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For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism
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Post by Professor Dan Fullerton on March 27, 2015

Correct link:

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

Transcription: 1998 AP Practice Exam: Free Response Questions

Hello, everyone, and welcome back to

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

Make it a great day everybody.1794