AP Physics C: Electricity and Magnetism Practice Test 9 – 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. Two small charged spheres carry charges of 6 μC and 8 μC and are held 0.3 m apart in a vacuum. What is the magnitude of the electrostatic force between them?
- (A) 2.4 N
- (B) 4.79 N
- (C) 9.59 N
- (D) 7.19 N
Show Answer
Answer: B – F = kq1q2/r2 = 4.79 N.
Q2. A point charge of 5 μC is fixed in space. What is the magnitude of the electric field it produces at a point 0.3 m away?
- (A) 9.99 × 105 N/C
- (B) 2.5 × 105 N/C
- (C) 4.99 × 105 N/C
- (D) 7.49 × 105 N/C
Show Answer
Answer: C – E = kQ/r2 = (8.99 × 109)(5 × 10-6)/(0.3)2 = 4.99 × 105 N/C.
Q3. A closed Gaussian surface encloses a net charge of 18 μC. What is the total electric flux through the surface?
- (A) 1.02 × 106 N·m2/C
- (B) 4.07 × 106 N·m2/C
- (C) 3.05 × 106 N·m2/C
- (D) 2.03 × 106 N·m2/C
Show Answer
Answer: D – Gauss’s law: Φ = Qenc/ε0 = 2.03 × 106 N·m2/C.
Q4. A solid conducting sphere of radius 0.1 m carries a total charge of 20 μC uniformly on its surface. What is the electric field magnitude at a point 0.6 m from its centre?
- (A) 4.99 × 105 N/C
- (B) 9.99 × 105 N/C
- (C) 2.5 × 105 N/C
- (D) 7.49 × 105 N/C
Show Answer
Answer: A – Outside a spherical charge distribution, E = kQ/r2 = (8.99 × 109)(20 × 10-6)/(0.6)2 = 4.99 × 105 N/C.
Q5. A nonconducting sphere of radius 0.5 m carries a total charge of 15 μC uniformly distributed through its volume. What is the electric field magnitude at a point 0.2 m from its centre?
- (A) 4.32 × 105 N/C
- (B) 2.16 × 105 N/C
- (C) 1.08 × 105 N/C
- (D) 3.24 × 105 N/C
Show Answer
Answer: B – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R3 = 2.16 × 105 N/C.
Q6. An infinitely long straight wire carries a uniform linear charge density of 8 μC/m. What is the electric field magnitude at a perpendicular distance of 0.1 m from the wire?
- (A) 2.88 × 106 N/C
- (B) 7.19 × 105 N/C
- (C) 1.44 × 106 N/C
- (D) 2.16 × 106 N/C
Show Answer
Answer: C – For an infinite line, Gauss’s law gives E = λ/(2πε0r) = 2kλ/r = 1.44 × 106 N/C.
Q7. An infinite nonconducting sheet carries a uniform surface charge density of 11 μC/m2. What is the electric field magnitude near the sheet, on either side?
- (A) 3.11 × 105 N/C
- (B) 1.24 × 106 N/C
- (C) 9.32 × 105 N/C
- (D) 6.21 × 105 N/C
Show Answer
Answer: D – For an infinite nonconducting sheet, E = σ/(2ε0) = 6.21 × 105 N/C.
Q8. A thin spherical shell of radius 0.15 m carries a total charge of 8 μC on its surface. What is the electric field magnitude at a point inside the shell, 0.075 m from its centre?
- (A) 0 N/C
- (B) 1.28 × 107 N/C
- (C) 3.2 × 106 N/C
- (D) 6.39 × 106 N/C
Show Answer
Answer: A – Inside a uniformly charged spherical shell the enclosed charge is zero, so by Gauss’s law E = 0.
Q9. A thin rod of length 0.4 m carries a uniform linear charge density of 8 μC/m. What is the electric field magnitude at a point on the rod’s axis, 0.1 m beyond one end of the rod?
- (A) 1.15 × 106 N/C
- (B) 5.75 × 105 N/C
- (C) 2.88 × 105 N/C
- (D) 8.63 × 105 N/C
Show Answer
Answer: B – Integrating dE = kλdx/x2 from x = 0.1 m to x = 0.5 m gives E = kλL/[a(a + L)] = 5.75 × 105 N/C.
