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### Electric Potential & Electric Potential Energy

• The work done per unit charge in moving a charge between two points in an electric field is a scalar quantity known as the electric potential difference, or voltage.
• The work done in move this charge is equal to the change in the object's electric potential energy (U=qV)
• Equipotential lines show lines of equal electrical potential. They always cross electric field lines at right angles.
• The electric potential due to a point charge is given by kq/r. To find the potential difference due to multiple point charges, add up the potentials due to each individual charge.
• Equipotential lines show points of equal electric potential (similar to how topographic maps show points of equal altitude, or gravitational potential).
• As the distance between equipotential lines decreases, the steepness of the surface (or the gradient of the potential) increases.
• Electric permittivity is a material property describing a material's ability to store energy in an electric field.

### Electric Potential & Electric Potential Energy

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:08
• Electric Potential Energy 0:58
• Gravitational Potential Energy
• Electric Potential Energy
• Electric Potential
• Example 1 1:59
• Example 2 3:08
• The Electron-Volt 4:02
• Electronvolt
• 1 eV is the Amount of Work Done in Moving an Elementary Charge Through a Potential Difference of 1 Volt
• Conversion Ratio
• Example 3 4:52
• Equipotential Lines 5:35
• Topographic Maps
• Lines Connecting Points of Equal Electrical Potential
• Always Cross Electrical Field Lines at Right Angles
• Gradient of Potential Increases As Equipotential Lines Get Closer
• Electric Field Points from High to Low Potential
• Drawing Equipotential Lines 6:49
• E Potential Energy Due to a Point Charge 8:20
• Electric Force from Electric Potential Energy 11:59
• E Potential Due to a Point Charge 13:07
• Example 4 14:42
• Example 5 15:59
• Finding Electric Field From Electric Potential 19:06
• Example 6 23:41
• Example 7 25:08
• Example 8 26:33
• Example 9 29:01
• Example 10 31:26
• Example 11 43:23
• Example 12 51:51
• Example 13 58:12

### Transcription: Electric Potential & Electric Potential Energy

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

I'm Dan Fullerton, and today, we are going to talk about electric potential and electric potential energy.0003

Our objectives include determining the electric potential in the vicinity of 1 or more point charges.0009

Calculating the electrical work done on a charge.0014

Determining the direction and approximate magnitude of the electric field and various positions given a sketch of equal potentials.0018

Calculating the electrostatic potential energy of a system of 2 or more point charges.0025

Finally, stating a relationship between electric field and electric potential.0031

As we get into this one, please understand electric potential is one of the trickier concepts as we are talking about electricity and magnetism.0039

It is not one that typically makes a whole lot of intuitive sense, at least the first couple of times you see that.0046

Some of this may feel little bit nebulous and cloudy as you go through it the first time but that is not unusual at all.0050

Alright, electric potential energy, let us start there.0058

When an object is lifted against gravity by applying a force for some distance, we had to do work to give that object gravitational potential energy.0062

When the charged object is moved against an electric field by applying a force for some distance0070

we also have to do work to give that object electric potential energy.0075

The work done per unit charge in moving a charge between 2 points in an electric field is a scalar known as the electric potential.0080

More often times, you will hear that referred to, rather informally as voltage.0087

The units are known as V or 1 V is a J/ C.0094

The work done is equal to the change in the objects electric potential energy which0099

we are going to symbolize U for potential energy and e for electrical,0104

where the electrical potential energy is the charge × the electric potential d.0109

Taking a look at charge from work.0118

A potential difference of 10 V exist between 2 points A and B in an electric field,0123

what is the magnitude of charge that requires 2 × 10⁻² J of work to move it from A to B?0128

We have a potential difference between A and B of 10 V, we do not know charge but0135

we do know that the work done in moving it from A to B was 2 × 10⁻² J.0145

Work going from A to B = Q × electric potential.0153

Therefore, our charged Q is the work done ÷ the electric potential which is 2 × 10⁻² J ÷ 10 V.0159

Or charge is equal to 2 × 10⁻³ C, pretty straightforward problem to start us off.0176

Let us take a look at electric energy.0188

How much electrical energy is required to move a 4 micro C charge to a potential difference of 36 V?0190

We are looking for electric potential energy.0199

We know our charge is 4 micro C or 4 × 10⁻⁶ C and our potential difference is 36 V.0204

U = QV which is 4 × 10⁻⁶ C × 36 V which gives us the electric potential energy of 1.44 × 10⁻⁴ J,0216

another straightforward starter sort of problem.0238

When we talk about these different types of energy, oftentimes electrical energy into the work done is a very small portion of the Joule.0243

So small that talking about Joules does not always make a whole lot of sense.0252

A smaller alternate nonstandard unit of energy that is oftentimes much more convenient to use is known as the electron volt, given the abbreviation eV.0256

Where 1 eV is the amount of work done in moving an elementary charge through a potential difference of 1 V.0267

The charge on 1 electron move to a potential difference of 1 V is 1 eV.0273

The conversion ratio 1 eV is 1.6 × 10 ⁻19 J just like 1 elementary charge is 1.6 × 10 ⁻19 C.0279

