For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### 1998 AP Practice Exam: Free Response 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 0:00
- 1998 AP Practice Exam Link 0:14
- Free Response 1 0:22
- Free Response 2 10:04
- Free Response 3 16:22

### 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: Free Response Questions

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

*I'm Dan Fullerton, and in this final lesson we are going to continue our work on the 1998 practice exam,*0003

*this time focusing on the free response questions.*0010

*Let us dive right in and make sure you have got the test printed out.*0014

*I highly recommend that you take a shot at these questions before coming back to the video.*0017

*Starting with number 1, let us look at part A.*0022

*We have that sphere B1 and a point/ A and that B1 sphere is held in place.*0027

*We area asked to find the charge on sphere B1.*0034

*As I look at this, the first thing I think I'm going to do is draw a free body diagram for B1 to see all the forces that are on it.*0038

*Let us take a look here.*0046

*If there is our sphere, we will put our Y axis and X axis.*0048

*And labeling their forces, we have an electrical force repulsion to the right.*0057

*We have a force of tension from our string, we will call that T.*0062

*We have the weight of the sphere MG.*0069

*Using Newton’s second law, we can write that the net force in the Y direction is going to be T sin 70°.*0075

*If 70° here, if that is 20 to match what they have on our diagram would give us 4 θ, - MG = 0,*0088

*which implies then that T must equal MG/ sin 70° which is our mass 0.025 × the acceleration*0094

*due to gravity on the surface of the Earth 9.8 m / s² / the sin of 70° or about 0.26 N.*0111

*We try the same thing in the X direction.*0123

*The net force in the X direction, we have the electrical force to the right - T cos 70, the X component of that tension.*0127

*Those all have to be = to 0 because it is in equilibrium, it is not accelerating.*0139

*Which implies that the electrical force = T cos 70° is going to be, we just found T was 0.26 N cos 70° or 0.089 N.*0144

*If we know the electrical force, we should be able to back out that charge using in Coulomb’s law.*0162

*Our electrical force is K QA QB / R² which implies that the charge on sphere B is just going to be our electrical force × R² ÷ K QA.*0167

*Our electrical force we just found was 0.089 N.*0187

*The distance between the centers of our charged particles is 1.5 m, K 9 × 10⁹ N meters²/ C² which is the same as 1/ 4 π ε₀.*0196

*The charge on A is 120 µc.*0212

*Put that all into my calculator and I find the value for QB of 1.86 × 10⁻⁷ C.*0218

*Part A, check.*0231

*Part B asks, suppose this b1 is replaced by a second suspended sphere B2 that has the same mass but this one is conducting.*0235

*If we establish equilibrium again, what happens to that equilibrium angle?*0246

*That angle has going to have to be less, that angle θ is going to be less than 20°*0251

*because the charges are going to move in the conductor leaving more positive charge on the far side of your B2 sphere*0257

*and less on the near side near A.*0264

*Effectively, that distance between charges increases so the electrical force between them is going to decrease by Coulomb’s law.*0266

*Electrical force decreases, it is going to come down a little bit.*0274

*I would say that that is going to be less than 20°.*0277

*Here for part C, the sphere B2 is now replaced by a very long horizontal none conducting tube.*0282

*The tube is hollow with thin walls, gives us the radius and the uniform positive charge per unit length λ.*0290

*Use Gauss’s law to show that the electric field is given by that expression.*0298

*I’m really good at Gauss’s law, so we will try that.*0303

*Here is our tube and the Gaussian surface I'm going to choose is a cylinder around that.*0306

*Choose a Gaussian surface over here at some distance R from its center.*0317

*The radius of the little one inside this R, the radius of our Gaussian surface is r.*0324

*We can write Gauss’s law, integral / the closed surface of E ⋅ DA is equal to the total enclosed charge ÷ ε 0.*0333

*The left hand side electric field should be constant because we chose this with symmetry, that in mind.*0346

*And the area of our close surface is going to be 2 π R × its length L and*0353

*that is going to be equal to the enclosed charge which is the linear charge density × our length ÷ ε₀.*0361

