AP Physics 1 & 2 Electric Potential Difference

Hi everyone and welcome back to Educator.com.0000

This lesson is on electric potential difference, oftentimes called just electric potential or voltage.0003

Our objectives are going to be to find and calculate electric potential energy and electric potential difference, to determine the potential difference due to a series of point charges...0009

...finding the electric field strength between two charged parallel plates, finding the energy stored in a parallel plate capacitor and finally, calculating the capacitance of the parallel plate capacitor.0019

Let us start by talking about electric potential energy.0032

When we lifted an object against gravity by applying a force for some distance, work was done to give that object gravitational potential energy.0035

At the same token, when you take a charged object and it is moved against an electric field by applying a force for some distance, you have to do work to give that object electric potential energy.0043

Imagine we have a positive charge here; it is stuck in space and a long ways away, infinitely far away, I have a little point charge.0053

If I want to bring that point charge -- its positive -- close to this big positive charge, I have to do work -- I have to push harder and harder, and harder, and harder, and harder to get it closer and closer.0060

Once it is in this position it has a lot of potential energy -- electric potential energy.0070

It wants to be repelled because if I let go of it, it is going to flinging off that way; it is going to convert that electric potential energy into kinetic energy.0075

So we had to do work in bringing it from a long ways away until it was at this position.0083

If we do work on an object, we have given it energy and in this case, we gave electric potential energy.0089

Now the work done per unit charge, and moving that charge through that electric field, is a scalar and that is known as the electric potential or electric potential difference if you are talking about the potential between two different areas.0093

The units are volts which is equal to a joule per coulomb (J/C) and the work done is equal to the change in the object's electric potential energy.0109

So electric potential energy is charge times electric potential difference.0117

Let us start off with a problem.0127

If we have a potential difference of 10 volts between two points (A) and (B) in an electric field, what is the magnitude of charge that requires 2 × 10^-2 J of work to move it from (A) to (B)?0129

Well let us start off with our givens -- electric potential difference (V) is 10 volts; our electric potential energy is going to be 2 × 10^-2 J because that is the amount of work we had to do, and we are looking for charge.0142

If electric potential energy is Q × V, then that means (Q) must be electric potential energy divided by (V) or 2 × 10^-2 J/10 volts, which gives us a charge of 2 × 10^-3 C.0162

It is pretty straightforward.0187

Let us take a look at electric energy.0189

How much electric energy is required to move a 4 microcoulomb (MC) charge through a potential difference of 36 volts?0191

Well our charge now is 4 MC and MC is 4 × 10^-6 C; our potential difference is V = 36 volts, and we are looking for electrical energy.0199

Electrical energy is charge times voltage, which is going to be 4 × 10^-6 C × 36 volts (voltage)...0216

...which implies that our electric potential energy will be 0.00014 J.0227

Oftentimes, when we are dealing with very small charges -- something like a joule -- is not a very convenient form of energy.0242

We are talking about things × 10^-15, ^-16, ^-17, ^-18 J -- it is just not very convenient.0249

So there is another non-standard unit of energy that is very commonly used, it is called the electron-volt and it is given the symbol (eV) and it is a very small portion of a joule.0255

It is the amount of work done in moving an elementary charge through a potential difference of 1 volt.0268

So if you were to think about it, if electric potential energy is charge times voltage and your charge is 1 elementary charge and you move it through 1 volt, the electric potential energy is 1 eV.0273

If we were to do that in standard units, we would have said that this is the charge of 1e, which is 1.6 × 10^-19 C × 1 volt and that would have given us 1.6 × 10^-19 J.0290

These have to be the same, therefore, 1 eV = 1.6 × 10^-19 J.0306

It is just another unit of energy that you oftentimes use when you are dealing with very small charges.0313

Keep in mind though, if you are going to use these values for energy in other formulas, the SI unit, the standard unit that is going to make all your units work out, is still going to be joules. That is the standard.0319

A proton is moved through a potential difference of 10 V in an electric field. How much work in electron-volts was required to move this charge?0334