Q10. Two point charges, 4 μC and 8 μC, are fixed 0.4 m apart on the x-axis. What is the magnitude of the net electric field at the midpoint between them?
- (A) 1.8 × 106 N/C
- (B) 4.49 × 105 N/C
- (C) 8.99 × 105 N/C
- (D) 1.35 × 106 N/C
Show Answer
Answer: C – Each field at the midpoint is kq/(d/2)2; the fields point in opposite directions, so the net magnitude is |E2 – E1| = 8.99 × 105 N/C.
Q11. Two large parallel sheets carry surface charge densities of +8 μC/m2 and -3 μC/m2. What is the electric field magnitude in the region between the sheets?
- (A) 1.24 × 106 N/C
- (B) 3.11 × 105 N/C
- (C) 9.32 × 105 N/C
- (D) 6.21 × 105 N/C
Show Answer
Answer: D – Each sheet contributes σ/(2ε0) and the two contributions add between the sheets, giving E = (σ1 + σ2)/(2ε0) = 6.21 × 105 N/C.
Q12. An electric dipole has dipole moment 3 × 10-9 C·m. What is the approximate electric field magnitude on the dipole axis at a distance of 0.25 m from its centre, far from the dipole?
- (A) 3.45×103 N/C
- (B) 1.73×103 N/C
- (C) 6.9×103 N/C
- (D) 5.18×103 N/C
Show Answer
Answer: A – On the axis, far from the dipole, E ≈ 2kp/r3 = 3.45×103 N/C.
Unit 9: Electric Potential
Q13. What is the electric potential at a point 0.2 m from a point charge of 6 μC, taking the potential to be zero at infinity?
- (A) 5.39 × 105 V
- (B) 2.7 × 105 V
- (C) 1.35 × 105 V
- (D) 4.05 × 105 V
Show Answer
Answer: B – V = kQ/r = (8.99 × 109)(6 × 10-6)/0.2 = 2.7 × 105 V.
Q14. A charge of 5 μC is moved through a potential difference of 400 V. How much work is done by the electric field if the charge moves from the higher potential to the lower potential?
- (A) 4 × 10-3 J
- (B) 1 × 10-3 J
- (C) 2 × 10-3 J
- (D) -2 × 10-3 J
Show Answer
Answer: C – Work by the field is W = qΔV = (5 × 10-6)(400) = 2 × 10-3 J.
Q15. A conducting sphere of radius 0.25 m carries a charge of 10 μC. What is the electric potential at its surface, with zero potential at infinity?
- (A) 1.8 × 105 V
- (B) 7.19 × 105 V
- (C) 5.39 × 105 V
- (D) 3.6 × 105 V
Show Answer
Answer: D – V = kQ/R = 3.6 × 105 V.
Q16. A thin rod of length 0.8 m carries a uniform linear charge density of 2 μC/m. What is the electric potential at a point on the rod’s axis, 0.1 m beyond one end?
- (A) 3.95 × 104 V
- (B) 7.9 × 104 V
- (C) 1.98 × 104 V
- (D) 5.93 × 104 V
Show Answer
Answer: A – Integrating dV = kλdx/x from x = 0.1 m to x = 0.9 m gives V = kλ ln[(L + a)/a] = 3.95 × 104 V.
Q17. A ring of radius 0.3 m carries a total charge of 10 μC uniformly around its circumference. What is the electric potential on the ring’s axis at a distance of 0.4 m from its centre?
- (A) 3.6 × 105 V
- (B) 1.8 × 105 V
- (C) 8.99 × 104 V
- (D) 2.7 × 105 V
Show Answer
Answer: B – Every charge element is at distance √(R2 + x2) from the point, so V = kQ/√(R2 + x2) = 1.8 × 105 V.
Q18. Two point charges of 4 μC and 6 μC are held 0.3 m apart. What is the electric potential energy of this pair?
- (A) 1.438 J
- (B) 0.36 J
- (C) 0.719 J
- (D) 1.079 J
Show Answer
Answer: C – U = kq1q2/r = 0.719 J.