*Therefore, our electric field is going to be λ L / 2 π ε₀ RL.*0370

*Our L's make a ratio of 1 so that is just going to the λ / 2 π ε₀ R.*0379

*And if we start putting our values in here, λ is 0.1 × 10⁻⁶ C / m, 2 π ε₀ is 8.54 × 10 ⁻12 C²/ N-m².*0386

*And we also have our R down here.*0406

*This implies then that the electric field is going to be, I come up with about 1797/ R N-m/ C,*0410

*which is approximately 1800/ assuming R is in meters, N/ C with R in meters.*0422

*We prove that 1800/ R N/ C.*0437

*Very good, moving on to part B.*0443

*A small sphere A with charged 120 µc is now brought in the vicinity of the tube and held at distance of 1.5 m.*0448

*Find the force the tube exerts on the sphere.*0455

*That should be pretty easy now that we have the electric field strength.*0458

*The force is just charge × electric field which is going to be our 120 × 10⁻⁶ C × our 1800 N/ C / R 1.5 m or 0.144 N.*0463

*And part E, let us take a look at E here.*0486

*Calculate the work done against the electrostatic repulsion to move sphere A toward*0494

*the tube from a distance R = 1.5 m to a distance R= 0.3 m from the tube.*0500

*That should be a pretty straightforward calculation of work.*0506

*Work = the integral of F ⋅ DR, which is going to be the opposite of the integral as we go from R = 1.5 m to 0.3 m*0509

*throughout the work done against the electrostatic repulsion of, we have our 120 × 10⁻⁶ our charge × our 1800 N/ C ÷ R*0521

*to give us our force QE DR, which implies then that the work is going to be equal to -0.216, when I pull the constants out.*0539

*Integral from R = 1.5 to 0.3 of DR/ R.*0552

*Integral of DR/ R is nat log of R.*0559

*W = -0.216 log of R evaluated from 1.5 to 0.3, which is -0.216 log of 0.3 - log of 1.5,*0563

*which is when I plugged into my calculator about 0.348 J.*0583

*That finishes up free response problem number 1.*0595

*Let us move on to number 2, a circuit problem with the capacitor and inductor.*0598

*As we look here at number 2, we are going to start off with the circuit, the switches initially open,*0605

*the capacitor is uncharged, and there is a voltmeter but they are not showing the measure of potential difference across R1.*0612

*On the diagram above, draw the voltmeter with a proper connections for measuring the potential difference.*0618

*That was nice and simple, what you will only do is you take your drawing that you have there, there is your 20V.*0625

*We have R1 just put your voltmeter in parallel V with R1.*0629

*Taking a look at part B, at time T = 0, the switch is moved to position A so we now have the capacitor in the circuit.*0640

*Find the voltmeter reading for the time right after you do.*0649

*The trick here is the moment you close that, that capacitor initially is going to act like a wire.*0653

*Effectively, you have a circuit that looks like this.*0658

*There is R1, there is R2.*0662

*R1 is 10 ohms, R 2 is 20 ohms, and if we want to know the voltmeter reading,*0670

*we could do this as a voltage divider and probably not too tough to say that it is going to drop 1/3 of the voltage across,*0677

*1/3 of the resistance to 10 ohms which will be 1/3 of 20 or 6.67V.*0685

*But more formally, we can do this by saying the total current in the circuit is E/ R which is going to be 20 V/ your total resistance 30 ohms or 0.667 amps.*0690

*The voltage drop across R1 is just going to be current × resistor 1 which is 0.667 amps × our 10 ohms which is about 6.67 V.*0704

*Moving on to part C, after a long time the measurement of potential difference across R1 is again taken,*0721

*determine for this later time the voltmeter reading.*0731

*After a long time, the capacitor is going to act like an open.*0736

*That is going to break your circuit, you are not going to have any current flowing through R1.*0741

*Therefore, there is no voltage drop across R1 so the voltmeter reading is going to be equal to 0.*0747

*And C2, the charge on the capacitor.*0757

*The entire voltage will be across the capacitor at that point.*0759

*If C = Q/ V, that implies that Q = CV, which is going to be your capacitance 15 µf or 15 × 10⁻⁶ F ×*0763

