Free practice final
Physics 2 practice final (free, with answers)
A typical Physics 2 final covers electric forces and fields, electric potential, capacitors, DC and RC circuits, magnetic forces and fields, and electromagnetic induction, often with some optics at the end. Every constant you need is given in the question. Work each problem on paper, keeping track of units and directions, before opening the answer.
Topics
- Electric Force and Field
- Electric Potential
- Capacitors
- DC Circuits
- RC Circuits
- Magnetic Force
- Magnetic Fields of Currents
- Electromagnetic Induction
- Geometric Optics
15 questions
Try each one before you open its answer.
1. A +3.0 μC point charge and a −4.0 μC point charge are 0.20 m apart. What is the magnitude of the electric force between them, and is it attractive or repulsive? (k = 8.99 × 10⁹ N·m²/C²)
- A 2.7 N, repulsive
- B 0.54 N, attractive
- C 2.7 N, attractive
- D 0.54 N, repulsive
Show answer
Answer: C. 2.7 N, attractive
Coulomb's law gives F = k|q₁q₂|/r² = (8.99 × 10⁹)(3.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.20)² ≈ 2.7 N. Opposite charges attract, and forgetting to square the distance gives the wrong 0.54 N.
2. Point charges of +2.0 nC and −2.0 nC are 0.10 m apart. What are the magnitude and direction of the electric field at the midpoint between them? (k = 8.99 × 10⁹ N·m²/C²)
Show answer
Answer: About 1.4 × 10⁴ N/C, pointing toward the negative charge
Each charge is 0.050 m from the midpoint and makes a field of kq/r² = (8.99 × 10⁹)(2.0 × 10⁻⁹)/(0.050)² ≈ 7.2 × 10³ N/C there. The field points away from the positive charge and toward the negative charge, which is the same direction for both, so the two fields add to about 1.4 × 10⁴ N/C.
3. Two equal positive point charges are fixed a distance d apart. Taking V = 0 infinitely far away, which statement is true at the point midway between them?
- A E = 0 and V = 0
- B E = 0 and V is positive
- C E is nonzero and V = 0
- D E is nonzero and V is positive
Show answer
Answer: B. E = 0 and V is positive
The electric field is a vector, and the two equal fields at the midpoint point in opposite directions and cancel. Potential is a scalar, so the two positive contributions add to V = 2kq/(d/2) = 4kq/d, which is positive.
4. A proton (mass 1.67 × 10⁻²⁷ kg, charge 1.60 × 10⁻¹⁹ C) starts from rest and is accelerated through a potential difference of 1000 V. What is its final speed?
Show answer
Answer: About 4.4 × 10⁵ m/s
The electric potential energy the proton loses becomes kinetic energy, so qΔV = ½mv². Then v = √(2qΔV/m) = √(2 × 1.60 × 10⁻¹⁹ × 1000 / (1.67 × 10⁻²⁷)) ≈ 4.4 × 10⁵ m/s.
5. A parallel-plate capacitor is charged by a battery and then disconnected from it. The plates are then pulled apart to twice their original separation. What happens to the voltage V across the capacitor and the energy U it stores?
- A V doubles and U doubles
- B V doubles and U halves
- C V halves and U halves
- D V stays the same and U halves
Show answer
Answer: A. V doubles and U doubles
Once disconnected, the charge Q has nowhere to go, and doubling the separation halves C = ε₀A/d. Then V = Q/C doubles and U = Q²/(2C) doubles, the extra energy coming from the work done pulling the plates apart. If the battery stayed connected, V would stay fixed and U would halve instead.
6. Capacitors of 3.0 μF and 6.0 μF are connected in series across a 12 V battery. Once they are fully charged, how much charge is on each capacitor?
Show answer
Answer: 24 μC on each
In series, 1/C = 1/3.0 + 1/6.0 = 1/2.0, so the equivalent capacitance is 2.0 μF. Series capacitors carry the same charge, Q = CV = (2.0 μF)(12 V) = 24 μC, which puts 8.0 V across the 3.0 μF capacitor and 4.0 V across the 6.0 μF one.
7. An ideal 18 V battery is connected to a 6.0 Ω resistor in series with a parallel pair of 4.0 Ω and 12 Ω resistors. What is the current through the 4.0 Ω resistor?
- A 0.50 A
- B 1.5 A
- C 2.0 A
- D 4.5 A
Show answer
Answer: B. 1.5 A
The parallel pair is equivalent to (4.0 × 12)/(4.0 + 12) = 3.0 Ω, so the total resistance is 9.0 Ω and the battery current is 18/9.0 = 2.0 A. The voltage across the parallel pair is (2.0 A)(3.0 Ω) = 6.0 V, so the 4.0 Ω resistor carries 6.0/4.0 = 1.5 A and the 12 Ω resistor carries the other 0.50 A.
