AP Physics C: Electricity and Magnetism Practice Test 10 – 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 8 μC and 9 μC and are held 0.3 m apart in a vacuum. What is the magnitude of the electrostatic force between them?
- (A) 14.384 N
- (B) 3.596 N
- (C) 7.192 N
- (D) 10.788 N
Show Answer
Answer: C – Use Coulomb’s law F = kq1q2/r2 = (8.99 × 109)(8 × 10-6)(9 × 10-6)/(0.3)2 = 7.192 N.
Q2. A point charge of 6 μC is fixed in space. What is the magnitude of the electric field it produces at a point 0.2 m away?
- (A) 6.74 × 105 N/C
- (B) 2.7 × 106 N/C
- (C) 2.02 × 106 N/C
- (D) 1.35 × 106 N/C
Show Answer
Answer: D – E = kQ/r2 = 1.35 × 106 N/C.
Q3. A closed Gaussian surface encloses a net charge of 15 μC. What is the total electric flux through the surface?
- (A) 1.69 × 106 N·m2/C
- (B) 8.47 × 105 N·m2/C
- (C) 3.39 × 106 N·m2/C
- (D) 2.54 × 106 N·m2/C
Show Answer
Answer: A – Gauss’s law: Φ = Qenc/ε0 = 1.69 × 106 N·m2/C.
Q4. A solid conducting sphere of radius 0.2 m carries a total charge of 5 μC uniformly on its surface. What is the electric field magnitude at a point 0.4 m from its centre?
- (A) 5.62 × 105 N/C
- (B) 2.81 × 105 N/C
- (C) 1.4 × 105 N/C
- (D) 4.21 × 105 N/C
Show Answer
Answer: B – Outside a spherical charge distribution, E = kQ/r2 = (8.99 × 109)(5 × 10-6)/(0.4)2 = 2.81 × 105 N/C.
Q5. A nonconducting sphere of radius 0.4 m carries a total charge of 6 μC uniformly distributed through its volume. What is the electric field magnitude at a point 0.1 m from its centre?
- (A) 1.69 × 105 N/C
- (B) 4.21 × 104 N/C
- (C) 8.43 × 104 N/C
- (D) 1.26 × 105 N/C
Show Answer
Answer: C – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R3 = 8.43 × 104 N/C.
Q6. An infinitely long straight wire carries a uniform linear charge density of 4 μC/m. What is the electric field magnitude at a perpendicular distance of 0.2 m from the wire?
- (A) 7.19 × 105 N/C
- (B) 1.8 × 105 N/C
- (C) 5.39 × 105 N/C
- (D) 3.6 × 105 N/C
Show Answer
Answer: D – For an infinite line, Gauss’s law gives E = λ/(2πε0r) = 2kλ/r = 3.6 × 105 N/C.
Q7. An infinite nonconducting sheet carries a uniform surface charge density of 8 μC/m2. What is the electric field magnitude near the sheet, on either side?
- (A) 4.52 × 105 N/C
- (B) 9.04 × 105 N/C
- (C) 2.26 × 105 N/C
- (D) 6.78 × 105 N/C
Show Answer
Answer: A – For an infinite sheet, E = σ/(2ε0) = (8 × 10-6)/(2 × 8.85 × 10-12) = 4.52 × 105 N/C.
Q8. A thin spherical shell of radius 0.15 m carries a total charge of 6 μC on its surface. What is the electric field magnitude at a point inside the shell, 0.075 m from its centre?
- (A) 9.59 × 106 N/C
- (B) 0 N/C
- (C) 2.4 × 106 N/C
- (D) 4.79 × 106 N/C
Show Answer
Answer: B – Inside a uniformly charged spherical shell the enclosed charge is zero, so by Gauss’s law E = 0.
Q9. A thin rod of length 0.5 m carries a uniform linear charge density of 3 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) 4.5 × 105 N/C
- (B) 1.12 × 105 N/C
- (C) 2.25 × 105 N/C
- (D) 3.37 × 105 N/C
Show Answer
Answer: C – Integrating dE = kλdx/x2 from x = 0.1 m to x = 0.6 m gives E = kλL/[a(a + L)] = 2.25 × 105 N/C.
Q10. Two point charges, 6 μC and 3 μC, are fixed 0.6 m apart on the x-axis. What is the magnitude of the net electric field at the midpoint between them?
- (A) 5.99 × 105 N/C
- (B) 1.5 × 105 N/C
- (C) 4.5 × 105 N/C
- (D) 3 × 105 N/C
Show Answer
Answer: D – Each field at the midpoint is kq/(d/2)2; the fields point in opposite directions, so the net magnitude is |E2 – E1| = 3 × 105 N/C.
