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

### 1998 AP Practice Exam: Multiple Choice Questions

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
- 1998 AP Practice Exam Link
- Multiple Choice 36
- Multiple Choice 37
- Multiple Choice 38
- Multiple Choice 39
- Multiple Choice 40
- Multiple Choice 41
- Multiple Choice 42
- Multiple Choice 43
- Multiple Choice 44
- Multiple Choice 45
- Multiple Choice 46
- Multiple Choice 47
- Multiple Choice 48
- Multiple Choice 49
- Multiple Choice 50
- Multiple Choice 51
- Multiple Choice 52
- Multiple Choice 53
- Multiple Choice 54
- Multiple Choice 55
- Multiple Choice 56
- Multiple Choice 57
- Multiple Choice 58
- Multiple Choice 59
- Multiple Choice 60
- Multiple Choice 61
- Multiple Choice 62
- Multiple Choice 63
- Multiple Choice 64
- Multiple Choice 65
- Multiple Choice 66
- Multiple Choice 67
- Multiple Choice 68
- Multiple Choice 69
- Multiple Choice 70

- Intro 0:00
- 1998 AP Practice Exam Link 0:11
- Multiple Choice 36 0:36
- Multiple Choice 37 2:07
- Multiple Choice 38 2:53
- Multiple Choice 39 3:32
- Multiple Choice 40 4:37
- Multiple Choice 41 4:43
- Multiple Choice 42 5:22
- Multiple Choice 43 6:00
- Multiple Choice 44 8:09
- Multiple Choice 45 8:27
- Multiple Choice 46 9:03
- Multiple Choice 47 9:30
- Multiple Choice 48 10:19
- Multiple Choice 49 10:47
- Multiple Choice 50 12:25
- Multiple Choice 51 13:10
- Multiple Choice 52 15:06
- Multiple Choice 53 16:01
- Multiple Choice 54 16:44
- Multiple Choice 55 17:10
- Multiple Choice 56 19:08
- Multiple Choice 57 20:39
- Multiple Choice 58 22:24
- Multiple Choice 59 22:52
- Multiple Choice 60 23:34
- Multiple Choice 61 24:09
- Multiple Choice 62 24:40
- Multiple Choice 63 25:06
- Multiple Choice 64 26:07
- Multiple Choice 65 27:26
- Multiple Choice 66 28:32
- Multiple Choice 67 29:14
- Multiple Choice 68 29:41
- Multiple Choice 69 31:23
- Multiple Choice 70 31:49

### 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: 1998 AP Practice Exam: Multiple Choice Questions

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

*In this lesson, we are going to take an AP Practice Exam and go through the multiple choice portion.*0003

*The link of that is available right here.*0011

*I highly recommend that you take a minute, go through, and print it out, and try and take the test on your own first.*0013

*Then come back and we will go through the walk through this together.*0020

*As we take a look here, let us start on the E and M portion of this practice test from 1998.*0025

*In question 36 is where we are going to begin.*0031

*As we look at number 36, let us see.*0036

*We a resistor and capacitor acted in series to a battery.*0040

*I always like to draw these out.*0044

*There is V0 ±, we have got our resistor R, we got our capacitor C, and there is our circuit.*0048

*I will define the direction for positive current flow there I.*0059

*Which in the following questions relating the current and the circuit and the charge describes the circuit?*0065

*If we use Kirchhoff’s voltage law and we go around, let us start from here and go around clockwise.*0071

*I could have write my KVL that we have -V0 + IR + the voltage across our capacitor VC = 0.*0076

*But since capacitance is charge / voltage, voltage is charge/ capacitance so we could write this as - V0 + IR + Q/ C = 0.*0091

*Or V0 - IR -Q/ C = 0 and that will match the form of answer B.*0108

*37, which of those following combinations of 4 ohm resistors would dissipate 24 W connected to a 12V battery?*0126

*First, it would be nice know what that resistance is.*0135

*If power is V² / R, then resistance is V²/ our power which is going to be 12 V²/ 24W.*0139