Let us look at how easy this can be if we use electron-volts.0342

If our charge of a proton is +1 elementary charge, our potential difference is 10 volts, then the electric potential energy is charge times potential difference or 1e × 10 volts, which is 10 eV -- very straightforward.0346

If we wanted to do that in joules, we could have done 1.6 × 10^-19 C × 10 volts = 1.6 × 10^-18 J if we went through all that, or we could have converted electron-volts to joules and we are all done, knowing that 1 eV = 1.6 × 10^-19 J.0364

Thankfully, this problem made it nice and easy though in telling us that the answer was going to be in electron-volts.0384

All right. Let us talk about equipotential lines for a minute.0392

When you talk about topographic maps, if you have gone hiking or you were in some sort of surveying an organization, you have probably seen a topographic map where you have lines that show you areas of equal altitude.0397

Those are lines of equal gravitational potential energy.0411

We have the same sort of thing in the electrical world -- we have lines of equal electrical potential which we call equipotential lines.0415

Now equipotential lines always cross electric field lines at right angles and if you move a charged particle in space, as long as you stay on an equipotential line, you do not do any work.0423

As equipotential lines get closer together, the gradient of the potential increases.0435

You have a steeper slope of potentials -- kind of like if you have a topographic map and your equal altitude lines get closer together, you are looking at a steeper cliff, a steeper gradient.0440

So what I am showing here on the right is a positive charge -- we have the electric field lines which we have done before -- the black lines radiating away from the positive charge -- our equipotential lines must cross them at right angles.0451

Everywhere that an electric field line intersects an equipotential line, we have a 90 degree or a right angle, and equipotential lines show lines of constant potential.0464

If I were to take a charge -- let us put a charge right here -- and I want to move it over to here, somewhere else on the same equipotential line.0476

The net work done is going to be 0 because you end up at the same potential energy, the same potential because you have the same charge.0486

Let us take a look at how we could draw some equipotential lines.0496

Around positive point charges, they are pretty easy because it always intersects at a right angle, so these must look like circles.0499

I will do my best to draw a circle here -- that would be 1 equipotential line, and we could probably draw another equipotential line here -- pretty close -- and in a perfect world, all of those would intersect at 90 degrees.0508

Around a negative point charge, we would have the same idea -- crossing all the electric field lines at right angles, so we will get a circular pattern for our equipotential lines.0523

Over here on the bottom left where we have our dipole again of a positive and negative charge -- well now our equipotential lines get a little bit more complicated.0536

The one right here is pretty easy, if we draw an equipotential line right through the middle -- that is pretty straightforward -- it crosses all of those at 90 degrees.0544

But in order to cross all of these at 90 degrees as we move in here, we are going to have to adjust that a little bit and we are going to start to see some curve to it.0554

So there is an equipotential line on that side and on the other side, we are going to have the same sort of thing by symmetry.0565

I am trying to cross all these as best as I can at 90 degrees and we will get an equipotential line that looks something like that.0574

And over here on the right, now again we have two positive charges, so let us see what we get here.0584

Again, right down the middle, we are going to have that point where we do not have anything.0591

But as we go just a little bit off here, we are going to have to cross this at 90 degrees; we are going to have to get really, really steep here to cross at 90 degrees, and as we do that, it looks like we are going to come back around.0598

I am doing my best to draw that at 90 degree intersecting angles, let us try that again over here.0617

We will start from this side this time, crossing all of these at about 90 degrees and that must curve pretty steeply to come back there to give us our equipotential lines.0622

So the key point is: equipotential lines always cross electric field lines at right angles.0636

Let us talk about the potential due to a point charge just like we talked about the electric field due to a point charge.0646

To calculate the potential difference due to a point charge, the electric potential -- well if force is KQ1Q2/R² and we found electric field was going to be KQ/R², well potential is just going to be KQ/R.0652

The nice thing about potential is that it is a scalar.0668

We do not have to worry about direction, we can add them up in scalar form and save us a lot of work.0671