Q19. In a region of space the electric potential varies as V(x) = 40x2 + 10, where V is in volts and x is in metres. What is the x-component of the electric field at x = 0.5 m?
- (A) 40 V/m
- (B) -80 V/m
- (C) -20 V/m
- (D) -40 V/m
Show Answer
Answer: D – Ex = -dV/dx = -80x, so at x = 0.5 m, Ex = -40 V/m.
Q20. A thin conducting spherical shell of radius 0.2 m carries a charge of 18 μC. What is the electric potential at a point inside the shell, halfway to its centre?
- (A) 8.09 × 105 V
- (B) 4.05 × 105 V
- (C) 1.62 × 106 V
- (D) 1.21 × 106 V
Show Answer
Answer: A – Inside a conducting shell the potential is constant and equal to the surface value, V = kQ/R = 8.09 × 105 V.
Q21. A uniformly charged disk of radius 0.2 m carries a surface charge density of 6 μC/m2. What is the electric potential on the disk’s axis at a distance of 0.1 m from its centre?
- (A) 8.38 × 104 V
- (B) 4.19 × 104 V
- (C) 2.09 × 104 V
- (D) 6.28 × 104 V
Show Answer
Answer: B – Integrating rings gives V = 2πkσ( √(R2 + x2) – x ) = 4.19 × 104 V.
Unit 10: Conductors and Capacitors
Q22. A parallel-plate capacitor has plates of area 0.04 m2 separated by 4 mm in air. What is its capacitance?
- (A) 1.77 × 10-10 F
- (B) 4.43 × 10-11 F
- (C) 8.85 × 10-11 F
- (D) 1.33 × 10-10 F
Show Answer
Answer: C – C = ε0A/d = (8.85 × 10-12)(0.04)/(4 × 10-3) = 8.85 × 10-11 F.
Q23. A capacitor of capacitance 20 pF is filled completely with a dielectric of dielectric constant 3, with no change in plate separation. What is the new capacitance?
- (A) 120 pF
- (B) 30 pF
- (C) 70 pF
- (D) 60 pF
Show Answer
Answer: D – C = κC0 = 3 × 20 pF = 60 pF.
Q24. A 30 pF capacitor is charged to 30 V. How much energy is stored in the capacitor?
- (A) 1.35 × 10-8 J
- (B) 6.75 × 10-9 J
- (C) 2.7 × 10-8 J
- (D) 2.02 × 10-8 J
Show Answer
Answer: A – U = (1/2)CV2 = 0.5 × (30 × 10-12) × (30)2 = 1.35 × 10-8 J.
Q25. Capacitors of 12 μF and 3 μF are connected in series. What is their equivalent capacitance?
- (A) 4.8 μF
- (B) 2.4 μF
- (C) 1.2 μF
- (D) 3.6 μF
Show Answer
Answer: B – 1/C = 1/12 + 1/3, so C = 2.4 μF.
Q26. Capacitors of 6 μF and 6 μF are connected in parallel across a 24 V battery. What total charge is drawn from the battery?
- (A) 5.76 × 10-4 C
- (B) 1.44 × 10-4 C
- (C) 2.88 × 10-4 C
- (D) 4.32 × 10-4 C
Show Answer
Answer: C – Ceq = 12 μF, so Q = CeqV = 2.88 × 10-4 C.
Q27. The plates of a parallel-plate capacitor carry surface charge densities of magnitude 3 μC/m2. What is the electric field magnitude between the plates?
- (A) 1.69 × 105 N/C
- (B) 6.78 × 105 N/C
- (C) 5.08 × 105 N/C
- (D) 3.39 × 105 N/C
Show Answer
Answer: D – Between the plates, E = σ/ε0 = 3.39 × 105 N/C.
Q28. A spherical capacitor consists of concentric conducting spheres of radii 0.15 m and 0.25 m. What is its capacitance?
- (A) 4.17 × 10-11 F
- (B) 2.09 × 10-11 F
- (C) 8.34 × 10-11 F
- (D) 6.26 × 10-11 F
Show Answer
Answer: A – C = 4πε0ab/(b – a) = 4.17 × 10-11 F.
