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### Capacitors

• Capacitors store electrical energy in electric fields.
• C=Q/V
• You can calculate capacitance by assuming a charge of Q on each conductor; finding the electric field between the conductors; calculating V by integrating the electric field; and solving for C using C=Q/V.
• The amount of energy stored per unit volume in a capacitor is known as the field energy density.

### Capacitors

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
• What is a Capacitor? 0:42
• Electric Device Used to Store Electrical Energy
• Place Opposite Charges on Each Plate
• Develop a Potential Difference Across the Plates
• Energy is Stored in the Electric Field Between the Plates
• Capacitance 1:22
• Ratio of the Charge Separated on the Plates of a Capacitor to the Potential Difference Between the Plates
• Units of Capacitance
• Formula
• Calculating Capacitance 1:59
• Assume Charge on Each Conductor
• Find the Electric Field
• Calculate V by Integrating the Electric Field
• Utilize C=Q/V to Solve for Capitance
• Example 1 2:44
• Example 2 5:30
• Example 3 10:46
• Energy Stored in a Capacitor 15:25
• Work is Done Charging a Capacitor
• Solve For That
• Field Energy Density 18:09
• Amount of Energy Stored Between the Plates of a Capacitor
• Example
• Dielectrics 20:44
• Insulating Materials Place Between Plates of Capacitor to Increase The Devices' Capacitance
• Electric Field is Weakened
• The Greater the Amount of Polarization The Greater the Reduction in Electric Field Strength
• Dielectric Constant (K) 22:30
• Formula
• Net Electric Field
• Key Take Away Point
• Example 4 24:00
• Example 5 25:50
• Example 6 26:50
• Example 7 28:53
• Example 8 30:57
• Example 9 32:55
• Example 10 34:59
• Example 11 37:35
• Example 12 39:57

### Transcription: Capacitors

Hi, everyone, I'm Dan Fullerton and I like to welcome you back to www.educator.com.0000

In this lesson, we are going to talk about capacitors.0005

Our objectives for this lesson include relating stored charge voltage and stored energy for capacitor.0008

Understanding the physics of a parallel plate capacitor.0014

Deriving expressions for the energy stored in a parallel plate capacitor.0018

Describing the electric field inside cylindrical and spherical capacitors.0022

Deriving expressions for capacitance of cylindrical and spherical capacitors.0027

Finally, describing how insertion of a dielectric between the plates of a charged parallel plate capacitor affects its electrical characteristics.0032

So let us start by talking about what a capacitor is.0041

A capacitor is an electric device that is used to store electrical energy.0044

Typically, we think of it as 2 conducting plates.0048

They do not always have to be parallel plates, you can actually make capacitors0051

that are rolled up cylindrical but you have 2 different conducting plates with an insulating material between them.0056

It is known as a dielectric.0059

That insulating material can be anything from paper to air to a vacuum to even different types of oil0061

depending on the type of capacitor you are looking at.0068

You placed opposite charges on each plate which develops a potential difference across the plates so0070

the energy is actually stored as electric field between those plates.0077

Capacitance, which gets a symbol C is going to be the ratio of the charge separated on the plates of the capacitor to the potential difference across the plates.0083

And the units of capacitance are C/V also known as farads which get the symbol F.0092

A farad is a very big amount of capacitance.0100

More often than not, we will be dealing with things like mf, µf, nf, even pf.0103

The formula for capacitance, capacitance = charge ÷ potential, C = Q/ V.0112

To calculate capacitance, what we are going to do is follow these four basic steps.0121

We are going to first assume a charge of + Q and –Q on each conductor of the plate.0126

Find the electric field between the conductors, you may have a lot of practice doing that recently.0132

Oftentimes, that is going to be Gauss’s law.0137

Calculate the electric potential by integrating the electric field.0141

Usually, this is going to involve V = the opposite of the integral of E ⋅ DL.0145

Finally, once we have the electric field and we have the potential, we can use C = Q/ V with the charge that we have assumed to solve for the capacitance.0153

Let us do an example here with parallel plates.0164

Determine the capacitance between identical parallel plates of area A, separated by some distance D.0167

If we assume +Q and -Q on each of those plates, we are going to develop an electric field from the positive side to the negative side.0175

There is our electric field E.0186

Our first step, number 1, assume + Q and –Q on each plate.0190

Check, we have done that.0200

Step 2, determine the electric field.0203

The electric field we know due to parallel plates, due to a plane of charge, our electric field between those is going to be σ/ ε₀.0206

