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

### Related Articles:

### Electric Charge & Coulomb's Law

- Electric charge (q) is a fundamental property of certain particles. The smallest amount of isolatable charge is the elementary charge (e), equal to 1.6×10
^{-19}coulombs. Charge can be positive or negative. - Protons have a charge of +1e. Electrons have a charge of −1e. Neutrons are neutral. Atoms with an excess of protons or electrons are known as ions.
- Charges can move freely in conductors. Charges cannot move freely in insulators.
- Like charges repel, opposite charges attract. Coulomb’s Law describes the magnitude of the electrostatic force between charges.
- Charging by contact is known as conduction. If a charged conductor is brought into contact with an identical neutral conductor, the net charge will be shared across the two conductors. Charging an object without placing it in contact with another charged object is known as induction.
- When a charged object is brought near a conductor, the electrons in the conductor are free to move. When a charged object is brought near an insulator, the electrons are not free to move, but they may spend a little more time on one side of their orbit than another, creating a net separation of charge in a process known as
**polarization**. The distance between the shifted positive and negative charges, multiplied by the charge, is known as the**electric dipole moment**.

### Electric Charge & Coulomb's Law

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
- Objective
- Electric Charges
- Matter is Made Up of Atoms
- Most Atoms are Neutral
- Ions
- Coulomb
- Elementary Charge
- Law of Conservation of Charge
- Example 1
- Example 2
- Conductors and Insulators
- Conductors Allow Electric Charges to Move Freely
- Insulators Do Not Allow Electric Charges to Move Freely
- Resistivity
- Charging by Conduction
- Example 3
- The Electroscope
- Charging by Induction
- Bring Positive Rod Near Electroscope
- Ground the Electroscope
- Sever Ground Path and Remove Positive Rod
- Example 4
- Polarization and Electric Dipole Moment
- Coulomb's Law
- Electrostatic Force, Also Known as Coulombic Force
- How Force of Attraction or Repulsion Determined
- Formula
- Coulomb's Law: Vector Form
- Example 5
- Example 6
- Example 7
- Example 8

- Intro 0:00
- Objective 0:15
- Electric Charges 0:50
- Matter is Made Up of Atoms
- Most Atoms are Neutral
- Ions
- Coulomb
- Elementary Charge
- Law of Conservation of Charge
- Example 1 2:39
- Example 2 3:42
- Conductors and Insulators 4:41
- Conductors Allow Electric Charges to Move Freely
- Insulators Do Not Allow Electric Charges to Move Freely
- Resistivity
- Charging by Conduction 5:32
- Conduction
- Balloon Example
- Charged Conductor
- Example 3 6:28
- The Electroscope 7:16
- Charging by Induction 7:57
- Bring Positive Rod Near Electroscope
- Ground the Electroscope
- Sever Ground Path and Remove Positive Rod
- Example 4 9:39
- Polarization and Electric Dipole Moment 11:46
- Polarization
- Electric Dipole Moment
- Coulomb's Law 12:38
- Electrostatic Force, Also Known as Coulombic Force
- How Force of Attraction or Repulsion Determined
- Formula
- Coulomb's Law: Vector Form 14:18
- Example 5 16:05
- Example 6 18:25
- Example 7 19:14
- Example 8 23:21

### 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 Charge & Coulomb's Law

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

*I am Dan Fullerton, and today we are going to start a series of lessons for AP Physics C, *0003

*specifically the Electricity and Magnetism course.*0008

*Let us tag in by talking about electric charge and Coulomb’s law.*0011

*Some of our objectives are going to be to calculate the charge on an object and explain the law of conservation of charge.*0016

*That is going to be a big idea throughout the course.*0022

*Describe differences between conductors and insulators.*0025

*Explain the difference between conduction and induction.*0029

*Talk about polarization and inducing charges or induction, and use Coulomb’s law in the principle of super position, *0033

*which is going to come up a lot in this course to solve for the force of the charge particle due to other point chargers.*0040

*With that, why do not we dive in and start talking about electric charges.*0047

*First off, matter is made up of atoms, and atoms contain protons which are positive, *0052

*electrons which are negative, and neutrons in the nucleus which are neutral.*0057

*Most atoms are neutral, they have the same number of protons and the same number of electrons.*0062

