Electricity and Magnetism Practice Exam for AP Physics C

Free-Response Questions

Time: 45 minutes. You may refer to the Constants sheet and Equations sheet.  You may also use a calculator on this portion of the exam.

E&M 1

A metal sphere of radius R₁ carries charge +Q. A concentric spherical metal shell, of inner radius R₂ and outer radius R₃, carries charge +2Q.

1. Let r represent the distance from the center of the spheres. Calculate the electric field as a function of r in each of the following four regions:
   1. between r = 0 and r = R₁
   2. between r = R₁ and r = R₂
   3. between r = R₂ and r = R₃
   4. between r = R₃ and r = O
2. How much charge is on each surface of the outer spherical shell? Justify your answer.
3. Determine the electric potential of the outer spherical shell.
4. Determine the electric potential of the inner metal sphere.

E&M 2

A 1 MΩ resistor is connected to the network of capacitors shown above. The circuit is hooked to a 10-V battery. The capacitors are initially uncharged. The battery is connected and the switch is closed at time t = 0.

1. Determine the equivalent capacitance of C₁, C₂, and C₃.
2. Determine the charge on and voltage across each capacitor after a long time has elapsed.
3. On the axes below, sketch the total charge on C₃ as a function of time.



4. After the capacitors have been fully charged, the switch is opened, disconnecting C₁ and C₂ from the circuit. What happens to the voltage across and charge on C₃? Justify your answer.

E&M 3

In the laboratory, far from the influence of other magnetic fields, the Earth's magnetic field has a value of 5.00 × 10^–5 T. A compass in this lab reads due north when pointing along the direction of Earth's magnetic field.

A long, straight current-carrying wire is brought close to the compass, deflecting the compass to the position shown above, 48° west of north.

1. Describe one possible orientation of the wire and the current it carries that would produce the deflection shown.
2. Calculate the magnitude B_wire of the magnetic field produced by the wire that would cause the deflection shown.
3. The distance d from the wire to the compass is varied, while the current in the wire is kept constant; a graph of B_wire vs. d is produced. On the axes below, sketch the shape of this graph.



4. It is desired to adjust this plot so that the graph becomes a straight line. The vertical axis is to remain B_wire, the magnetic field produced by the wire. How could the quantity graphed on the horizontal axis be adjusted to produce a straight line graph? Justify your answer.
5. The current carried by the wire is 500 mA. Determine the slope of the line on the graph suggested in part (d).

STOP. End of Physics C—Electricity and Magnetism Practice Exam—Free-Response Questions