We have done that before, check.0215

Step 3, calculate the electric potential between the plates.0219

If V is equal to the opposite of the integral of E ⋅ DL, that is going to be - E0224

which is constant between the plates × the integral of DL is just going to give us D, because we have that constant electric field.0233

This implies then, since we already know that our electric field is σ/ ε₀, we can write then that our electric potential is equal to σ/ ε₀ × T.0242

But we also know that surface charge density σ is charge ÷ area.0265

We can write this as V = QV/ ε₀ A.0284

Alright, there is our third step. Going to our fourth step then, use C = Q/ V to solve for capacitance.0284

That is going to be Q/, we have for V which is QD/ ε₀ A, which gives us ε₀ A/ D.0293

The formula for the capacitance of a parallel plate capacitor.0308

There is no dielectric in here, it is air.0312

We are going to assume the dielectric consonant of 1.0314

We will talk about what that means here in a little bit.0317

But the capacitance due to parallel plate capacitor ε₀ × the area of the plates ÷ the distance between those plates, ε₀ A/ D.0319

Let us do an example with a cylindrical capacitor.0331

Determine the capacitance of a long thin hollow conducting cylinder of radius RB, surrounding along solid conducting cylinder of radius RA.0334

First step, assume charge of +Q and –Q, that is easy enough to do.0345

Let us assume we have got +Q here and we have got –Q over there.0351

We also need to talk about the charge on here.0356

Let us define a linear charge density λ which is going to be Q/ L, the charge per length.0359

Q would be equal to λ L, of course.0367

We are going to use Gauss’s law to find the electric field.0372

We will try and be relatively wise in our choice of our Gaussian surface.0375

Let us pick one that is in between here where we want to know the electric field.0381

Our Gaussian surface will be a cylinder going back that way.0385

The integral over the closed surface of E ⋅ DA is our enclosed charge ÷ ε 0,0389

which implies then if we look at our left hand side that is going to be E × our area 2 π R × some length L has to equal our enclosed charge,0401

which is going to be enclosed charge Q is going to be λ L ÷ ε₀, which implies then that our electric field is going to be λ/ 2 π ε₀ R between our conductors.0417

We have done that before as well.0439

Step 3 then involves finding the potential from the electric field.0442

V = the opposite of the integral of E ⋅ DL is going to be the opposite0448

of the integral from our radius = RA to RB of our electric field which we just determine as λ/ 2 π ε₀ R × DR.0456

A little bit of math, let us pull our constants outside the integral sign.0475

That is going to be - our constants λ/ 2 π ε₀ integral from R = RA to RB of DR/ R.0479

The integral of DR/ R is going to be the nat log of R.0497

As we continue with our math here, this implies that our potential is going to be - λ/ 2 π ε₀ × the nat log of R evaluated from RA to RB.0503

Going a step further, that is - λ/ 2 π ε₀.0524

We will have this multiplied by the log of RB - the log of RA.0531

We are going to use the property that the difference of the logs is the log of the quotient of the 2.0539

That will be - λ/ 2 π ε₀ × the log of RA/ RB.0546

If I go make that RA/ RB, I can get rid of my negative sign out here.0556

We also still know λ is Q/ L.0564

We determine that way over here.0567

We can substitute that in using λ = Q/ L to say that V = Q/ 2 π ε₀ L log of RA/ RB.0569

There is our potential which leads us to our fourth step, solve for the capacitance using C = Q/ V.0590

If C is Q/ V that is going to be Q/, we just determined V that is Q/ 2 π ε₀ L × log of RA/ RB.0598

Let me do a little bit of algebra.0620

It will be 2 π ε₀ L ÷ the nat log of RA/ RB.0622

Derivation for the capacitance of a cylindrical capacitor.0636

What if we had a spherical capacitor?0643

Let us take a look here.0647

Determine the capacitance of a thin hollow conducting shell of radius RB concentric around the solid conducting sphere of radius RA.0648

We are going to follow the same steps we did before.0658

We are first going to assume charges of +Q on one of our conductors and -Q on the other.0660

Then, we are going be use Gauss’s law to find the electric field.0667

We really want to know the electric field, that is going to be between our two conductors.0671