*your potential difference across your capacitor 20 V or 3 × 10⁻⁴ C.*0775

*There is part C, let us take a look at D.*0787

*At a still later time t = T, the switch is moved to position B.*0795

*Determine the voltmeter reading right after that happens.*0800

*Now we have got a circuit that looks kind of like this, for 2D we got out 20V.*0803

*At that next position for our switch, there is R1.*0812

*Here is our inductor and we have R2 in the mix, that is a 2 H conductor.*0816

*Initially, the inductor poses current flowing and acts like an open so potential across R1 is going to be 0.*0827

*Let us take a look at 2 E, a long time after t = T, the current in R1 reaches some constant final value I final.*0839

*Determine what that is going to be.*0848

*After a long time, that inductor just acts like a wire so I final is going to be equal to the potential ÷ the resistance in our circuit.*0851

*Which is again, 20 V/ 30 ohms or 0.667 amps.*0861

*Part 2, determine the energy stored in the inductor.*0872

*The energy in the inductor is given by ½ LI² and it is going to be ½ × our inductance 2 H × the square of our current 0.667 amps² or about 0.445 J.*0875

*We have got a part F, write but do not solve the differential equation for the current in R1 as a function of time.*0896

*We can use Faraday’s law to do this, the integral/ the closed loop of E ⋅ DL = - the time rate of change of the magnetic flux.*0908

*As we go around here, remember no electric field in there so we have got - E + I R1.*0921

*We have got the voltage drop across I R2 and all of that has to equal their change in flux - L DI DT.*0932

*There is no electric field here so no contribution to the left hand side from your inductor.*0944

*We got ID IDT so I suppose we have already written that differential equation.*0949

*But just to clean it up a touch, let us call that E – I R1 + R2 - L DI DT = 0.*0954

*There is the differential equation and we do not have to solve it this time.*0971

*That is a straightforward number 2 and I think we are going to make up for here in number 3.*0976

*This is one of the trickier free response problems, more involved free response problems I have seen on AP exam.*0982

*Here we have a conducting bar of mass M that is placed on some conducting rails*0992

*at distance L apart that is at an angle with respect to the horizontal.*0997

*We have got a magnetic field coming out of that ramp.*1001

*We are asked to determine the current on the circuit when the bar has reached the constant final speed, so 3A.*1005

*First thing I'm going to do is I'm going to make a free body diagram for that bar and I'm going to tilt my axis*1013

*so it is a little easier for me to see what is going.*1021

*On our bar we are going to have a magnetic force up the ramp and we are going to have gravitational force pulling it down.*1029

*We know that that force is going to be up the ramp, we can use the right hand rule Q × V × B*1040

*to find the force on the charges in the bar creating the current.*1046

*And then the charges in the bar move to find the force on the bar which is going to be up the incline.*1049

*As I do this, as I take a look here, determine the current in the circuit when the bar is reached a constant final speed.*1054

*Let us see, net force in the X direction, Newton’s second law point our free body diagram*1065

*is going to be the X component of its weight MG sin θ - FB the magnetic force, all has to equal MA.*1072

*Since V is constant, A = 0.*1085

*Therefore, we can say that the magnetic force has to equal MG sin θ, its constant final speed.*1089

*The magnetic force, my current flowing through the wire is going to be I LB.*1100

*Therefore, we can state that MG sin θ must equal I LB or the current must be equal to MG sin θ/ LB.*1107

*For part B, we are asked to determine the constant final speed of the bar.*1130

*For B, if current is potential / resistance which is – D φ B DT/ R.*1138

*Let us see here, we could do that and all of that must equal MG sin θ/ LB.*1151

*It would be helpful to know what this D φ B term is.*1162

*Φ B is going to be B × A or in this case B LX are geometry.*1167

*D φ B with respect to time is just going to be BL and X is the only thing that changes with time.*1177

*That is DX DT which is going to B BL × its velocity.*1185

*I can put that in up there for that.*1189

*This implies then that, we have B LV/ R = MG sin θ/ LB.*1194

*And solving just for B 4V, V would then equal, we will have MG R sin θ/ L² B².*1212

*Plugging away through this, part C.*1234

*Part C says, determine the rate at which energy is being dissipated when its reach its constant final speed.*1241

