For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

### 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
- Objectives
- What is a Capacitor?
- 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
- Ratio of the Charge Separated on the Plates of a Capacitor to the Potential Difference Between the Plates
- Units of Capacitance
- Farad
- Formula
- Calculating Capacitance
- 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
- Example 2
- Example 3
- Energy Stored in a Capacitor
- Field Energy Density
- Dielectrics
- 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)
- Example 4
- Example 5
- Example 6
- Example 7
- Example 8
- Example 9
- Example 10
- Example 11
- Example 12

- 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
- Farad
- 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

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### 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 capacitors *0051

*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 oil *0061

*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 so *0070

*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 - E *0224

*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 opposite *0448

*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 and *0775

*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 stored *1012

*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 stored*1043

*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 plates *1092

*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 energy*1201

*and write this as electric potential energy stored per volume = ½ ε₀ × the square of our electric field *1211

*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 here *1452

*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 capacitance*1470

*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 capacitance*1518

*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

*Let us start with what we know.*1622

*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 charged *1742

*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

*let us say that the radii R2 is about twice R1.*2123

*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 angles *2427

*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

1 answer

Last reply by: Professor Dan Fullerton

Thu Feb 11, 2016 10:43 AM

Post by Daniel Jansson on February 11, 2016

What happened to the minus sign in example 1 when you integrated the E-field?

Thanks!

/Danny (engineering physics)

1 answer

Last reply by: Professor Dan Fullerton

Fri Oct 2, 2015 6:22 AM

Post by Michael Norton on October 1, 2015

I thought V = E*d not... V = E/d

1 answer

Last reply by: Professor Dan Fullerton

Thu Apr 23, 2015 12:21 PM

Post by Sauvik Banik on April 23, 2015

On example 10, I understand you use c=(Q/V) and re-write V=(Q/C) but then how did you come up with (4*pi*E*r2) as a value of C?

3 answers

Last reply by: John X

Mon May 18, 2015 3:11 AM

Post by Thadeus McNamara on April 9, 2015

10:36, can you please write the final answer in terms of Ke? because i got exactly that answer in terms of Epsilon knot, but my hw is asking for it in terms of Ke. I'm assuming Ke = 1/4piKe , because my hw didnt say what Ke meant