For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

### Electric Fields & Forces

- Matter has a property called resistivity, which relates to how easily electric charges move through the material.
- Electric charge is conserved. Net charge is the sum of all charges of all objects in a system.
- The smallest observed unit of charge that can be isolated is the elementary charge.
- Electric force results from the interaction of two objects with electric charge.
- The magnitude of the electric force (F) exerted on an object with charge q by an electric field E is F=qE.
- The magnitude of the electric field vector is proportional to the net electric charge of the object(s) creating the field.
- The electric field outside a spherically symmetric charged object is radial and follows an inverse square law as a function of the distance from the center of the object.
- The electric field around dipoles and other systems of electrically charged objects is radial and follows an inverse square law as a function of the distance from the center of the object.
- The electric field inside a conductor at equilibrium is zero.

### Electric Fields & Forces

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 Charges
- Matter is Made Up of Atoms
- Protons Have a Charge of +1
- Electrons Have a Charge of -1
- Most Atoms Are Neutral
- Ions
- Fundamental Unit of Charge is the Coulomb
- Like Charges Repel, While Opposites Attract
- Example 1: Charge on an Object
- Example 2: Charge of an Alpha Particle
- Conductors and Insulators
- Conductors Allow Electric Charges to Move Freely
- Insulators Do Not Allow Electric Charges to Move Freely
- Resistivity is a Material Property
- Charging by Conduction
- Example 3: Charging by Conduction
- The Electroscope
- Charging by Induction
- Example 4: Electrostatic Attraction
- Coulomb's Law
- Charged Objects Apply a Force Upon Each Other = Coulombic Force
- Force of Attraction or Repulsion is Determined by the Amount of Charge and the Distance Between the Charges
- Example 5: Determine Electrostatic Force
- Example 6: Deflecting an Electron Beam
- Electric Fields
- The Property of Space That Allows a Charged Object to Feel a Force
- Electric Field Strength Vector is the Amount of Electrostatic Force Observed by a Charge Per Unit of Charge
- The Direction of the Electric Field Vector is the Direction a Positive Charge Would Feel a Force
- Example 7: Field Between Metal Plates
- Visualizing the Electric Field
- Electric Field Lines Point Away from Positive Charges and Toward Negative Charges
- Electric Field Lines Intersect Conductors at Right Angles to the Surface
- Field Strength and Line Density Decreases as You Move Away From the Charges
- Electric Field Lines
- E Field Due to a Point Charge
- Electric Fields Are Caused by Charges
- Electric Field Due to a Point Charge Can Be Derived From the Definition of the Electric Field and Coulomb's Law
- To Find the Electric Field Due to Multiple Charges
- Comparing Electricity to Gravity
- Example 8: E Field From 3 Point Charges
- Example 9: Where is the E Field Zero?
- Example 10: Gravity and Electricity
- Example 11: Field Due to Point Charge

- Intro 0:00
- Objectives 0:10
- Electric Charges 0:34
- Matter is Made Up of Atoms
- Protons Have a Charge of +1
- Electrons Have a Charge of -1
- Most Atoms Are Neutral
- Ions
- Fundamental Unit of Charge is the Coulomb
- Like Charges Repel, While Opposites Attract
- Example 1: Charge on an Object 2:22
- Example 2: Charge of an Alpha Particle 3:36
- Conductors and Insulators 4:27
- Conductors Allow Electric Charges to Move Freely
- Insulators Do Not Allow Electric Charges to Move Freely
- Resistivity is a Material Property
- Charging by Conduction 5:05
- Materials May Be Charged by Contact, Known as Conduction
- Conductors May Be Charged by Contact
- Example 3: Charging by Conduction 5:38
- The Electroscope 6:44
- Charging by Induction 8:00
- Example 4: Electrostatic Attraction 9:23
- Coulomb's Law 11:46
- Charged Objects Apply a Force Upon Each Other = Coulombic Force
- Force of Attraction or Repulsion is Determined by the Amount of Charge and the Distance Between the Charges
- Example 5: Determine Electrostatic Force 13:09
- Example 6: Deflecting an Electron Beam 15:35
- Electric Fields 16:28
- The Property of Space That Allows a Charged Object to Feel a Force
- Electric Field Strength Vector is the Amount of Electrostatic Force Observed by a Charge Per Unit of Charge
- The Direction of the Electric Field Vector is the Direction a Positive Charge Would Feel a Force
- Example 7: Field Between Metal Plates 17:58
- Visualizing the Electric Field 19:27
- Electric Field Lines Point Away from Positive Charges and Toward Negative Charges
- Electric Field Lines Intersect Conductors at Right Angles to the Surface
- Field Strength and Line Density Decreases as You Move Away From the Charges
- Electric Field Lines 20:09
- E Field Due to a Point Charge 22:32
- Electric Fields Are Caused by Charges
- Electric Field Due to a Point Charge Can Be Derived From the Definition of the Electric Field and Coulomb's Law
- To Find the Electric Field Due to Multiple Charges
- Comparing Electricity to Gravity 23:56
- Force
- Field Strength
- Constant
- Charge/ Mass Units
- Example 8: E Field From 3 Point Charges 25:07
- Example 9: Where is the E Field Zero? 31:43
- Example 10: Gravity and Electricity 36:38
- Example 11: Field Due to Point Charge 37:34