We will draw our Gaussian sphere right in there.0679

The integral / the close surface of E ⋅ DA = Q enclosed/ ε₀.0684

The left hand side becomes E × the area of our Gaussian sphere 4 π R² must equal our enclosed charge +Q ÷ ε₀.0695

Or as we have done many × before, E = Q/ 4 π ε₀ R² in the electric field between those two.0708

Now to find the potential, let me go back to V = the opposite of the integral of E ⋅ DL,0720

which is - the integral from RA to RB of our electric field Q/ 4 π ε₀ R² DR.0730

Once again, we will pull the constants out of the integral sign.0747

We will have -Q/ 4 π ε₀ integral from RA to RB of 1/ R squared is what we have there.0751

I will write it as R⁻² DR.0765

Let us get to work here.0773

This implies then than V = -Q/ 4 π ε₀ the integral of R⁻² is just going to be -1/ R and0775

that is evaluated from RA to RB, which is -Q/ 4 π ε₀ -1/ RB - -1/ RA which is going to be +1/ RA.0787

Or +Q = 4 π ε₀, just distributing that negative sign through, × 1/ RB -1/ RA.0808

Let us see if we can simplify this by putting in a way common denominator RA RB.0824

We can say that our potential is equal to Q/ 4 π ε₀ ×, they give us RA - RB/ RA RB.0829

Since we know that RA is less than RB, RA being less than RB, when we do this, we get a negative.0846

Let us take that the magnitude of the potential will be Q/ 4 π ε₀ RB- RA/ RA RB.0854

Because the sign of capacitance does not make sense.0867

And finally step 4, find the capacitance by using C =Q/ V.0871

If C = Q/ V that is going to be Q/, what we just determined for our potential Q/ 4 π ε₀ RB - RA/ RA RB.0878

A little simplification comes out to 4 π ε₀ RA RB all ÷ RB – RA.0900

The derivations for different capacitor configurations.0921

Let us take a look at the energy stored in a capacitor.0926

Work is done in charging a capacitor which is what allows that capacitor to store energy.0929

If you consider 2 uncharged conductors enclosed proximity,0934

the potential difference in moving some amount of charge Q from the negative plate to the positive plate is going to be Q ÷ that capacitance.0937

Moving more charge has to increase the potential.0945

Therefore, the electric potential energy of the charge and the capacitor also has to go up.0948

How would you solve for that, how would put that numerically, quantify that?0954

The electric potential energy stored in a capacitor is going to be, let us integrate from some charges,0959

let us start with no charge until we get to our final charge in the capacitor Q of Q/ C DQ.0967

U = QV VQ/ C × our charge.0977

That is going to be, I can pull that capacitance out of there, that is a constant for the purposes of our changing charge.0982

That will be 1/ C integral from 0 to Q of q DQ which is 1/ C × Q² / 2 evaluated from 0 to Q which will be ½ Q²/C.0988

But we also know that C = Q/ V, therefore we could write this as the potential energy stored1012

in that capacitor is ½ Q² ÷, C is Q/ V which is s ½ QV in other form.1021

We could put in and get another form by recognizing that C = Q/ V.1037

Therefore, Q = C × V by replacing Q with C × V, determine that the potential energy stored1043

in my capacitor is ½ × CV from our Q, × the V that we already have there or ½ CV².1051

We have a couple different formulas for the energy stored in the capacitor.1066

½ CV², ½ QV ½, Q²/ C, they will all give you the same answer.1070

Just use whichever one happens to be most expedient for the problem you are working with.1081

Let us take a look now at the amount of energy stored in the capacitor.1087

The amount of energy stored as electric field per unit volume of your capacitor between those plates1092

is known as the field energy density which gets the symbol u_e.1097

Let us consider a parallel plate capacitor of plate area A and some plate separation B.1103

And we can think of the energy stored in the capacitor is the work that was required to create the electric field between those plates.1111

Therefore, the capacitor stores energy by creating the electric field and when it does that,1118

the amount of energy stored as electric field per unit volume is the field energy density U.1123

Our total stored electric potential energy which we just determine is ½ CV².1130

We can rewrite, knowing that our capacitance is ε₀ A/ D.1138

Our electric field between for a parallel plate capacitor is the potential/ separation B/ D.1144

