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 Fields

- 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

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 Fields
- Property of Space That Allows a Charged Object to Feel a Force
- Detect the Presence of an Electric Field
- Electric Field Strength Vector
- Direction of the Electric Field Vector
- Example 1
- Visualizing the Electric Field
- Electric Field Lines
- E Field Due to a Point Charge
- Derived from the Definition of the Electric Field and Coulomb's Law
- Finding the Electric Field Due to Multiple Point Charges
- Comparing Electricity to Gravity
- Example 2
- Example 3
- Example 4
- Example 5
- Charge Densities
- Example 6
- Example 7
- Example 8
- Example 9
- Example 10
- Example 11

- Intro 0:00
- Objectives 0:09
- Electric Fields 1:33
- Property of Space That Allows a Charged Object to Feel a Force
- Detect the Presence of an Electric Field
- Electric Field Strength Vector
- Direction of the Electric Field Vector
- Example 1 3:00
- Visualizing the Electric Field 4:13
- Electric Field Lines 4:56
- E Field Due to a Point Charge 7:19
- Derived from the Definition of the Electric Field and Coulomb's Law
- Finding the Electric Field Due to Multiple Point Charges
- Comparing Electricity to Gravity 8:51
- Force
- Field Strength
- Constant
- Charge Units vs. Mass Units
- Attracts vs. Repel
- Example 2 10:06
- Example 3 17:25
- Example 4 24:29
- Example 5 25:23
- Charge Densities 26:09
- Linear Charge Density
- Surface Charge Density
- Volume Charge Density
- Example 6 27:26
- Example 7 37:07
- Example 8 50:13
- Example 9 54:01
- Example 10 1:03:10
- Example 11 1:13:58

### 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 Fields

*Hi, everyone, and welcome back to www.educator.com.*0000

*I'm Dan Fullerton and in this lesson we are going to talk about electric fields.*0004

*Our objectives are going to be to calculate the electric field due to 1 or more point charges.*0007

*To analyze electric field diagrams.*0014

*To calculate the electric field by various continuous charge distributions by integration and this is going to get a little bit involved for our second lesson.*0017

*This is one of the trickier parts of the course.*0026

*Take some time here, do not always expect to get that on the first try.*0028

*And then apply Gauss’s law to find the electric field for planar, spherical, and cylindrical symmetric charge distributions.*0032

*We are going to set some of the base line up in this lesson and then on the following lesson on Gauss’s law specifically.*0039

*We are really going to dive to that with a bunch of examples and finish up that once.*0045

*We will not finish this fourth objective here in this lesson itself.*0049

*As we get started, I would like to mention that a lot of this work in AP Physics C is fairly complicated especially the first time you see it.*0054

*The notation is a little tricky, there is a lot to think about, a lot to do.*0061

*You are not going to get all of these the first time through.*0065

*Doing this has to be an active pursuit.*0068

*As we come to the example problems, I recommend hitting the pause button and seeing if you can do the problems.*0070

*If you get stuck, play it a bit more, and if you can get through it, do so and then check your answer with what you see in the video.*0075

*There is a lot more active learning that needs to be done to really ingrain some of these concepts.*0081

*Please take the time to do that as you work through our example problems.*0087

*Moving on, what are electric fields? if the electrostatic force is a non contact or field force, it operates across some distance.*0092

*The property of space that allows a charged object to feel a force is called the electric field.*0101

*It is a made up concept to help us understand how that force works.*0107

*You can detect the presence of an electric field by placing a positive test charge, *0111

*add various points in space and then measuring the resulting force on that test charge.*0116

*The electric field strength factor which we symbolize with a E and you will see it *0121

*in textbooks sometimes with bold or with an arrow over top of it, it is known as a vector.*0126

*It is the amount of electrostatic force observed by a charge per unit of charge.*0131

*The electric field vector is the force it feels ÷ the amount of charge.*0135

*The direction of the electric field vector is a direction of a positive test charge would feel a force.*0141

*As we talk about electric fields, it is important to remember that these have units and that they are going to have some meaning.*0147

*The standard units of the electric field are units of force N/ charge N/ C.*0153

*We are going to find out later that is also equivalent to volt per meter but standard units you are going to see most of the time are N/ C.*0163

*Let us start off with a simple example and then we will work our way into more and more complex examples.*0175

*2 oppositely charged parallel metal plates 1 cm apart exert a force of 3.6 × 10 ⁻15 N on electron placed between plates.*0182

*Calculate the magnitude of the electric field strength, we want to know how strong the electric field is between the plates.*0192

*We can do this in a pretty straightforward manner saying that the electric field is = electrostatic force ÷ the amount of charge.*0199

*That is going to be 3.6 × 10 ⁻15 N ÷ our charge, it is the charge that we have on an electron*0210

*because it is an electron between those plates, the magnitude of the charge on electron is an elementary charge or 1.6 × 10⁻¹⁹ C.*0221

*When I go through and do this with my calculator, I come up with about 2.25 × 10⁴ N/ C.*0232

*Hopefully, a pretty straightforward example to get us started.*0247

*Let us talk about how we could visualize this electric field.*0250

*Since, you can actually see the electric field you can visualize it by drawing *0254

*what we call field lines to show the direction of the electric force on the positive test charts that we would put somewhere in space.*0257

*Electric field lines always point away from positive charges and toward negative charges, *0264

*away from the positives and toward the negatives.*0270

*Electric field lines never cross each other and they always intersect conductors at right angles to the surface.*0274

*Stronger fields have closer and denser lines, weaker field the lines are further apart.*0281

*The field strength and line density decreases as you move away from the charges.*0286

*To do just a simple one to begin with.*0291

*If we were to take a look at a positive point charge, we have electric field lines moving away from it, moving radially away.*0293

*Electric field is the same at all points that are in equal distance from the center of that point charge so you have the same density of lines.*0303

*You have the symmetry.*0309

*You can almost think of a positive charge when we draw electric field lines as kind of a magic blow dryer.*0313

*It is blowing air out from that point in all directions and what you are drawing is the vectors of the air coming out of it.*0319

*On the other hand, a negative charge works the same way but all the lines are coming into it.*0325

*A positive charge would be sucked in this direction.*0331

*You can almost think of these as the magic vacuum cleaner in space, it is sucking all the air in and you can draw just the wind vectors there.*0334

*Away from positive charges in the negative charges.*0343

*Also note that, as you get further away the distance between these increases quickly showing that the electric field strength drops off quickly.*0348

*We have another example of an inverse square law here, just like we had for the electrostatic force.*0356

*If we put these charges together to make what is called the dipole to charges, positives go toward negatives, *0366

*the electric field lines begins at the positive and ends at the negative.*0372

*They are symmetric and if we were to have a particle hearsay, positive test charge somewhere sitting right there for example, *0376