An example with energy eV, we have a proton move through a potential difference of 10 V in an electric field,0293

how much work in eV was required to move this charge?0300

Our charge is + 1 elementary charge, our potential difference is 10 V.0305

Work done on this charge × V which is 1 elementary charge × 10 V or 10 eV.0312

This is straightforward when you are dealing with these tiny units and elementary charges.0328

It makes it much more convenient.0332

Oftentimes, we you think about topographic maps, maps of showing elevation, they show you lines of equal altitude or equal gravitational potential.0336

We have the same sort of thing in the electricity world.0347

The lines connecting points of equal electric potential are known as equal potential lines.0350

Equal potential lines always cross electrical field lines at right angles.0356

They are always perpendicular.0359

If you move the charge particle in space and stay on equal potential line, you cannot do any work.0362

You have to change the equal potential line you are on.0368

You have to change your potential in order to do work.0370

If you where to start over here for example, and wander through any path and come up back to the same equal potential line, the work done is 0.0374

As equal potential lines get closer together, the gradient of the potential increases, equivalent kind of to a steeper slope of potentials.0387

An electric field points from high to low potential.0396

And because you had this analogy with slopes, they actually even call this the gradient as you get deeper into physics.0399

Let us take a look at some equal potential lines.0407

If you wanted to draw an equal potential lines on these electric field diagrams,0410

we have to remember that they always intersect field lines at right angles.0414

They are perpendicular.0419

If I were to do it over here, I would probably draw an equal potential line something like that, especially if I had better art skills.0420

And you can draw another equal potential line out here but notice it is always crossing these at right angles.0432

Those should be perfect circles but not so good at circles.0440

The same idea over here, we would draw equal potential lines they would be circles around these charges,0445

with the charges at the center always crossing at 90° angles.0453

Down here however, an equal potential line would go like that so it is always crossing at right angles.0459

Over here, you might have it looking something more like that.0467

Down here, same idea, right in between you might have something like a D shape that is going to,0472

When it is curling like that, let me have to draw them so they always intersect at right angles.0485

My art skills are not so great, but I think you get the general idea.0495

The electric potential energy due to a point charged, find the work required to take a point charge Q2 from infinity,0501

some long ways away where its electric potential energy is 0 to some point it is a distance r away from the point charge Q1.0509

Assuming these are both positive, if we have this charge due to a long ways away, we have got to do some work.0517

We got to push it to get all the way over here to point Q2.0521

How much work was done in doing that?0525

I’m first going to define that we have an electrical force in this direction along the line from Q1 to Q2 we will call that fe, the electrical force.0528

The work done is going to be the integral going from some position infinity to r =R of our force.0540

If we are going this way and the force is going one way, the force is going the opposite direction,0552

they will have to be - fe ⋅ dl from our definition of work in mechanics force with displacement.0556

That is going to be equal to,0567

I do not like that minus sign so I'm just going to switch these directions here and say0569

we are going to go from r = R to infinity of fe ⋅, instead of dl, I’m going to write it in terms of our variable dr,0572

which is going to be the integral from R to infinity of,0587

the electrical force we know from Coulomb’s law, that is going to be 1/ 4 π ε₀ Q1 Q2 / R² and we also have dr here.0593

I’m going to pull my constants out of the integration, 1/ 4 π ε₀ is all a constant in this case and0608

Q1 Q2 their charges are not changing in this problem so they can be pulled out.0616

This becomes Q1 Q2/4 π ε₀ integral from R to infinity of r⁻² dr.0620

The work done we just have to integrate that, work = we still have our Q1 Q2/4 π ε₀ × the integral of r⁻² is just -1/ r.0638

-1/ r evaluated from r to infinity which implies then that the work done is going to be Q1 Q2/4 π ε₀ -1/ infinity is 0, - -1/ r is just going to be 1/ R.0653

If that was the work that we did, it implies that the potential energy that it now has must be the same, whatever work we did on,0675

if it is starting with 0 now must be its potential energy.0683

That is going to be 1/ 4 π ε 0 Q1 Q2/ R.0686

Note here that this is independent of path because the Coulombic force, the electrical force is a conservative force.0701

It does not matter what path you take to get there, you can push it straight there, you can wobble all around.0709

But only the starting and ending points are going to matter.0714

Let us see if we can find electric force from electric potential energy.0719

Using that relationship for conservative force f = - du dl, in our case that is going to be - d/dr0724

of our electric potential energy 1/4 π ε₀ Q1 Q2 / R.0736

We will pull our constants out of the derivation so that is -1/ 4 π ε₀ Q1 Q2 × the derivative with respect to r of 1/ r.0749

Therefore, the force is going to be 1/ 4 π ε₀, the derivative of 1/ r is going to be -1/ r² so we end up with Q1 Q2 / r², Coulomb’s law.0764

We can go in the other direction as well which should make sense.0781

How about electric potential due to a point charge?0786

Electric potential known as voltage is the work per unit charge required to bring a charge from infinity to some point r in an electric field.0790

Potential is a work done per unit charge which is going to be 1/4 π ε₀ Q1 Q2/ R our work ÷ our charge q.0801