Therefore, our voltage is ED.1153

We can rewrite this as U = ½, replace C with ε₀ A/ D.1157

Or replace our V² with E squared D².1166

Then we can state that our potential energy stored there is ½ ε₀ A E² D.1174

Let us look at this a little bit further.1186

What is A × D, cross sectional area × the separation?1188

That is the volume between the plates of the capacitor.1195

The volume where that electric field is stored.1199

If A × D is the volume, we can pull A × D out over under our electric potential energy1201

and write this as electric potential energy stored per volume = ½ ε₀ × the square of our electric field1211

and the potential energy per unit volume is what we call u_e, the field energy density.1223

So far we have been talking about capacitors that have air or vacuum between the plates.1235

What happens if we put another material between the plates instead?1240

Let us take a look here, as we talk about dielectrics.1244

Dielectrics are insulating materials which are placed between the plates of the capacitor to increase the devices capacitance.1248

When the dielectric is placed between the plates of the capacitor, the electric field between those plates is weaken.1261

The reason it does that, is if we have some electric field, let us draw in this direction.1265

And we happen to have a fairly neutral atom, the molecule of a dielectric, we draw it like that.1272

When the electric field is pointing in one direction, the electrons in this molecule want to go in the direction opposite the electric field.1278

They try and spend a little bit more time over here and a little bit less over here.1289

And what you really do to your molecule is you are slightly polarize it, where this side is negative,1293

this side is positive, and you set up a small induced electric field from + to -, positive to negative,1302

which opposes the over all electric field.1312

You have made it weaker, you have created an opposing electric field.1315

The greater the amount of polarization, the greater the reduction in the electric field strength.1318

Therefore, for a fixed charge on the plates, let us call that Q, the voltage that decreases increasing the capacitance because C = Q/ V.1323

Q is fixed, our potential is going to go down therefore, our capacitance goes up when you insert that dielectric.1334

Something that has a greater amount of polarization.1342

Now one way we can quantify the amount of polarization is what we call the dielectric constant which is going to get the symbol κ.1349

The amount by which the capacitance is increased when the dielectric is introduced is known as the dielectric constant.1358

That corresponds to the amount of the electric field strength is reduced due to that polarization effect.1365

The more the molecules of the atom of the dielectric are polarized, the greater the dielectric constant,1371

the stronger the weakening of an electric field becomes.1376

If we talk about our capacitance as being ε₀ A/ D, once we insert a dielectric in there, we have to throw in this κ effect, where ε is going to be κ ε₀.1381

This is the permittivity of free space, if we multiply that by the dielectric constant we get our effective permittivity.1399

We could write this now as C being equal to ε A/ D.1405

That includes our κ in and our ε₀ factor.1410

The net electric field E is going to be the electric field strength we had without the dielectric ÷ κ.1415

That relationship shows you how the κ reduces the electric field strength.1426

Key takeaway point here is that ε = κ ε₀.1431

Now there are times when you may have more than one capacitor in a configuration.1440

You may have 2 capacitors in series for example, one lined up right after the other.1444

How do you deal with that?1449

Let us assume we have a charge +Q on the top of that capacitor which means you must have -Q here1452

and by conservation of charge, that means you have +Q here and must have -Q down there.1458

We will call this potential 1 across capacitor 1, potential 2 across capacitor 2.1464

If we wanted to find the equivalent capacitance of the setup, our equivalent capacitance1470

is going to be our effective charge ÷ our total potential.1478

The charge is still +Q ÷ our total potential difference is V1 + V2, which with a little bit of manipulation that is Q ÷, if C = Q/ V, V = Q /C.1482

V1 becomes Q/C1 + V2 which is Q/C2.1499

We could rewrite this as the equivalent capacitance is going to be, if I divide those Q out, 1/ 1/ C1 + 1/ C2.1507

Or to make this a little bit easier to read, it is typically written as 1/ the equivalent capacitance1518

is equal to 1/ C1 + 1/ C2, if you have two capacitors.1525

And if you have more, you just keep going in similar fashion.1531

Capacitors in series 1/ the equivalent capacitance is 1/ the first capacitance + 1/ the second capacitance and on and on,1535

for however many capacitors you have in series.1544

We can do the same sort of thing for capacitors in parallel.1548

Let us assume that we have a charge Q1 on C1 and some charge Q2 on C2.1553

Our voltage across our capacitors must be the same, that means this must be -Q1 and that will be -Q2.1560

And our equivalent capacitance again, it is going to be our total charge ÷ our potential difference.1567