Q29. The electric field in a region of space has magnitude 5000 N/C. What is the electric energy density in that region?
- (A) 5.53 × 10-5 J/m3
- (B) 1.11 × 10-4 J/m3
- (C) 2.21 × 10-4 J/m3
- (D) 1.66 × 10-4 J/m3
Show Answer
Answer: B – u = (1/2)ε0E2 = 1.11 × 10-4 J/m3.
Unit 11: Electric Circuits
Q30. A 24 V battery is connected across a 8 Ω resistor. What current flows through the resistor?
- (A) 6 A
- (B) 1.5 A
- (C) 3 A
- (D) 4.5 A
Show Answer
Answer: C – I = V/R = 24/8 = 3 A.
Q31. Resistors of 40 Ω and 60 Ω are connected in parallel. What is their equivalent resistance?
- (A) 12 Ω
- (B) 48 Ω
- (C) 36 Ω
- (D) 24 Ω
Show Answer
Answer: D – 1/Req = 1/R1 + 1/R2, so Req = 24 Ω.
Q32. A resistor of 24 Ω is connected across a 24 V source. What power is dissipated in the resistor?
- (A) 24 W
- (B) 48 W
- (C) 12 W
- (D) 36 W
Show Answer
Answer: A – P = V2/R = (24)2/24 = 24 W.
Q33. A battery of emf 12 V and internal resistance 2 Ω is connected to an external resistor of 4 Ω. What is the terminal voltage of the battery?
- (A) 16 V
- (B) 8 V
- (C) 4 V
- (D) 12 V
Show Answer
Answer: B – Current I = emf/(R + r) = 2 A; terminal voltage V = IR = 8 V.
Q34. An RC circuit has resistance 4700 Ω and capacitance 10 μF. What is its time constant?
- (A) 0.094 s
- (B) 0.023 s
- (C) 0.047 s
- (D) 0.07 s
Show Answer
Answer: C – τ = RC = (4700)(10 × 10-6) = 0.047 s.
Q35. A capacitor of 10 μF, initially uncharged, is connected in series with a 4000 Ω resistor across a 20 V battery. What is the voltage across the capacitor after one time constant has elapsed?
- (A) 25.285 V
- (B) 6.321 V
- (C) 18.964 V
- (D) 12.642 V
Show Answer
Answer: D – After one time constant, V = V0(1 – e-1) = 0.632 × 20 = 12.642 V.
Q36. A capacitor charged to 14 V discharges through a 2500 Ω resistor. Its capacitance is 50 μF. What is the voltage across the capacitor after two time constants?
- (A) 1.89 V
- (B) 0.947 V
- (C) 3.79 V
- (D) 2.84 V
Show Answer
Answer: A – τ = RC = 0.125 s; V = V0e-t/τ = 14e-2 = 1.89 V.
Q37. A capacitor discharges through a 500 Ω resistor. Its capacitance is 25 μF. How long does it take for the charge on the capacitor to fall to 0.25 of its initial value?
- (A) 0.035 s
- (B) 0.017 s
- (C) 8.66 × 10-3 s
- (D) 0.026 s
Show Answer
Answer: B – q = Q0e-t/RC; setting q/Q0 = 0.25 gives t = -RC ln(0.25) = 0.017 s.
Q38. A battery of emf 24 V is connected in series with resistors of 6 Ω and 12 Ω. What is the voltage across the 12 Ω resistor?
- (A) 32 V
- (B) 8 V
- (C) 16 V
- (D) 24 V
Show Answer
Answer: C – Series current I = 24/(6 + 12) = 1.333 A; V = IR2 = 16 V.
Q39. A current of 5 A flows through a resistor of 8 Ω. What power is dissipated as heat in the resistor?
- (A) 100 W
- (B) 400 W
- (C) 300 W
- (D) 200 W
Show Answer
Answer: D – P = I2R = (5)2(8) = 200 W.
Q40. A total current of 12 A splits between two parallel resistors of 4 Ω and 12 Ω. What current flows through the 4 Ω resistor?