*144/ 24 will be 6 ohms.*0149

*We need a combination of resistors that will give us 6 ohms.*0152

*If those are all 4 ohms resistors, it looks to me like E gives us a 4 ohm resistor in series with 2/ 4.*0156

*The 2/ 4 in parallel gives us 2 ohms.*0164

*4 + 2 is 6, correct answer must be E.*0166

*38, we have 2 uncharged conductors 1 and 2 on insulating stands, they are in contacts.*0175

*Let us draw those and we are going to bring a negatively charged rod near them, there it is.*0181

*When we do that, the electrons are closest to the rod and they are screaming away.*0187

*They are repelled so they are all going to hang out over here as much as possible, leaving this one slightly positively charged.*0192

*We separate those spheres, what is now true of conductor number 2, this one over here.*0199

*It is negatively charged so the answer must be C.*0206

*Taking a look at 39, we have a couple of charges in a test charge + Q, *0212

*what is the direction of force on the test charge due to the other two charges?*0223

*We are going to get a force from the top one that is down the force F.*0227

*We are going to get a force to the left F from the one on the right because they repel.*0232

*The net is going to be down into the left or E.*0237

*40, relates to the same question, if F is the magnitude of the force on the test charge due to 1, what is the net force acting on it?*0244

*We can figure this out with the Pythagorean Theorem.*0253

*That is going to be in that direction, the combination of those two.*0257

* The total is going to be √ F² + F² which is √ 2F² or √ 2 × F.*0260

*The answer is D. *0274

*Taking a look at 41, Gauss’s law provides a convenient way to calculate the electric field outside and each of the following accept what?*0278

*A large plate, we did that.*0298

*A sphere, we did that.*0301

*A cubed, it is going to be difficult with Gauss’s law because that is symmetry piece.*0303

*The long solid and a long hollow cylinder, our lectures on Gauss’s law we did A, B, D, and E.*0307

*The only one that is going to be difficult to use Gauss’s law for that purpose is going to be C.*0314

*Let us take a look, moving on here to 42, have a wire resistance R dissipating some power when the current passes through it.*0321

*The wire is replaced by another wire with resistance 3R, what is the power dissipated with the same current?*0332

*Power is I² R so our final power, once we triple our resistance is going to be I² × 3R, which is just going to be 3 I² R or 3 × our initial power.*0338

*The answer must be D, 3P.*0354

*Moving on 43, we got a narrow beam of protons producing that current of 1.6 ma.*0361

*They are 10⁹ proton in each meter, which is the best estimate of the average speed of the protons in the beam?*0369

*I like it, a little more challenging.*0375

*Current is 1.6 × 10⁻³ amps.*0378

*We have got 10⁹ protons / m and we are trying to find the average speed.*0383

*The way I might go about this is, I’m first going to recognize the current as charge per unit time.*0394

*We need to also somehow get velocity in here so I'm going to look at velocity being distance /time.*0400

*Therefore, time is distance / velocity and I'm going to plug that in for T here to get that my current is going to be Q/ D × V.*0406

*Let us also look and see if we can figure out the charge / m, since we got a Q/ D factor here.*0421

*We have 10⁹ elementary charges, the charge on each proton / m.*0427

*Let us convert that into Coulomb’s, that is 1.6 × 10 ⁻19 C/ elementary charge, which gives us 1.6 × 10 ⁻10 C / m.*0433

*Putting all of this together now, we said current is Q/ D × V.*0447

*Therefore, V must be current × D/ Q which is going to be our current was 1.6 × 10⁻³ amps Q/ D, that is what we have here C/ m.*0454

*Q/ D in the denominator would be 1.6 × 10 ⁻10 C/ m, which is going to leave us with 10⁷ m/ s.*0469

*Is it one of our answer? It is, answer D.*0482

*Very good, let us take a look at 44.*0487

*Which of the following describes the line of magnetic field in the vicinity of the beam due to the beam's current?*0493