And to find the potential difference due to multiple point charges, we just take the sum of the electric potentials due to each individual point charge -- again not worrying about any vector nature.0676

Electric potential energy then can be found by multiplying the electric potential by the charge.0686

So the electric potential energy due to a point charge is (QV) or Q × KQ/R, therefore electric potential energy is going to be (K) times the product of your two charges divided by the distance between them.0691

Let us take a look at a sample problem here.0712

Find the electric potential at point (P) which is located 3 m away from a -2 C charge.0715

What is the electric potential energy of a half-coulomb charge situated at point (P)?0721

Let us start by finding the electric potential at point (P).0726

(V) at point (P) is going to be KQ/R where K = 9 × 10⁹ N-m²/C², our Q = -2 C, and our distance = 3 m, so that is going to give us about -6 × 10⁹ V.0729

What is the electric potential energy of a half-coulomb charge situated at that point?0751

Well the electric potential energy is charge times voltage -- our potential -- which is going to be 0.5 C × -6 × 10⁹ V or -3 × 10⁹ J.0756

A nice, straightforward applications of those formulas.0779

Let us try one that is a little more involved, kind of mirroring what we did with the electric field.0783

Let us find the electric potential at the origin due to the three charges shown in the diagram and at the end it says if we place an electron at the origin, what electric potential energy does it possess?0787

Now this is awfully similar to what we did when we were finding the net electric field at the origin, but now we are looking for potential.0797

So we are going to do it with the same basic strategy -- let us find the potential at the origin due to each of the three individual charges and then we will add them up.0803

So the potential due to the green charge, that is going to be KQ/R = 9 × 10⁹N-m²/C², our charge is 2 C and our distance from the origin is 8.0812

That is going to be just 2.25 × 10⁹ V.0828

Now let us do the red charge -- V = KQ/R, which is going to be 9 × 10⁹ × -2 C/8, which is going to be -2.25 × 10⁹ volts.0836

And finally, let us find the potential due to our 1 C charge -- V = KQ/R or 9 × 10⁹ × 1/R.0857

We said last time that if we make a right triangle here, that side is 2, that side is 2, therefore the hypotenuse must be the square root of 2² + 2²...0871

...so (R) is going to be square root of 2² + 2² or square root of 8, which is 3.18 × 10⁹ volts.0881

Now to find the total -- the potential here at the origin -- we have to sum up the potentials to each of those three charges.0894

The total is just going to be: 2.25 × 10⁹ + -2.25 × 10⁹, so we are going to cancel this out, that is going to be 0, and all we are left with is this 3.18 × 10⁹ volts.0902

That is the potential due to those three point charges when you are here at the origin.0917

Finally, we are asked to find what happens to the electric potential energy if we place an electron at the origin; what is its electric potential energy?0922

Well, let us do that.0931

The electric potential energy due to that electron is just going to be charge times voltage and our charge is -1.6 × 10^-19 C, because it is an electron; it is a negative...0934

Our voltage is 3.18 × 10⁹ volts, which is going to give us a potential energy of about -5.1 × 10^-10 J.0947

Let us point out one other item here.0961

If instead we did this in electron-volts, this would have been -1e × 3.18 × 10⁹ volts, which would have given us -3.18 × 10⁹ eV.0963

So you can see where using electron-volts could be a lot more convenient, but since it does not specify which units it wants for our answer, we will circle both of them -- they will both be correct.0979

All right. Let us talk a little bit about parallel plate configurations.0994

Configurations in which we have two parallel plates of opposite charge that are situated a fixed distance from each other are very common in physics because this is how we make a capacitor, an electrical device used to store charge.0997

And the general look of these is, we will take one plate, put another one just a little bit of distance from it, they have some cross-sectional area (A), and we will situate them at some distance from each other (D), and then typically we are going to put some charge like (+Q) up here and (-Q) there -- there is the basics of a capacitor.1009

Now the electric field due to two parallel plates -- as long as you are away from the edges of those plates -- is constant.1034

