AP Physics C: Electricity and Magnetism Practice Test 5 – 60 Questions with Answers
Test yourself with 60 original AP Physics C: Electricity and Magnetism practice questions covering all 6 units of the course: electric charges, fields, and Gauss’s law, electric potential, conductors and capacitors, electric circuits, magnetic fields and electromagnetism, and electromagnetic induction. Every answer includes the full working. All questions are calculus-based.
Unit 8: Electric Charges, Fields, and Gauss’s Law
Q1. An infinitely long straight wire carries a uniform linear charge density of 82 nC/m. What is the magnitude of the electric field 44 cm from the wire?
- (A) 1.68 × 103 N/C
- (B) 3.35 × 103 N/C
- (C) 6.7 × 103 N/C
- (D) 1.34 × 104 N/C
Show Answer
Answer: B – By Gauss’s law, E = 2kλ/r = 3.35 × 103 N/C.
Q2. A very large flat sheet carries a uniform surface charge density of 24 μC/m2. What is the magnitude of the electric field near the sheet, away from its edges?
- (A) 6.78 × 105 N/C
- (B) 1.36 × 106 N/C
- (C) 2.71 × 106 N/C
- (D) 3.39 × 105 N/C
Show Answer
Answer: B – For one sheet, E = σ/(2ε0) = 1.36 × 106 N/C.
Q3. A solid insulating sphere of radius 14 cm carries a total charge of 5 μC uniformly distributed through its volume. What is the magnitude of the electric field at a point 34 cm from its centre?
- (A) 1.94 × 105 N/C
- (B) 7.78 × 105 N/C
- (C) 3.89 × 105 N/C
- (D) 1.56 × 106 N/C
Show Answer
Answer: C – Outside, the sphere acts as a point charge, so E = kQ/r2 = 3.89 × 105 N/C.
Q4. A thin ring of radius 23 cm carries a total charge of 25 μC uniformly around its circumference. What is the magnitude of the electric field on the ring’s axis 36 cm from its centre?
- (A) 5.19 × 105 N/C
- (B) 2.08 × 106 N/C
- (C) 1.04 × 106 N/C
- (D) 2.59 × 105 N/C
Show Answer
Answer: C – Integrating around the ring, E = kQx/(R2 + x2)3/2 = 1.04 × 106 N/C.
Q5. A thin rod of length 25 cm carries a total charge of 10 nC uniformly along its length. What is the magnitude of the electric field on the rod’s axis 11 cm beyond one end?
- (A) 2.27 × 103 N/C
- (B) 1.14 × 103 N/C
- (C) 4.54 × 103 N/C
- (D) 9.08 × 103 N/C
Show Answer
Answer: A – Integrating dq contributions gives E = kQ/[d(d + L)] = 2.27 × 103 N/C.
Q6. A closed surface encloses a net charge of 15 μC. What is the net electric flux through the surface?
- (A) 8.47 × 105 N·m2/C
- (B) 3.39 × 106 N·m2/C
- (C) 4.24 × 105 N·m2/C
- (D) 1.69 × 106 N·m2/C
Show Answer
Answer: D – Gauss’s law: Φ = Qenc/ε0 = 1.69 × 106 N·m2/C.
Q7. A particle of mass 12 mg and charge 7 μC is placed in a uniform electric field of 1136 N/C. Ignoring gravity, what is the magnitude of its initial acceleration?
- (A) 331 m/s2
- (B) 1.33 × 103 m/s2
- (C) 663 m/s2
- (D) 166 m/s2
Show Answer
Answer: C – a = qE/m = 663 m/s2.
Q8. An electric dipole consists of charges +11 μC and -11 μC separated by 16 mm. Estimate the magnitude of the electric field on the dipole axis 37 cm from its centre, where the distance is much larger than the separation.
- (A) 6.25 × 104 N/C
- (B) 3.12 × 104 N/C
- (C) 1.25 × 105 N/C
- (D) 2.5 × 105 N/C
Show Answer
Answer: A – On the axis far away, E is approximately 2kp/r3 = 6.25 × 104 N/C.