*Either way, concentric circles around the beam.*0499

*Absolutely, that is got to be at 44 A.*0502

*45, we have got a couple of charges on X axis and we want to know where the electric field strength could be 0.*0508

*That is got to be between the two so that they cancel out.*0518

*And where is it going to be between the two?*0525

*That has got to be a little bit closer to that + 2Q charge so that the distance factor *0526

*to make up for the fact that the charge on the right has less charge, so less force.*0532

*I would say it has to be between 2 and 3, answer C.*0536

*And 46, relating to the same problem.*0544

*The electric potential is negative at some points on the line in which of the following ranges?*0546

*That does not make sense, with only positive charges we are not going to have negative electric potentials.*0551

*I'm going to go with E on that one.*0557

*There are no places where you are going to have a negative potential at least in that current charge configuration.*0560

*Let us take a look here at number 47, we have got a graph showing the electric potential and region of space as a function of position.*0570

*Where would the charged particle experience the force of greatest magnitude?*0580

*You are going to have the greatest force where you have the greatest electric field strength.*0584

*Remember that the electric field is - DV DR so if that relates to the slope, *0592

*we are going to have the strongest field where we have the strongest, the greatest slope on our voltage position graph.*0598

*That to me looks like it is going to be position D, the greatest slope is going to have the greatest electric field.*0604

*Therefore, the greatest force on a charged particle.*0612

*48, work that must be done by an external agent to move a point charge of 2 µc from the origin 2, 3 m away as 5 J.*0619

*Find the potential difference.*0628

*Potential difference is work per unit charge, that is going to be 5 J / 2 mc or 0.002 C.*0631

*It is just 2500 V, the answer is C.*0639

*49, we have got a wire loop, we have got a wire, and we are looking for net force on the loop.*0648

*We have done similar things here in our lectures.*0655

*Here is our current I2, let us draw our loop up here.*0658

*Where this is I1 going that direction and the first thing I'm going to do if I want to know the net force on the loop is I'm going to figure out the direction of the magnetic field due to I2.*0667

*From the right hand rule, it looks like that is going to be into the plane of the page *0677

*so I'm just going to draw that with some axis up here to help me remember that.*0680

*They should be uniform but I have poor drawing skills here.*0687

*The direction of the force, by the right hand rule, if we have positive current, *0695

*point the fingers of my right hand in the direction current is flowing, bend in the direction of the magnetic field.*0700

*You are going to see a force that is up from that section of wire.*0705

*Let me use red here.*0708

*Up from that section of wire, over here it is going to be to the right.*0710

*Here it is going to be down and here it is going to be the left.*0715

*To where it is going to be that these two are going to cancel, these are not however *0719

*because the closest wire is going to be in the stronger magnetic field because it is closer to the source.*0723

*The smaller distance from the source of that magnetic fields.*0728

*This magnetic field is stronger here than here.*0731

*We will have a greater force down so the answer must be A, toward the wire.*0733

*Number 50, one of the trickier problems on the test.*0746

*A uniform magnetic field B is parallel to the XY plane as shown and we have got a proton initially moving with velocity V at that angle θ.*0750

*How will that follow that kind of path?*0758

*In here you have really got to think and visualize in 3 dimensions.*0760

*How the right hand rule is going to work?*0764

*Point your fingers in the direction it is moving, bend in the direction of the magnetic field, *0766

*and you are going to see an evolving, twisting path as you get a force but keep bringing it around in the helical path.*0772

*That is going to be D, A helical path with its axis parallel to the Y axis in the direction of the magnetic field.*0780

*51, taking a look here, we have a parallel plate capacitor.*0791

*I like drawing parallel plate capacitors, let us do that.*0798

*Parallel plate capacitor + Q on one plate -Q on the other and that is separated by some distance D.*0805

*Good so far.*0816

*Let us see, if each plate has an area A.*0818

*A single proton of charge + E is released from rest of the surface of the positively charged plate.*0822

*There is our proton charge + E. *0829

*What kinetic energy be or what will it be proportional to when it hits the other plate?*0831