Solutions

Multiple-Choice Solutions

1. B—An electric field exists regardless of the amount of charge placed in it, and regardless of whether any charge at all is placed in it. So both experimenters must measure the same field (though they will measure different forces on their test charges).
2. D—You could use Gauss's law to show that the field outside the sphere has to decrease as 1/r 2, eliminating choices B and E. But it's easier just to remember that an important result of Gauss's law is that the electric field inside a conductor is always zero everywhere, so D is the only possibility.
3. B—While in the region between the plates, the negatively charged electron is attracted to the positive plate, so bends upward. But after leaving the plates, there is no more force acting on the electron. Thus, the electron continues in motion in a straight line by Newton's first law.
4. C—This is a Newton's third law problem! The force of A on B is equal (and opposite) to the force of B on A. Or, we can use Coulomb's law: The field due to A is k(2Q)/(4 m)². The force on B is QE = k2QQ/(4 m)². We can do the same analysis finding the field due to B and the force on A to get the same result.
5. A—The charge is in equilibrium, so the horizontal component of the tension must equal the electric force. This horizontal tension is 0.1 N times sin 30° (not cosine because 30° was measured from the vertical), or 0.05 N. The electric force is qE, where q is 0.002 C. So the electric field is 0.050 N/0.002 C. Reduce the expression by moving the decimal to get 50/2, or 25 N/C.
6. D—The answer could, in principle, be found using the integral form of Coulomb's law. But you can't do that on a one-minute multiple-choice problem. The electric field will point down the page—the field due to a positive charge points away from the charge, and there's an equal amount of charge producing a rightward field as a leftward field, so horizontal fields cancel. So, is the answer B or D? Choice B is not correct because electric fields add as vectors. Only the vertical component of the field due to each little charge element contributes to the net electric field, so the net field must be less than kQ/R².
7. A—Use the symmetry of the situation to see the answer. Because the infinitely large plane looks the same on the up side as the down side, its electric field must look the same, too—the field must point away from the plane and have the same value.
8. C—Another way to look at this question is, "where would a small positive charge end up if released near these charges?" because positive charges seek the smallest potential. The positive charge would be repelled by the +2Q charge and attracted to the –Q charges, so would end up at point C. Or, know that potential due to a point charge is kq/r. Point C is closest to both –Q charges, so the r terms will be smallest, and the negative contribution to the potential will be largest; point C is farthest from the +2Q charge, so the r term will be large, and the positive contribution to the potential will be smallest.
9. A—A positive charge is forced from high to low potential, which is generally to the left; and the force on a positive charge is in the direction of the electric field. At point P itself the electric field is directly to the left because an electric field is always perpendicular to equipotential surfaces.
10. C—The charge on the metal sphere distributes uniformly on its surface. Because the non-conducting sphere also has a uniform charge distribution, by symmetry the electric fields will cancel to zero at the center. Outside the spheres we can use Gauss's law: E·A = Q_enclosed/ε_o. Because the charge enclosed by a Gaussian surface outside either sphere will be the same, and the spheres are the same size, the electric field will be the same everywhere outside either sphere. But within the sphere? A Gaussian surface drawn inside the conducting sphere encloses no charge, while a Gaussian surface inside the non-conducting sphere does enclose some charge. The fields inside must not be equal.
11. A—That's what Gauss's law says: Net flux through a closed surface is equal to the charge enclosed divided by ε_o. Though Gauss's law is only useful when all charge within or without a Gaussian surface is symmetrically distributed, Gauss's law is valid always.
12. B—The electric field at the center of the ring is zero because the field caused by any charge element is canceled by the field due to the charge on the other side of the ring. The electric field decreases as 1/r² by Coulomb's law, so a long distance away from the ring the field goes to zero. The field is nonzero near the ring, though, because each charge element creates a field pointing away from the ring, resulting in a field always along the axis.
13. C—Capacitance of a parallel plate capacitor is ε_oA/d, where A is the area of the plates, and d is the separation between plates. To halve the capacitance, we must halve the area or double the plate separation. The plate separation in the diagram is labeled c, so double distance c.
14. E—We are told that the capacitors are identical, so their capacitances must be equal. They are hooked in parallel, meaning the voltages across them must be equal as well. By Q = CV, the charge stored by each must also be equal.
15. E—First determine the voltage of the battery by Q = CV. This gives V = 600 μC/2 μF = 300 V. This voltage is hooked to the three series capacitors, whose equivalent capacitance is 6 μF (series capacitors add inversely, like parallel resistors). So the total charge stored now is (6 μF)(300 V) = 1800 μC. This charge is not split evenly among the capacitors, though! Just as the current through series resistors is the same through each and equal to the total current through the circuit, the charge on series capacitors is the same and equal to the total.
16. B—The charge does reside on the surface, and, if the conductor is alone, will distribute evenly. But, if there's another nearby charge, then this charge can repel or attract the charge on the sphere, causing a redistribution.
17. D—The voltage must stay the same because the battery by definition provides a constant voltage. Closing the switch adds a parallel branch to the network of resistors. Adding a parallel resistor reduces the total resistance. By Ohm's law, if voltage stays the same and resistance decreases, total current must increase.
18. C—The time constant for an RC circuit is equal to RC. The resistance used is the resistance encountered by charge that's flowing to the capacitor; in this case, 40 Ω. So RC = 20 s.
19. E—Assume that the current runs clockwise in the circuit, and use Kirchoff 's loop rule. Start with the 7-V battery and trace the circuit with the current: + 7V – I(3Ω) + 3V – I(2Ω) = 0. Solve for I to get 2.0 A.
20. C—The intrinsic property of the light bulb is resistance; the power dissipated by a bulb depends on its voltage and current. When the bulbs are connected to the 100-V source, we can use the expression for power P = V ²/R to see that the bulb rated at 50 watts has twice the resistance of the other bulb. Now in series, the bulbs carry the same current. Power is also I ²R; thus the 50-watt bulb with twice the resistance dissipates twice the power, and is twice as bright.
21. E—The electron bends downward by the right-hand rule for a charge in a B field—point to the right, curl fingers into the page, and the thumb points up the page. But the electron's negative charge changes the force to down the page. The path is a circle because the direction of the force continually changes, always pointing perpendicular to the electron's velocity. Thus, the force on the electron is a centripetal force.
22. A—This is one of the important consequences of Ampére's law. The magnetic field inside the donut is always along the axis of the donut, so the symmetry demands of Ampére's law are met. If we draw an "Ampérean Loop" around point P but inside r₁, this loop encloses no current; thus the magnetic field must be zero.
23. C—The magnetic field due to the wire at the position of the electron is into the page. Now use the other right-hand rule, the one for the force on a charged particle in a magnetic field. If the charge moves down the page, then the force on a positive charge would be to the right, but the force on a (negative) electron would be left, toward the wire.
24. C—A positive charge experiences a force in the direction of an electric field, and perpendicular to a magnetic field; but the direction of a force is not necessarily the direction of motion.
25. A—The electric field due to any finite-sized charge distribution drops off as 1/r ² a long distance away because if you go far enough away, the charge looks like a point charge. This is not true for infinite charge distributions, though. The magnetic field due to an infinitely long wire is given by