What we are going to get is 1/ 4 π ε₀ q/ r.0819

If you happen to have more than 1 charge, just add up the electric potentials due to each of those individual charges.0829

Since, electric potential is a scalar not a vector, you do not have to worry about direction.0835

It makes it a whole lot simpler.0842

Electric potential is the sum over all i for how many charges you happen to have Qi, Q2, Q3, Q4, however many happen you have of 1/ 4 π ε₀ Qi/ ri.0844

Since that is a constant, that can come out as 1/ 4 π ε₀ sum /i of their charges ÷ the distance.0863

Let us see if we can do a couple samples to go along with these.0878

Find the electric potential at point P located 3 m from the -2 C charge.0883

What is the electric potential energy of ½ C charge situated at point P?0888

Let us find its potential first.0894

The potential point P has 1 /4 π ε₀ Q/ r which is 1/4 π ε₀.0898

Our charge is -2 C ÷ our distance 3 m which is -6 × 10⁻⁹ V.0909

To find the electric potential energy of ½ C charged at that point, that is going to be charge ×0925

our potential which is 0.5 C × our potential which we just said was -6 × 10⁻⁹ V,0933

for a total of -3 × 10⁻⁹ J.0947

How about the potential due to point charges?0959

In this problem should look fairly familiar.0961

Last time we were using this to find electric field, now we want to find the electric potential at the origin due to the 3 charges shown in the diagram.0963

And then finally, if an electron is placed at the origin what electric potential energy does it possess?0971

Let start out with the electric potential and we will start out with electric potential from our green charge up here.0976

The potential due to that green charge is 1/4 π ε₀ × Q/ r.0982

Q is 2 C, our distance from the origin is going to be 8 m or 2.25 × 10⁹ V.0990

Doing the same thing for our red charge down here, our potential is 1/4 π ε₀ × -2 C / 8 m or -2.25 × 10⁹ V.1002

We have our blue charge, potential due to the blue charge is 1/4 π ε₀ × our charged 1 C/ the distance,1020

these are Pythagorean theorem here, 2 and 2 , √2² + 2².1032

We have √2² + 2² which is going to give us about 3.18 × 10⁹ V.1039

T get the total then, all we have to do is add these up.1051

2.25 × 10⁹ -2.25 × 10⁹, that is 0.1057

It leaves us with just 3.18 × 10⁹ V, for the second part of the question.1062

If we place an electron at the origin, there it is, what electric potential energy does it possess?1075

Electric potential energy is charge × voltage, it is going to be -1.6 × 10 ⁻19 C × our voltage, our electric potential 3.18 × 10⁹ V or -5.1 × 10 ⁻10 J.1086

We could also do that in eV if we wanted to make our math a little simpler.1112

In that case, this would be -1 e, this would be the same, and we would come up with.1116

Let me write that in 3.18 × 10⁹ V, we will multiply those to come up with -3.18 × 10⁹ eV.1124

A little bit simpler math but note that this is non standard units.1139

Finding the electric field from electric potential, we can do that as well.1147

If our potential is 1/4 π ε₀ Q/ R, we can take the derivative of both sides to say that the derivative of the potential with respect to R1152

must equal the derivative of the right hand side with respect to R which is 1/4 π ε₀ Q / R.1169

Or pulling those constants out, Q is not going to change, 4 π ε₀ is not going to change, this becomes Q/4 π ε₀ × the derivative with respect to R of 1/ R.1181

Which implies then that dv dr = we still have Q/ 4 π ε₀.1200

The derivative of 1/ R is just -1/ R² so that is equal to -Q/4 π ε₀ R².1212

If we multiply both sides in this equation, the dv dr and our answer over here by the unit vector and the R direction r ̂ ,1228

This implies then the dv dr r ̂ = -2 /4 π ε₀ R² r ̂.1237

If you recall our definition for electric field due to that point charge is Q/4 π ε₀ R², the direction of r ̂.1251

Therefore, we can write then this is the opposite of the electric field or electric field = - dv dr in the direction of r ̂.1266

There is a relationship we can use between the electric field and the electric potential.1281

If we go that way, we should be able to find electric potential from electric field and the answer of course is yes.1290

Let us do that.1299

Let us start with electric potential V is W/ Q which is going to be 1/Q × our definition of the work done which is the integral from r to infinity of fe ⋅ dr.1301

We did that previously, which is going to be the integral from r to infinity of the force ÷ the charge × ⋅dr.1318

Which implies then given that the electric field is the force per unit charge, that we now have the potential is the integral from r to infinity of E ⋅ dr.1335

True, but usually you see this written a little bit differently.1356

Let us go ahead to show that.1360

Potential difference which is what we are really talking about is the potential at some point B - the potential at some point A,1364

potential difference from A to B, is equal to the opposite of the integral from A to B of E ⋅ dl1371

which is how you normally going to see that written and probably is a little bit more useful.1382

Both are correct but this one is probably the one that is much more common as opposed to going from here1386

where you may see this written as - the integral from infinity to r of E ⋅ dr.1393

This by the way is the change in electric potential energy per unit charge as well.1399

