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

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### 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
- Objectives
- Electric Potential Energy
- Example 1
- Example 2
- The Electron-Volt
- 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
- Equipotential Lines
- 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
- E Potential Energy Due to a Point Charge
- Electric Force from Electric Potential Energy
- E Potential Due to a Point Charge
- Example 4
- Example 5
- Finding Electric Field From Electric Potential
- Example 6
- Example 7
- Example 8
- Example 9
- Example 10
- Example 11
- Example 12
- Example 13

- 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

### 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: 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 distance*0070

*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 which*0099

*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 but*0135

*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 say*0569

*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 and*0608

*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/dr*0724

*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 R*1152

*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 ⋅ dl*1371

*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 here*1386

*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 x*1523

*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

*You can find that on your formula sheet table of contents for the exam.*1713

*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

*I get about 0.18 J.*1793

*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 and*1864

*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 so*2247

*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 ahead*2339

*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 and*2608

*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 4*2747

*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 so*2902

*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 configuration*2939

*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 potential*3177

*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 field*3395

*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 and*3510

*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 that*3538

*the electric field × that area 4 π R² = our total enclose charge is just Q ÷ ε₀*3551

*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 R*3587

*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/ r*3609

*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 us*4339

*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

1 answer

Last reply by: Professor Dan Fullerton

Fri Apr 8, 2016 6:12 AM

Post by Ayberk Aydin on April 7, 2016

For example 8, isn't the electric field directed opposite the displacement? So there should be a negative multiplier? And isn't the charge of the electron at the end negative also? Or did you just realize those two negative would cancel out at the end?

1 answer

Last reply by: Professor Dan Fullerton

Mon Jan 18, 2016 7:20 AM

Post by Shehryar Khursheed on January 17, 2016

Hello Mr. Fullerton, I have a question about the 2003 FRQ, part e.

You used an integral to find the charge enclosed within the gaussian region. When I attempted this problem on my own, I did not use an integral. This is what I did:

1) q=(rho)V

2) Because we are interested in a specific "r", we can just plug in the values for an arbitrary "r" within the spherical cloud.

3) A specific region enclosed will have a volume of (4/3)(pi)r^3, therefore...

4)Qenc= (rho not)(1-(r/R))*((4/3)(pi)r^3)

5)I plugged this into Gauss's Law and ended up with a similar answer, except I did not have a 3/4 in the fraction of r/R

I was wondering, why didn't this approach work. Even if the charge density isn't uniform, I can still plug in the function as long as I have the same function of the volume, can't I? I hope you understand what I'm trying to ask because I'm not sure if I can explain my problem best in writing. Thanks!

1 answer

Last reply by: Professor Dan Fullerton

Sun Apr 5, 2015 6:42 AM

Post by Luvivia Chang on April 4, 2015

hello Professor Dan Fullerton,

why is Electric force F=-du/dl insiead of du/dl?

why is the "-" there ?

1 answer

Last reply by: Professor Dan Fullerton

Sun Mar 29, 2015 9:39 AM

Post by holly song on March 28, 2015

why I keep getting network error message and couldn't watch your lessons? Thanks!