*it is going to feel a force in the direction of the field lines which we do not show all of them there but it would feel a force in roughly that direction.*0384

*Or if we put a positive charge over here, it would feel a force in that direction.*0391

*What if you have a negative charge?*0397

*Let us take and do an example with a negative charge.*0400

*If we put a negative charge, let us say we put it right over here that purple spot, *0403

*it is going to go in the opposite direction of the field line that would feel a force that way which only makes sense,*0408

*it would be attracted by that positive test charge and move to the right.*0413

*If we happen to have 2 positive charges or 2 negative charges, you get this other pattern as well, *0417

*where there is a point in the middle of those, where there is no force or everything cancels out.*0423

*Some examples of how you can draw electric field lines.*0429

*We did a couple point charge diagrams, let us talk about the electric field due to point charges in a little bit more detail.*0434

*Electric fields are caused by electrical charges, of course.*0441

*The electric field due to a point charge can be derived from the definition of the electric field and we just learned about in our previous lesson, Coulomb’s law.*0445

*If the electric field is the force ÷ the charge and the magnitude of the force is given by Coulomb’s law here.*0453

*Or if you want it to, we can go to our other notation and say that the force is 1/4 π E₀, remember that is equivalent to the K × Q1 Q2/ R².*0462

*We can start to put these together and say for example the electric field here is the force K Q1 Q2/ R² ÷ Q.*0476

*We end up with one of the Q cancels out makes a ratio of 1, we end up with KQ/ R².*0489

*Or if you are looking at it in this notation here, the electric field would be 1/ 4π E₀ × Q/ R².*0498

*We have our formulas for the electric field due to a point charge.*0510

*To find the electric field due to a multiple point charges, what we do is we take the vector sum of *0517

*all of the electric fields due to the individual point charges.*0522

*We use the law of super position to just add all of those up.*0526

*You will notice a bunch of comparisons electricity to gravity.*0531

*The formula electric forces K Q1 Q2/ R² compared to gravity G M1 M2/ R².*0535

*The only major difference mass vs. charge and the constant, the fudge factor.*0542

*The electric field strength is the force ÷ the charge.*0548

*With gravity, the gravitational field strength was the gravitational force ÷ the mass.*0552

*We can also write this, our formula is KQ/ R² for electricity, for gravity GM/ R², see they are parallels here.*0559

*The constants are different but that is just because we are trying to make the math workouts so our units come out to something that makes sense.*0568

*The charge units are Coulomb’s, the mass units are kilograms.*0575

*The math has tons of similarities.*0578

*One that we have not pointed out yet though that is a very important, gravity only attracts.*0581

*The electrostatics can attract and repel.*0589

*This is very important.*0598

*Let us do a problem where we have got the electric field from a couple of point chargers.*0600

*Find the electric field at the origin due to the three charges shown in the diagram.*0607

*We got a charge here at 0, 8 m which is 2 C, a huge amount of charge.*0611

*Over here at 2, 2 we have got a + 1 C charge and here at 8, 0, 8 m on the X axis we have -2 C charge.*0618

*How can we do that?*0628

*What I'm going to do is I'm going to find the electric field of the origin due to each of these 3 charges individually and then take their vector sum.*0630

*First thing is let us find the electric field at the origin due to do this green charge up here.*0639

*Since, this is a positive charge we should be able to see that down here at the origin, *0645

*the field lines go away from positive charges our electric field should be in that direction.*0649

*The electric field due to that + 2 C point charge at the origin is going to be KQ/ R² where K is 9 × 10⁹.*0657

*Our charge is 2 C ÷ our distance from the origin 8 m that we have to square or I get about 2.81 × 10⁸ N/ C down which is in vector form 0, -2.81 × 10⁸ N/ C.*0670

*There is the electric field at the origin due to this green charge up here.*0700

*Let us do it now for the red charge.*0704

*We can see if this is a -2 C charge, the electric field is going to go toward the -2 C charge so they should feel that the origin is going to go toward the right.*0709

*We already know the direction.*0719

*The magnitude then is given by KQ/ R² which is 9 × 10⁹ Nm²/ C² × 2 C charge.*0722

*I’m not going to worry about the magnitude since I already know the direction, ÷ the square of the distance 8 m between those.*0733

*Or again, we end up with 2.81 × 10⁸ N/ C to the right or in vector notation that is going to be 2.81 × 10⁸ N/ C in the x and nothing in the y, 0.*0742

*We have got to figure out our portion due to this blue charge is 1 C charge at 2, 2.*0762

*By inspection, we should be able to see that the field lines are going to get away from the positive charge so *0770

*the origin that is going to be going in that direction.*0776

*To find its magnitude however, our same formula E = KQ / R² where K is 9 × 10⁹.*0780

*Our charge now is 1 C and the square of this distance.*0791

*To do that, we got to do a little bit of math.*0795

*I'm going to make a right triangle realizing that this is 2, this is 2, so I can use that the Pythagorean Theorem to find out that distance that is 2 and 2.*0799

*This is going to be the √2² + 2² so R² is √2² + 2² is just √2² + 2²² or 8.*0809

*That is 9 × 10⁹/ 8 which will be 1.13 × 10⁹ N/ C down into the left in a 45° angle.*0825

*What we have been doing this in vector notation so what we do with that?*0838

*We do not have a straight x and y component right from our answer yet.*0843

*We got to do a little bit of math here and realize that that is going to be equal to.*0847

*The x component is going to the left so that is going to be -1.13 × 10⁹ N/ C in the x component cos 45°.*0852

*Our y component is going to be going down as well so that will be negative.*0866

*I end up with -1.13 × 10⁹ N/ C sin 45° or when I do the math there that is = -7.95 × 10⁸ N/ C in the x and -7.95 × 10⁸ N/ C in the y.*0871

*Just taking a minute to step back.*0899

*We now have the magnitude and the broken down vector notation of the electric fields due to each of these 3 charges.*0900

*To find the total electric field, we need to add these keeping in mind of their vectors.*0908

*We cannot just add their straight magnitudes, we have to add vector components.*0912

*To find the total electric field, we are going to add up all of the x components and all of the y components.*0917

*Our x components we have a 0 from the green one.*0926

*We have 2.81 × 10⁸ from our red charge and we have a -7.95 × 10⁸ N/ C from our blue charge.*0930

*We have these three to add up for our x and for the y component we are going to do the same basic thing.*0949

*We have -2.81 × 10⁸ N/ C due to our green charge.*0958

*We have no contribution in the y direction due to our red charge and we have -7.95 × 10⁸ contribution from our blue charge.*0965

*We add up the x, we add up the y to find out that our total electric field is going to be, *0981

*when we add all those up we are going to get about -5.14 × 10⁸ N/ C in the x direction and about -1.08 × 10⁹ N/ C in the y.*0990

*Our total electric field is going down into the left, you can figure out the angle if we wanted to.*1013