A couple conversions that you go from potential to electric field and potential energy and how to put all of that wonderful stuff together.1408

Let us do a couple more samples now.1419

Find the electric potential of the origin due to the following charges.1422

We have a 2 micro C charge at 3, 0 and -5 micro C charge at 0, 5 and 1 micro C charge at 4, 4.1426

We are going to assume that all of these are given as points on the axis where that is 3 m, 5 m, and so on.1436

The electric potential is 1/ 4 π ε₀ and then we just have to add up for all i, our charges ÷ their distance.1446

It is going to be 1/ 4 π ε₀ × our first point is 2 × 10⁻⁶ C of 2 micro C and it is 3 m away from the origin.1457

Then we have a second charge -5 × 10⁻⁶ C and it is 5 m away from the origin.1470

And then finally, we have 1 micro C charge that is at 4, 4.1478

Its distance from the origin is going to be √ 4² + 4².1483

I do not have to worry about direction again since electric potential is a scalar, just plug through the math here and find that you are going to get about -1410 V.1490

Another example, get lots of practicing.1508

Given an electric potential function, V is a function of x as 5 x² -7 x, find the magnitude and direction of the electric field at x = 3 m.1511

Our electric field we just said was –dv dr in the direction of r ̂ so that is going to be - the derivative with respect to x1523

we are in 1 dimension here of 5x² -7x in the i ̂ direction is going to be - the derivative is going to be 10x - 7 in the i ̂ direction or that is 7 -10x (i.) ̂1534

But we know that x = 3 m so we can solve for it specifically and say that the electric field is going to be 7 -10 × 3 m in the direction of i ̂.1558

7 -30 is just going to be -23 V/ m in the i ̂ direction, one way to solve that.1574

Let us put this together with a little bit of energy concerns.1590

An electron was released from rest in a uniform electric field of 500 N/ C.1594

What is its velocity after its traveled 1 m?1600

Our change in kinetic energy is going to be equal to the opposite of the change in potential energy from conservation of energy.1606

We also know here that our change in potential energy is going to be -Q integral from A to B of E ⋅ dr.1614

Then change in kinetic energy = Q integral from A to B of E ⋅ dr.1630

Which implies then that our change in kinetic energy is going to be how the electric field is constant,1646

it is a uniform electric field, so we can pull that out.1652

QE integral from A to B of dr which is just going to be QE × the change in R.1655

Our change in kinetic energy, if it starts from rest is just going to be ½ MV², we have ½ MV² = QE Δr.1666

Therefore, solving for V we can say that V = √2Q E Δr/ M.1678

Substituting in with our known values that is going to be 2 × 1.6 × 10 ⁻19 C × 500 N/ C × 1 m ÷ mass of our electron 9.11 × 10 ⁻31 kg.1689

There is our square root and I come up with about 1.33 × 10⁷ m/ s.1719

Let us keep doing some more examples and make sure we really have these down.1735

I will get into a couple of practice AP problems.1738

The work required to establish a charge system, fairly common type of question.1742

2 point charges 1, 5 micro C and one that is 2 micro C are placed ½ m apart, how much work was required to establish this charge system?1747

We had these 2 charges to get them ½ m apart and had you have done some work because they do not want to be there beside each other, they repel each other.1758

How much work did you have to do?1766

That is the potential energy of the system which is 1/4 π ε 0 Q1 Q2/ R which is 1/4 π ε 0 × 5 × 10⁻⁶ × 2 × 10⁻⁶ ÷ ½ m.1768

What is the electric potential halfway between the 2 charges?1801

To find that, we will go with our formula for electric potential that is 1/4 π ε₀ × Q1 / R + Q2.1806

Let us call that R1, R2, even though R1 and R2 are both going to be the same ¼ m.1824

That is going to be 1/4 π ε₀ Q1 5 × 10⁻⁶ C/ R1 .25 m + the second charge 2 × 10⁻⁶ C ÷ ¼ m.1830

I come up with about 252,000 V or 252 kV.1851

Let us get into some of these AP style questions that are pretty challenging type questions and1864

quite likely on the AP exam you are going to see, at least in the free response question.1870

question that has some electric potential derivation.1875

On most years, you are going to see Gauss’s law question and electric potential question.1878

Sometimes, they are even involved in the same questions as we are going to see here.1883

We will start off with a 2013 APC E and M exam free response number 1.1887

It is highly recommend you pause the video, take a minute, download it and print in out,1893

look it over for a few minutes before you come back to the video and hit play again.1896

Let me pull the question out as well and we are looking at 2013 free response number 1 on the E and M exam,1902

where we have got this very long solid none conducting cylinder.1909

The first question it asks us, using Gauss’s law derive an expression for the magnitude of the electric field inside the edges of that cylinder.1913

Draw an appropriate Gaussian surface on the diagram.1925

First thing, let us give us a representation of that diagram here.1927

I will put the center line, the axis, here in green.1932

We have this cylinder that goes around it.1943

I will put that in blue.1946

Here is our cylinder, there we go.1948

Using Gauss’s law, derive an expression for the electric field inside that.1960

If I'm looking for the most symmetric Gaussian surface I can come up with, I think I’m going to draw a cylinder inside that.1964