*Varying at which energy is dissipated.*1247

*A lots of fancy way of saying find the power.*1249

*We know I, we know R, so power = I² R is just going to be this² multiplied by R.*1251

*We will have M² G² R sin² θ / L² B².*1260

*Not too bad on C.*1279

*D, express the speed of the bar as a function of time T from the time it is released at T equal 0.*1283

*This one looks a bit more involved.*1292

*Let us see, we know MG sin θ - the magnetic force = MA Newton’s second law.*1297

*We said the magnetic force was I LB, we also know that A is the time rate of change of velocity DV DT.*1310

*MG sin θ – I LB must equal M DV DT.*1320

*But we also know that I = B LV / R, therefore, MG sin θ -, substituting in here for FB, B LV/ R.*1332

*We have still got our LB again, LV LB = M DV DT.*1355

*Let us move that M by all sides by M.*1370

*I get on the left hand side G sin θ - L² B² V/ MR = DV DT.*1375

*Which implies then, it is time to do our separation of variables.*1394

*This might get a little bit ugly, DV/ L² B² V/ MR - G sin θ = – DT.*1398

*We are going to try and integrate this.*1419

*We are going to integrate, the composite.*1422

*The integral from V equal 0 to some final value of V of, we have got DV / L² B² V / MR -G sin θ.*1425

*If I want that = the integral from t equal 0 to T of DT with a negative sign, I want this to fit the formula DU/ U.*1445

*I have got DV, I got V here.*1459

*I have got all this other stuff there as well.*1461

*I’m going to have to have that at the top to make that fit the form so that is L² B² / MR.*1464

*If I multiplying the left by L² B²/ MR, I have to multiply the right by L² B²/ MR × negative integral from 0 to T of DT.*1473

*That fits the form of DU/ U.*1486

*This implies then that this should be a nat log of our U which is down here,*1491

*L² B² V/ MR -G sin θ all evaluated from V = 0 to V, must equal.*1499

*This thankfully is little bit easier to integrate, - L² D² T/ MR.*1514

*We can expand this in the left as we substitute in our 0 and V.*1525

*The left hand side will be, we get L² B² V/ MR - G sin θ - the log of - G sin θ = - L² B² T/ MR.*1529

*We can compress this right hand side.*1557

*The difference of logs is the log of the quotient.*1558

*That is going to imply that the log of, we have L² B² V/ M R - G sin θ/ -G sin θ = - L² B² T/ MR.*1562

*Where do we go from here?*1594

*Let us see, it looks like we are going to have to do some simplifications.*1597

*I think we can get rid of that G sin θ or at least incorporate it a little bit better.*1601

*Which implies then that the log of - L² B² V/ MR G sin θ + 1 = - L² B² T/ MR, raising both sides to E*1606

*to get rid of that natural log, which implies in the left hand side becomes - L² B² V/ M R G sin θ +1*1632

*must equal E ^- L² B² T/ MR, which implies then, moving on to the next page.*1646

*L² B² V/ MR G sin θ = 1 - E ^- L² B² T/ MR.*1660

*Since we are solving for V, we can do one last step.*1678

*Get VL by itself and say that V =, we will have MR G sin θ/ L² B² × (1 - E ^- L² B² T/ MR).*1683

*Let us get that a nice big 3D box because good heaven knows we certainly earned that on that problem.*1714

*A lot involved on part D there.*1721

*Quite a tricky bit of work and a lot of math.*1725

*Let us finish off with part E, so 3 even.*1729

*Suppose the experiment is performed again, this time with the second identical resistor connecting the rails at the bottom.*1735

*Will this affect the final speed and how? Justify your answer.*1741

*If you have 2 resistors in parallel, your total equivalent resistance is going to go down.*1747

*We just said that the final velocity is MG R sin θ/ B² L².*1758

*If R decreases, our final velocity must go down.*1770

*I would say that the final speed increases because R is decreasing there.*1774

*Hopefully that gets you a great start to AP Physics C Electricity and Magnetism.*1785

*Thank you so much for joining us at www.educator.com.*1792

*Make it a great day everybody.*1794

0 answers

Post by Professor Dan Fullerton on March 27, 2015

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