*But this gives us the vector sum of these 3 electric field contributions to give us our total.*1019

*Let us make sure we highlight that after all that work.*1027

*There is our total electric field at 0, 0 due to those 3 point charges.*1031

*It is kind of complicated but if you take it step by step, it is really not bad.*1039

*Let us take a look at an example where we are looking to find out where the electric field is 0.*1046

*Here on this line, we have 2 point charges, we have a + 1 C charge at x = -6 m and the + 2 C charge at x = 5 m.*1051

*The electric fields from both of these go away from positive charges so we got an electric field to the right here and to the left on that side of them.*1063

*Due to the red charge, we have an electric field to the right, electric field to the left.*1072

*There should be some point on this number line where the red and the blue exactly cancel out and there is no electric field.*1078

*By inspection, as I look at this, we have a smaller charge here so I would expect that *1085

*we would have the point at 0, maybe a little bit to the left of the middle of those 2 charges.*1091

*For 5 and 6, about -.5 would probably be halfway between them so I'm expecting we would probably get an answer that is somewhere over in this region.*1098

*Before we even go into the math, let us qualitatively take a look, make something of a guess and *1108

*then we will go see if everything matches up when we go through our math here.*1113

*What we are going to do to find this out is, let us say that this is our point and they do not know exactly where it is *1118

*but we will define the distance between our 1 C charge and our point where the electric field is 0, let us call that distance R.*1123

*If we call that R, then the distance from here to our red charge must be 11 -R because the total distance between them is 11 so R + 11 – R is 11 m.*1133

*We can go in and we can start doing our math to find out exactly where that point resides where the electric field is 0.*1148

*Looking at the blue charge first, the electric field due to that 1 C charge is K Q1/ R².*1155

*If we look at our red charge over here, the electric field due to that is going to be K Q2/ our distance is going to be 11 - R².*1165

*The total electric field we know has to be 0 at this point.*1178

*The total electric field which is going to be our electric field due to the blue charge *1183

*which is K × 1 C ÷ R² - this one was going to the left our red electric field contribution *1188

*K × 2 ÷ 11 - R² and we know all of that has to = 0.*1203

*We have got an algebra problem.*1217

*We have got K on the left, K on the right, 0 over here, so what we can do is we can factor out the K's *1220

*which implies 1/ R² from the left, after we cancel out our K’s must be equal to,*1226

*I’m going to move this portion to the right hand side.*1240

*We have 2 ÷ this 11- R it is kind of awkward to use that way.*1242

*I’m going to multiply that out 11 - R² is 11 -R × 11 - R or 121 11² -11 R -11 R is -22 R + R².*1250

*I’m trying to solve this for R so what I might do here is maybe I will cross multiply this and just make it look a little simpler,*1267

*to say then that we will have 2 R² = we still have our R² -22 R + 121 *1275

*which implies then if I subtract R² from both sides, I get R² = and then we have R² =.*1293

*Let us do it a little bit differently.*1310

*I think we can save a step here.*1311

*We are going to put this into a quadratic equation so I could bring all of this over to the left, subtract all of this and say that R² + 22 R -121 = 0.*1313

*And that is a little bit slicker.*1325

*I'm taking 2 R² - R² to get R² -22 R or adding 22 R to both sides and then subtracting the 121 to rearrange.*1326

*We got a system that we can put in a quadratic formula because I do not see a very easy way to factor that in any other direction.*1338

*If we use the quadratic formula, we can then say that our R must be equal to, quadratic formula -B ± √B² - 4 AC/ 2 A.*1345

*In this case, -B is going to be our -22 ± √22² -4 A × C 1 × 121 is just going to be that -121/ 2 A and A is 1 so that is ÷ 2.*1364

*A little bit of calculator work and you should come up with 2 possible answers, either R = 4.56 or R = -26.6.*1390

*In here, we got to use a little bit of common sense and say which one of these make sense?*1402

*If R is 4.56 that means this distance over here, pretty close to our guess a bit, that make sense.*1406

*If R was -26.6 we are way to the left of the screen and that is not going to make any sense.*1413

*We have got a left vector from this electric field, left there and they are going to add together.*1419

*You are not going to get 0.*1423

*We can cross that one out as not making sense for our problem.*1425

*Therefore, we know that R must equal 4.56.*1428

*Where on this diagram is that 0 point?*1434

*If we start at -6 and we go 4.56 to the right that tells us that our x position*1438

*must be -6 + 4.56 or about -1.44 m on our number line.*1445

*The exact answer should be right around there.*1454

*Finding the electric field due to a couple of point charges we can use the law of super position.*1461

*Let us take a look at a little bit more straight forward example for a moment, stepping back.*1468

*A distance of 1 m separates the centers of 2 small charge sphere, they exert a gravitational force FG and electrostatic force FE on each other.*1473

*If the distance between the center of the spheres is increased to 3 m, we are tripling that distance, *1485

*what happens to the gravitational force and electrostatic force?*1491

*If we triple the distance, they get further apart, of course the force has to get smaller so we can get rid of those choices.*1495

*Again, this is an inverse square law relationship for both the gravity and the electrostatic force.*1503

*Inverse square law, triple the distance 1/9 of the force.*1512

*The correct answer here must be A.*1517

*The direction of the electric field due to a point charge.*1523

*In the diagram here below, P is a point that is near a sphere with charge -2 C.*1527

*What is the direction of the electric field at point P?*1532

*If you remember our rules for electric field lines, electric field lines go in to negative charges.*1535

*Over here at P, we must have a force to the left.*1545

*The electric field there must be pointing to the left, assuming that would be a positive test charge.*1550

*To the left would be the correct answer.*1557

*Moving on, let us see if we can get into a little bit deeper derivations.*1565

*To do that, we are going to have to talk for a moment about charge density.*1570

*Let us assume we have something like a line that is uniformly charged.*1574

*The total charge ÷ the length of that line would give you the linear charge density which *1578

*we are going to give the abbreviation the symbol λ Λ, it is amount of charge per unit length.*1583

*In other cases, we may be looking at a distribution of charge that is on some sort of surface.*1590

*A linear charge density does not make sense but instead we are going to look at total charge per unit area *1595

*that we are going to give the symbol sigma Σ, a surface charge density.*1602

*We may also be looking at objects that are three dimensional that have some volume.*1608

*When we have charged uniformly distributed through a volume, it can have some volume charge density taking the total charge ÷ the unit volume.*1616

*And that we are going to give the symbol ρ Ρ.*1625

*We are going to use these definitions fairly regularly throughout the rest of the course.*1627

*A linear charge density, charge ÷ length.*1631

*An area charge density, charge ÷ area.*1635

*The volume charge density, charge ÷ volume.*1638

*Let us put this in to play as we try and find the electric fields due to a semicircle of charge.*1644

*We have a thin insulating semicircle of charge total charge Q at some radius R centered around at point C.*1652