I will put that in here in purple so you can see it.1970

Here is our Gaussian surface and now we are trying to find the magnitude of the electric field there.1973

For part A, we will start by writing Gauss’s law integral / the close surface of E ⋅ da, it is our total charge enclosed ÷ ε 0.1985

The left hand side and we have done this a couple of times already, with Gauss’s law it is just going to be E × the area of our Gaussian surface.1999

The right hand side, our charge enclosed is going to be that volume charge density ρ × the volume that is enclosed by ÷ ε₀.2008

Our left hand side becomes E and that area 2 π R × some length l, which we will define as the length of our cylinder must equal ρ × our volume.2019

We have got π R² × the length ÷ ε₀.2040

Solving for the electric field then, E is going to be equal to, we have got ρ × π R² l / 2π Rl ε₀.2048

A couple of cancellation simplifications we can make here.2067

We got a π, we have got a π, and l and l, and R and R², and I think that I will do it and come up with ρ × r / 2 ε₀.2069

The electric field inside that long solid, none conducting cylinder.2087

For part B, we are asks to use Gauss’s law to derive an expression for the magnitude of the electric field outside of that.2095

At some r that is greater than the radius R of the blue cylinder.2104

We will go back to Gauss’s law again, integral/ the close surface of E ⋅ da = our enclosed charge ÷ ε 0,2109

which implies then the left hand side Ea = the charge enclose is just going to be the volume charge density ρ × V ÷ ε₀.2122

Our left hand side E × our area 2 π R × length must equal the right hand side, ρ × our volume which is now π R² l ÷ ε₀,2138

Because the charge stops when you get to the edges of that R cylinder, when you get to that radius.2155

Then a little simplification again E = ρ × π R² l/ 2 π Rl ε₀.2163

We have got π and π, an R and l, we do not have an R, we still have that R down there.2179

Let me redraw him in R.2188

That gives us ρ R²/ 2 ε₀ R.2190

Next up, move in on to C.2206

On the axis sketch a graph of the electric field as a function of radial distance from r=0 to 2R.2209

Let us draw our axis in here first.2217

Here we have electric field, here we have r, we have got R in here, and we have got 2R.2233

We need to know the electric field.2244

As we go from 0 to R, we have already found out that that is a ρ R/ 2 ε₀ so2247

we can draw our nice linear fit there and after that it looks like we fall off with respect to 1/ R.2255

R there is a constant so then this would be sort of this shape and things I would label right there at that point where r = R.2267

We got ρ R/ 2 ε₀.2281

We know that because we can plug in R up here for little r or we can plug R down here for little r and get the same thing either way.2286

And then we get over here to 2r, we also have another important point, that little r =2R.2296

We end up with ρ R/ 4 ε₀.2304

I would use that as my graph, therefore part C.2310

I think that covers C pretty well.2317

Let us take a look now at D, derive an expression for the magnitude of the potential difference from r = 0 to R.2321

Alright, let us start with E = - dv dr in the direction of r ̂.2332

There are other ways you can do this, you got to know your formulas, you can jump in a few steps ahead2339

But it is like doing this every now and then starting from the simpler expressions.2345

Then dv dr = - E which implies then that V = - the integral of E ⋅ dr which is going to be - the integral from r = 0 to R.2350

We already found that our electric field that is ρ r/ 2 ε₀ dr which implies then that our potential V = let us pull other constants ρ, 2, and ε₀.2369

We will have - ρ/ 2 ε₀ integral from 0 to R of r dr.2387

Or - ρ/ 2 ε₀ × R²/ 2 evaluated from 0 to R, which is going to be - ρ/ 2 ε₀ × R²/ 2 -0,2397

which implies then that our potential is going to be - ρ R²/ 4 ε₀.2416

Or the magnitude of the potential difference is just going to be ρ R²/ 4 ε₀.2427

Part 2 D2, is a potential wire at r =0 or r = R?2441

As we went from 0 to r, we obtained a negative potential difference.2453

Let VB - VA was – ρ R²/ 4 ε₀.2459

If we said VA - VB must be ρ R² / 4 ε₀, A must be bigger or a potential is higher where A is what we are going to call r = 0 and B is r = R.2466

I would say then that we are higher at r = 0.2491

And finally, they ask us to do another graph here.2504

The non conducting cylinder is replaced with a conducting cylinder of the same shape and same linear charge density.2510

Sketch the electric field from r = 0 to 2r.2516

Alright, let us go to the next page to do that.2520

We will draw our axis again and here we have r, here we have our electric field, and we will mark off R and capital 2R.2525

If it is a conductor, we know the electric field inside a conductor is 0.2550

This piece becomes nice and simple E = 0 inside the conductor and outside it is the same as when we did part C.2554

It does not matter once you are outside that, the enclosed charge is the same.2568

We have something that looks like that.2572

Where again, we have the same key points ρ r / 2 ε₀.2577

I guess that is not drawn to scale very well.2587

Or/ 4 ε₀, same as part C.2590

That should cover that problem, the 2013 free response number 1.2596

But let us keep going, let us take a look at the 2006 E and M free response number 1.2601