Q9. A thin rod of length 46 cm carries a uniform linear charge density λ. Which integral gives the magnitude of the electric field at a point on the rod’s axis 31 cm beyond one end, where x is the distance from a charge element dq to the point?
- (A) ∪ kλ dx / x
- (B) ∪ kλ dx / x3
- (C) ∪ kλ dx / x2
- (D) ∪ λ dx / x2
Show Answer
Answer: C – Each element contributes dE = k dq/x2 = kλ dx/x2, and the contributions add along the axis.
Q10. Two charged beads carry charges 11 μC and 21 μC and are 56 cm apart in vacuum. What is the magnitude of the electrostatic force between them?
- (A) 6.62 N
- (B) 3.31 N
- (C) 13.2 N
- (D) 1.66 N
Show Answer
Answer: A – F = kq1q2/r2 gives 6.62 N.
Q11. A point charge of -29 μC is fixed in vacuum. What is the magnitude of the electric field at a point 11 cm from the charge?
- (A) 2.15 × 107 N/C
- (B) 1.08 × 107 N/C
- (C) 4.31 × 107 N/C
- (D) 8.62 × 107 N/C
Show Answer
Answer: A – E = k|q|/r2 gives 2.15 × 107 N/C.
Q12. Two positive point charges 6 μC and 15 μC lie on the x-axis 43 cm apart. A point P lies on the x-axis 22 cm beyond the second charge, on the side away from the first. What is the magnitude of the net electric field at P?
- (A) 1.46 × 106 N/C
- (B) 5.83 × 106 N/C
- (C) 2.91 × 106 N/C
- (D) 7.28 × 105 N/C
Show Answer
Answer: C – Both fields point away from P’s sources, so E = kq1/r12 + kq2/r22 = 2.91 × 106 N/C.
Unit 9: Electric Potential
Q13. A charge is moved slowly along an equipotential surface held at 258 V, from one point to another on the same surface. How much work is done by the electric field during this move?
- (A) 258 J
- (B) 129 J
- (C) 516 J
- (D) 0 J
Show Answer
Answer: D – Moving along an equipotential means ΔV = 0, so W = qΔV = 0.
Q14. A point charge of -22 μC is fixed in vacuum, with zero potential at infinity. What is the electric potential 68 cm from the charge?
- (A) -2.91 × 105 V
- (B) -1.45 × 105 V
- (C) -5.82 × 105 V
- (D) 2.91 × 105 V
Show Answer
Answer: A – V = kq/r = -2.91 × 105 V.
Q15. A charge of 14 μC is moved slowly from a point at potential 370 V to a point at potential 620 V. How much work is done by an external agent, ignoring any change in kinetic energy?
- (A) 0.00175 J
- (B) 0.007 J
- (C) 0.0035 J
- (D) 0.000875 J
Show Answer
Answer: C – W = qΔV = 0.0035 J.
Q16. Two point charges 2 μC and -15 μC are held 40 cm apart. What is the electric potential energy of this pair?
- (A) -0.337 J
- (B) -0.674 J
- (C) -1.35 J
- (D) 0.674 J
Show Answer
Answer: B – U = kq1q2/r = -0.674 J; the sign follows the sign of q2.
Q17. In a region, the electric potential varies with position as V(x) = 13x2 + (-37)x + 20, where V is in volts and x is in metres. What is the x-component of the electric field at x = 1.0 m?
- (A) 5.5 N/C
- (B) 22 N/C
- (C) 11 N/C
- (D) -11 N/C
Show Answer
Answer: C – Ex = -dV/dx = -(26x + -37); at x = 1.0 m this gives 11 N/C.
Q18. A thin ring of radius 25 cm carries a total charge of 33 μC uniformly around its circumference. What is the electric potential on the ring’s axis 43 cm from its centre, with zero potential at infinity?