*Let us go back to my definition of potential is work / charge, which implies that the work done which is QV.*0841

*When we get down here that is going to be the kinetic energy, 100% efficient in our problem.*0851

*But we also know that if C = Q/ V, then V = Q/ C.*0857

*We can write this as our charge of our proton × a charge on one of our plates ÷ the capacitance is equal to QV.*0862

*Since capacitance is proportional to area/ the separation of the plates,*0876

*we can say that the change in kinetic energy is going to be, not equal to, proportional to EQ/ C, which is A/ D.*0882

*DQ D/ A, do we have anything that looks like that, the answer is A.*0896

*Moving on to 52, in which of the following case does there exist a non 0 magnetic field that can be conveniently determined using Ampere’s law.*0906

*Outside the point charge if it is at rest.*0919

*It cannot be A, that is not going to work.*0921

*You do not get a magnetic field from a charge at rest.*0923

*Inside a stationary cylinder carrying current, no it is not moving again.*0926

*No charge, no magnetic field.*0929

*How about C, inside a very long current carrying solenoid?*0931

*We actually did that in the Ampere’s law video.*0935

*We know the answer is C, but let us take a look at the next 2 just for kicks.*0938

*At the center of the current carrying loop of wire, we did that but not with Ampere’s law.*0942

*That was pretty good, be pretty tough to integrate.*0947

*We did that with the Biot-Savart law and outside the square current carrying loop of wire.*0950

*That would be pretty tough to integrate too.*0954

*C is going to be the nice simple easy one.*0956

*53, positive beam of protons moves parallel to the X axis in the positive X direction as shown.*0961

*The magnetic field is pointed in the positive Y direction, in what direction must the electric field be pointed?*0970

*But let us se, the magnetic field is pointing in the positive Y and it is going to go with 0 deflection.*0976

*Then the magnetic force has to exactly balance the electrical force.*0983

*By the right hand rule then, that means that the direction of the electric field must be in the -Z direction *0986

*in order to balance the force from the magnetic field.*0994

*That is got a be E, negative Z direction.*0998

*Taking a look at 54, a vertical length of copper wire moves to the right with a steady velocity V *1005

*in the direction of constant horizontal magnetic field B.*1012

*To the following describes the induce charges on the ends of the wire.*1016

*54, V cross B is going to be 0, they are in the same direction.*1020

*I'm going to go with E on that one.*1025

*55, suppose an electron charge -E orbit a proton in a circular orbit of constant radius R.*1031

*If we assume the proton stationary, we are only worried about the electrostatic forces.*1043

*Find the kinetic energy of the two particle system which is going to be the kinetic energy of the electron.*1048

*The force on the electron, the electrical force is going to be the centripetal force, *1056

*which implies then our electrical force by Coulomb’s law is 1/ 4 π ε₀, the charge on each of these proton and electron.*1062

*The magnitude is E so that is going to be E² ÷ R and that has to equal our centripetal force MV²/ R.*1070

*Q1 Q2/ R, get rid of square there.*1087

*We got to solve then for MV² is going to be equal to, we multiply that by R.*1093

*We are going to get a E²/ 4 π ε₀ R.*1101

*And MV² is awfully close to the kinetic energy.*1111

*If we want the kinetic energy, that is kinetic energy so I just divide that by 2.*1114

*Our kinetic energy then is going to be E²/ 8 π ε₀ R.*1120

*Because that is one of our choices, 1/ 8 π ε₀ R.*1133

*Choice B gives us the correct answer.*1139

*56, we have got a square loop of wire with side L and resistance R.*1148

*It is held at rest in a uniform field but the field decreases with time according to that formula.*1154

*Find the induced current in the loop.*1161

*Faraday’s law, E = –D φ BDT which is - the derivative of BA because our area is not changing.*1164

*Which is going to be - the derivative with respect to time of, our magnetic field strength B is given by the formula A – BT.*1175