Another interesting question, this one focusing more on the electric potential and2608

some of the inner relationships with the electric field as opposed to Gauss’s law.2613

Take a moment, print that out, look at it, give it a shot, come back here and we will see what we have got.2617

Here we have 4s² with a point P in the middle and 4 charges on those points.2626

We have got point P right here and around that we have this square where we have 4 different charges.2633

We will call this corner 1, corner 2, corner 3, and corner 4.2644

On the diagram, indicate with an arrow the direction of the net electric field at point P.2652

Due to charge 4 it is going to be up into the right because they repel.2657

Due to Q, it is going to be up in the right because it is attracted.2661

And these 2 both negative are going to cancel out.2664

Our net electric field at point P got to go that way.2667

Moving on to part B, derive an expression for the magnitude of the electric field at point P.2674

By symmetry, the electric field at 1 + electric field at .3 has to be equal to 0.2681

We are going to worry about 2 and 4.2687

The electric field at 0.2 is going to be Q/4 π ε₀ R².2691

The electric field at .4 is going to be Q/4 π ε₀ R².2699

But what is R² here?2708

If this whole distance is a then that would be a/ 2, that is a/ 2.2710

If we want to find R², use the Pythagorean theorem that is a/ 2² + a/ 2² which is 2a/ 2².2718

And therefore, R² = 2a²/ 4 which is a²/ 2 and then r must equal a/ √2.2732

The electric field at point P is the electric field at point P due to 2 + the electric field at point P due to 42747

which is going to be 2Q /4 π ε₀ R² which is 2Q/4 π ε₀ and R² in the denominator is going to give us a² and 2 up in the numerator.2755

2 × 2 and a 4 down there, I end up with Q/ π ε₀ a².2777

That will work, how about the electric potential at point P?2794

Part 2 of this question of B, find the electric potential at point P.2800

That should be a little more straightforward.2805

Electric potential at point P just 1/4 π ε₀ × the sum of all those little charges of that Q/ R for each of those charges,2809

which is going to be 1/4 π ε₀.2821

The r are all the same, that is a constant because they are all the same distance from point P which is going to be a/ √2 × we have +Q and 3 –Q.2827

-Q -Q -Q gives us -2Q √2/ 4a π ε₀.2841

With a little simplification here, that is going to be -Q/ √2 and we have got π ε₀ a.2856

How about C, a positive charge is placed at point P it is then moved from P to point R which is at the midpoint of the bottom side of the square.2875

It is going to end up over here.2883

If the charge is moved, is the work done by the electric field positive, negative, or 0?2888

You got to explain it.2893

For part C here, say that has to be negative, why?2897

We only know the electric field is up into the right, the charge is being moved against that field so2902

we are doing the work against the field, the field is not doing the work.2911

That the charges moved against the field so the work done is less than 0.2923

The work done on it by the electric field is negative because we are doing the work.2930

We are giving the system energy.2934

Moving on to part D, describe one way to replace a single charge in this configuration2939

that would make the electric field at the center of the square equal to 0 and justify your answer.2951

Let us draw this again, just so that we do not lose track of what we are doing.2956

We have got these here, we have got P in the middle, and I think they call these 1, 2, 3,and 4.2963

The first thing it asks us is what would we do in order to make the electric field at the center of the square =0?2973

The way I would do that is I would replace the charge here at 2 that is negative with +Q.2982

If we did that, we have +Q here, +Q here, -Q here and -Q here.2989

The fields all cancel at point P.2997

I would say something like replace Q2 with +Q.3000

Fields cancel at point P.3010

For part 2, describe a way to replace a single charge in this configuration so electric potential of the center of the square is 0 but the electric field is not.3020

We start off again with, we put these charges back -Q -Q -Q and + Q.3030

If we replace Q1 over here with + Q, let us make that a +.3038

When we do that, the electric potential is going to sum to 0.3048

Remember it does not care about direction, it is the scalar piece.3052

I have a potential of 0 but the field no longer cancels.3056

We are going to have a net electric field to the right.3060

I would say that we can replace Q1 with +Q charge.3062

V which is the sum of Q/ 4 π ε₀ R = 0 but field no longer cancels at point P.3072

The electric field at point P instead would point to the right.3093

Another AP style question down, good practice.3098

Let us do another one, now we are going to go to 2006 E and M.3102

I do not think we needed that page.3110

Let us go to the 2005 E and M free response number 1 and then you can print out from the same place, take a minute to work through it,3111

and come back here and un pause it, once you have a minute to look it over.3119

We are given a graph that shows us an xy plot along with field lines on there,3127

that says points A, B, and C are all located at y = 0.06 m and you can see that right from the graph.3135

A1, what should these 3 points is the magnitude of the electric field the greatest?3142

It is got to be the greatest at point C, since the field lines are the densest there.3148