*We want to find the electric field at point C due to the semicircle of charge.*1660

*Before I even begin the math, the first thing I’m going to do is observe for a minute.*1665

*If this is symmetrical around C and it is uniformly charged, there should not be any horizontal component of electric field.*1669

*That should all cancel out.*1677

*We should only have a vertical component that we need to deal with.*1679

*By symmetry, we can simplify the problem and work in just the horizontal direction.*1683

*As we do this, the first thing I’m going to do is I'm going to define some linear charge density where our linear charge density, *1688

*the charge per unit length on this, is the amount of charge on that length ÷ its length,*1695

*which in this case is going to be its total charge Q ÷ the length of that semicircle.*1703

*If the distance around an entire circle is its circumference 2 π R, half of that for a semicircle must be π R.*1710

*Our linear charge, hence the Λ must be charged ÷ π R.*1719

*As we do this, we are also going to break this up and our strategy is going to be to take and look at little tiny pieces of the semicircle.*1725

*And as we go through it, we are going to find the electric field due to each of those pieces and then add them all up.*1733

*I'm going to define this little piece and draw a couple of lines here just to make this a little bit clear.*1740

*It is not quite as clear as we wanted there.*1751

*It looks something like that.*1755

*What we are going to do is we are going to call this R Θ so that this angle right there is a little differential of Θ, a little tiny piece of Θ.*1757

*The amount of charge that we are enclosing up here in this little piece, that length is going to be RD Θ.*1769

*The amount of charge that is enclosed there which we are going to call DQ, a little tiny bit of the charge *1778

*is going to be that length RD Θ × the linear charge density in that area which we just defined as Λ.*1783

*The differential of charge, the tiniest little bit of charge and that little bit is RD Θ × Λ.*1794

*To find then the electric field due to that just that little bit of charge, we will find the differential. *1802

*The little tiny piece of electric field which we are only looking in the y direction is going to be, we could use KQ/ R².*1808

*I’m going to use 1/ 4π E₀ version, 1/ 4π Ε 0 is the same as K × our charge DQ ÷ the distance between them R² *1817

*and we are looking at just the y components, so sin Θ.*1832

*Once we have that we can start to get set up and do all of or math.*1838

*The hardest part of these problems is just getting into the habit learning how to set these up.*1841

*There is no better way than to just do a problem after problem after problem and check as you go on.*1846

*Let us take this a little bit further.*1854

*We have our RD Θ here, our differential of the angle if that is Θ and we just said that the differential of the electric field *1858

*in the y direction due to this little bit of charge was 1/ 4π Ε 0 DQ/ R² sin Θ which implies then since we know that DQ = Λ RD Θ. *1866

*We just said that on our previous screen that the differential EY = 1/ 4π E₀ and replace DQ with Λ RD Θ.*1881

*Λ RD Θ we will make that stand out ÷ R² × sin Θ.*1903

*To add up all these little bits of the electric field what we are going to do is an operation called integration.*1918

*It is adding all of these tiny little pieces up to get a whole.*1925

*We will integrate on both sides, do the same thing to both sides you get the same thing.*1929

*We are going to integrate, to get the entire electric field we need to integrate from Θ = 0 all the way across until we get to Θ = π.*1937

*Because once around an entire circle is 2 π, halfway around the semicircle would be π.*1946

*We are going to integrate from Θ = 0 to π.*1952

*What we can do, what is nice over here is you can pull constants out of the integration.*1959

*1/ 4π E₀ that is all a constant anywhere in our problem, that is not going to change.*1964

*Let us rewrite this again and say that this then is, here I see a simplification.*1972

*R and R² we can simplify there.*1979

*We can write this as integral of DEY = the integral from Θ= 0 to π.*1983

*We have 1/ 4π E₀.*1991

*We got R in the denominator, we have got Λ sin Θ D Θ.*1995

*We write it that way before we really get into the math, it is like simplifying as much as possible.*2005

*What we can do is we can pull all of the constants out of the integral.*2011

*The left hand side is the integral of all these little pieces of tiny pieces of the y give us the total EY.*2015

*The electric field in the y direction or the y component of that is going to be equal to, pulling other constants,*2023

*Λ is a constant 4π E₀ and R are all constants for the purposes of this problem.*2029

*That will be Λ / 4π E₀ R integral from Θ = 0 to π of sin Θ D Θ = Λ/ 4π E₀ R, the integral of the sin is the opposite of the cos.*2034

*We have the opposite of that cos of Θ evaluated from 0 to π which implies then the y component of the electric field *2061

*that is going to be Λ/ 4π E₀ R and then we have – cos π - cos of 0.*2071

*Cos of π is -1 so --1 is going to give us a 1 here, 1 – cos 0 is 1.*2089

*Two negatives we will get 1 there.*2098

*What I end up with is the y component of my electric field is going to be Λ/ 4π E₀ R.*2100

*All of this is just 2.*2112

*Therefore, the y component of our electric field becomes 2 Λ / 4π E₀ R.*2119

*2/4 I can reduce to 1/2 or Λ/ 2 π E₀ R.*2128

*There we have our y component of our electric field but we can even take that a bit further.*2138

*That is a good answer but we know the total charges so we can substitute that and get rid of that Λ knowing that *2144

*we already defined previously Λ = Q/ π R to say then that our y component of the electric field is Q / π R × whatever we had before 2 π E₀ R, *2151

*which implies then that the electric field, the y component is going to be, we got a π² and R², so that is going to be our total charge Q ÷ 2 E₀.*2177

*We have got π² and then R².*2193

*Quite a bit of math involved but our strategy is pretty straightforward.*2203

*Find and break up our little bit of charge.*2208

*Find the electric field due to that little bit of charge and then we just add that up for all of those little bits of charge to make the whole.*2212

*That is what we are really did with all of this math.*2219

*Probably worth doing another couple of these.*2225

*Thin, straight, uniform line of charge, find the electric field at distance D from a long straight and slating rod of length L at a point P *2229

*which is perpendicular to the wire and equidistance from the end of the wire, assuming the wire is uniformly charged.*2239

*We have got this point P lined up right with the center of this length L of an insulating rod that is uniformly charged.*2245

*Our strategy again is going to be very similar.*2253

*Let us even write that down to make sure we have got it nice and clear.*2257

*First thing we are going to do is divide the total charge Q into smaller charges Δ Q.*2261

*We are going to find the electric field due to each Δ Q.*2275

*Once we have done that, we are going to add up the electric field due to each Δ Q to get our total electric field.*2288

*And finally, as I look at this problem it looks like we can make another symmetry argument and should be pretty obvious that *2307

*if this is centered on our wire then we are not going to have to worry about any vertical components of the electric field.*2313

*It is all going to be horizontal.*2320

*We are going to use symmetry to show that the y component of the electric field is 0 or therefore we are only going to worry about EX, *2321