### AP Physics 1 & 2 Exam Online Course

### Transcription: Electric Fields & Forces

*Hi everyone! We are thrilled to have you back with us here at Educator.com.*0000

*Today we are going to start a lesson on electric fields and forces which is the beginning of our unit on electricity and magnetism.*0003

*Our objectives are going to be to calculate the charge on an object and explain the Law of Conservation of Charge, describe differences between conductors and insulators, and explain the difference between conduction and induction.*0010

*We will also use Coulomb's Law to solve for the force on a charged particle due to other point charges, calculate the electric field due to one or more point charges and finally to analyze electric field diagrams.*0023

*Let us start by talking about electric charges.*0034

*As you know matter is made up of atoms, and those atoms even have smaller particles -- subatomic particles such as protons, electrons and neutrons.*0037

*Now protons have a charge of +1 and +1e is a +1 elementary charge, the smallest stable single charge, that is equal to 1.6 × 10 ^{-19} coulombs (C), the SI unit of charge.*0045

*Electrons on the other hand have a charge of -1 elementary charge and neutrons are neutral.*0060

*Now most atoms are neutral -- they have equal numbers of protons and electrons.*0066

*The positives and the negatives balance out for a net charge of 0; it is neutral.*0070

*But if an atom loses its electrons, loses an electron or two, it is going to become positively charged, or if it gains an electron or two, it is going to become negatively charged.*0075

*We call those charged atoms 'ions'.*0085

*Now the fundamental unit of charge in the SI system is the coulomb (C), and that is a big amount of isolated charge.*0089

*The smallest isolated unit of electric charge is the elementary charge, 1e on a proton, -1e on an electron and it's magnitude is 1.6 × 10 ^{-19} C.*0098

*Now like charges repel each other, opposites attract, and of course electric charge is conserved -- that is called the Law of Conservation of Charge.*0110

*You cannot spontaneously have just a (+) charge appear.*0119

*If you start off with 0 net charge, later on in a closed system you must still have 0 net charge.*0122

*If you start off with 0 net charge you could have a +1e and a -1e so that your net is still 0, but you cannot spontaneously get a +3e and a -1e for a net of 2 -- charge has to be conserved.*0127

*Now let us take a look at charge on an object.*0143

*Mittens the cat possesses an excess of 6,000,000 electrons.*0146

*Let us find the net charge on Mittens in coulombs SI units.*0150

*Well charge gets the letter (Q) and that is going to be 6 × 10 ^{6} electrons, and since they are negative I will put that there so we have -6 × 10^{6} elementary charges...*0154

*...and we are going to try and convert that to coulombs and the way we are going to do that is by multiplying by 1 again -- that old trick.*0168

*We want elementary charges to go away so I will put (e) down here and I want units of coulombs, so I will put that up here, and then I have to write numbers in here to make a ratio of 1.*0175

*Well 1 elementary charge = 1.6 × 10 ^{-19} C.*0186

*Now when I go through and do the math, my elementary charge units are going to cancel out and I will get -6 × 10 ^{6} × 1.6 × 10^{-19} C for a total of -9.6 × 10^{-13} C as my answer. Great!*0192

*Let us take a look at the charge of an alpha particle.*0216

*An alpha particle, which is also known as a helium nucleus, consists of 2 protons and 2 neutrons, no electrons.*0219

*So what is the charge of an alpha particle?*0225

*Well if an alpha particle has 2 protons, those are the only charged particles, it must have a charge of +2 elementary charges.*0228

*But let us convert that into coulombs -- +2 elementary charges, and we need to multiply that by 1, I will write 1 as 1e = 1.6 × 10 ^{-19} C...*0239

*...so I get a charge of 3.2 × 10 ^{-19} C when I put that into SI units -- 3.2 × 10^{-19}.*0254

*All right. As we talk about materials -- conductors are materials that allow charges to move freely.*0268

*They have a very low resistivity, so charges can go through them very, very easily.*0274

*Insulators, on the other hand, do not allow charges to move freely.*0279

*They have what is known as a very high resistivity, and resistivity is a material property.*0282

*If you look up the resistivity of gold, it has a certain value; if you look up the resistivity of glass, it has a specific value.*0288

*It is a material property measured in ohm meters and it typically gets the symbol ρ, a squirrely little P.*0296