- (A) 2.98 × 105 V
- (B) 1.19 × 106 V
- (C) 1.49 × 105 V
- (D) 5.96 × 105 V
Show Answer
Answer: D – Every element is the same distance away, so V = kQ/√(R2 + x2) = 5.96 × 105 V.
Q19. A thin spherical shell of radius 26 cm carries a total charge of 23 μC uniformly on its surface. What is the electric potential at a point 5 cm from its centre, with zero potential at infinity?
- (A) 3.98 × 105 V
- (B) 1.59 × 106 V
- (C) 3.18 × 106 V
- (D) 7.95 × 105 V
Show Answer
Answer: D – Inside a charged shell the potential is constant and equal to its surface value, V = kQ/R = 7.95 × 105 V.
Q20. A particle of mass 45 μg and charge 50 nC starts from rest and accelerates through a potential drop of 554 V. What is its final speed, ignoring gravity and other forces?
- (A) 17.5 m/s
- (B) 35.1 m/s
- (C) 70.2 m/s
- (D) 8.77 m/s
Show Answer
Answer: B – qΔV = (1/2)mv2, so v = √(2qΔV/m) = 35.1 m/s.
Q21. Point charges 5 μC and 10 μC are fixed in vacuum. At a certain point P, the distances to the two charges are 34 cm and 35 cm respectively. What is the electric potential at P, with zero potential at infinity?
- (A) 1.95 × 105 V
- (B) 7.78 × 105 V
- (C) -3.89 × 105 V
- (D) 3.89 × 105 V
Show Answer
Answer: D – Potentials add as scalars: V = kq1/r1 + kq2/r2 = 3.89 × 105 V.
Unit 10: Conductors and Capacitors
Q22. Capacitors of 6 μF, 17 μF and 4 μF are connected in parallel. What is their equivalent capacitance?
- (A) 13.5 μF
- (B) 54 μF
- (C) 27 μF
- (D) 20.2 μF
Show Answer
Answer: C – In parallel, Ceq = C1 + C2 + C3 = 27 μF.
Q23. A 39 μF capacitor is charged to 95 V. How much energy is stored in the capacitor?
- (A) 0.176 J
- (B) 0.088 J
- (C) 0.352 J
- (D) 0.044 J
Show Answer
Answer: A – U = (1/2)CV2 = 0.176 J.
Q24. A 7 μF capacitor is connected across a 21 V battery. What charge is stored on each plate?
- (A) 7.35 × 10-5 C
- (B) 0.000294 C
- (C) 0.000147 C
- (D) 0.000588 C
Show Answer
Answer: C – Q = CV = 0.000147 C.
Q25. A spherical capacitor consists of two concentric conducting shells of radii 7 cm and 17 cm. What is its capacitance?
- (A) 6.62 × 10-12 F
- (B) 1.32 × 10-11 F
- (C) 2.65 × 10-11 F
- (D) 3.31 × 10-12 F
Show Answer
Answer: B – C = 4πε0ab/(b – a) = 1.32 × 10-11 F.
Q26. The voltage across a parallel-plate capacitor is 352 V and the plate separation is 8 mm. What is the magnitude of the uniform electric field between the plates?
- (A) 4.4 × 104 N/C
- (B) 2.2 × 104 N/C
- (C) 8.8 × 104 N/C
- (D) 1.76 × 105 N/C
Show Answer
Answer: A – E = V/d = 4.4 × 104 N/C.
Q27. A parallel-plate capacitor has plates of area 129 cm2 separated by 8 mm in vacuum. What is its capacitance?
- (A) 7.14 × 10-12 F
- (B) 2.85 × 10-11 F
- (C) 5.71 × 10-11 F
- (D) 1.43 × 10-11 F
Show Answer
Answer: D – C = ε0A/d = 1.43 × 10-11 F.
Q28. A parallel-plate capacitor has capacitance 176 pF in vacuum. The gap is then completely filled with a dielectric of constant 2.5 while the plates remain isolated. What is the new capacitance?
- (A) 440 pF
- (B) 220 pF
- (C) 880 pF
- (D) 110 pF
Show Answer
Answer: A – C = κC0 = 440 pF.