not proportional to 1/r ²; the magnetic field produced by a wire around a torus is zero outside the torus by Ampére's law.

26. C—The magnetic field produced by a single loop of wire at the center of the loop is directly out of the loop. A solenoid is a conglomeration of many loops in a row, producing a magnetic field that is uniform and along the axis of the solenoid. So, the proton will be traveling parallel to the magnetic field. By F = qvB sinθ, the angle between the field and the velocity is zero, producing no force on the proton. The proton continues its straight-line motion by Newton's first law.
27. E—By the right-hand rule for the force on a charged particle in a magnetic field, particle C must be neutral, particle B must be positively charged, and particle A must be negatively charged. Charge B must be more massive than charge A because it resists the change in its motion more than A. A proton is positively charged and more massive than the electron; the neutron is not charged.
28. E—The force on the positive charge is upward; the force on the negative charge is downward. These forces will tend to rotate the dipole clockwise, so only A or E could be right. Because the charges and velocities are equal, the magnetic force on each = qvB and is the same. So, there is no net force on the dipole. (Yes, no net force, even though it continues to move to the left.)
29. D—The centripetal force keeping the electrons in a circle is provided by the magnetic force. So set qvB = mv ²/r. Solve to get r = (mv)/(qB). Just look at orders of magnitude now: r = (10^–31 kg) (10⁷m/s)/(10^–19 C)(10^–5 T). This gives r = 10²⁴/10²⁴ = 10⁰ m ~1m.
30. E—An element of current produces a magnetic field that wraps around the current element, pointing out of the page above the current and into the page below. But right in front (or anywhere along the axis of the current), the current element produces no magnetic field at all.
31. D—A long way from the hole, the magnet produces very little flux, and that flux doesn't change much, so very little current is induced. As the north end approaches the hole, the magnetic field points down. The flux is increasing because the field through the wire gets stronger with the approach of the magnet; so, point your right thumb upward (in the direction of decreasing flux) and curl your fingers. You find the current flows counterclockwise, defined as positive. Only A or D could be correct. Now consider what happens when the magnet leaves the loop. The south end descends away from the loop. The magnetic field still points down, into the south end of the magnet, but now the flux is decreasing. So point your right thumb down (in the direction of decreasing flux) and curl your fingers. Current now flows clockwise, as indicated in choice D. (While the magnet is going through the loop, current goes to zero because the magnetic field of the bar magnet is reasonably constant near the center of the magnet.)
32. A—You remember the equation for the induced EMF in a moving rectangular loop, ε = Blv. Here l represents the length of the wire that doesn't change within the field; dimension a in the diagram. So the answer is either A or C. To find the direction of induced current, use Lenz's law: The field points out of the page, but the flux through the loop is increasing as more of the loop enters the field. So, point your right thumb into the page (in the direction of decreasing flux) and curl your fingers; you find the current is clockwise, or left-to-right across the resistor.
33. D—Start by finding the direction of the induced current in the wire using Lenz's law: the magnetic field is out of the page. The flux increases because the field strength increases. So point your right thumb into the page, and curl your fingers to find the current flowing clockwise, or south in the wire. Now use the right-hand rule for the force on moving charges in a magnetic field (remembering that a current is the flow of positive charge). Point down the page, curl your fingers out of the page, and the force must be to the west.
34. C—There is clearly nonzero flux because the field does pass through the wire loop. The flux is not BA, though, because the field does not go straight through the loop—the field hits the loop at an angle. So is the answer BA cos 30°, using the angle of the plane; or BA cos 60°, using the angle from the vertical? To figure it out, consider the extreme case. If the incline were at zero degrees, there would be zero flux through the loop. Then the flux would be BA cos 90°, because cos 90° is zero, and cos 0° is one. So don't use the angle of the plane, use the angle from the vertical, BA cos 60°.
35. C—The inductor resists changes in current. But after a long time, the current reaches steady state, meaning the current does not change; thus the inductor, after a long time, might as well be just a straight wire. The battery will still provide current I, of which half goes through each equal resistor.