*All right. Materials may be charged by contact known as conduction.*0306

*You might have tried this trick before -- rub a balloon against your hair someday.*0310

*Some electrons from the atoms in your hair get transferred to the balloon.*0313

*The balloon now has a net negative charge and your hair, because it lost some electrons, now has a net positive charge.*0317

*You can also charge conductors by contact.*0324

*For example, if you bring a charge conductor into contact with an identical neutral conductor, you will share the charge across those two conductors.*0327

*Let us take a look and see how that would work.*0335

*A conductor carrying a net charge of 8 elementary charges is brought into contact with an identical conductor with no net charge.*0338

*When they are brought into contact because their conductors and charges can move freely, those charges -- those 8e -- they are all positive; that is a positive charge; they are going to repel; they want to get as far apart from each other as they can.*0346

*So what do they do? They split up so you get 4e on each conductor.*0357

*If those conductors are identical, then you get the exact same charge on each.*0361

*Then if you split them apart, you now have 4 elementary charges on each object.*0365

*So what would the charge be?*0370

*4e on each one, is going to be 4 times the charge on an elementary charge -- 1.6 × 10 ^{-19} C, which is 6.4 × 10^{-19} C.*0372

*Now on a conductor, the charge is always going to sit on the outside surface.*0387

*So the electric field inside the conductor, as we talk about fields here soon, is always going to be 0.*0392

*Something to remember is that an electric field inside a conductor is 0.*0398

*Let us talk about the electroscope.*0405

*The electroscope is a really cool tool that is used to detect small electric charges based on conduction.*0407

*It consists of a conducting rod in a beaker that is then insulated from the outside world, except for this metal knob that sticks out the top.*0413

*If you were to bring something like a positively charged rod over to it to share the charge, you are going to get a net positive charge on your metal rod and as that is distributed throughout the rod, at the bottom you have these two very thin leaves.*0421

*If you have positive charges on each of the leaves, they are going to repel and you get a spreading of the leaves, which indicates that you have a charge.*0434

*You could have the same basic thing happen if you were trying to do this with a negative charge.*0442

*If we draw just the metal part of our electroscope here and we bring a negatively charged rod near it...*0446

*...well now the electrons that are already on that metal rod without even touching, they are going to be repelled from the negative charge there and they are going to try and hang out as far away as possible from that negative charge -- leaving this end positive.*0456

*Well once again down here on those leaves you have negative charges on each leaves they will repel, so this is really a charge detector.*0470

*You could also charge by induction -- that is charging a conductor without actually coming into contact with another charged object.*0480

*So if we start off with a neutral electroscope here and bring a positive rod near it, we will have the electrons all want to gather near the positive charge -- opposites attract -- leaving a net positive charge down near the leaves of the electroscope; they will spread apart.*0487

*Now what we are going to do though is while we hold that rod in place, we are going to connect this metal bar to ground and by some connection to ground -- to the earth through a conductor -- the earth acts as an infinite sync or source of electrons.*0502

*If you need some electrons to balance things out you could pull them straight from the earth; we have tons of extras.*0517

*If you need to get rid of some, you can throw them into the earth very easily.*0522

*So now when you connect this to ground it sees all the positive charges here and you will start sucking some electrons up from the ground in order to make that balance nice and happy.*0526

*Now if you disconnect the ground connection, those charges are stuck on the metal conductor, you have a net negative charge on the electroscope and of course down here where you once again have negative charges on the leaves, the light charges repel each other and you see the spreading of the scope.*0535

*You have therefore charge to the electroscope by induction -- you have not touched it specifically to the source of charges to that glass rod.*0553

*Let us talk about electrostatic attraction.*0564

*A positively charged glass rod attracts object X. What can you say about the net charge of object X?*0566

*Well before we can answer that, let us think about what could happen here.*0574

*If we have a positively charged glass rod, it is pretty easy to see that if you have some other negatively charged object that it will attract.*0577

*So without a doubt, you could say that you could attract a negatively charged object.*0590

*That is easy, but what if the object is neutral?*0594

*Well if it is a conductor and it is a neutral object, you will actually have some of the electrons in the object who will want to hang out over near the positive charge.*0598

*It remains electrically neutral, but because you have a negative closer to the positive, then the positive-positive you have a net attractive force, so you can actually attract neutral objects.*0607

*What on the other hand would happen if you had that positive rod and instead of a conducting sphere like we had there let us talk about an insulating sphere.*0618

*Well the atoms actually in this insulating sphere are made up of a positive nucleus and a negative electron orbiting it or multiples.*0629

*When you bring the positive charge near it, the electrons are going to want to be attracted to that positive rod.*0642

*They will want to spend just a little bit more time in their orbits over toward that positive rod side.*0647