Q11. Two large parallel sheets carry surface charge densities of +10 μC/m2 and -3 μC/m2. What is the electric field magnitude in the region between the sheets?
- (A) 7.34 × 105 N/C
- (B) 3.67 × 105 N/C
- (C) 1.47 × 106 N/C
- (D) 1.1 × 106 N/C
Show Answer
Answer: A – Between the sheets the fields add: E = (σ1 + |σ2|)/(2ε0) = 7.34 × 105 N/C.
Q12. An electric dipole has dipole moment 5 × 10-9 C·m. What is the approximate electric field magnitude on the dipole axis at a distance of 0.3 m from its centre, far from the dipole?
- (A) 6659.259 N/C
- (B) 3329.63 N/C
- (C) 1664.815 N/C
- (D) 4994.444 N/C
Show Answer
Answer: B – On the axis, far from the dipole, E ≈ 2kp/r3 = 3329.63 N/C.
Unit 9: Electric Potential
Q13. What is the electric potential at a point 0.4 m from a point charge of 4 μC, taking the potential to be zero at infinity?
- (A) 1.8 × 105 V
- (B) 4.5 × 104 V
- (C) 8.99 × 104 V
- (D) 1.35 × 105 V
Show Answer
Answer: C – V = kQ/r = (8.99 × 109)(4 × 10-6)/0.4 = 8.99 × 104 V.
Q14. A charge of 4 μC is moved through a potential difference of 100 V. How much work is done by the electric field if the charge moves from the higher potential to the lower potential?
- (A) 8 × 10-4 J
- (B) 2 × 10-4 J
- (C) -4 × 10-4 J
- (D) 4 × 10-4 J
Show Answer
Answer: D – Work by the field is W = qΔV = (4 × 10-6)(100) = 4 × 10-4 J.
Q15. A conducting sphere of radius 0.1 m carries a charge of 16 μC. What is the electric potential at its surface, with zero potential at infinity?
- (A) 1.44 × 106 V
- (B) 2.88 × 106 V
- (C) 7.19 × 105 V
- (D) 2.16 × 106 V
Show Answer
Answer: A – Outside and at the surface, V = kQ/R = 1.44 × 106 V.
Q16. A thin rod of length 0.8 m carries a uniform linear charge density of 3 μC/m. What is the electric potential at a point on the rod’s axis, 0.2 m beyond one end?
- (A) 8.68 × 104 V
- (B) 4.34 × 104 V
- (C) 2.17 × 104 V
- (D) 6.51 × 104 V
Show Answer
Answer: B – Integrating dV = kλdx/x from x = 0.2 m to x = 1 m gives V = kλ ln[(L + a)/a] = 4.34 × 104 V.
Q17. A ring of radius 0.4 m carries a total charge of 12 μC uniformly around its circumference. What is the electric potential on the ring’s axis at a distance of 0.3 m from its centre?
- (A) 1.08 × 105 V
- (B) 4.32 × 105 V
- (C) 2.16 × 105 V
- (D) 3.24 × 105 V
Show Answer
Answer: C – Every charge element is at distance √(R2 + x2) from the point, so V = kQ/√(R2 + x2) = 2.16 × 105 V.
Q18. Two point charges of 5 μC and 4 μC are held 0.5 m apart. What is the electric potential energy of this pair?
- (A) 0.18 J
- (B) 0.719 J
- (C) 0.539 J
- (D) 0.36 J
Show Answer
Answer: D – U = kq1q2/r = 0.36 J.
Q19. In a region of space the electric potential varies as V(x) = 60x2 + 5, where V is in volts and x is in metres. What is the x-component of the electric field at x = 0.4 m?
- (A) -48 V/m
- (B) -24 V/m
- (C) -96 V/m
- (D) -72 V/m
Show Answer
Answer: A – Ex = -dV/dx = -120x, so at x = 0.4 m, Ex = -48 V/m.
Q20. A thin conducting spherical shell of radius 0.1 m carries a charge of 9 μC. What is the electric potential at a point inside the shell, halfway to its centre?
- (A) 1.62 × 106 V
- (B) 8.09 × 105 V
- (C) 4.05 × 105 V
- (D) 1.21 × 106 V
Show Answer
Answer: B – 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 0.3 m carries a surface charge density of 5 μC/m2. What is the electric potential on the disk’s axis at a distance of 0.1 m from its centre?