*the X component due to symmetry.*2332

*It is always nice to simplify these wherever you can because they are complicated on their own.*2337

*We have some uniform charge on this insulating rod, if they are uniformly charged we can talk again about a linear charge density.*2342

*Λ was going to be our total charge Q divided by our total length of the wire L.*2352

*Before we did the math, we are going to set this up as carefully as possible.*2357

*If this total length is L, I'm going to call this Y = - L / 2 and we will call this end L/ 2.*2362

*I have got our distance D toward point P and let us take just a little bit of charge, a little bit of length of that wire and *2372

*we will find out what that is going to look like as far as the electric field contribution.*2379

*I'm going to draw a line.*2385

*There is our electric field due to that little bit I, we are only going to be worrying about the x component of that.*2392

*If we do this, we will define this angle as Θ to that little piece I.*2399

*We will have some y component, some y value, y distance from our center line which will call yi to that little piece of line i with charge Qi or DQ.*2404

*As we do this, the length of this is probably going to be important.*2416

*The distance from our charge due to a point where we want to know the electric field.*2419

*We will call that Ri and by the Pythagorean theorem I can see that that is going to be equal to the √D² + yi².*2423

*All right, that looks pretty good.*2438

*Finally, let us write our equation for the x component of the electric field due to this little bit I before I get heavy into the math.*2440

*The electric field due to this little portion I in the x direction is going to be 1/ 4π E₀, our K value again × the charge DQ ÷ RI² *2449

*the distance from our charge to our point × the cos of Θ I because we are after the x component.*2466

*I think we have got everything we need for our setup.*2474

*It is time to start getting into the math and plugging through this thing.*2478

*Let us dive in.*2484

*The electric field in the x direction = 1/ 4π E₀ × Δ Q in that little piece ÷ RI² × cos of Θ I which implies then *2487

*we know that RI = √y I² + D² and the cos of Θ I, as we look here cos SOHCAHTOA.*2509

*Adjacent / hypotenuse is going to be D/ RI.*2522

*cos Θ = D/ RI and we can write this as EI in the x direction is 1/ 4π E₀ Δ Q/ y I² + D² is RI².*2527

*We can get rid of the square root sign × cos Θ I which is D/ RI which is y I² + D² √ ½.*2546

*Let us get rid of that Δ Q now.*2560

*Our Δ Q we know is Q/ L × our little bit of differential of y up and down that length.*2565

*Then the x component due to that little piece I is 1/ 4π E₀.*2575

*We have our Q, we have our D, we have our Dy, be careful because D and Dy are different.*2585

*D does not mean D multiplied Dy.*2595

*Dy here is differential of y, little tiny piece of Dy.*2597

*Q D Dy/ L, capital L is still down there.*2601

*We still have our YI² + D² but because we have y² + d² × y² + d² ^½.*2606

*I can write that as 2³/2, which implies then that the EI in the x direction.*2616

*Let us do the whole thing, let us integrate both sides.*2628

*Let us say that just by looking at that little piece, the x component is going to be the integral from y = - L / 2 to L / 2.*2632

*We are going to add up from here all the way to the top, all of those little pieces of 1/ 4π E₀ × we have got Q × D/ L.*2641

*We also have multiplied by our DY/ y I² + D²³/2,*2657

*which implies then that the x component of our electric field is, we can pull other constants.*2671

*Q and D are not going to change.*2680

*Our L is not going to change and 4π E₀ is not going to change.*2682

*Those are all constants for the purposes of our problems so they can come out of the integral sign.*2686

*We have QD / 4π E₀ L × the integral from - L/ 2 to L/ 2 of DY/ yI² + D²³/2.*2691

*This is a tricky integral, not one that I expect you to be able to do with that.*2718

*When we get to an integral like this where I like to go is the back of a calculus book, the front.*2722

*It is a great place to go and look up some of those integration formulas.*2727

*If you can find a form that this fits where it solved for you, let somebody else do the work.*2732

*Use the formula that gives you the integration and that is exactly what I'm going to do here.*2736

*I have looked this up already.*2741

*This implies then using the formula that I looked up, the integral of some Dx/ A² + x²³/2 = x ÷ a² × √a² + x² + constant of integration.*2744

*This fits that same format.*2771

*Instead of Dx, we have got Dy, our x² we got yi², a² we got D².*2773

*We can use this formula to do our integration.*2780

*Taking the next step using this formula, then our electric field x component is Q D/ 4π E₀ L.*2785

*I'm very careful to make sure that my solution is using the same piece as the same correlation with this formula.*2798

*I use green to help highlight what we have done here.*2807

*Instead of x we have y/, instead of a² we have our constant D², then we have got √a² + x² which for us will be √y² + D² ^½ .*2810

*We are evaluating this whole integration as for - L/ 2 to L/ 2.*2830

*Evaluated from - L/ 2 to L/ 2 which is equal to,*2836

*Our left hand side over here still stays the same Q × D/ 4π E₀ L.*2844

*We have got D² here and the D up here, we can do some simplification and leave us with just a D in the denominator.*2855

*We will put a D down here and that one can go way.*2865

*We also now have to deal with all of this mess.*2872

*Let us just go right in, I will do this over here to give us some room.*2878

*L/ 2/ L / 2² + D² ^½ - -L/ 2 ÷ -L / 2² + D² ^½.*2883

*A little bit more simplification, that is = Q/ 4π E₀ DL this piece, × L/ 2 - -L/ 2 that is going to be the same denominator when you square that.*2909

*It is just going to give us one total L so that is × L ÷ L/ 2² + D² ^½.*2925

*Putting all of this together, we got an L down here and L up here.*2939

*We can simplify that, I end up with the x component of the electric field at point P due to this line of charge as Q/ 4π E₀ D × L/ 2² + D²¹/2.*2944

*We will make it nice and pretty and put in the square root sign there.*2969

*There is our answer.*2973

*Almost feel I need a nap after that.*2978

*There is a lot involved, step by step.*2981

*Each little step is not that difficult but there is a lot of bookkeeping, a lot of keeping track of the details.*2984

*When you see these problems, take your time, do not let them intimidate you but it is all about practice.*2990

*Just doing it again and again and again, nobody gets it on the first time.*2995

*It is scary the first couple of times you look at it and it is supposed to be, stick with it.*3000

*Let us take the solution and do a little bit more with a lower here though.*3008

*Let us see what would happen if this line was infinitely long?*3012

*In that case, the electric field in the x direction we will find that out by taking the limit as our length of that wire *3018

*approaches infinity of our answer from before, Q/ 4π E₀ D × L/ 2² + √D².*3028

*That can give us the answer.*3048

*You got to be careful as we did the limit because as L gets very big you would think that this is just going to dominate.*3051

*You also have to remember that Q increases and goes toward infinity as L does.*3057