8. A battery with an emf of 9.0 V and an internal resistance of 0.50 Ω is connected to a 4.0 Ω resistor. What is the voltage across the battery's terminals?
Show answer
Answer: 8.0 V
The current is I = emf/(R + r) = 9.0/(4.0 + 0.50) = 2.0 A. The internal resistance drops Ir = (2.0 A)(0.50 Ω) = 1.0 V, so the terminal voltage is 9.0 − 1.0 = 8.0 V.
9. At t = 0, an uncharged 50 μF capacitor is connected in series with a 2.0 kΩ resistor and a 10 V battery. What is the voltage across the capacitor at t = 0.10 s?
- A 3.7 V
- B 5.0 V
- C 6.3 V
- D 10 V
Show answer
Answer: C. 6.3 V
The time constant is τ = RC = (2.0 × 10³ Ω)(50 × 10⁻⁶ F) = 0.10 s, so t = 0.10 s is exactly one time constant. A charging capacitor follows V = 10(1 − e^(−t/τ)), which gives 10(1 − e^(−1)) ≈ 6.3 V, about 63% of the battery voltage.
10. In a right-handed x, y, z coordinate system, an electron moves in the +x direction through a uniform magnetic field that points in the +y direction. In which direction is the magnetic force on the electron?
- A In the +z direction
- B In the +y direction
- C There is no force, because v and B are perpendicular
- D In the −z direction
Show answer
Answer: D. In the −z direction
The force is F = qv × B, and (+x) × (+y) = +z, which would be the direction of the force on a positive charge. The electron's negative charge reverses it to the −z direction, and the force is largest, not zero, when v and B are perpendicular.
11. A straight 0.50 m wire carrying 4.0 A lies in a uniform 0.30 T magnetic field, at an angle of 30° to the field. What is the magnitude of the magnetic force on the wire?
Show answer
Answer: 0.30 N
The force on a current-carrying wire is F = ILB sin θ, with θ the angle between the wire and the field. So F = (4.0)(0.50)(0.30) sin 30° = 0.60 × 0.50 = 0.30 N.
12. A long straight wire carries a current of 10 A. What is the magnitude of the magnetic field 5.0 cm from the wire? (μ₀ = 4π × 10⁻⁷ T·m/A)
- A 4.0 × 10⁻⁵ T
- B 8.0 × 10⁻⁵ T
- C 1.3 × 10⁻⁴ T
- D 2.5 × 10⁻⁴ T
Show answer
Answer: A. 4.0 × 10⁻⁵ T
For a long straight wire, B = μ₀I/(2πr) = (4π × 10⁻⁷)(10)/(2π × 0.050) = 4.0 × 10⁻⁵ T. Using the formula for the center of a loop, μ₀I/(2R), instead gives the wrong 1.3 × 10⁻⁴ T.
13. A flat coil of 200 turns, each of area 0.010 m², sits in a uniform magnetic field perpendicular to its plane. The field drops steadily from 0.50 T to zero in 0.10 s. What is the magnitude of the induced emf?
- A 0.050 V
- B 0.10 V
- C 5.0 V
- D 10 V
Show answer
Answer: D. 10 V
Faraday's law gives |emf| = NΔΦ/Δt = NAΔB/Δt = (200)(0.010)(0.50)/0.10 = 10 V. Leaving out the number of turns gives 0.050 V, the emf of a single loop.
14. A bar magnet is held on the axis of a flat circular wire loop, with its north pole facing the loop, and is then pushed toward the loop. Viewed from the magnet's side, what is the direction of the current induced in the loop?
- A Clockwise
- B Counterclockwise
- C There is no current, because the magnet does not touch the loop
- D There is no current until the magnet passes through the loop
Show answer
Answer: B. Counterclockwise
The magnet's field points away from its north pole, so it passes through the loop away from the viewer, and that flux grows as the magnet approaches. By Lenz's law the induced current makes a field pointing back toward the magnet, which by the right-hand rule is a counterclockwise current as seen from the magnet's side. The loop's near face acts as a north pole that repels the approaching magnet.
15. An object stands 30 cm in front of a thin converging lens with a focal length of 10 cm. Where is the image, and what kind of image is it?
- A 15 cm on the far side of the lens; real and inverted
- B 15 cm on the same side as the object; virtual and upright
- C 7.5 cm on the far side of the lens; real and inverted
- D 20 cm on the far side of the lens; real and inverted
Show answer
Answer: A. 15 cm on the far side of the lens; real and inverted
The thin-lens equation 1/do + 1/di = 1/f gives 1/di = 1/10 − 1/30 = 2/30, so di = +15 cm. A positive image distance means a real image on the far side of the lens, and the magnification m = −di/do = −0.5 means it is inverted and half the size of the object.