*So you are going to actually get the atoms, polar atoms or molecules to polarize just a little bit the ones at the edge, and as they do that you have a slight attractive force that is stronger between the positive and negative than the repulsive force from the positive to the positive.*0652

*You have created a slight polarization of those molecules electrically and you can attract a neutral insulator that way as well.*0670

*So what is the answer?*0679

*The net charge of object X could be neutral or it could be negative to get attracted.*0680

*Same thing would happen if you have a negative glass rod, it could attract something that was either positive or neutral.*0686

*The only way to prove that two objects are charged is by repulsion.*0693

*You can attract neutral objects with one charged object; you cannot repel unless both objects are charged.*0698

*Let us talk about Coulomb's Law as we try and quantify this electrical force of attraction or repulsion.*0706

*As we know like charged objects repel and opposites attract.*0712

*Charged objects therefore must apply a force upon each other -- that is known as the electrostatic force or the coulombic force.*0717

*And similar to gravity, the force of attraction or repulsion is determined by the amount of charge instead of mass and the distance between the charges.*0723

*So the electric force is equal to some constant (k) -- that is called the electrostatic constant, it has the value 9 × 10 ^{9}N-m^{2} per C^{2} × the charge on the first object × the charge on the second object divided by the square of the distance between them.*0732

*It is another inverse square law -- and notice how similar that is to the force equation we had for gravity.*0749

*Universal gravitational force was (g) -- m1m2/r ^{2}.*0756

*Notice what we have here. Instead of (k), we had (g) for gravity and instead of charge we have masses and the distance between their centers.*0765

*It is almost the same thing, just in a different regime.*0777

*The biggest difference is that gravity cannot repel, while the electrostatic force can attract or repel.*0781

*So determining the electrostatic force -- three protons are separated from a single electron by a distance of 1 micron or 1 × 10 ^{-6} m.*0790

*Find the electrostatic force between them. Is that attractive or repulsive?*0800

*Well let us answer the easy question first.*0805

*If we have 3 protons, that is going to be a charge of +3 elementary charges, the electron will be -1 elementary charge, and opposite charges are going to attract.*0807

*Now, however, let us find the force between them.*0819

*Let us call our protons our first charge (q1), so that is going to be 3e or 3 × +1.6 × 10 ^{-19} C, which is 4.8 × 10^{-19} C.*0823

*Our second charge (q2) is going to be -1 × our elementary charge (1.6 × 10 ^{-19} C) or -1.6 × 10^{-19} C.*0836

*And the distance between them? R = 10 ^{-6} m.*0850

*So now we can apply Coulomb's Law.*0854

*The electric force = k (the electrostatic constant) times the product of our charges divided by the square of the distance between them.*0857

*So that is going to be 9 × 10 ^{9}N-m^{2}/C^{2} (k) × q1 (4.8 × 10^{-19} C) × q2 (1.6 × 10^{-19} C) all over the square of the distance between them -- 10^{-6} m^{2}.*0866

*Now I want you to notice something here -- I did not worry about the signs when I am using Coulomb's law.*0889

*Although you can, it is usually a lot simpler to calculate the magnitude of the force and then use what you know about positive - negative charges, attraction or repulsion to figure out the direction of the force.*0896

*Typically, do not worry about the signs of the charges.*0908

*It is a lot easier to just figure it out using common sense once you have done your calculations.*0911

*This implies then that the electric force, when I put all of this into my calculator, comes out to be right around 6.9 × 10 ^{-16} and the units of force of course are Newton's (N) and we already determined that that was an attractive force.*0917

*Let us take a look at a deflecting electron beam.*0937

*If we have a beam of electrons, a bunch of electrons all negative -- e-, e-, my symbol for electron (e-) -- all moving to the right.*0940

*It is directed into the electric field between two oppositely charged parallel plates.*0951

*What is the direction of the electrostatic force exerted on the electrons by the electric field?*0956

*Well even though we have not talked about electric field formally yet, it is pretty easy to see what is going to happen to our electron beam.*0960

*If these are negative charges, they are going to get attracted by this side so we are going to have a force that way to do attraction here, and they are going to be repelled by the negatives here so we will have a force that way.*0967

*So our net force -- both of these are going up, it is going to deflect it up, so we have a net force upwards.*0978

*Electric fields -- The electrostatic force, just like the gravitational force is a non-contact or field force -- we cannot see it.*0989

*The way we can detect it is by placing a charge somewhere in space and then seeing what force it has acting upon it.*0997

*The property of space that allows a charged object to feel a force is what we call the electric field -- that mental construct to help us understand what is going on.*1005

*You detect the presence of an electric field by placing a positive test charge at various points in space and measuring the force the test charge feels.*1012

*What that means is the electric field is always going to point in the direction of the force on a positive charge, not on a negative charge.*1021