*Their positive charges that are exactly balanced by the negative charges.*0067

*Atoms that have lost or gained electron have a net charge or no longer neutral, so they are called ions.*0071

*Also, the fundamental unit of charge is known as a Coulomb with the abbreviation C.*0079

*The Coulomb is a very large amount of electric charge.*0084

*Typically, we are going to be talking about units much smaller than the Coulomb.*0088

*Not always, but the vast majority of the time.*0092

*The smallest isolated unit of an electric charge is called the elementary charge. *0095

*It is the charge on a single electron or a single proton.*0100

*1 elementary charge is equal the 1.6 × 10⁻¹⁹ C.*0103

*You are going to see that come up again and again in the course, this fundamental unit.*0110

*It is also in your formula sheet for the AP Physics C test.*0115

*Like charges repel while opposites attract, you probably knew that already.*0119

*Electric charges conserve this law of conservation of charge is a huge idea again in AP Physics C.*0123

*What that really means is if we have any close system, whatever charge we start with, *0132

*is the same charge we our going to have when we our done.*0136

*If we start with a neutral atom and we pull off 2 electrons, we have -2 elementary charges over here, *0138

*we our left with must be 2 positive, 2 elementary charges, so that the sum total of the charge is still 0. *0144

*It is still neutral, that law of conservation of charge.*0149

*Let us take a look at an example or 2 and see how this works out.*0156

*First off, Mitt is a cat that possesses an axis of 6 × 10⁶ electrons, what is the net charge on N?*0160

*The way that I will go about doing that is let us find the charge abbreviated with a Q =6 × 10⁶ elementary charges and *0168

*the charge on each one of those electrons is -1.6 × 10⁻¹⁹ C because electrons *0181

*are negatively charged for every elementary charge or electron.*0191

*Therefore, when we multiply through we are going to have our e cancel out *0196

*6 × 10⁶ × -1.6 × 10⁻¹⁹ give us an answer of right around -9.6 × 10⁻¹³ C.*0201

*Pretty straightforward, let us do another one.*0218

*Let us find a charge on Α particle.*0221

*An Α particle also known as a helium nucleus consists of 2 protons and 2 neutrons, what is the charge of an Α particle?*0224

*The first thing we have to recognize here is that the neutrons are neutral, they have no charge.*0234

*They do not contribute any charge to the answer to our question.*0239

*Instead, we need to focus on these 2 protons.*0242

*The charge Q is +2 elementary charges which is going to be 2 × 1.6 × 10⁻¹⁹ C for a total of 3.2 × 10⁻¹⁹ C.*0245

*Again, notice our answer how much smaller than it is than the Coulomb, 10⁻¹⁹ C.*0266

*Typically, we are going to be dealing with quantities of charge that are smaller than Coulomb’s. *0270

*Not always, but quite a bit of time.*0276

*Let us move on for a moment and talk about conductors and insulators.*0279

*Conductors allow electric charges to move freely.*0283

*They have a very low resistivity, another term it is going to be good to know.*0286

*Insulators, on the other hand, do not allow electric charge to move freely.*0291

*Therefore, they have a very high resistivity.*0295

*Resistivity is a material property.*0298

*It tells you how freely that material allows electrons to move and it gets the symbol ρ ρ.*0301

*The Greek letter ρ, it is like a squiggly P.*0306

*Materials that have a high resistivity things like a glass, plastic, my flannel blanket at home, they are all good insulators.*0310

*They are all of a very high resistivity.*0319

*Materials with a very low resistivity are good conductors, things like gold and silver.*0322

*Most of the metals typically are very good conductors.*0327

*Let us talk about how we charge something and we will start by talking about charging by conduction.*0333

*When things are charged by contact we call that conduction.*0337

*Here is a trick you can try, rub a balloon against your hair for a little bit, *0341

*some electrons from the atoms in your hair are transferred to the balloon, an example of conduction.*0345

*The balloon therefore, picks up some extra electrons it becomes negatively charged, *0350

*while your hair, by the law of conservation of charge must become positively charged.*0355

*We can test by now by taking the balloon and going and trying to stick it onto a wall or something like that.*0359

*Oftentimes, you can get it to a stick which is showing that it is charged in the lowest charge and by common sense*0366