And the electric field, as we know, runs from positive to negative, so anywhere between these parallel plates -- as long as we stay a little bit away from the edges -- is constant throughout this entire region; it is given by the voltage, the potential difference across the plates divided by their distance.1040

A nice and easy way to calculate a uniform electric field, and that is true -- equals V/D here, here, here, here, here -- anywhere between those plates, it is constant.1056

The magnitude of that electric field strength is given by the potential difference divided by the plate separation and the units are going to be Newton's per coulomb (N/C) again for an electric field or volts over distance (V/m).1066

So we are proving that we have the same units (N/C) as equivalent to a (V/m)1082

All right. A capacitor is an electrical device used to store charge.1090

It consists of two conducting plates separated by some sort of insulator.1094

That insulator could be air, it could be vacuum, it could be paper, it could be Jell-O; anything that is an insulator you could put in between the plates and you would still have a capacitor.1100

Now once the plates are charged, they are disconnected.1111

The charges then stuck on the plates until the plates are reconnected.1113

So if you put a charge on each plate, you create an electric field between the plates.1116

If you disconnect them, then the charge is stuck there, and you have all that energy that is stored in the electric field.1120

The amount of charge a capacitor can store for some given amount of potential difference across it is known as the device's capacitance (C).1126

Now notice (C) is capacitance, but it is also used as a unit of charge, coulombs, so we have to be careful with our (C)'s.1134

The units of capacitance are coulombs per volt also known as a farad (F), and a farad is a very large amount of capacitance.1142

So much more often we will be talking about millifarads, microfarads, nanofarads, even picofarads.1151

Our key formula for capacitance -- capacitance is equal to the charge divided by the potential difference between the plates or C = Q/V.1158

Capacitors store energy -- because the charges are on opposite plates of the capacitor, they exert forces on each other; it becomes an energy storage device where that energy is stored in the electric field.1170

You can find the energy stored in a capacitor by the formula: potential energy in a capacitor is 1/2 times the capacitance times the square of the potential difference.1180

If we write this as 1/2 CV² -- well we also just learned that C = Q/V, so I could write this as 1/2 and I am going to replace (C) with Q/V and I still have a V².1190

I can do a little bit of simplification here -- (V) -- and the squared goes away and we will come up with 1/2 QV, which is also equal to (V), potential energy stored in a charged capacitor.1208

Let us do an example with capacitance.1228

A capacitor stores 3 microcoulombs of charge -- Q = 3 × 10^-6 C -- with a potential difference of 1.5 volts across the plates, V = 1.5.1231

What is its capacitance and how much energy is stored in it?1244

Well its capacitance is C = Q/V or 3 × 10^-6 C/1.5volts = 2 × 10^-6 F which is 2 microfarads.1248

How much energy is stored in the capacitor?1270

We could do this a couple of different ways, but let us start off with 1/2 CV² -- that is going to be 1/2 times...1273

...we just found our capacitance, 2 microfarads or 2 × 10^-6 F, and our potential difference, 1.5 volts ², gives us 2.25 × 10^-6 J).1280

Or we could have used u = 1/2 QV, which will be 1/2 × 3 × 10^-6 C (charge) × 1.5 volts (potential difference), which amazingly is 2.25 × 10^-6 J.1296

And of course those have to be the same; we said the formulas were equivalent.1315

All right. Looking at the charge on a capacitor -- How much charge sits on the top plate of a 200 nanofarad capacitor when charged to a potential difference of 6 volts?1323

Well let us start there.1334

Capacitance is C = 200 nanofarads or 200 × 10^-9 F and potential difference is V = 6 volts.1337

So if C = Q/V, then that means our charge (Q) must be (CV) or 200 × 10^-9 F × 6 volts (potential difference), which is going to be about 1.2 × 10^-6 C.1347

How much energy is stored in the capacitor when it is fully charged?1369

Well when it is fully charged, u = 1/2 CV² which is 1/2 our capacitance, 200 × 10^-9 or 200 nanofarads × 6 volts², which is going to give us about 3.6 × 10^-6 J.1373