*The electric field points the direction of a force on a positive charge.*1029

*We quantify this with the electric field strength vector (E) and that is the amount of electrostatic force observed by a charge per unit of charge.*1034

*So the electric field strength (E) is the electric force divided by the amount of charge; it is the force per unit charge.*1044

*Its units are going to be -- Well, force is Newton's (N), (q) charge is in coulombs so that is Newton's per coulomb (N/C) and we will later find out that that is equivalent to volts per meter.*1052

*Newton's per coulomb or volts per meter is the same thing.*1066

*And again, the direction of the electric field vector is in the direction a positive charge would feel a force.*1070

*So let us take a look at another example -- finding the field between some metal plates.*1078

*Two oppositely charged parallel metal plates at 1 cm apart exert a force of 3.6 × 10 ^{-15}N on an electron placed between the plates.*1083

*Calculate the magnitude of the electric field strength between the plates.*1094

*Well let us write down what we know so far.*1098

*First off, our parallel plates are 1 cm apart -- so the distance between our plates is 0.01 m and there is a force (an electric force) of 3.6 × 10 ^{-15}N on an electron.*1100

*What do we know about an electron?*1115

*Its charge is -1e or -1.6 × 10 ^{-19}C.*1117

*We want the magnitude of the electric field strength, so we are looking for (E).*1124

*Well from our definition, the electric field strength is the electric force divided by the charge, which is 3.6 × 10 ^{-15}N/-1.6 × 10^{-19} C...*1130

*...which gives me an electric field strength of 2.25 × 10 ^{4} N/C and that would be negative, but since it is asking us for the magnitude, our answer is just 2.25 × 10^{4} N/C.*1145

*Let us talk a little bit more about the electric field and how we might be able to visualize it.*1167

*Since you cannot actually see it, you can visualize it by drawing what we call field lines that show the direction of the electric force on a positive test charge.*1172

*The electric field lines always point away from positive charges and toward negative charges.*1180

*Electric field lines never cross each other; they intersect conductors at right angles to the surface of the conductor; and stronger fields have closer or denser lines; and finally field's strength and line density decreases as you move away from the charges.*1186

*So having giving you some of these basic rules for our electric field lines, let us take a look and see what they actually look like.*1203

*Here I have four different examples.*1209

*In the upper left I am showing a positive charge and the electric field lines radiating outward from that charge and because it is a positive charge, electric field lines go away from it.*1212

*You have a radial pattern and here you have a more dense lines in closer to the charge, so you have a stronger electric field.*1222

*The further away you get, you have less dense lines, you have a less strong electric field -- which we know of course because the electric field or electric force follows an inverse square law relationship.*1231

*Around a negative charge, you also have this radial pattern but now the field lines point in.*1245

*Lines point away from positive charges and end at negative charges.*1251

*Now down here I have a couple of dipoles -- two charges near each other.*1256

*If we find the electric field, what we are going to do is vector addition of the individual fields.*1260

*But if you have a positive and a negative beside each other -- well electric field lines go from positive to negative charges.*1266

*So we are going to get a pattern that looks kind of like this, and you could fill in more lines if you wanted to as you interpolate between them, and so on.*1273

*The key is that they go from positive to negative -- they never cross -- and they show the direction of the force on a positive test charge.*1284

*If I were to go take and put a positive charge right there in space, it is pretty easy to see that the net force it is going to feel is probably going to be somewhere in that direction.*1291

*If I were to go take a positive charge and put it right there, that yellow positive charge is going to feel a force to the right -- toward that negative charge.*1302

*The electric field line is telling you the direction of the force on a positive test charge.*1309

*Now over here on the bottom right, we have two positive charges making a dipole.*1316

*While electric field lines go away from positive charges and they do not cross, you end up with a pattern that looks kind of like this.*1321

*What is interesting here is if you were to take a positive test charge and put it exactly in the middle of those two, the forces on it are going to balance out so you get a net force of 0.*1327

*Right in the middle, you actually have a dead spot while you are exactly at that point.*1340

*So that is a little help to get you going on visualizing electric field lines.*1345

*Let us talk about calculating the electric field due to a point charge.*1352

*Electric fields are caused by charges.*1356

*The electric field due to a point charge can be derived from what we know about the electric field and Coulomb's Law.*1358

*We already know that the electric field strength is the force divided by the charge and Coulomb's Law says that the electric force on two charged objects is this constant (k) times the product of the charges divided by the square of the distance between them.*1363

*We can put those together to find a definition, or a formula for the electric field due to a point charge.*1379

*If the electric field strength is force divided by charge and force is (k)q1q2/r ^{2}...*1385

*...and I still have this (q) in the denominator there, well we can cancel out or make a ratio of 1 from one of our sets of charges and we find that the electric field strength is kq/r ^{2}.*1397