*you know it is probably the balloon.*0370

*Conductors may be charged by contact.*0373

*The charge conductor brought into contact with an identical neutral conductor will share the charge across the conductors.*0375

*We will show example of that here right quick.*0385

*Here we have a conductor carrying a net charge of 8 elementary charges and *0389

*it spread into contact with an identical conductor that has no net charge.*0393

*What is the charge on each conductor after we bring them together and then separate them?*0397

*Once we touch them together, if they are identical spheres they are going to share the charge across them.*0402

*Each one is going to have 4 elementary charges and they split them apart to each have 4.*0407

*What is that in Coulomb’s?*0413

*We have 4 elementary charges which is 4 × 1.6 × 10⁻¹⁹ C or 6.4 × 10⁻¹⁹ C on each of those conducting spheres.*0415

*A useful tool for detecting charges known as the electroscope, basically what it is, is an insulating bottle of some sort *0436

*that has a metal rod in it and down at the bottom you have a couple of conducting foil leaves that are very easy to move.*0444

*If you bring something like a positively charged rod near the metal head, *0451

*the electrons are free to move in this metal rod trying to get as close as they can, opposites attract.*0457

*When that happens, as the electrons move to the top, you are left with a net positive charge here on the bottom *0464

*and then the leaves at the bottom, because they have the same charge, tend to spread apart.*0470

*It is kind of a charge detector.*0474

*We could also charge something by induction, *0479

*charging a conductor with out coming into contact with another charged object.*0481

*Here is an example of how we might do that with an electroscope.*0485

*If we start out and bring a positive rod near the head of the electroscope, the negative charges try and congregate near that positive rod.*0489

*It attracts the negative charges leaving a net positive charge down here at the bottom of the electroscope and the leaves spread apart.*0497

*What we do though, is we connect that metal rod to ground.*0506

*You can think of connecting the ground as connecting to the earth.*0511

*The earth is like this giants sink or source of charges.*0514

*It is kind of like a magic purse for any electrical charge.*0519

*If you need more negative charges connect it to ground, it will be pulled up.*0522

*If you have extra charges to get rid off, put it in the ground.*0525

*It is a great place to go store them for a while.*0528

*What we do is while we have a positive rod here but not touching, we attach the metal rod to ground and *0530

*because we have a positive charge, we want to have negative charges here.*0537

*The ground wire sucks up all these negative charges so we have a bunch of negative charges now in the head of the metal.*0541

*Then we disconnect the ground wire and the negative charges are stuck on the negative rod, on the metal rod.*0547

*Therefore, the leaves spread apart again but it is because they are charged negatively as we bring that rod away.*0556

*We ended up having a rod that was neutral to begin with, over here on the right when it is done it is negatively charged.*0561

*But we never touched our positively charged rod to it all we did was connect it to ground.*0569

*That would be an example of charging by induction.*0575

*Let us take a look at some examples.*0580

*A positively charged glass rod attracts object x.*0582

*The net charge of object x maybe, our choices are A it may be 0 or negative, B it may be 0 or positive, *0586

*C it must be negative, or D it must be positive.*0594

*Let us think about that.*0599

*If we take a positively charged glass rod and we bring it near some object x, *0600

*if it is a negatively charged they will definitely be attracted, opposites attract, so that could be true.*0609

*But also, if we have a positively charged rod and we bring it near a neutral object *0617

*what can happen is if we look closely, the entire thing is negative, or it is neutral.*0624

*When we bring the positively charged rod near it, the atoms that are nearest to the rod, *0633

*the electrons on those atoms are attracted to the positive rod.*0638

*What they will do is actually spend a little bit more time orbiting the nucleus over towards the positive rod *0642

*so with just a little tiny bit of a shift in the material.*0649

*But the electrons spending a little bit more time on this side of the nucleus than this side, *0654

*that gives us a negative charge over on this side of the object and we can still get attraction.*0658

*A charged object can attract a neutral object so we know that the answer must be 0 or negative.*0665

*The only way to prove 2 objects are charged is repulsion.*0674

*You can charge a neutral object with a charged object.*0678

*If you want to prove that they are both charged they have to repel.*0682

*This, by the way, is known as polarization of those atoms.*0687

*The whole thing is electrically neutral but because electrons spend a little bit more time over here *0693

*you get a little bit stronger force of attraction than you do repulsion in the net attraction occurs.*0698