*We cannot do it quite that simply.*3061

*Instead, we can look at this and say you know as L approaches infinity, L/ 2² is going to get so much bigger than D² that this D² really is not going to matter.*3063

*The √L/ 2² is going to simplify to L/ 2 in the limit.*3073

*We would get Q/ 4π E₀ D L/ 2 but we also know if you recall Λ is Q/ L or Q = Λ L.*3079

*As L gets very big having a set charge is going from infinity does not make sense.*3098

*We got to put it in terms of a linear charge density.*3103

*We would write this as Λ L/, we have here L / 2 × 4 is just going to give us L/ 2 L.*3106

*We will have 2 π E₀ DL and our L make a ratio of 1.*3120

*We can state then that the electric field in the x direction due to that infinitely long line of charge would be Λ/ 2 π E₀ D.*3129

*That looks a whole lot nicer.*3142

*We are going to come back to a problem like this and solve it in a different way, *3145

*a much smoother way in our next lesson when we talk about Gauss’s law.*3149

*How about the electric field if the distance D is infinite?*3154

*Instead of an infinite line, we are an infinite distance away.*3157

*We can follow the same basic strategy, the horizontal component of an electric field *3161

*is going to be the limit, as D approaches infinity of Q/ 4π E₀ D × √L/ 2² + D².*3168

*As D approaches infinity now, this L/ 2 piece is not going to matter.*3189

*The √D² is going to be D so we are going to get Q/ 4π E₀.*3195

*We had 1 D, we got a D from this piece, D².*3204

*That look pretty familiar, that looks like it is a point charge.*3209

*Instead of D, think of R, how far you are.*3213

*As you get infinitely far away from this wire, it is going to start looking like a point charge.*3216

*Imagine you are zillions of miles away, you cannot tell it is a line.*3221

*Electrically you cannot tell either, it starts to act like a point charge.*3224

*That answer should make sense as well.*3227

*You are very long distance away that piece of wire starts to act like a point charge.*3229

*Let us do another derivation here.*3239

*I think we need all the practice we can get.*3241

*We are going to try and find the electric field at a point on the axis here P that is perpendicular to a ring, *3243

*a thin insulating ring that is uniformly charge at some radius R.*3249

*We are going to use the same strategies again, all of the real physics work is in the setup.*3255

*By symmetry as I look at this, I can see that the only electric field is going to be in z direction.*3261

*Everything else along the x and the y cancel out so that will help a little bit.*3267

*We can also define for this problem a linear charge density Λ which is Q/ L.*3272

*It is going to be Q/ the length of our ring is a circumference 2 π R.*3280

*Or Δ Q as we look at some portion of the ring is going to = our linear charge density × our radius × the arc length that we go through here.*3289

*I'm going to call this as we go up to here, let us call this direction φ Φ.*3306

*Times some little bit of Φ, differential of Φ, D Φ.*3314

*Once we have done that, we can start looking at some of the geometry of the problem here.*3320

*We need to know the electric field at point P due to this little bit of charge.*3325

*I will draw my vector distance here being nice happy green color.*3330

*This will be our electric field due to that little bit of charge.*3342

*This distance we will call Ri which in this case we can use the Pythagorean Theorem, it is going to be we have got 1 piece of the triangles Z, the other piece R.*3353

*This will be the √z² here is the component + R² the constant, and the radius of our hoop here, our ring of charge.*3368

*We will call our angle here Θ i.*3379

*What else do we need to define?*3385

*Usually we can get that little piece of electric field defined before we really dive in.*3387

*Let us say that the electric field due to that little piece we are looking at just as z component 1/ 4π E₀.*3394

*We have got our charge Δ Q ÷ our distance Ri².*3403

*It looks like we are dealing with the cos of Θ I, the horizontal component as we have set this up currently.*3409

*We can do a little bit more work here.*3420

*We know that RI is z² + R² we already defined that.*3421

*RI = √z² + R² and cos of Θ i.*3426

*We also look at the cos that is the adjacent / the hypotenuse that is going to be, our adjacent side is our z component and our hypotenuse is Ri again.*3434

*Rewriting this, Ei in the z direction is 1/ 4π E₀ × our charge Δ Q and pulls up there ÷ Ri² *3448

*is going to be z² + R² × cos Θ i which is going to be z / √z² + R².*3463

*If you want to simplify that before moving on which is probably a good idea, we have 1/ 4π E₀.*3481

*We will throw in our z, we have got z/ z² + R²³/2 × our Δ Q.*3491

*Before you get too nervous, this one is actually a little simpler than the last one.*3510

*Let us see if we can dive into the math now and solve this one.*3514

*Electric field due to a little bit i in the z direction is going to be 1/ 4π E₀ z/ z² + R²³/2 Δ Q.*3522

*But we know what Δ Q is.*3541

*Δ Q is just going to be our linear charge × Λ × the radius × D Φ.*3546

*Then we get that E in the z direction is going to be the integral from Φ = 0 all the way around the circle this time so from Φ= 0 to 2 π of 1/ 4π E₀.*3557

*We have our z/ z² + R²³/2 Λ R D Φ.*3576

*Next up, electric field in the z direction = let us pull all of our constants out of here.*3590

*4π E₀ a constant in our problems z constant, z² must be a constant, R a constant, Λ is a constant, R again is a constant.*3597

*We can pull most of these out of the integral sign so we have Λ R/ 4π E₀ z / z² + R²³/2 integral from 0 to 2π of D Φ.*3609

*That looks so much friendlier.*3632

*The reason we can do that is none of these depend on our variable of integration here Φ.*3634

*They are all constants with respect to any changes in Φ.*3639

*The integral of DΦ is just going to be Φ.*3644

*Therefore, we can say that Ez = the integral of DΦ is Φ from 0 to 2 π that is just going to be 2 π.*3647

*We have 2 π from this × Λ R ÷ 4π E₀ × z ÷ z² + R²³/2.*3656

*This is much better.*3682

*Our Λ we know what that is equal to, so this implies then knowing that Λ we already defined as Q/2 π R.*3684

*We can say that the z component of our electric field is = we have 2 π Λ was Q/ 2 π R.*3695

*We still have R up here and the 4π E₀ down here, a z/ z² + R²³/2.*3714

*It looks like we can make a couple of simplifications here.*3729

*2 π/ 2 π does make a ratio of 1.*3732

*We have got R and R, those make a ratio of 1 and that looks pretty good.*3735

*We can rewrite this then the electric field in the z direction is going to be 1/ 4π E₀.*3742

*We got a Q and a z ÷ we are left with z² + R²³/2.*3753

*Hopefully, you are starting to see a pattern in these.*3772

*The same sort of strategy applied again and again and again.*3775

*If we take a single ring, we just solved for the electric field there but what happens if we start solving for electric field due to a bunch of rings?*3780

*That is going to be our next problem.*3789

*A uniformly charged disk, define the electric field due to a uniformly charged *3792