- (A) 9 A
- (B) 18 A
- (C) 4.5 A
- (D) 13.5 A
Show Answer
Answer: A – By current division, I1 = ItotalR2/(R1 + R2) = 9 A.
Q41. A series RC circuit with resistance 6 Ω and capacitance 10 μF is connected to a 24 V battery and left for a very long time. What is the final voltage across the capacitor?
- (A) 12 V
- (B) 24 V
- (C) 6 V
- (D) 48 V
Show Answer
Answer: B – After a long time the charging current falls to zero, so the full battery voltage appears across the capacitor: 24 V.
Unit 12: Magnetic Fields and Electromagnetism
Q42. A particle of charge 4 μC moves at 4000 m/s perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the magnetic force on the particle?
- (A) 0.013 N
- (B) 3.2 × 10-3 N
- (C) 6.4 × 10-3 N
- (D) 9.6 × 10-3 N
Show Answer
Answer: C – F = qvB = (4 × 10-6)(4000)(0.4) = 6.4 × 10-3 N.
Q43. A straight wire of length 0.6 m carries a current of 8 A perpendicular to a uniform magnetic field of 0.6 T. What is the magnitude of the magnetic force on the wire?
- (A) 1.44 N
- (B) 5.76 N
- (C) 4.32 N
- (D) 2.88 N
Show Answer
Answer: D – F = ILB = (8)(0.6)(0.6) = 2.88 N.
Q44. A long straight wire carries a current of 10 A. What is the magnetic field magnitude at a distance of 0.1 m from the wire?
- (A) 2 × 10-5 T
- (B) 4 × 10-5 T
- (C) 1 × 10-5 T
- (D) 3 × 10-5 T
Show Answer
Answer: A – B = μ0I/(2πr) = 2 × 10-5 T.
Q45. A circular loop of radius 0.05 m carries a current of 4 A. What is the magnetic field magnitude at the centre of the loop?
- (A) 1.01 × 10-4 T
- (B) 5.03 × 10-5 T
- (C) 2.51 × 10-5 T
- (D) 7.54 × 10-5 T
Show Answer
Answer: B – B = μ0I/(2R) = 5.03 × 10-5 T.
Q46. A solenoid has 500 turns per metre and carries a current of 4 A. What is the magnetic field magnitude inside the solenoid, far from its ends?
- (A) 5.03 × 10-3 T
- (B) 1.26 × 10-3 T
- (C) 2.51 × 10-3 T
- (D) 3.77 × 10-3 T
Show Answer
Answer: C – B = μ0nI = (4π × 10-7)(500)(4) = 2.51 × 10-3 T.
Q47. An electron (mass 9.11 × 10-31 kg, charge magnitude 1.60 × 10-19 C) moves at 2000000 m/s perpendicular to a uniform magnetic field of 0.02 T. What is the radius of its circular path?
- (A) 1.14 × 10-3 m
- (B) 2.85 × 10-4 m
- (C) 8.54 × 10-4 m
- (D) 5.69 × 10-4 m
Show Answer
Answer: D – r = mv/(qB) = (9.11 × 10-31)(2000000)/[(1.60 × 10-19)(0.02)] = 5.69 × 10-4 m.
Q48. A rectangular coil of 250 turns and area 0.015 m2 carries a current of 2 A in a magnetic field of 0.6 T. What is the maximum torque on the coil?
- (A) 4.5 N·m
- (B) 2.25 N·m
- (C) 9 N·m
- (D) 6.75 N·m
Show Answer
Answer: A – τmax = NIAB = (250)(2)(0.015)(0.6) = 4.5 N·m.
Q49. In a velocity selector, a charged particle passes undeflected through crossed electric and magnetic fields of 600 V/m and 0.3 T. What is the speed of the particle?
- (A) 1×103 m/s
- (B) 2×103 m/s
- (C) 4×103 m/s
- (D) 3×103 m/s
Show Answer
Answer: B – For undeflected passage, qE = qvB, so v = E/B = 600/0.3 = 2×103 m/s.