*That is the formula for the electric field due to a point charge.*1412

*To find the electric field due to multiple point charges, take the vector sum of the electric fields due to each of the individual point charges-- you just keep adding them up in a vector fashion.*1425

*So let us take a look at comparing electricity to gravity again, there are so many similarities.*1436

*Force is F = (k)q1q2/r ^{2} for electricity and for gravity it is (g)m1m2/r^{2}.*1442

*We have swapped the (k) for (g) and the (q) for (m), charge for mass, but the same basic pattern though.*1449

*The field strength -- electric field strength is the electric force divided by the charge, the gravitational field strength (g) was the gravitational force divided by the mass -- again a nice parallel.*1456

*The field strength due to a point charge was kq/r ^{2} and for gravity that was gm/r^{2} -- the same parallel.*1467

*Now they have different constants and notice (k) here on the left is 8.99 × 10 ^{9}?*1477

*I tend to just make that a nice, simple 9 × 10 ^{9}, so if you see that written differently somewhere, do not worry, it is close enough.*1482

*On the right side the gravitational constant is 6.67 × 10 ^{-11}N-m^{2}/kg^{2}.*1490

*Again, these are really just fudge factors to make the units work out and the charge units were coulombs; the mass units were kilograms -- a lot of parallels.*1497

*Let us take a look at a little bit more in-depth problem solving.*1508

*Find the electric field at the origin due to the 3 charges shown in the diagram.*1513

*We have a +2 C charge here at (0,8), a +1 C charge here at (2,2) and a -2 C charge at (8,0).*1517

*Well what we are going to do -- our strategy is going to be to find the electric field at the origin to each of the individual charges and then add those up in vector fashion.*1527

*So let us start by looking at our green charge up here at (0,8).*1536

*If that is a +2 charge the electric field is going to point away from it, so we are going to have the electric field going in all these directions and at the origin, that must be pointing down.*1541

*So right away we know that we only have to worry about the (y) component.*1551

*Now the electric field strength is kq/r ^{2} where (k) is 9 × 10^{9} and our (q) is 2 C and the distance from the origin to our charge is 8 m.*1556

*So that is going to be 8 ^{2} or 2.81 × 10^{8} N/C down.*1573

*And if I want to write that in bracket notation, my (x) component is 0 and my (y) component is going to be -2.81 × 10 ^{8} N/C because we know that must be pointing down.*1582

*So there is the electric field due to our green charge.*1599

*Let us do the red charge next.*1603

*The electric field due to that red charge is again kq/r ^{2} where (k) is 9 × 10^{9}, (q) is -2C, but again we are not going to worry about signs at the moment -- let us get the magnitude and we will use common sense to figure out its direction...*1605

*...and its distance is 8 units from the origin, so 8 ^{2} is going to give us the exact the same thing -- 2.81 × 10^{8} N/C is the magnitude.*1624

*But if this is a negative charge, electric field lines point in to negative charges.*1634

*So from the origin, we must be going toward the right due to that -2 C charge, so this would be 2.81 × 10 ^{8} N/C toward the right or in the positive (x) direction.*1641

*In bracket notation, I am going to write that as 2.81 × 10 ^{8} N/C and it has no (y) component.*1653

*So we have the electric field due to the green charge, the electric field due to our red charge, so now let us see what it is due to our blue charge.*1663

*If this is +1 C, it should be pretty obvious to see that the electric field at the origin will be pointing away from it -- that direction.*1672

*So we will have components in the (-x) and the (-y) direction here.*1679

*Let us find its magnitude -- E = kq/r ^{2} which is 9 × 10^{9}, our charge is 1 C and now we have to get the distance from the origin to that charge.*1684

*Well to do that, I am going to make a right triangle here and if that is 2 units and that is 2 units, then by the Pythagorean Theorem, this must be the square root of 2 ^{2} + 2^{2}.*1699

*So that is going to be over square root of 2 ^{2} + 2^{2} squared.*1711

*Or 9 × 10 ^{9}/8, which is about 1.13 × 10^{9} N/C, but that is going to be down and to the left at an angle of 45 degrees because it has equal (x) and (y) components, which we can see just from the symmetry of the situation.*1719

*So that means that if I want to break this into (x) and (y) components, this is going to be equal to...*1740

*...Well we are going to have to have its magnitude 1.13 × 10 ^{9} N/C and the (x) component, we will multiply by the cos(45 degrees) and I know that is going to be negative.*1745

*And for the (y), it is also going to be -1.13 × 10 ^{9} × the sin(45 degrees), the (y) component.*1760

*And those are going to be the same -- sin(45) and cos(45) are the same thing.*1770

*When I run that through my calculator, I find that that is going to be about -7.95 × 10 ^{8} for the (x) and -7.95 × 10^{8} for the (y) -- all in Newton's per coulomb.*1774