*and slating disk of some radius R to point B perpendicular to the disk as shown on the diagram.*3797

*To do this when we are going to rely heavily on what we just did.*3804

*Realizing we already talked about how you find a ring of charge, our strategy is going to be to take our ring that we just solve for.*3808

*We know the electric field due to that ring and just say let us start with a little ring at the center, *3823

*find its electric field and make a bigger ring and add it, and a bigger ring.*3828

*We are going to integrate from the radius is 0 all the way to R, of all of these little rings to get the total electric field at point B.*3834

*Relying heavily on what we have done previously.*3843

*Again, we can make a symmetry argument that we only have to deal with the z component because we are centered on this disk.*3846

*This time we are going to define a surface charged area of Σ which is total charge Q ÷ A.*3853

*In this case, our area of the disk is π R².*3861

*If this is our ring here in purple, the charge of that ring is Δ Q on our ring there.*3869

*We will call the radius of our ring some Δ,*3878

*Let me make that a lowercase just to help differentiate what we are doing.*3883

*A Δ R, as we go out, we will call this little r, this variable that we are going to change.*3888

*We can look at the charge contained Δ Q.*3895

*Δ Q is going to be that surface charge density × the area of this ring.*3900

*To find the area of the ring, we have got to consider here is our ring, what happened if we cut it 1 point and then spread it out to make a line.*3905

*We get a little skinny rectangle.*3913

*The length of that rectangle is going to be Σ × the area, the length of that is going to be 2 π × Ri and the thickness of that rectangle is going to be Δ R.*3915

*With that in place, we can start to find our electric field due to some little bit of our disk, our little bit being that single ring.*3934

*The electric field in the z direction due to that little ring is 1/ 4π E₀ again × we have our z Δ Q due to that little piece of i/ z² + Ri²³/2.*3944

*Coming from the solution to our previous problem, we just did that work.*3973

*This implies then that Δ Q, using Δ Q = Σ × 2 π Ri Δ R that the electric field due to that ring i in the z direction*3978

*is 1/ 4π E₀ × z × Δ Q i which we just define over hear in green.*3997

*Σ 2π Ri Δ R ÷ z² + Ri²³/2.*4007

*There is our starting point then we are going to integrate this, *4024

*we are going to add these all up as we go from a little tiny ring here in the center*4027

*and expand it outwards and upwards until we get to the entire width of disk.*4031

*I think we are ready for the math.*4038

*Electric field in the z direction is going to be the integral adding all these up of 1/ 4π E₀ × z Σ 2π Ri DR / z² + Ri²³/2.*4041

*Let us pull out our constant and see if we can simplify this a little bit.*4069

*Which implies that Ez = Σ and z can both come out, those are constants.*4072

*Our 2 E₀ can come out and this 2 and 4, we can do a little bit division there and say that we end up with Σ z/ 2 E₀.*4079

*A π in a π cancel out, integral from r = 0 all the way to the R, the entire radius of our disk.*4092

*We are left with r DR / z² + R²³/2.*4100

*To integrate this I'm going to do a little bit of a substitution trick.*4112

*This is not quite so obvious to me how to do so, I like to simplify things as much as possible.*4116

*I'm going to say, I’m going to define a new variable U and say that u = z² + R².*4121

*This would be u³/2.*4129

*If I do that the differential of u, our only variable here is the R so that is going to be 2 RDR.*4132

*To make all of this fit that form, I have got u down here.*4141

*I almost have Du up here.*4146

*If I had a 2 here this would be Du/ u.*4149

*Let us multiply it to here and change the values.*4153

*I have to divide by a 2 as well out there.*4155

*What I have is a constant × the integral of Du/ u³/2 and that I know how to integrate.*4159

*Not Du/ u, Du/ u³/2 and that I know how to integrate.*4171

*We can rewrite this electric field in the z direction, we have Σ z/ 4 E₀ integral.*4177

*I have changed my variable of integration from R to u.*4191

*Now we went from r = 0 corresponds if r is 0, u is z².*4194

*We are integrating from u = z² to r = R so this must be u = x² + R².*4201

*Our simplified version of what we are integrating now becomes u⁻³/2 Du, much easier for me to integrate.*4214

*That becomes Σ z/ 4 E₀ × -2/ √u¹/2 evaluated from z², 2z² + R².*4227

*There is the integration I have to substitute in here.*4249

*We end up with Ez = Σ z/ 4 E₀ - 2/ z² + R²¹/2 - -2/ u ^½ which in this case is z² ^½ or z.*4252

*A little bit of algebra here that is equal to Σ z and we can pull out a 2 here from both sides ÷ 4 E₀ × that is going to be 1/ z-1/ √ z² + R².*4283

*If I have a 2 here and the 4 here, we can simplify there the electric field in the z direction will be 2 there, 2 there, Σ/ 2 E₀.*4309

*I can take out the z and say that we have 1 - z/ z² + √R².*4328

*This is just a little bit of algebra going from here to here and simplifying that out.*4346

*There is our answer.*4350

*I think I’m happy with that, not really feel like simplifying any further.*4354

*What if that disk or an infinite disk?*4359

*What if you expanded it and expanded it until it is infinitely big?*4362

*We almost have the solution to an infinite plane of charge here, *4367

*let us do that by taking the limit as R approaches infinity of our answer which should be Σ/ 2 E₀ × 1 - z/ √z² + √R².*4371

*As I do this, R gets really big we are going to have 1 – 0.*4392

*The R is going to dominate over here so we end up with Σ/ 2 E₀.*4402

*If that disk is infinitely big as it becomes an infinite plane, there is no dependence on how far you are from that plane at all.*4412

*Which makes sense, if it is an infinite plane, that distance from it really have any tangible practical meaning any more.*4420

*We get this nice simple Σ/ 2 E₀, it is all a function of your surface charge area.*4429

*Practice is good, let us keep going.*4438

*Hit the break and by all means do so but comeback in we will finish this up here with one last problem.*4440

*Find the electric field due to a finite uniformly charged rod of length L on its side at some distance D away from the end of the rod.*4448

*Uniformly charged rod we are some distance from it, let us find the electric field here.*4456

*By symmetry arguments, it looks like you are only going to need the x component.*4462

*We will do the same thing, let us take and break this up.*4467

*Let us call this little piece of that Δ Q and close there.*4470

*We will define some x distance and distance from P to our Δ Q, there is our x.*4476

*This Δ Q is going to be = to some linear charge density × whatever that little piece of x is.*4484

*Δ x or we could say that the differential of Q that tiny little piece of charge enclosed is the linear charge density × Dx.*4495

*Where Λ is linear charge density is Q/straight forward L.*4507

*The electric field due to this little tiny I, bit of charged in the x direction is 1/ 4π E₀ Δ Q/ distance².*4514