Q50. Two long parallel wires 0.2 m apart carry currents of 5 A and 10 A in opposite directions. What is the magnetic field magnitude at the midpoint between the wires?
- (A) 6 × 10-5 T
- (B) 1.5 × 10-5 T
- (C) 3 × 10-5 T
- (D) 4.5 × 10-5 T
Show Answer
Answer: C – Each wire contributes μ0I/(2πr) with r = 0.1 m, and the contributions add at the midpoint, giving B = 3 × 10-5 T.
Unit 13: Electromagnetic Induction
Q51. A flat loop of area 0.02 m2 lies perpendicular to a uniform magnetic field of 0.5 T. What is the magnetic flux through the loop?
- (A) 0.02 Wb
- (B) 5 × 10-3 Wb
- (C) 0.015 Wb
- (D) 0.01 Wb
Show Answer
Answer: D – Φ = BA cosθ = (0.5)(0.02) × 1 = 0.01 Wb.
Q52. A coil of 500 turns and area 0.01 m2 experiences a magnetic field change of 0.4 T in 0.2 s, with the field perpendicular to the coil. What is the magnitude of the induced emf?
- (A) 10 V
- (B) 20 V
- (C) 5 V
- (D) 15 V
Show Answer
Answer: A – |ε| = N A ΔB/Δt = (500)(0.01)(0.4)/0.2 = 10 V.
Q53. A conducting rod of length 0.4 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.3 T, sliding on conducting rails. What is the motional emf induced between the ends of the rod?
- (A) 0.48 V
- (B) 0.24 V
- (C) 0.12 V
- (D) 0.36 V
Show Answer
Answer: B – ε = BLv = (0.3)(0.4)(2) = 0.24 V.
Q54. A loop of resistance 12 Ω experiences an induced emf of 6 V. What is the magnitude of the induced current in the loop?
- (A) 1 A
- (B) 0.25 A
- (C) 0.5 A
- (D) 0.75 A
Show Answer
Answer: C – I = ε/R = 6/12 = 0.5 A.
Q55. A solenoid has 1000 turns per metre, cross-sectional area 0.04 m2, and length 0.4 m. What is its self-inductance?
- (A) 0.04 H
- (B) 0.01 H
- (C) 0.03 H
- (D) 0.02 H
Show Answer
Answer: D – L = μ0n2Al = 0.02 H.
Q56. An LR circuit has inductance 3 H and resistance 60 Ω. What is its inductive time constant?
- (A) 0.05 s
- (B) 0.025 s
- (C) 0.1 s
- (D) 0.075 s
Show Answer
Answer: A – τ = L/R = 3/60 = 0.05 s.
Q57. An inductor of 2 H carries a current of 3 A. How much energy is stored in its magnetic field?
- (A) 18 J
- (B) 9 J
- (C) 4.5 J
- (D) 13.5 J
Show Answer
Answer: B – U = (1/2)LI2 = 0.5 × 2 × (3)2 = 9 J.
Q58. An LC circuit has inductance 30 mH and capacitance 20 μF. What is its natural frequency of oscillation?
- (A) 103 Hz
- (B) 411 Hz
- (C) 205 Hz
- (D) 308 Hz
Show Answer
Answer: C – f = 1/[2π√(LC)] = 1/[2π√((30 × 10-3)(20 × 10-6))] = 205 Hz.
Q59. A coil of 50 turns and area 0.02 m2 rotates at angular speed 50 rad/s in a uniform magnetic field of 0.2 T. What is the maximum induced emf in the coil?
- (A) 20 V
- (B) 5 V
- (C) 15 V
- (D) 10 V
Show Answer
Answer: D – εmax = NBAω = (50)(0.2)(0.02)(50) = 10 V.
Q60. In an LC circuit, the maximum current is 2 A. The inductance is 20 mH and the capacitance is 10 μF. What is the maximum voltage across the capacitor?
- (A) 89.443 V
- (B) 178.885 V
- (C) 44.721 V
- (D) 134.164 V
Show Answer
Answer: A – Energy conservation gives (1/2)LI2max = (1/2)CV2max, so Vmax = Imax√(L/C) = 89.443 V.
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