*Now I want the total charge or the total electric field at the origin due to those three.*1793

*So for the total, all I am going to do is I am going to add up the (x) components and add up the (y) components.*1800

*My (x) components, I have 0 from the green charge, I have +2.81 × 10 ^{8} from the red charge and I have -7.95 × 10^{8} from my blue charge -- so that is the (x) component.*1808

*For the (y) component, I have -2.81 × 10 ^{8} (green) + 0 (red) - 7.95 × 10^{8} (blue), and again all of that is N/C.*1830

*So I will pull out my calculator and add up all my (x)'s and add up all my (y)'s and I end up with a total electric field of about -5.14 × 10 ^{8} for the (x) and -1.08 × 10^{9} for the (y) and all of that is in N/C.*1850

*A little bit more math here to get through it but really the same basic concept -- find the electric field due to each of those points and then sum them up to get the total electric field over here at the origin.*1874

*We have a component to the left, we have a component that is down so we are going to have something where we have a net electric field down in that sort of southwestern quadrant (Quadrant 3 of the graph).*1888

*Let us take a look at another problem.*1903

*Determine the (x) coordinate here on this line where the electric field is 0 using the diagram below.*1906

*We have a +2 C charge here on the right and then +1 C charge here on the left.*1912

*We are going to follow the same strategy again.*1918

*Let us take a look at the electric field due to that blue charge and we will call that E1 and that will be kq1/r ^{2}.*1921

*When we do this, let us try and figure out roughly where our answer is going to be.*1933

*As I look down here, if I have a +1 C charge to the right, it is going to give me an electric field to the right; +2 C over here to the left is going to give me an electric field to the left.*1938

*And because +2 is stronger, I would expect that my 0-point is probably going to be somewhere a little closer to the +1 charge than it is to the +2 charge.*1948

*So I am going to make a guess and say that we are probably going to be somewhere over in that sort of region.*1958

*If we call this distance between our +1 C charge there at -6 and our point (r), then that means that over here, this distance must be 11 - r.*1965

*If I want to draw the electric field due to E2, due to this red charge, we will call that E2 is kq2 over -- and our distance now will be 11 - r ^{2}.*1987

*We want to find out the point where those two -- where the net electric field is 0, so where those are going to be exactly equal in magnitude.*2001

*Our total electric field is going to be -- well we have our blue electric field over here -- (k) × q1 (1 C) divided by r ^{2} and we will have to subtract because this is going to the left -- the electric field due to the red charge.*2011

*That will be -(k) and the charge is 2, divided by 11 - r ^{2}, and we want the point where that all is equal to 0.*2036

*First thing I can do here as I look at my problem is I see a simplification right away -- we can divide (k) out of all of that.*2049

*I could rewrite this then as 1/r ^{2} and I will move the red part to the right-hand side, which is going to be equal to 2/(11 - r)^{2}.*2059

*Let us extend that out, that will be 121 - 11r - 11r = -22r + r ^{2}.*2074

*Now with a little bit of math here, if I multiply both sides by r ^{2}, on the left-hand side I am going to get 1 = 2r^{2}/121 - 22r + r^{2}...*2084

*...or with some cross-multiplication -- 2r ^{2} = 121 - 22r + r^{2}.*2100

*It is starting to look at a quadratic equation, so let us get it in that form.*2110

*This implies then that 2r ^{2} - r^{2} = r^{2} + 22r - 121 = 0.*2115

*Now we can apply our quadratic formula.*2126

*Some of you have calculators that may do that automatically or you can go through the work there and what I find is that I get a value for r of 4.56.*2130

*It means if this is 4.56 then this distance, 11 - r, must be 11 - 4.56 = 6.44.*2143

*So determine the x-coordinate where the electric field is 0 using the diagram below.*2154

*If we want the x-coordinate, well now I just need to go back and finish this problem up.*2160

*To get the x-coordinate, that is going to be -- well if we start here at -6 and we add this distance r to it (4.56), I get an x-coordinate of about -1.44.*2165

*So it looks to me like our actual point for the correct answer should be at about -1.44 right about there and we will make our guesstimate go away.*2179

*We have found the x-coordinate where the electric field is 0.*2191

*Let us take a look at an example where we look at gravity and electricity.*2199

*A distance of 1 m separates the centers of two small charged spheres.*2203

*The spheres exert gravitational force (Fg) and electrostatic force (FE) on each other.*2207

*If the distance between the sphere's center is increased to 3 m -- so we are tripling the distance -- the gravitational force and electrostatic force, respectively, may be represented as...*2213

*This is an inverse square law problem again -- we have tripled the distance and because distance is squared, we are going to have a change by a factor of 9.*2223

