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
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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 acceleration0094
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 sphere0257
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 and0353
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 toward0494
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 m0509
throughout the work done against the electrostatic repulsion of, we have our 120 × 10⁻⁶ our charge × our 1800 N/ C ÷ R0521
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 rails0992
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 axis1013
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 × B1040
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 diagram1065
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 E1606
to get rid of that natural log, which implies in the left hand side becomes - L² B² V/ M R G sin θ +11632
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