*And our area is just L², it is a square.*1188

*L² can come out, it is going to be L² × the derivative of A – BT, which implies that induced EMF then is just going to be BL².*1191

*The current then is E/ R just going to be BL² / R.*1207

*And how about direction, it looks like with time but that magnetic field is getting weaker.*1218

*Dense law as we wanted to oppose that change that will give us a counterclockwise current by the right hand rule.*1224

*BL²/ R counterclockwise, looks like that is going to be choice E.*1230

*57, a negatively charged particle in a uniform magnetic field moves in a circular path as shown,*1240

*which of the following graphs to picks have the frequency of revolution F depends on the radius R.*1249

*Let us start, we know that the velocity is distance /time or in this case 2 π R ÷ T, the time for once around is the period.*1258

*We can also look at this from a centripetal force perspective.*1268

*The centripetal force is MV² / R which has to be equal to whatever is causing that force.*1272

*What is causing the centripetal force is the magnetic force here, which is going to be Q VB.*1281

*Therefore, we could solve for V and say that that is going to be Q BR/ M.*1287

*Putting these together, Q BR/ M is our velocity has to equal 2 π R / T.*1297

*But by the way, 1/ the period is the frequency so I'm going to write that as 2 π R F.*1309

*And just solve for frequency is going to be Q BR/ M × 2 π R or Q B/ 2 π M.*1315

*Notice here that we have no dependence on the radius.*1331

*The correct answer must be A, does not depend on radius at all.*1334

*Let us take a look at 58, the only force acting on an electron is due to a uniform electric field.*1343

*The electron moves at constant.*1353

*It has got to be a constant acceleration because it is a constant force.*1356

*The direction is opposite that of the field because the field points in the direction of the force a positive charge would feel.*1363

*58 right away, A.*1369

*59, in a region of space we got a spherical symmetric electric potential given by the function V of R = KR², where K is a constant.*1375

*What is the magnitude of the electric field when your distance R from the origin?*1387

*The electric field from potential is - DV DR which is just going to be -2 KR.*1392

*Therefore, the magnitude of the electric field is just going to be 2K and we are looking at R₀ instead of R.*1400

*2 K R0 is going to be our answer C.*1408

*In the following question 60, what is the direction of the electric field at RO and the direction of the force on electron.*1414

*Notice we had negative here, so that means that the electric field is going to be toward the origin and *1422

*because it is a force on the electron, it is going to the opposite direction of the field.*1428

*The answer there B, it is toward the origin with the force, the electric field toward the origin of the force in the opposite direction.*1433

*The answer is B because of this -2 KR that we just found.*1440

*61, 2 charged particles each with a charge of +Q are located along the X axis at X = 2 and X =4.*1450

*Which of the following shows the graph of the magnitude of the electric field along the X axis?*1459

*As I look at 61, it is pretty obvious right in between that has to be 0 so that gets rid of choice C and choice B.*1464

*D and E do not make any sense, it is going to be A.*1475

*62, positive electric charge moved at constant speed between two locations in an electric field*1481

*with no work done by or against the field, how can that occur?*1490

*Right away, I’m thinking that the only way that happens is if you are on equal potential.*1493

*D, the charge is moved along the equal potential line.*1498

*There it is, nice and simple.*1501

*63, we got a non conducting hollow sphere of radius R carrying a large charge + Q.*1507

*A small charge + Q is located at point P and what must be the work done in moving the charge +Q *1513

*from P through the hole to the center of the sphere?*1521

*We are only going to have to do work to get to the surface and that amount of work is going to be charge × the potential difference, *1525

*which is going to be charge × the final potential - the initial potential, which is going to be our charge × *1534

*our final K Q/ R the radius - K Q/ r, we can do some factoring Kq Q 1 / R -1 / r.*1543

*It looks like it is going to be the choice E.*1561

*64, we have got a couple capacitors altering µf connected to the circuit with a 12 V battery.*1568

*Find the equivalent capacitance between X and Z.*1577

*The first thing I'm going to do is see that those 2, 3 µf capacitor is in parallel.*1580