Now our total charge that was just Q1 + Q2 ÷ our potential difference which is Q1/ V + Q2 / V which is just C1 + C2.1573

Or more generally, the equivalent capacitance for capacitors in parallel, you just add up their individual capacitances.1589

C1 + C2 and on and on, for however many capacitors you happen to have handy in this parallel configuration.1596

Let us do a couple numerical examples, see how those go for us.1607

A capacitor stores 3 mc of charge with a potential difference of 1.5 v across the plates.1612

Find its capacitance and the amount of energy stored in the capacitor.1618

The charge is 3 mc 3 × 10⁻⁶ C.1625

Our potential difference V 1.5 v and we are trying to find capacitance.1631

Capacitance is charged ÷ voltage which is 3 × 10⁻⁶ C ÷ 1.5 v, which comes out to about 2 × 10⁻⁶ F or 2 µf.1640

However you prefer to write it.1661

Now for how much energy is stored in the capacitor, we can go about that a couple of different ways but let us start with U = ½ CV².1665

That will be ½ × our capacitance 2 × 10⁻⁶ F × our potential 1.5 v² or about 2.25 × 10⁻⁶ J.1677

Or just as easily, you could have done this from the perspective of our stored potential energy is ½ QV.1693

Especially, since we are giving QV right away.1701

That will be ½ × our charge 3 × 10⁻⁶ C × our voltage 1.5 v, our potential which is 2.25 × 10⁻⁶ J.1703

Of course, they are the same, that is a little assuring.1719

That will be really troubled if they were not.1722

There is our energy stored in our capacitor.1727

How much charge sits on the top plate of a 200 nf capacitor when charged to a potential difference of 6 v?1735

We have got some follow on questions about how much energy is stored when it is charged1742

and how much energy is stored when the voltage across it is 3 v?1746

Let us start with what we know, our capacitance is 200 nf, 200 × 10⁻⁹ F, our potential is 6 v.1750

How much charge?1765

I would use C = Q/ V which implies that Q = C × V which is 200 × 10⁻⁹ F × our voltage 6 v, for a total of about 1.2 × 10⁻⁶ C.1766

How much energy is stored in the capacitor when it is fully charged?1788

We can do that using U = ½ CV² which is ½ × our capacitance 200 × 10⁻⁹ F × our potential 6 v² or about 3.6 × 10⁻⁶ J.1792

What happens when we cut that voltage in half? When we go from 6 to 3 v?1815

Can you just cut that energy in half?1819

Not that simple, let us go to our formula again.1822

U = ½ CV² that will be ½ × our capacitance 200 × 10⁻⁹ F × 3 v², which is 9 × 10⁻⁷ J which is 1/4 of what we had here,1826

because we have that square factor in potential.1844

Alright let us see little bit about designing a capacitor.1854

How far apart should the plates in an air gap capacitor B if the area of the top plate is 5 × 10⁻⁴ m² and capacitor must store 50 mj of charge and an operating potential of 100 v.1859

We will start again with what we know.1873

Area is 5 × 10⁻⁴ m² must hold 50 mj or 0.05 J of energy and the potential is going to be 100 v.1875

I would start with potential energy in our capacitor is ½ CV² but we also know for a parallel plate air gap capacitor, C = ε₀ A/ D.1892

Then our potential energy is ½, or replace C with ε₀ A/ D × square of our potential.1909

Solving then for the separation of the plates D, we find that D is equal to ε₀ A × square of our potential ÷ 2 ×.1921

The stored potential energy is going to be ε 8.85 × 10 ⁻12 × our area 5 × 10⁻⁴ m² 100 v² ÷ 2 × 0.05 J.1934

0.1 J which gives us the separation of about 4.43 × 10 ⁻10 m.1956

Those things are pretty close.1970

Let us do another calculation problem.1973

Find the capacitance of 2 parallel plates of length 1 mm and width 2 mm, if they are separated by 3 µm of air?1977

Let us start there, C = ε₀ A/ D which will be our ε₀ 8.854 × our area is going to be 2 mm or 0.002 m,1986

× its length 0.001 m, 1 mm ÷ the separation 3 microns or 3 × 10⁻⁶ m, which gives us if ε₀ is 8.5 × 10 ⁻12 and it is,2003

a capacitance value of 5.9 × 10 ⁻12 F or 5.9 pf.2021

What would the devices capacitance be if the 3 µm of air were replaced by 3 µm of glass, which has a dielectric constant of 3.9.2037