*With that I think we are set to start doing our math.*4531

*Electric field due to that little bit i in the x direction is 1/ 4π E₀ Δ Q/ x² which implies then *4538

*that the total electric field in the x direction will be the integral of 1/ 4π E₀ R DQ / x².*4549

*Which we already said DQ is Λ Dx so we can write that as pulling 1/ 4π E₀ out of the integral sign, integral of Λ Dx/ x².*4563

*Where our limits of integration, we are going to integrate from x = that distance d to x = D + L to get us that total length of our line of charge.*4582

*This implies then that the electric field in the x direction is Λ/ 4π E₀, our Λ can come out as well, it is a constant.*4594

*The integral from D to D + L of x², x⁻² Dx which is Λ/ 4π E₀ × -1/ x from D to D + L,*4607

*which is Λ/ 4π E₀ substituting in we have -1/ D + L - -1 / D which implies then that the electric field in the x direction is Λ/ 4π E₀.*4633

*And doing a bit of rearranging here, putting a common denominator D × D + L we end up with.*4659

*Let us write that down here first, D × D + L we end up with D - D + D + L.*4667

*It is little bit easier to deal with which is Λ/ 4π E₀ - D + D is going to be 0 that is just going to leave an L / D × D + L.*4677

*We also have said again that Λ = Q/ L so that is going to be = I will replace Λ with Q/ L.*4694

*We have still got our 4π E₀ down here × L/ D × D + L.*4706

*L in the bottom and L to the top make a ratio of 1 electric field in the x direction.*4719

*1/ 4π E₀ × Q/ D × D + L.*4725

*That looks like enough practice problems for now.*4742

*Please take your time, if you need to do these again go back practice some and see if you can do them on your own or see how far you can get, *4744

*then check the video again, pause and rewind our wonderful things.*4751

*Thank you so much for watching www.educator.com.*4754

*We will see you in the next lesson when we talk about Gauss’s law.*4758

*Thanks and make it a great day.*4761

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 13, 2017 6:18 AM

Post by Mark Sim on March 13 at 06:15:30 AM

I find example V11 very difficult to understand. Can you teach me how to do?

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 13, 2017 6:16 AM

Post by Mark Sim on March 13 at 06:13:21 AM

Dear Prof,

I have a question for you from my physics workbook.

Which of the following provides evidence for the quantization of charge?

(a) Electron diffraction experiments

(b)Millikan'oil drop experiment

(c) Photoelectric effect experiments

(d) The deflection of electron beams by electric and magnetic fields

2 answers

Last reply by: Professor Dan Fullerton

Thu Mar 2, 2017 6:50 AM

Post by Harsh Ranawat on March 2 at 05:00:20 AM

Sir,

why L/2 and -L/2? Is it because we are considering the point "p" at a distance d from the midpoint of "l"?

1 answer

Last reply by: Professor Dan Fullerton

Sun Sep 18, 2016 7:09 AM

Post by Mohsin Alibrahim on September 17, 2016

Hello Mr F,

For example 7, are we assuming that the rod is positively charged ?

1 answer

Last reply by: Professor Dan Fullerton

Mon Oct 12, 2015 1:10 PM

Post by QAIS ATTARWALA on October 10, 2015

Hi professor, I don't understand why for example six you used sin theta instead of cos theta. Can you clarify?

1 answer

Last reply by: Professor Dan Fullerton

Sat Aug 15, 2015 2:35 PM

Post by Micheal Bingham on August 14, 2015

Hello Mr. Fullerton, if I place a positive charge and a negative charge at the end of a spring and a problem asks me to calculate the initial attracting force, do I have to take in account hooke's law? It says the spring was uncompressed.

3 answers

Last reply by: Professor Dan Fullerton

Mon Sep 21, 2015 8:53 AM

Post by Mohsin Alibrahim on June 6, 2015

Hello Mr F.

For example 5, what if the charge of P was negative ?

Thanks for the wonderful lecture, I'm watching it for the second time and things starts to clearer than before.

1 answer

Last reply by: Professor Dan Fullerton

Tue Jun 2, 2015 7:22 PM

Post by Mohsin Alibrahim on June 2, 2015

Hello Mr F

Would you please elaborate on why symmetry cancel out charges in the x and y directions not in the z ?

1 answer

Last reply by: Professor Dan Fullerton

Sat May 30, 2015 6:26 PM

Post by Mohsin Alibrahim on May 30, 2015

Hello Professor Fullerton,

I understand from the notes that the electric filed point away from the positive charges toward negative, so why in ex III there are electric filed in both left and right ?

1 answer

Last reply by: Professor Dan Fullerton

Sat May 30, 2015 6:24 PM

Post by Mohsin Alibrahim on May 30, 2015

Hello Professor Fullerton,

Why is q positive in ex1 ? Should it negative since we're dealing with an electron not a proton ?

1 answer

Last reply by: Professor Dan Fullerton

Wed May 13, 2015 6:14 AM

Post by Arjun Srivatsa on May 12, 2015

How often have you seen them ask for an integral derivation for electric field on an FRQ? I was wondering how well I needed to know this

2 answers

Last reply by: Professor Dan Fullerton

Tue Apr 28, 2015 5:59 AM

Post by Arshin Jain on April 28, 2015

Hello Professor Fullerton,

Thanks for the fundamentally overwhelming yet helpful lecture!

~ John

===============================================================

Note: I resolved my concern while completing the last sentence of my question. I am leaving the question (and answer) for others, should a similar concern arise.

During example X (i.e. @ 71:00), I am unable to understand, on the second line, the reason for (-2) in the numerator. Could you please explain?

My approach led me to replacing the numerator prior to integration with du = 2rdr, which leaves the integral (after u-substitution) to u^(-3/2).

Ah, I notice that during integral, 1/(-1/2) is the multiplying factor for the power rule.

It seems that I was a bit rusty on my Integration. The concern has been alleviated, Thanks regardless.

1 answer

Last reply by: Thadeus McNamara

Tue Mar 3, 2015 9:16 PM

Post by Thadeus McNamara on March 3, 2015

@ around 51:30, how does rad((L/2)^2+d^2) simplify down to (L/2)?

1 answer

Last reply by: Jingwei Xie

Fri Jan 9, 2015 6:50 PM

Post by Professor Dan Fullerton on January 9, 2015

Hi Jingwei. In example 6 you are correct, the electric field is exerted from all directions, but by observation we can see that all the horizontal components are going to cancel out at point C. The only thing we'll have to worry about are the vertical components. So, we only look at the y-component of the electric field (which is the entire electric field multiplied by sin(theta) in this problem.

0 answers

Post by Jingwei Xie on January 9, 2015

Hello Mr. Fullerton,

In example 6, I don't really understand why the sine was multiplied. What do you mean by vertical here? Do you mean the electric field above point C? But isn't the field strength exerted from all directions?

Thank you!