ring here in the center4027

and expand it outwards and upwards until we get to the entire width of disk.4031

I think we are ready for the math.4038

Electric field in the z direction is going to be the integral adding all these up of 1/ 4π E₀ × z Σ 2π Ri DR / z² + Ri²³/2.4041

Let us pull out our constant and see if we can simplify this a little bit.4069

Which implies that Ez = Σ and z can both come out, those are constants.4072

Our 2 E₀ can come out and this 2 and 4, we can do a little bit division there and say that we end up with Σ z/ 2 E₀.4079

A π in a π cancel out, integral from r = 0 all the way to the R, the entire radius of our disk.4092

We are left with r DR / z² + R²³/2.4100

To integrate this I'm going to do a little bit of a substitution trick.4112

This is not quite so obvious to me how to do so, I like to simplify things as much as possible.4116

I'm going to say, I’m going to define a new variable U and say that u = z² + R².4121

This would be u³/2.4129

If I do that the differential of u, our only variable here is the R so that is going to be 2 RDR.4132

To make all of this fit that form, I have got u down here.4141

I almost have Du up here.4146

If I had a 2 here this would be Du/ u.4149

Let us multiply it to here and change the values.4153

I have to divide by a 2 as well out there.4155

What I have is a constant × the integral of Du/ u³/2 and that I know how to integrate.4159

Not Du/ u, Du/ u³/2 and that I know how to integrate.4171

We can rewrite this electric field in the z direction, we have Σ z/ 4 E₀ integral.4177

I have changed my variable of integration from R to u.4191

Now we went from r = 0 corresponds if r is 0, u is z².4194

We are integrating from u = z² to r = R so this must be u = x² + R².4201

Our simplified version of what we are integrating now becomes u⁻³/2 Du, much easier for me to integrate.4214

That becomes Σ z/ 4 E₀ × -2/ √u¹/2 evaluated from z², 2z² + R².4227

There is the integration I have to substitute in here.4249

We end up with Ez = Σ z/ 4 E₀ - 2/ z² + R²¹/2 - -2/ u ^½ which in this case is z² ^½ or z.4252

A little bit of algebra here that is equal to Σ z and we can pull out a 2 here from both sides ÷ 4 E₀ × that is going to be 1/ z-1/ √ z² + R².4283

If I have a 2 here and the 4 here, we can simplify there the electric field in the z direction will be 2 there, 2 there, Σ/ 2 E₀.4309

I can take out the z and say that we have 1 - z/ z² + √R².4328

This is just a little bit of algebra going from here to here and simplifying that out.4346

There is our answer.4350

I think I’m happy with that, not really feel like simplifying any further.4354

What if that disk or an infinite disk?4359

What if you expanded it and expanded it until it is infinitely big?4362

We almost have the solution to an infinite plane of charge here,4367

let us do that by taking the limit as R approaches infinity of our answer which should be Σ/ 2 E₀ × 1 - z/ √z² + √R².4371

As I do this, R gets really big we are going to have 1 – 0.4392

The R is going to dominate over here so we end up with Σ/ 2 E₀.4402

If that disk is infinitely big as it becomes an infinite plane, there is no dependence on how far you are from that plane at all.4412

Which makes sense, if it is an infinite plane, that distance from it really have any tangible practical meaning any more.4420

We get this nice simple Σ/ 2 E₀, it is all a function of your surface charge area.4429

Practice is good, let us keep going.4438

Hit the break and by all means do so but comeback in we will finish this up here with one last problem.4440

Find the electric field due to a finite uniformly charged rod of length L on its side at some distance D away from the end of the rod.4448

Uniformly charged rod we are some distance from it, let us find the electric field here.4456

By symmetry arguments, it looks like you are only going to need the x component.4462

We will do the same thing, let us take and break this up.4467

Let us call this little piece of that Δ Q and close there.4470

We will define some x distance and distance from P to our Δ Q, there is our x.4476

This Δ Q is going to be = to some linear charge density × whatever that little piece of x is.4484

Δ x or we could say that the differential of Q that tiny little piece of charge enclosed is the linear charge density × Dx.4495

Where Λ is linear charge density is Q/straight forward L.4507

The electric field due to this little tiny I, bit of charged in the x direction is 1/ 4π E₀ Δ Q/ distance².4514

With that I think we are set to start doing our math.4531

Electric field due to that little bit i in the x direction is 1/ 4π E₀ Δ Q/ x² which implies then4538

that the total electric field in the x direction will be the integral of 1/ 4π E₀ R DQ / x².4549

Which we already said DQ is Λ Dx so we can write that as pulling 1/ 4π E₀ out of the integral sign, integral of Λ Dx/ x².4563

Where our limits of integration, we are going to integrate from x = that distance d to x = D + L to get us that total length of our line of charge.4582

This implies then that the electric field in the x direction is Λ/ 4π E₀, our Λ can come out as well, it is a constant.4594

The integral from D to D + L of x², x⁻² Dx which is Λ/ 4π E₀ × -1/ x from D to D + L,4607

which is Λ/ 4π E₀ substituting in we have -1/ D + L - -1 / D which implies then that the electric field in the x direction is Λ/ 4π E₀.4633

And doing a bit of rearranging here, putting a common denominator D × D + L we end up with.4659

Let us write that down here first, D × D + L we end up with D - D + D + L.4667

It is little bit easier to deal with which is Λ/ 4π E₀ - D + D is going to be 0 that is just going to leave an L / D × D + L.4677

We also have said again that Λ = Q/ L so that is going to be = I will replace Λ with Q/ L.4694

We have still got our 4π E₀ down here × L/ D × D + L.4706

L in the bottom and L to the top make a ratio of 1 electric field in the x direction.4719

1/ 4π E₀ × Q/ D × D + L.4725

That looks like enough practice problems for now.4742

Please take your time, if you need to do these again go back practice some and see if you can do them on your own or see how far you can get,4744

then check the video again, pause and rewind our wonderful things.4751

Thank you so much for watching www.educator.com.4754

We will see you in the next lesson when we talk about Gauss’s law.4758

Thanks and make it a great day.4761