*Let us talk about that a little bit more.*0706

*When a charged object is brought in your conductor, the electrons and the conductor are free to move. *0709

*But when a charged object is brought in your insulator, the electrons are not free to move.*0713

*They spend a little more time on one side of their orbit trading a net separation of charge known as polarization.*0717

*The distance between the shifted positive and negative charges multiplied by the charge is known as the electric dipole moment.*0726

*We just have a couple of diagrams showing that with the conductor at charged rod in a conductor.*0734

*The electrons spend time here, it is easy to see the attraction with the charged rod and insulator,*0740

*where the charges are not free to move inside the atoms and set themselves.*0746

*The electron spends a little bit more time toward the positive side and you still get that net force of attraction.*0750

*We have been talking about this force of attraction, repulsion between charges, *0758

*let us go formalize that a little bit and see if we can calculate it using what is known as Coulomb’s law.*0762

*Charged objects apply a force upon each other known as the electrostatic force or sometimes called the Coulombic force.*0768

*Similar to gravity, the force of attraction or repulsion is determined by the amount of charge and the distance between the charges.*0775

*It is actually another inverse square law.*0781

*It is the inverse square of the distance between the centers of those charges.*0784

*If we have charge 1 Q1 over here, Q2 over here, the magnitude of the force between them can be found *0788

*by taking some constant, some fudge factor K × the first charge in Coulomb’s × *0795

*the second charge in Coulomb’s ÷ the square of the distance between their centers.*0801

*Be careful here, we are using R is the distance between the centers of the object.*0805

*In this case, we our not talking about our radius itself, R is the distance between their centers.*0810

*When we do this, we end up with K constant of roughly 9 × 10⁹ Nm²/ C².*0816

*Notice, how similar this is to the force of gravity.*0828

*Universal gravitational constant × the first mass × second mass × the square of the distance between them.*0832

*The K and G are different fudge factor to make the units work out right.*0840

*You have the same form and they work in a very similar manner.*0844

*The other big difference between gravity and the Coulombic force is that the Coulombic force can repel whereas gravity only ever attracts.*0847

*Let us take a look at Coulomb’s law now in vector form.*0859

*Here we have charge 1 and charge 2, R 1 2 is the vector from 1 to 2 and *0862

*we our also going to define a unit vector in the direction from 1 to 2 which we our going to call R ̂ 1 to 2.*0871

*These are both positive what we expect is a force of 1 on 2, recalling it over to the right.*0878

*Writing that mathematically, the force of 1 on 2 is equal to our fudge factor K 9 × 10⁹ Nm²/ C² × the magnitude of the first charge, *0885

*the magnitude of the second charge ÷ the square of the distance between them and the R ̂ 1 2 was just saying the direction.*0897

*It is from 1 to 2 on the direction of that line.*0904

*Another way to write this is we can take this K and write it as 1/ 4π Ε0, the permittivity of free space.*0908

*9 × 10⁹ or 1/ 4π E₀ are the same thing.*0917

*We are going to use this form of that fudge factor quite regularly in the E and M course*0924

*because it is going to help us with some derivations later.*0929

*But really they are the same thing.*0932

*This is a constant and this is a constant, they have the same value.*0934

*Where, if we write it this way Ε 0 is 8.85 × 10⁻¹² C²/ Nm².*0937

*K= 1/ 4π E₀.*0945

*If you are doing these calculations and you see this, easy enough to just swap in K for it if you want.*0948

*9 × 10⁹ or it looks like you can maybe cancel a π or E₀.*0954

*Go from K that way, they are the exact same thing.*0959

*Determining the electrostatic force, we have 3 protons separated from a single electron by a distance of 1 micron.*0966

*3 protons + 3 elementary charges separated from a single electron -1 elementary charge by a distance of 1 micron or 1 × 10⁻⁶ m.*0973

*Find the force between them and is it attractive or repulsive.*0986

*Let us do the easy part first, attractive or repulsive, they are opposites they must attract.*0990

*Let us find the magnitude of that force.*0999

*Our first charge, Q1 is 3 elementary charges which is 3 × 1.6 × 10⁻¹⁹ C or 4.8 × 10⁻¹⁹ C.*1001

*Our second charge, Q2 is the charge of our single electron which is -1 elementary charge *1020