Q29. Capacitors of 4 μF and 7 μF are connected in series. What is their equivalent capacitance?
- (A) 1.27 μF
- (B) 5.09 μF
- (C) 10.2 μF
- (D) 2.55 μF
Show Answer
Answer: D – 1/Ceq = 1/C1 + 1/C2, so Ceq = 2.55 μF.
Unit 11: Electric Circuits
Q30. A current of 4 A flows in a wire of cross-sectional area 2 mm2. The free-electron density is 8.5 × 1028 m-3. What is the drift speed of the electrons?
- (A) 7.35 × 10-5 m/s
- (B) 0.000294 m/s
- (C) 0.000588 m/s
- (D) 0.000147 m/s
Show Answer
Answer: D – vd = I/(neA) = 0.000147 m/s.
Q31. Resistors of 58 Ω and 85 Ω are connected in series across a 25 V battery. What is the voltage across the 85 Ω resistor?
- (A) 14.9 V
- (B) 7.43 V
- (C) 29.7 V
- (D) 3.72 V
Show Answer
Answer: A – V2 = emf × R2/(R1 + R2) = 14.9 V.
Q32. A current of 1 A flows uniformly through a wire of cross-sectional area 1 mm2. What is the current density in the wire?
- (A) 5 × 105 A/m2
- (B) 1 × 106 A/m2
- (C) 2 × 106 A/m2
- (D) 2.5 × 105 A/m2
Show Answer
Answer: B – J = I/A = 1 × 106 A/m2.
Q33. A wire of resistivity 16.0 × 10-8 Ω·m has length 4 m and cross-sectional area 1.5 mm2. What is its resistance?
- (A) 0.213 Ω
- (B) 0.853 Ω
- (C) 0.107 Ω
- (D) 0.427 Ω
Show Answer
Answer: D – R = ρL/A = 0.427 Ω.
Q34. Resistors of 59 Ω, 117 Ω and 21 Ω are connected in series. What is their equivalent resistance?
- (A) 98.5 Ω
- (B) 394 Ω
- (C) 197 Ω
- (D) 148 Ω
Show Answer
Answer: C – In series, Req = R1 + R2 + R3 = 197 Ω.
Q35. Resistors of 60 Ω and 20 Ω are connected in parallel. What is their equivalent resistance?
- (A) 7.5 Ω
- (B) 15 Ω
- (C) 30 Ω
- (D) 60 Ω
Show Answer
Answer: B – 1/Req = 1/R1 + 1/R2, so Req = 15 Ω.
Q36. A resistor of 47 Ω is connected across a 60 V source. How much power is dissipated in the resistor?
- (A) 76.6 W
- (B) 38.3 W
- (C) 153 W
- (D) 19.1 W
Show Answer
Answer: A – P = V2/R = 76.6 W.
Q37. A battery of emf 29 V and internal resistance 4 Ω is connected to an external resistor of 26 Ω. What is the current in the circuit?
- (A) 0.483 A
- (B) 0.967 A
- (C) 1.93 A
- (D) 3.87 A
Show Answer
Answer: B – I = emf/(R + r) = 0.967 A.
Q38. A resistor of 32 kΩ and a capacitor of 9 μF are connected in series. What is the time constant of this RC circuit?
- (A) 0.288 s
- (B) 0.144 s
- (C) 0.576 s
- (D) 0.072 s
Show Answer
Answer: A – τ = RC = 0.288 s.
Q39. An initially uncharged capacitor is connected in series with a 29 kΩ resistor to a battery. The RC time constant is not given directly. How long after connection will the capacitor reach 75% of its maximum charge? The capacitance is 90 μF.
- (A) 1.81 s
- (B) 3.62 s
- (C) 7.24 s
- (D) 0.905 s
Show Answer
Answer: B – τ = RC = 2.61 s, and q = Qmax(1 – e-t/τ), so t = -τ ln(1 – 0.75) = 3.62 s.