*Now is that going to be 9 times greater force or 1/9 the force?*2233

*Well we are going to have a smaller force because they are getting further away.*2237

*Because they are both inverse square laws, our correct answer must be A -- 1/9 the gravitational force and 1/9 the electrical force.*2241

*We will have a last problem here.*2254

*In the diagram below, (P) is a point near a sphere that has a charge of -2 C.*2256

*What is the direction of the electric field at point (P)?*2262

*The way I would start this sort of problem is if we have a negative charge, let us draw some electric field lines.*2266

*Electric field lines point in to negative charges.*2272

*This line is just going to keep going so it looks like over here at point (P), we would have an electric field pointing that direction.*2277

*So the direction of the electric field at point (P) would be to the left as shown right there.*2287

*Hopefully that gets you a good start on electric fields and forces.*2297

*Thank you so much for your time. Look forward to talking to you soon.*2300

*Make it a great day everyone!*2303

1 answer

Last reply by: Professor Dan Fullerton

Mon Feb 29, 2016 10:23 AM

Post by Sarmad Khokhar on February 29, 2016

In 6:02 how did you figured out that there would be positive charge ?

1 answer

Last reply by: Professor Dan Fullerton

Mon Feb 29, 2016 10:13 AM

Post by Sarmad Khokhar on February 29, 2016

Do we need to know questions that involve vector addition for coulomb's law in AP Physics 2 course ?

1 answer

Last reply by: Professor Dan Fullerton

Sun Feb 7, 2016 11:31 AM

Post by Gabrielle Martinez on February 6, 2016

I wonder why I can't find the " Electric Field Due to a Line of Charge" section. I've attempted to search online on various sites for explanations and have not been successful.

1 answer

Last reply by: Professor Dan Fullerton

Mon Nov 23, 2015 7:17 AM

Post by Parth Shorey on November 16, 2015

Is the Q&A active?

1 answer

Last reply by: Professor Dan Fullerton

Sat Aug 15, 2015 2:35 PM

Post by Anh Dang on August 13, 2015

In example 5, so if it was repulsion instead, would the electrostatic force be negative?

1 answer

Last reply by: Professor Dan Fullerton

Thu Aug 13, 2015 5:24 PM

Post by Anh Dang on August 13, 2015

For the electroscope, why is it that when you touch the knob with a negative charge rod, you get a negative charge at the bottom of the leaves and the positive charge at the knob part? Where did the positive charges come from?

And why , if reversed, when you put a positive charged rod to the knob, the whole metal part is positive without a negative charge at the knob?

1 answer

Last reply by: Professor Dan Fullerton

Sun Apr 12, 2015 4:49 PM

Post by Geoffrey Miller on April 12, 2015

Great lecture, Professor Fullerton. I have a question regarding example 5. Why is the answer positive and not negative 6.9*10^6? Is it because the force is a scaler in this instance?

2 answers

Last reply by: Hassan BIn Mazhar

Sat Oct 25, 2014 1:30 AM

Post by Hassan BIn Mazhar on October 23, 2014

Hello sir,I would like to ask u a question:If human body is a conductor and Earth is an infinite source of charge then would we have an electrostatic attraction from Earth as well?

1 answer

Last reply by: Professor Dan Fullerton

Sun May 25, 2014 11:05 AM

Post by Matej Neumann on May 25, 2014

Could the example 8 also be solved by using vectors? with the equation

E=k*q*r/r^3 and then adding all the j and i components together?

0 answers

Post by Madina Abdullah on May 5, 2014

Thank you Sir

1 answer

Last reply by: Professor Dan Fullerton

Sat Sep 7, 2013 5:48 PM

Post by Sam Mukau on September 7, 2013

for example 7, does the distance between the plates not matter? you did not use it in the calculation

1 answer

Last reply by: Professor Dan Fullerton

Sat Aug 31, 2013 12:50 PM

Post by Jude Nawlo on August 31, 2013

Also, in example 9, since both of the points are positive charges, why do they move in a direction towards each other if we know that like charges repel each other?

1 answer

Last reply by: Professor Dan Fullerton

Sat Aug 31, 2013 9:11 AM

Post by Jude Nawlo on August 31, 2013

For Example 3, I understand that the two conductors will distribute their charges equally, which result in both of them having four elementary charges. I understand the mathematical procedure of multiplying by 1.6 x 10^19 to get the charge of each. But I am confused about signs. If we are denoting the elementary charge as "e," then isn't that referring to an electron, and if so, why isn't the final charge of each conductor negative (since electrons are negative?)Or does the elementary charge refer to a positive or negative state or charge?

1 answer

Last reply by: Professor Dan Fullerton

Fri May 3, 2013 6:21 AM

Post by help me on May 2, 2013

Wonderful videos :) Thank you very much for starting your own Educator course. Very helpful, indeed.