*or -1.6 × 10⁻¹⁹ C and the distance between them is 10⁻⁶ m.*1027

*Therefore, when we go to find the electrostatic force that is going to be K Q1 Q2/ R².*1039

*K is 9 × 10⁹ Nm²/ C² × Q1 4.8 × 10⁻¹⁹ C × Q2 -1.6 × 10⁻¹⁹ C ÷ the square of the distance between them which is 10⁻⁶ m².*1049

*When I go through and do my math, plug it in my calculator, *1082

*I come up with something like the electrostatic force is about equal to 6.9 × 10⁻¹⁶ N.*1085

*Of course that must be attractive.*1095

*Moving on, a beam of electrons is directed into the electric field between 2 oppositely charged parallel plates.*1105

*What is the direction of the electrostatic force exerted on the electrons by the electric field?*1113

*Let us see, our electron beam is actually going to be just a bunch of electrons traveling to the right.*1119

*As the electrons travel to the right, their negative charges they see the positive charge up here, *1127

*they are going to feel a force towards the positive due to this plane.*1132

*And down here they see a negative, their negative they are going to be repelled *1139

*that is also going to push them in the upper directions. Our net force must be up.*1143

*Let us see if we can put this together into a slightly more involved problem.*1146

*We have 3 point charges located at the corners of a right triangle as shown where Q1 and *1149

*Q2 are both 3 micro Coulomb’s, 3 × 10⁻⁶ C and Q3 is – 4 micro Coulomb’s.*1160

*If Q1 and Q2 are each 1 cm from Q3, find the net force on Q3.*1167

*Let us start by looking at the force of 1 on 3, we will call that F 1, 3 and that is going to be equal to *1175

*K Q1 Q3 / R² which is 9 × 10⁹ Nm²/ C² × our charge on Q1 3 micro Coulomb’s *1187

*or 3 × 10⁻⁶ C × Q3 which we said was -4 × 10⁻⁶ C.*1200

*I’m just going to work with magnitudes for now and figure out the directions when we are done ÷ the square of the distance between them.*1210

*1 cm is 0.01 m and we have to square that.*1218

*I end up with about 1080 N when I plug that into my calculator.*1223

*We are looking at the force of 1 on 3.*1229

*If this is negative and this is positive, we are going to have a force upward of on 3 due to 1.*1231

*That is 1080 N up or I could write that in unit vector notation 1080 N in the J ̂ direction.*1240

*Let us take a look at the force on Q3 due to Q2.*1253

*The force of 2 on 3 is going to be K Q3 Q2/ R² which is going to be 9 × 10⁹ × our first charge 3 micro C, *1260

*3 × 10⁻⁶ × our second charge 4 × 10⁻⁶ / the square of the distance between them again 0.01 m² or 1080 N.*1277

*As we look at this one, we have a positive charge here *1293

*and negative charge here so the force on 3 due to 2 is going to be to the right.*1296

*1080 N to the right or 1080 N in the I ̂ direction, in the x direction unit vector notation.*1302

*If we want the total force, we will use the law of super position and just add these together that is going to be, *1309

*we will have this component 1080 N to the right in the i ̂ direction + we have our 1080 N in the J ̂ direction.*1319

*Or if you wanted to write this in component form that is just 1080 N in the x, 1080 N in the y.*1336

*If we wanted to know the magnitude of that force, we have 1080 N up, 1080 N to the right which makes a right triangle.*1348

*We can use the Pythagorean Theorem to figure that out.*1354

*A total force then, the magnitude of that total force is going just be √1080² + 1080² *1358

*or about 1527 N and the direction it is going to be an angle of about 45° NE something like that.*1370

*1527 N, one to the right, one up the combination.*1387

*Let us take a look at one last example here in this lesson and see how it goes.*1395

*Putting all of this stuff together, we have 2 identically charged balls of mass 5 mg hanging from the ceiling by a light string of length 20 cm.*1399

*When we say light string in Physics that means you can neglect the mass of the string, makes life a whole lot simpler.*1411

*The total angle between them is 12° find the magnitude of the charge on each ball.*1416

*The first thing I'm going to do is let me draw in a little bit here, *1424

*I'm going to draw a center line between these just so that I can see what is going on here.*1428