Free-Response Solutions

For answers that are numerical, or in equation form:

*For each part of the problem, look to see if you got the right answer. If you did, and you showed any reasonable (and correct) work, give yourself full credit for that part. It's okay if you didn't explicitly show EVERY step, as long as some steps are indicated and you got the right answer. However:

*If you got the WRONG answer, then look to see if you earned partial credit. Give yourself points for each step toward the answer as indicated in the rubrics below. Without the correct answer, you must show each intermediate step explicitly in order to earn the point for that step. (See why it's so important to show your work?)

*If you're off by a decimal place or two, not to worry—you get credit anyway, as long as your approach to the problem was legitimate. This isn't a math test. You're not being evaluated on your rounding and calculator-use skills.

*You do not have to simplify expressions in variables all the way. Square roots in the denominator are fine; fractions in non-simplified form are fine. As long as you've solved properly for the requested variable, and as long as your answer is algebraically equivalent to the rubric's, you earn credit.

*Wrong, but consistent: Often you need to use the answer to part (a) in order to solve part (b). But you might have the answer to part (a) wrong. If you follow the correct procedure for part (b), plugging in your incorrect answer, then you will usually receive full credit for part (b). The major exceptions are when your answer to part (a) is unreasonable (say, a car moving at 105 m/s, or a distance between two cars equal to 10–100 meters), or when your answer to part (a) makes the rest of the problem trivial or irrelevant.

For answers that require justification:

*Obviously your answer will not match the rubric word-for-word. If the general gist is there, you get credit.

*But the reader is not allowed to interpret for the student. If your response is vague or ambiguous, you will NOT get credit.

*If your response consists of both correct and incorrect parts, you will usually not receive credit. It is not possible to try two answers, hoping that one of them is right. (See why it's so important to be concise?)

CE&M 1

1.  

1 pt: Inside a conductor, the electric field must always be zero. E = 0.

1 pt: Because we have spherical symmetry, use Gauss's law.

1 pt: The area of a Gaussian surface in this region is 4πr ². The charge enclosed by this surface is Q.

1 pt: So, E = Q_enclosed/ε_oA = Q/4π ε_or ².

1 pt: Inside a conductor, the electric field must always be zero. E = 0.

2 pts: Just as in part 2, use Gauss's law, but now the charged enclosed is 3Q. E = 3Q/4π ε_or ².

2.  

1 pt: –Q is on the inner surface.

1 pt: +3Q is on the outer surface.

1 pt: Because E = 0 inside the outer shell, a Gaussian surface inside this shell must enclose zero charge, so –Q must be on the inside surface to cancel the +Q on the small sphere. Then to keep the total charge of the shell equal to +2Q, +3Q must go to the outer surface.