How much energy is stored in the capacitor when the voltage across its plate is 3 volts?1398

Well when we get to 3 volts, u = 1/2 CV² again -- that is going to be 1/2 × 200 × 10^-9 × 3 volts² = 9 × 10^-7 J.1404

That will be 1/4 of that value, so if we cut the voltage in half, we have 1/4 the value, and that is because of that V² relationship -- so pretty good at calculating charge on a capacitor and the energy stored in a charged capacitor.1422

Let us talk about the design of capacitors.1437

What determines how much charge a capacitor can store?1440

Well the area of the plates -- as you have bigger plates you can store more charge, the separation of the plates plays a role in the capacitance and the insulating material between them.1444

In short, the capacitance is given by this value, ε -- that is called the permittivity and that is a constant that has to do with the material between the plates -- cross-sectional area in square meters and the separation of the plates (D).1462

Now if you have an air gap capacitor or a vacuum capacitor, our baseline ε, our baseline permittivity is 8.85 × 10^-12 C²/N-m².1481

If you put something other than air or vacuum between the plates, then you have to go from the permittivity of free space and multiply it by what is known as the dielectric constant, where your permittivity is going to be your dielectric constant (K) × ε0 and materials have larger (K)'s.1495

For air or vacuum, that is going to be 1, so ε is equal to that.1517

If you have something that is a little bit more resistive, a better insulator for example, that is going to be a bigger number for (K), so that will increase your capacitance as you put a different insulator in there.1521

Let us see how that works.1535

How far apart should the plates of an air gap capacitor be if the area of the top plate is 5 × 10^-4 m² and the capacitor must store 50 mJ of charge and an operating potential difference of 100 volts.1537

It should say it must store 50 mJ of energy -- that makes more sense.1554

Well if our area is 5 × 10^-4 m² and our potential energy is 0.05 J, we would have a potential of 100 volts.1559

Let us see what we can determine here -- u = 1/2 CV², but we also know that C = ε0 A/D and because it is an air gap capacitor, I can leave that ε0, because I do not have any dielectric constant that I have to put in front of it there.1571

So that means that u = 1/2 × ε0 A/D × V².1591

If I rearrange this a little bit to find how far apart or the distance between them, I would say that the distance then solving for (D) is going to be...1605

...ε0 × (A) × V² divided by 2 × 8.85 × 10^-12 (potential energy) × 5 × 10^-4 m² (area)...1613

...× 100 volts² (potential)/2 × 0.05 J.1631

When I do all that, I come up with a separation distance of about 4.43 × 10^-10 m, so just combining our formulas in order to solve for the unknown.1642

Let us do some more capacitance calculations.1658

Find the capacitance of two parallel plates of length 1 mm and with 2 mm if they are separated by 3 micrometers of air.1662

All right. To begin with, capacitance is ε0 A/D; it is an air gap capacitor so I do not have to worry about a dielectric constant there.1673

That will be 8.85 × 10^-12 × our area, which is length × width, or 0.001 m × 0.002 m/3 micrometers or 3 × 10^-6 m (the separation of the plates).1682

If I put that in my calculator, and I come up with about 5.9 × 10^-12 F which is also 5.9 picofarads.1703

Now what would the device's capacitance be if we replace that air with glass which has a dielectric constant of 3.9?1719

Well the only thing we have to do there is, if before, (C) was ε0 A/D -- because we are no longer using air or vacuum because we have a dielectric constant of 3.9 -- 3.9 ε0 becomes my permittivity.1727

So now, all I have to do -- all of this is the same -- is multiply my answer by 3.9.1747

That is going to be 3.9 × 5.9 × 10^-12 F or 2.3 × 10^-11 F which is 23 picofarads.1752

By inserting a dielectric material with a higher dielectric constant, I increased my capacitance because the dielectric constant was 3.9 times larger than the original, so my capacitance is 3.9 times larger than the original.1768