Q40. A capacitor charged to 81 V discharges through a resistor. The RC time constant is 1 s. What is the voltage across the capacitor 0.5 s after discharge begins?
- (A) 24.6 V
- (B) 98.3 V
- (C) 49.1 V
- (D) 12.3 V
Show Answer
Answer: C – V = V0e-t/τ = 49.1 V.
Q41. An uncharged capacitor is connected to a 58 V battery through a resistor. The RC time constant is 1.5 s. What is the voltage across the capacitor 2 s after connection?
- (A) 42.7 V
- (B) 21.4 V
- (C) 85.4 V
- (D) 10.7 V
Show Answer
Answer: A – V = V0(1 – e-t/τ) = 42.7 V.
Unit 12: Magnetic Fields and Electromagnetism
Q42. An electron (mass 9.11 × 10-31 kg, charge 1.60 × 10-19 C) moves perpendicular to a uniform magnetic field of 30 mT. What is the period of its circular motion?
- (A) 5.96 × 10-10 s
- (B) 1.19 × 10-9 s
- (C) 2.38 × 10-9 s
- (D) 2.98 × 10-10 s
Show Answer
Answer: B – T = 2πm/(qB) = 1.19 × 10-9 s.
Q43. Two long parallel wires 17 cm apart carry currents of 13 A and 14 A in the same direction. What is the magnitude of the magnetic force per unit length on either wire?
- (A) 0.000214 N/m
- (B) 0.000107 N/m
- (C) 0.000428 N/m
- (D) 0.000856 N/m
Show Answer
Answer: A – F/L = μ0I1I2/(2πd) = 0.000214 N/m.
Q44. A rectangular loop of area 31 cm2 has 50 turns and carries a current of 3 A in a uniform magnetic field of 47 mT. The angle between the loop’s normal and the field is 45°. What is the magnitude of the torque on the loop?
- (A) 0.00773 N·m
- (B) 0.0309 N·m
- (C) 0.00386 N·m
- (D) 0.0155 N·m
Show Answer
Answer: D – τ = NIAB sinθ = 0.0155 N·m.
Q45. A long solid cylindrical wire of radius 10 cm carries a current of 28 A uniformly distributed over its cross-section. What is the magnitude of the magnetic field inside the wire at a distance 6 cm from its axis?
- (A) 1.68 × 10-5 T
- (B) 3.36 × 10-5 T
- (C) 6.72 × 10-5 T
- (D) 0.000134 T
Show Answer
Answer: B – By Ampere’s law inside, B = μ0Ir/(2πR2) = 3.36 × 10-5 T.
Q46. A long straight wire carries a current of 16 A. What is the magnitude of the magnetic field 28 cm from the wire?
- (A) 5.71 × 10-6 T
- (B) 2.29 × 10-5 T
- (C) 4.57 × 10-5 T
- (D) 1.14 × 10-5 T
Show Answer
Answer: D – B = μ0I/(2πr) = 1.14 × 10-5 T.
Q47. A circular loop of radius 13 cm has 20 turns and carries a current of 9 A. What is the magnitude of the magnetic field at the centre of the loop?
- (A) 0.00087 T
- (B) 0.000435 T
- (C) 0.00174 T
- (D) 0.000217 T
Show Answer
Answer: A – B = μ0NI/(2R) = 0.00087 T.
Q48. A solenoid of length 80 cm has 378 turns and carries a current of 1 A. What is the magnitude of the magnetic field inside the solenoid, away from its ends?
- (A) 0.000297 T
- (B) 0.00119 T
- (C) 0.00238 T
- (D) 0.000594 T
Show Answer
Answer: D – B = μ0nI, with n = 472 turns/m, giving 0.000594 T.
Q49. A straight wire of length 32 cm carries a current of 15 A in a uniform magnetic field of 51 mT. The angle between the wire and the field is 90°. What is the magnitude of the magnetic force on the wire?
- (A) 0.122 N
- (B) 0.245 N
- (C) 0.49 N
- (D) 0.0612 N
Show Answer
Answer: B – F = ILB sinθ = 0.245 N.