3.  

1 pt: Because we have spherical symmetry, the potential due to both spheres is 3Q/4π ε_or, with potential equal to zero an infinite distance away.

1 pt: So at position R₃, the potential is 3Q/4π ε_oR₃. (Since E = 0 inside the shell, V is the same value everywhere in the shell.)

4.  

1 pt: Integrate the electric field between R₁ and R₂ to get V = Q/4π ε_or + a constant of integration.

1 pt: To find that constant, we know that V(R₂) was found in part (c), and is 3Q/4π ε_oR₃. Thus, the constant is

1 pt: Then, potential at R₁ = Q/4π ε_oR₁ + the constant of integration.

CE&M 2

1.  

1 pt: The series capacitors add inversely,

so C_eq for the series capacitors is 3 μF.

1 pt: The parallel capacitor just adds in algebraically, so the equivalent capacitance for the whole system is 5 μF.

2.  

1 pt: After a long time, the resistor is irrelevant; no current flows because the fully charged capacitors block direct current.

1 pt: The voltage across C3 is 10 V (because there's no voltage drop across the resistor without any current).

1 pt: By Q = CV the charge on C₃ is 20 μC.

1 pt: Treating C₁ and C₂ in series; the equivalent capacitance is 3 μF, the voltage is 10 V (in parallel with C₃).

1 pt: The charge on the equivalent capacitance of C₁ and C₂ is 30 μC; thus the charge on C₁ = 30 μC, and the charge on C₂ is also 30 μC (charge on series capacitors is the same).

1 pt: Using Q = CV, the voltage across C₁ is 7.5 V.

1 pt: Using Q = CV, the voltage across C₂ is 2.5 V.

3.  

1 pt: For a graph that starts at Q = 0.

1 pt: For a graph that asymptotically approaches 20 μC (or whatever charge was calculated for C₃ in part b).

1 pt: For calculating the time constant of the circuit, RC = 5 s.

1 pt: For the graph reaching about 63% of its maximum charge after one time constant.

4.  

1 pt: For recognizing that the voltage does not change.

1 pt: For explaining that if voltage changed, then Kirchoff 's voltage rule would not be valid around a loop including C₃ and the battery (or explaining that voltage is the same across parallel components, so if one is disconnected the other's voltage is unaffected)

CE&M 3

1.  

1 pt: For placing the wire along a north–south line.

1 pt: The wire could be placed above the compass, with the current traveling due north. (The wire also could be placed underneath the compass, with current traveling due south.) (Points can also be earned for an alternative correct solution: for example, the wire could be placed perpendicular to the face of the compass (just south of it), with the current running up.)

2.  

1 pt: The B field due to Earth plus the B field caused by the wire, when added together as vectors, must give a resultant direction of 48° west of north.

1 pt: Placing these vectors tail-to-tip, as shown below, tan 48° = B_wire/B_Earth.



1 pt: So B_wire = B_Earthtan 48° = 5.6 × 10^–5 T.

3.  

1 pt: The magnetic field due to a long, straight, current-carrying wire is given by

where r is the distance from the wire to the field point, represented in this problem by d.

1 pt: So B is proportional to 1/d; this results in a hyperbolic graph.

1 pt: This graph should be asymptotic to both the vertical and horizontal axes.

4.  

1 pt: Place 1/d on the horizontal axis.

2 pts: The equation for the field due the wire can be written

Everything in the first set of parentheses is constant. So, this equation is of the form y = mx, which is the equation of a line, if 1/d is put on the x-axis of the graph. (1 point can be earned for a partially complete explanation. On this problem, no points can be earned for justification if the answer is incorrect.)

5.  

1 pt: The slope of the graph, from the equation above, is

1 pt: For plugging in values correctly, including 0.5 A or 500 mA.

1 pt: For units on the slope equivalent to magnetic field times distance (i.e., T·m, T·cm, mT·m, etc.).

1 pt: For a correct answer, complete with correct units: 1.0 × 10^–7 Tm, or 1.0 × 10^–4 mT·m.