*I'm going to call the angle write there θ.*1434

*If the whole angle between them is 12°, this θ is 6° and that θ must be 6°.*1440

*Let us call this ball 1 and that ball 2.*1448

*If we start from there, I always like to look at free body diagrams to help me identify forces and what is going on.*1452

*Starting with a free body diagram let us look at the ball on the left, ball number 1.*1459

*We have our object which I will draw as a dot, our forces on it we have its weight mg down, *1464

*we have the tension from the string T, and we have the force of electrostatic repulsion to the left FE.*1474

*I have a force in an angle and that always makes things more complicated.*1486

*What I want to do is break up this tension into components along the x and y axis.*1490

*The way I’m going to do that is I'm going to redraw my free body diagram to make a pseudo free body diagram *1495

*where I still have MG down, I have the electrostatic force to the left but now I'm going to break up this tension into a component *1501

*that is vertical and the component that is horizontal.*1511

*And looking at my angle here or if I were to look at it here as well, it is pretty easy to see *1515

*that the component to the right is going to be T sin θ that is opposite R θ and this component, *1522

*the vertical component will be T cos θ because the horizontal is adjacent to our angle as we have defined θ.*1530

*I can start writing Newton’s second law equation to solve the problem.*1539

*Let us start with a net force in the x direction, I can say net force in the x direction *1543

*and I just look at my pseudo free body diagram and write in all the forces in the x direction.*1550

*I have T sin θ in the positive x direction and I have the electrostatic force in the negative x direction *1555

*and because this is an equilibrium it is not accelerating, I know that the total must be 0.*1567

*Therefore, I could write then that T sin θ = electrostatic force.*1575

*There is one equation we can use, let us try the same thing with the y direction, net force in the y direction *1586

*and I look at my pseudo free body diagram again, I have T cos θ in the positive y direction – MG.*1595

*It is not accelerating in the y direction so that = 0.*1607

*Therefore, I can state that T cos θ must be equal to MG.*1611

*I got my 2 equations, looks like I can probably do some simplifications here.*1621

*I'm going to pull a little trick, as I look at this, if that is equal to that and that is equal to that, *1626

*if I divide the first equation by the second, that also still has to be equal.*1632

*T sin θ/ T cos θ = FE/ MG.*1636

*Let us write that, T sin θ/ T cos θ must be equal to our electrostatic force ÷ gravity.*1643

*T/ T is 1, sin θ/ cos θ is the tan θ so this implies then that tan θ =FE/ MG.*1661

*or I could rewrite this to say that the electrostatic force equals MG tan θ *1677

*but the electrostatic force we can get from Coulomb’s law that is equal to K Q1 Q2/ R².*1684

*Q1 and Q2 are the same, these are identical spheres so this is just Q and Q.*1698

*The trick then is this R is the distance between them, we could figure that out implies then *1703

*we know the length of the string and we know the angle, we can use a little trig *1712

*to realize that distance from here to the center point must be L sin θ.*1715

*This 20 cm sin θ.*1721

*The distance between the first and the second sphere must be twice that.*1724

*Our R must be 2 × the length of the string sin θ.*1728

*If I rewrite this again, we can say that K Q1 × Q2 if they are the same thing, *1734

*just Q that is K Q² / R² is going to be 4 L² sin² θ must be equal to MG 10 θ *1741

*which implies then that our charge itself, rearranging this and solving for just Q must be equal to the √4 MG L² sin² θ tan θ / K.*1761

*That is quite the equation.*1782

*Let us substitute in what we know to say that Q = 4 × 5 g, 0.005 kg × G 9.8 m /s² on the surface of the earth × *1784

*our length 0.2 m² × the square of the sin of our angle which is 6° × the tan of angle 6°/ 9 × 10⁹ K N m²/ C².*1800

*Take the square root of that and I come up with about 3.16 × 10⁻⁸ C, the charge on each of those spheres.*1821

*Hopefully, that gets you a great start on electric fields and Coulomb’s law, electric charges and Coulomb’s law.*1837

*Come back for the next one and thank you very much for watching www.educator.com.*1843

*I will see you again real soon, make it a great day.*1847

1 answer

Last reply by: Professor Dan Fullerton

Mon Jan 23, 2017 4:46 AM

Post by El Einstein on January 23 at 12:44:42 AM

On example 5, how did you know that the electrostatic force was "attractive" and not "repulsive"?