Dense field lines, strongest field.3163

For A2, at which of these 3 points is the electric potential the greatest?3169

That has got to be at point A.3174

I would say that it is A because the field points from high to low potential3177

which in this case is left to right, A is farthest left.3194

Therefore, A is at highest potential.3206

Moving on to question B, an electron is released from rest at point B.3218

Qualitatively describe the electrons motion in terms of direction, speed, and acceleration.3224

If we release an electron at point B, it is a negative charge so it is going to go in the direction opposite the field line.3230

First thing is the electronic accelerates toward the left that moves with increasing speed because it is accelerating.3236

But it is going to keep accelerating at smaller and smaller rates, the speed increases by smaller and smaller amounts.3261

Let us write that.3267

Smaller and smaller amount per unit time as it moves because acceleration is decreasing.3282

For B2, calculate the electron speed after it is moved through a potential difference of 10 V.3306

Another energy question.3313

For B2, we would say that the W = QV.3315

The work done in this case is going to be the change in kinetic energy.3323

½ MV² = Q × B potential, be careful not to mix up v velocity and potential.3327

Therefore, velocity² = 2Q × electric potential ÷ mass or V = √2Q B/ M.3337

It gives us some values so we can actually solve for an exact speed.3351

V= √2 × 1.6 × 10 ⁻19 C × our voltage 10 ÷ our mass 9.11 × 10 ⁻31 kg.3356

I come up with about 1.87 × 10⁶ m/ s.3373

Taking a look here at part C, points B and C are separated by a potential difference of 20 V.3382

Estimate the magnitude of the electric field midway between them and state any assumptions that you make.3390

The electric field is roughly the potential ÷ the distance assuming you are roughly uniform electric field3395

which should be 20 V / it looks like we are about 1 cm, .01 m apart there which is 2000 V/ m.3403

We have to write what we are assuming and say assuming a roughly uniform e field in that region.3416

Onto D, on the diagram draw an equal potential line it passes through point B and intersects at least 3 electric field lines.3434

I’m just going to sketch this very quickly, I think you can get the idea without too much,3446

The graph looks kind of like that and we have some points coming in there, something kind of like that.3455

We have got to be down here somewhere.3467

What we need to do is we need to draw an equal potential line so that crosses all of these at right angles and goes through that point.3469

Something like that, you have to look at your graph to see exactly how to do it.3477

Draw these equal potential lines perpendicular to the field lines and make sure it goes through D.3481

And that should cover that 2005 question.3486

Finally, let us take a look at one last free response question in this unit then I will you a rest.3493

Let us take a look at the 2003 E and M free response number 1.3498

Another one that say, a little bit of a more unique sort of question.3503

You do not see this very often.3507

They give us a spherical cloud of charge of radius R with a total charge +Q and a non uniform volume charge density and3510

they gave us the equation for where that the charges distributed.3517

Determine the following as a function of r for when you are outside that sphere.3523

The magnitude E of the electric field that is just a straightforward Gauss’s law question by now.3529

Let us make sure we get that one right, still good practice.3535

A1 the integral / the close surface of E ⋅da = the enclose charge ÷ ε₀ which implies then that3538

the electric field × that area 4 π R² = our total enclose charge is just Q ÷ ε₀3551

Or the electric field = Q/4 π ε₀ R².3562

By now, quite a few of you could probably just look at that and say what the answer is because we have done it so much.3570

It is same as if it was that + Q point charge in the center.3574

However, take the time to go through this and get all the points for showing your work.3578

How about the electric potential V in part 2?3583

There is part 1, part 2, define the electric potential of V the opposite of the integral from infinity to R3587

of E ⋅ dl is going to be - the integral from infinity to R of Q/4 π ε₀ R² dr.3596

We will pull other constants -Q /4 π ε₀ integral from infinity to R of r⁻² dr which is -Q/4 π ε₀ × -1/ r3609

evaluated from infinity to R which is as we have done before is well Q /4 π ε₀ r.3629

You might have been able even to say that from a number of times we have done similar problems, go ahead show the work.3640

Get every single possible point here.3646

For part B, we have got a proton and we are going to place it at point P as shown and release it.3649

Describe its motion for a long time after it is released.3655

If we put it at P, the proton is going to move away from the sphere to the right and it is going to accelerate,3659

it is going to be moving in increasing velocity but with a decreasing acceleration because the force is going to get smaller the further way it gets.3665

I would write something along the lines of, the proton will move away from the sphere to the right, probably important to mention.3672

An increasing velocity with a decreasing acceleration.3690

Part C, an electron of charged magnitude E is now place at point P which is a distance r from the center of the sphere it is released,3710

find the kinetic energy of the electron as a function of R as it strikes the cloud.3721

This is a conservation of energy question and we have seen things like this before by now as well.3726

Where the potential energy that r= to the potential energy that R + the kinetic R implies that the kinetic R = the potential energy of r - the electric potential energy R.3733

Thankfully we know the formula for the electric potential energy U is Q1 Q2/4 π ε₀ R.3752

Therefore, this implies then the potential energy at R is going to be - EQ/4 π ε₀ r - -E capital Q/4 π ε₀ R.3767

Which implies in that the kinetic energy, if we want to simplify this and combine a little bit is just going to be EQ/4 π ε₀ × 1/ R -1/ r.3790