*Those add up so that is going to be 6 µf.*1585

*Then we have got to go through A 3 µf capacitor.*1590

*Those two in series, if we found that equivalent should tell us our answer.*1596

*We can do that a bunch of ways but right away we know that that is going to be less than the smallest so less than 3.*1601

*As I do that, 1/ C equivalent is 1/6 µf + 1/3 µf is going to be, 1/ C equivalent is going to being 1/6 µf + 2/6 micro F is 3/6 µf.*1615

*Therefore, C equivalent is 2 µf.*1634

*Our choice there for 64 must be B.*1640

*And 65, the follow on question.*1646

*Potential difference between Y and Z is going to be what?*1650

*We are looking for the potential difference if we can put a vault meter right there between Y and Z.*1654

*If C = Q/ V then potential = Q/ C, we know our charge is going to be VC, is going to be 12V × 2 µf or 24 µc.*1662

*The charge on each plate is 24 µc.*1685

*Once we know that, the potential between Y and Z is just going to be Q / C or we will have 24 µc on the plate ÷ that capacitance 3 µf is 8V.*1689

*65 should be D. *1705

*66, in the figure contains two identical light bulbs in series with a battery.*1713

*At first they are at equal brightness when switch S is closed, which of the following occurs to the bulbs?*1720

*Once switch S is closed, right away bulb 2 is not going to have any electricity going through.*1727

*All the electricity is going to go right through S, it has shorted out.*1734

*Our choice is either B or E.*1737

*If we short out bulb 2, all of that potential is now going to go through bulb 1 so it is going to expand more power, *1740

*the answer has to be B, bulb 1 gets brighter and bulb 2 goes out.*1748

*67, the bar magnet and a wire loop current carrying I are arranged in which direction is the force on the current loop due to the magnet?*1755

*As I look at that one, let us say that that current loop creates a south pole*1765

*but nears the magnet due to the right hand rule so that is going to attract it.*1770

*Your force then has to be A, toward the magnet.*1774

*68, a wire loop of area A is placed in a time varying that is spatially uniform magnetic field perpendicular to the plane of the loop.*1782

*The induced EMF was given by that function.*1791

*The time varying magnetic field could be given by, EMF is – D φ B DT is going to be B AT ^½.*1794

*Or which implies that E = - the derivative of the integral of B ⋅ DA which is - D/ DT of BA or - A DB DT.*1807

*Which implies then, we have – B T ^½ must equal DB DT, which implies that the integral of -BT ^½ DT must equal the integral of DB.*1828

*Or integral of DB is just B integral of - BT ^½ is going to be – B T³/2 / 3/2 + some constant, *1837

*implies then that B is going to be equal to -2/3 B T³/2, which is we are looking for that and not worry about direction.*1859

*2/3 BT³/2, the answer choice E.*1875

*That is more involved multiple choice in there. *1880

*69, talking about a capacitor with two identical conducting plates parallel to each other separated by a distance D,*1884

*the stand is connected if you ignore effects what is the electric field between the plates?*1892

*Electric field between the plates are parallel capacitor is constant as long as you are not near the edges.*1897

*Just B/ D, the answer is B.*1904

*70, an insulating plastic material is inserted between the plates without otherwise disturbing the system.*1909

*What does that do to the capacitance?*1916

*Capacitance is ε A/ D.*1919

*If you increase ε by putting in an insulator, your capacitance has to go up because A and D are changing.*1923

*Capacitances must increase, the answer has to be A.*1930

*Alright that is the end of the multiple choice section of this practice test.*1937

*Hopefully that gives you a good feel for where you are strong and some opportunities for some more work.*1940

*Thank you so much for watching www.educator.com.*1946

*Come on back in the next lesson and we will do the free response portion of this test.*1948

*Make it a great day everyone.*1952

0 answers

Post by Professor Dan Fullerton on March 27, 2015

Correct link: http://apcentral.collegeboard.com/apc/public/courses/211623.html