1 answer

Last reply by: Professor Dan Fullerton

Tue Aug 30, 2016 10:12 PM

Post by Samatar Farah on August 30, 2016

Hello Mr. Fullerton. On example 5, I kept getting a negative answer instead. Is it supposed to be negative or not? I might be missing something here.

1 answer

Last reply by: Professor Dan Fullerton

Thu Mar 31, 2016 4:33 PM

Post by Ayberk Aydin on March 31, 2016

On the last example, shouldn't the mass be 5*10^-6 kg and not 5*10^-3 kg since the units given are in milligrams and not grams?

2 answers

Last reply by: Jessie Sun

Fri Feb 5, 2016 10:15 AM

Post by Jessie Sun on January 25, 2016

Hello Professor,

When using the principle of superposition, how do you know whether to add the two forces or subtract them to get the net force?

In your example you added them, in what case would you subtract the two forces instead? And how do you determine the force that will be subtracted from the other in order to find the net force? Is it just the larger one?

2 answers

Last reply by: Professor Dan Fullerton

Tue Jan 12, 2016 12:59 PM

Post by Shehryar Khursheed on January 12, 2016

I noticed that you do not have real ap problems included in this course as you have in the mechanics course. Is there a particular reason for this becuase I found those examples very helpful?

1 answer

Last reply by: Professor Dan Fullerton

Mon Dec 21, 2015 6:21 AM

Post by Akilah Miller on December 20, 2015

Hello Professor,

In example 8, since we are given that the mass of the objects is 5mg, is the mass we use in the equation in the end supposed to be 5E-6 kg instead of 5E-3 kg?

1 answer

Last reply by: Professor Dan Fullerton

Mon Nov 23, 2015 7:33 AM

Post by Jeffrey Tao on November 10, 2015

Would it be beneficial to take multivariable calculus before starting this course, or is having taken AP Calculus BC just fine? I know Maxwell's equations include topics like surface integrals, which are not covered in single variable calculus, so I wasn't sure whether I would be able to understand them completely without taking multivariable calculus.

1 answer

Last reply by: Professor Dan Fullerton

Sat Sep 5, 2015 6:56 AM

Post by richmond mensah on September 4, 2015

but how did you know it was going towards the positive direction instead of the negative. I thought the charges go from positive to negative. giving a negative direction in the I and j. so -1080i-1080j

1 answer

Last reply by: Professor Dan Fullerton

Sun Jun 28, 2015 1:30 PM

Post by Derek Boutin on June 28, 2015

Professor Fullerton, what is the difference between this segment of lectures and the segment of lectures on AP Physics 1 & 2? I watched your videos on electricity and magnetism in AP Physics 1 & 2. Do you suggest I also watch these? I feel like some of them contain the same content.

2 answers

Last reply by: richmond mensah

Fri Sep 4, 2015 9:27 PM

Post by Richard Scafidi on June 11, 2015

Apologies if I missed it, but how did you find the direction of the particle in example 7?

1 answer

Last reply by: Professor Dan Fullerton

Thu May 28, 2015 5:19 PM

Post by Mohsin Alibrahim on May 28, 2015

Hello Professor,

How did you determine r in ex 5 ?

Thanks for the wonderful lecture.

2 answers

Last reply by: Professor Dan Fullerton

Fri Feb 20, 2015 5:11 PM

Post by Thadeus McNamara on February 20, 2015

this whole lecture is basically all ap physics b right ? (well now its called ap physics 1 and 2). when is new material going to be introduced? next lecture? I'm pumped

1 answer

Last reply by: Professor Dan Fullerton

Fri Nov 21, 2014 1:20 PM

Post by Shih-Kuan Chen on November 21, 2014

Hello Professor,

What is the difference between your courses of AP Physics C and Professor Jishi's? Why are his lectures much longer?

1 answer

Last reply by: Professor Dan Fullerton

Thu Nov 20, 2014 6:04 AM

Post by Shih-Kuan Chen on November 19, 2014

Hello Professor, sincere greetings.

I only have a basic understanding of electricity and magnetism. By watching all the videos in this AP electromagnetism course, do you think I will be well prepared enough to take the AP electromagnetism exam by the end of the school year?