Q50. A proton (mass 1.67 × 10-27 kg, charge 1.60 × 10-19 C) moves at 4 × 105 m/s perpendicular to a uniform magnetic field of 31 mT. What is the radius of its circular path?
- (A) 0.0673 m
- (B) 0.269 m
- (C) 0.135 m
- (D) 0.539 m
Show Answer
Answer: C – r = mv/(qB) = 0.135 m.
Unit 13: Electromagnetic Induction
Q51. An inductor of 122 mH carries a current of 1 A. How much energy is stored in its magnetic field?
- (A) 0.0305 J
- (B) 0.122 J
- (C) 0.244 J
- (D) 0.061 J
Show Answer
Answer: D – U = (1/2)LI2 = 0.061 J.
Q52. An inductor of 94 mH is connected in series with a resistor of 45 Ω. What is the time constant of this LR circuit?
- (A) 0.00104 s
- (B) 0.00209 s
- (C) 0.00418 s
- (D) 0.000522 s
Show Answer
Answer: B – τ = L/R = 0.00209 s.
Q53. An LR circuit with time constant 2.5 s is connected to a 15 V battery through a 26 Ω resistor. What is the current 1.5 s after the switch is closed, starting from zero current?
- (A) 0.13 A
- (B) 0.521 A
- (C) 0.0651 A
- (D) 0.26 A
Show Answer
Answer: D – I = (V/R)(1 – e-t/τ) = 0.26 A.
Q54. An inductor carrying an initial current of 1 A is disconnected from its battery and discharges through a resistor. The LR time constant is 1.5 s. What is the current 0.5 s after discharge begins?
- (A) 0.358 A
- (B) 1.43 A
- (C) 0.717 A
- (D) 2.87 A
Show Answer
Answer: C – I = I0e-t/τ = 0.717 A.
Q55. An LC circuit consists of an inductor of 125 mH and a capacitor of 70 μF. What is the frequency of oscillation of the circuit?
- (A) 26.9 Hz
- (B) 53.8 Hz
- (C) 108 Hz
- (D) 13.5 Hz
Show Answer
Answer: B – f = 1/(2π√(LC)) = 53.8 Hz.
Q56. A flat loop of area 136 cm2 is placed in a uniform magnetic field of 30 mT. The angle between the field and the normal to the loop is 60°. What is the magnetic flux through the loop?
- (A) 0.000204 Wb
- (B) 0.000102 Wb
- (C) 0.000408 Wb
- (D) 5.1 × 10-5 Wb
Show Answer
Answer: A – Φ = BA cosθ = 0.000204 Wb.
Q57. A coil of 50 turns and area 55 cm2 is perpendicular to a magnetic field that increases uniformly from 37 mT to 107 mT in 48 ms. What is the magnitude of the induced emf in the coil?
- (A) 0.201 V
- (B) 0.802 V
- (C) 0.401 V
- (D) 0.1 V
Show Answer
Answer: C – emf = N|ΔΦ/Δt| = 0.401 V.
Q58. A conducting rod of length 85 cm moves at 1 m/s perpendicular to a uniform magnetic field of 38 mT, with its length perpendicular to both the velocity and the field. What is the motional emf between its ends?
- (A) 0.0161 V
- (B) 0.0646 V
- (C) 0.0323 V
- (D) 0.129 V
Show Answer
Answer: C – emf = BLv = 0.0323 V.
Q59. An emf of 0.5 V is induced in a coil of resistance 10 Ω. What is the induced current in the loop?
- (A) 0.025 A
- (B) 0.1 A
- (C) 0.0125 A
- (D) 0.05 A
Show Answer
Answer: D – I = emf/R = 0.05 A.
Q60. A solenoid has 1005 turns, a cross-sectional area of 64 cm2 and a length of 26 cm. What is its self-inductance?
- (A) 0.0156 H
- (B) 0.0312 H
- (C) 0.0625 H
- (D) 0.00781 H
Show Answer
Answer: B – L = μ0N2A/l = 0.0312 H.
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