All right. Let us go through a couple more problems to make sure we have everything down.1787

An electron sits on a equipotential line of 5 volts.1790

How much work is required to move the electron to an equipotential line of -25 volts?1794

Let us make this one easy, let us do it in electron-volts.1799

As we do this, the potential energy is equal to charge times potential difference.1803

The charge on our electron is -1 elementary charge, and our potential difference, if we go from 5 to -25 volts -- well δV is final minus initial, so that is going to be -25 - 5 = -30 volts or 30 eV -- nice and straightforward.1809

Let us take a look at conservation of energy as we talk about these problems.1835

A proton is held at a fixed point in space where the electric potential is 500 kilovolts or 500,000 volts; the proton is then released.1840

Assuming no energy is lost to non-conservative forces, what is the speed of the electron at a point in space where the electric potential is 100,000 volts?1849

This is a conservation of energy problem.1859

It starts out initially with electric potential energy and that must be equal to its final electric potential energy.1861

Where did any excess energy go? That must be the final kinetic energy of our proton.1869

Well we know the initial is going to be (Q) times our initial voltage, so that must be equal to (Q) times our final voltage and our kinetic energy is going to be 1/2 mv².1875

So as I start solving to get v² all by itself, I could say then that v² = Q(Vi - Vf).1889

I will have the 2 in there, so 2 will factor out the Q (Vi - Vf) divided by (m), must be equal to V².1902

Therefore, v² = 2 × 1.6 × 10^-19 C (charge on our proton); Vi - Vf = 500,000 volts - 100,000 volts all divided by the mass.1912

If you have to look up the mass of a proton, you will find that it is about 1.67 × 10^-27 kg.1933

Plug all that into my calculator and I come up with something around 7.66 × 10¹³ m²/s².1941

So if I want just velocity, I take the square root of both sides and get that V = 8.75 × 10⁶ m/s -- converting electric potential energy into kinetic energy.1951

All right. Let us take a look at a problem with two conducting spheres.1970

These two conducting spheres, each with charge (Q) are connected by a wire as shown.1973

Do any charges flow between the spheres and how do their potentials compare?1979

Well the first thing we have to realize is when we connect them by a wire, all of a sudden they must be at equipotential -- anything that is connected by a conductor is going to be at equipotential.1984

So once we do that, these have to be at the same voltage.1993

That means that if we look over here at Q1, V1 = KQ1/R1.1997

Over here on the right-hand side, V2 = KQ2/R2, and because the potentials must be equal then, we could say that KQ1/R1 = KQ2/R2.2006

We have a nice simplification we can make there -- divide (K) out of both sides, therefore Q1/R1 = Q2/R2.2024

And since we know that R2 is going to be twice R1 -- measuring those lines, we could figure that out -- then Q1/R1 must equal Q2/2R,1...2036

...or with a little bit more rearrangement to say that Q2 = R1Q1/R1 or that is going to be just 2Q1.2054

If Q2 is equal to twice Q1, then we must have charge flowing from 1 to 2.2067

So our charge must be flowing that way -- we will have charge flowing from Q1 to Q2.2073

How did their potentials compare? They have to be equal.2082

Equipotential lines for a capacitor -- Draw the equipotential lines for the parallel plate capacitor below.2089

The first thing that I am going to do is it is probably easier to draw the electric field first, going from positive to negative.2095

So I will put in some electric field lines first, in green here -- electric field is constant between those.2101

And then we know that equipotential's are always crossing electric field lines perpendicularly, at right angles, so I could draw my equipotential's that way.2108

And if this is 0 volts, that will be maybe 1, 2, 3 volts, 4 volts, 5 volts.2127

We have a constant electric field between these but we do not have a constant potential.2136

Potential is going to have a linear gradient from 5 volts down to 0 volts.2140

Hopefully that gets you a good start on electric potential and electric potential difference.2146

We talked about electric potential energy and even some capacitors in there.2151

Thanks so much for your time, and make it a great day everyone!2155