All we have to do is change this a little bit, C = ε A/ D, instead of ε₀ A/ D which is going to be 3.9 × ε₀ 8.854 × 10 ⁻12.2049

There is our total ε × our area 0.002 × 0.001 m.2063

We have the same separation 3 microns, which gives us 2.3 × 10 ⁻11 F or about 23 pf.2070

The insertion of that dielectric may give us about almost a 4 × larger capacitance.2088

Let us take a look at an example with 2 conducting spheres.2098

We have 2 spheres here, each with charge Q and they are connected by a wire.2102

Do any charges flow between the spheres and how do their potentials compare?2107

The first thing I'm going to do as I look at this, is just to make this a little simpler.2112

We know Q2 is bigger than Q1, I'm going to say just for the purposes of qualitatively doing this problem,2117

As I look at these, we can say that the potential of the second sphere is going to be Q2/ 4 π ε₀ R2.2129

And in a similar fashion, the potential of 1 V1 is going to be Q1/ 4 π ε₀ R1.2143

Since they are connected by a wire, we already know that means they are going to be at the same potential.2153

If V1 = V2, we can write that Q1/ 4 π ε₀ R1 must equal Q2/ 4 π ε₀ R2.2158

And factoring out this 4 π ε₀, Q1/ R1 must equal Q2/ R2.2177

But if we said R2 is equal to 2 R1, just for the purposes of getting a qualitative feel for the problem,2186

then we can say then that Q1/ R1 is equal to Q2/ 2 R1.2193

Which implies then that Q2 = 2 Q1 R1 ÷ R1, cross multiplying or Q2 = 2 Q1, this has twice as much charge as that one does.2201

In order for that to happen, the charge must be flowing from Q1 to Q2.2223

That must have happen to get to the state.2228

They started with the same amount of charge, each had Q.2230

What do we know about the potentials?2234

That is easy, since we connected them by a wire and they are both conducting spheres, the potentials must be equal.2236

Let us take a look at an example where we are going to insert the dielectric into a capacitor.2252

We have an air gap parallel plate capacitor connected to a source of constant potential difference,2259

typically, a voltage source or a battery.2263

Inserting a dielectric between the plates of the capacitor increases which of these quantities?2266

Before I even go to these, I'm going to think about what happens when we do this, when we insert that dielectric.2271

We know that V has to remain constant because we have got a set source of potential difference, it tells us that in the problem.2278

If C = ε A/ D and ε is increasing, it went from ε₀ to some number bigger than as we put a dielectric in there,2287

that implies that our capacitance must be going up.2297

And we also know that C = Q/ V which implies that Q = CV.2302

Our C is going up, our V is remaining constant which implies then that the charge must be going up.2311

How about potential energy while we are here?2318

If U = ½ CV², while V is going up, or my V is constant, C is going up.2321

That implies that the energy stored must be going up.2330

Finally, let us take a look and say that potential we know between parallel plate capacitor is the electric fields ÷ the distance between the plates D,2335

which implies that the electric field = V × D.2344

V is constant and D is constant which implies then that the electric field must remain constant.2349

Let us see which ones of these increase?2357

Does charge on the capacitor increase?2359

We said over here charge increases so that one is true.2362

Does the voltage across the capacitor increase?2367

No, we said that is constant.2369

How about capacitance of the capacitor?2372

Absolutely, that increases.2374

Energy stored in the capacitor?2377

We already said that the energy stored is going up, that is a check.2380

And the electric field strength between the plates?2384

We said that was constant so our answers here are going to be A, C, and E.2387

Let us take a look at one last quick question.2395

Draw the equal potential lines for the parallel plate capacitor below.2400

The way I would do this to start off with, would be to draw the electric field.2404

The electric field, we know goes from positive to negative.2409

Let us draw some electric field lines here first.2412

If those are our electric field lines and we know equal potentials always cross at right angles2427

then we can draw our equal potential lines, something like this.2432

These will be our equal potential lines and I suppose probably we have one extra in here.2456

We go from 0v to something like 1v, 2v, 3v, 4v, and finally to 5v, up there.2462

Nice and straightforward, the big trick here making true that this always cross at right angles.2469

Thank you so much for watching www.educator.com.2476

Hopefully this gets you a great start on capacitors.2478

Thank you so much for your time and make it a great day everyone.2481