Two more parts to the question, the next 2 pieces are little bit more math intensive but hang with it.3813

If we look at part D, it says to derive an expression for ρ 0.3820

The way we do that is we realize that the total charge in that sphere is + Q, it gives us the total.3826

This is not really an electric fields question, so much as it is kind of a common sense to you, understand what the math and algebra means question.3833

Our total charge is Q, which is going to be the integral as we go from our radius = 0 to R,3842

the radius of that sphere of our volume charge density × the differential volume.3849

Adding up all those charges as we go from 0 to the radius R should give us capital Q.3856

We have got the formulas, we just have to back out what that ρ 0 is.3862

To do this it would be helpful to start off with some knowledge of dv.3867

Dv is going to be, as we go from 0 outwards, what we will take is we will take little tiny shells of area 4 π R² × some thickness dr.3873

dv will be4 π R² dr and we also know that our volume charge density function ρ =ρ₀ × 1 - r/ R.3885

Back to the math, Q = the integral from 0 to R of ρ which we said is ρ₀ × 1 - r/ R × 4 π R² dr.3907

Which implies then that Q = let us pull out our constants.3930

We have got a 4 π ρ₀, it can all come out.3938

Integral from 0 to R of, we got this R² in here, let us distribute that through r² - r³/ R dr.3944

Which will be 4 π ρ₀ the integral of r² is going to be r³/ 3 – r⁴/ 4R, remembering R is a constant,3959

and those are evaluate from 0 to R.3977

Which is going to be equal to 4 π ρ₀ × R³/ 3 – R⁴/ 4R.3982

The 0 is just going to 0 out, we will get anything from that piece which is going to be 4 π ρ₀ ×3998

let us see if we can get rid of the r⁴ and make that a cube.4009

Get rid of that and find any common denominator of 12 here, we can write this as 4 R³/12 -3 R³/ 12,4014

which implies then that Q = 4 π ρ 0 R³/ 12.4030

To find ρ₀ itself, ρ₀ then must be 12Q /4 π R³ or 12/ 4 gives us just 3Q/ π R³.4041

But we did a lot of work for that, let us put that in the 3D box.4063

We should be proud of ourselves, 3Q/ π R³.4066

One last piece, hang in there.4072

For part E, determine the magnitude E of the electric field as a function of r, for r is less than or equal to R.4076

When you are inside that sphere, it looks like you are probably going to be doing a bit more math again.4087

I'm going to go with Gauss’s law, the integral / the close surface of E ⋅ da = Q enclose/ ε₀ which implies then since A = 4 π R².4093

We know that the Q enclosed is going to be the integral from 0 to R of ρ dV.4117

We can write then that 4 π R² × the electric field = 1/ ε₀ × the integral from r= 0 to R of ρ dv.4126

We also can now fill in dv, just like before dv = 4 π R² dr.4147

This implies then that 4 π r² × our electric field is 1/ ε₀ integral from 0 to r of ρ is ρ₀ × 1 - r / R × 4 π r² dr,4158

which implies that then 4 π r² E, our left hand side = we will pull out the constants of the right hand side,4185

We have got for constants 4 π, we have got a ρ₀, we got an ε₀ out there already.4196

That is going to be 4 π₀/ ε₀ × the integral from 0 to r of r² - r³/ R dr.4202

We are getting the set up.4224

Let us see, we got a 4 π on both sides, we can maybe do some simplifications now.4228

We got to cancel 4 π and 4 π, and write that left hand side r² × electric field = ρ₀ / ε₀ integral of r² is r³/ 3.4233

We got r³/ 3 – r⁴/ 4R all evaluated from 0 to r, complies that r² capital E = ρ₀ /ε₀ × r²/ 3 – r⁴/ 4R.4250

The 0 does not give us any factors here that we need to worry about.4277

And we also now know from our part D problem that ρ₀ is equal to 3Q/ π R³.4282

Our electric field if I start solving for E all by itself, we are going to have on our right hand side, we have got ρ know which we just said was 3Q/ π R³.4298

We have still got the r² from the left hand side, we have got ε₀ there, and we also have this piece r³/ 3 – r⁴/ 4R.4314

And now it becomes an exercise in algebra.4329

Just trying to simplify this and pretty it up a little bit.4332

We will take a minute and clean it up.4335

That will be 3Q/ π ε₀ × R³, we will put the r² back in, in order to see what we can clean up there that will give us4339

r² in the denominator will give us r/ 3 - r²/ 4R which is also Q/ I ε₀ R³.4361

Let us see, if we multiply all of this by 3, factor out that 3 from there, we can get Q/ π ε₀ r³ 3.4381

Put the 3 in that will give us r -3r²/ 4r which implies that E = Q.4392

We can factor out a single lowercase r/ π ε₀ R³, 1 -3r/ 4R.4404

We might be will simplify that further but I think that is plenty.4418

There is our electric field as a function of r when you are inside that radius R.4425

Alright, hopefully that gets you going on electric potential and electric potential energy and how all of these starts to play together.4431

Thank you for watching www.educator.com.4438

We will see you again real soon.4440

Make it a great day everyone.4442