- (A) 1.22 × 105 V
- (B) 3.05 × 104 V
- (C) 6.11 × 104 V
- (D) 9.16 × 104 V
Show Answer
Answer: C – Integrating rings gives V = 2πkσ( √(R2 + x2) – x ) = 6.11 × 104 V.
Unit 10: Conductors and Capacitors
Q22. A parallel-plate capacitor has plates of area 0.05 m2 separated by 5 mm in air. What is its capacitance?
- (A) 1.77 × 10-10 F
- (B) 4.43 × 10-11 F
- (C) 1.33 × 10-10 F
- (D) 8.85 × 10-11 F
Show Answer
Answer: D – C = ε0A/d = (8.85 × 10-12)(0.05)/(5 × 10-3) = 8.85 × 10-11 F.
Q23. A capacitor of capacitance 10 pF is filled completely with a dielectric of dielectric constant 2, with no change in plate separation. What is the new capacitance?
- (A) 20 pF
- (B) 40 pF
- (C) 10 pF
- (D) 30 pF
Show Answer
Answer: A – C = κC0 = 2 × 10 pF = 20 pF.
Q24. A 40 pF capacitor is charged to 100 V. How much energy is stored in the capacitor?
- (A) 4 × 10-7 J
- (B) 2 × 10-7 J
- (C) 1 × 10-7 J
- (D) 3 × 10-7 J
Show Answer
Answer: B – U = (1/2)CV2 = 0.5 × (40 × 10-12) × (100)2 = 2 × 10-7 J.
Q25. Capacitors of 12 μF and 4 μF are connected in series. What is their equivalent capacitance?
- (A) 1.5×106 μF
- (B) 6×106 μF
- (C) 3×106 μF
- (D) 4.5×106 μF
Show Answer
Answer: C – 1/Ceq = 1/C1 + 1/C2, so Ceq = 3×106 μF.
Q26. Capacitors of 6 μF and 8 μF are connected in parallel across a 24 V battery. What total charge is drawn from the battery?
- (A) 6.72 × 10-4 C
- (B) 1.68 × 10-4 C
- (C) 5.04 × 10-4 C
- (D) 3.36 × 10-4 C
Show Answer
Answer: D – Ceq = 14 μF, so Q = CeqV = 3.36 × 10-4 C.
Q27. The plates of a parallel-plate capacitor carry surface charge densities of magnitude 8 μC/m2. What is the electric field magnitude between the plates?
- (A) 9.04 × 105 N/C
- (B) 4.52 × 105 N/C
- (C) 1.81 × 106 N/C
- (D) 1.36 × 106 N/C
Show Answer
Answer: A – Between the plates, E = σ/ε0 = 9.04 × 105 N/C.
Q28. A spherical capacitor consists of concentric conducting spheres of radii 0.05 m and 0.1 m. What is its capacitance?
- (A) 2.22 × 10-11 F
- (B) 1.11 × 10-11 F
- (C) 5.56 × 10-12 F
- (D) 1.67 × 10-11 F
Show Answer
Answer: B – C = 4πε0ab/(b – a) = 1.11 × 10-11 F.
Q29. The electric field in a region of space has magnitude 8000 N/C. What is the electric energy density in that region?
- (A) 1.42 × 10-4 J/m3
- (B) 5.66 × 10-4 J/m3
- (C) 2.83 × 10-4 J/m3
- (D) 4.25 × 10-4 J/m3
Show Answer
Answer: C – u = (1/2)ε0E2 = 2.83 × 10-4 J/m3.
Unit 11: Electric Circuits
Q30. A 24 V battery is connected across a 6 Ω resistor. What current flows through the resistor?
- (A) 8 A
- (B) 2 A
- (C) 6 A
- (D) 4 A
Show Answer
Answer: D – I = V/R = 24/6 = 4 A.
Q31. Resistors of 15 Ω and 30 Ω are connected in parallel. What is their equivalent resistance?
- (A) 10 Ω
- (B) 5 Ω
- (C) 20 Ω
- (D) 15 Ω
Show Answer
Answer: A – 1/Req = 1/R1 + 1/R2, so Req = 10 Ω.
Q32. A resistor of 18 Ω is connected across a 36 V source. What power is dissipated in the resistor?
- (A) 36 W
- (B) 72 W
- (C) 144 W
- (D) 108 W
Show Answer
Answer: B – P = V2/R = (36)2/18 = 72 W.
Q33. A battery of emf 24 V and internal resistance 2 Ω is connected to an external resistor of 4 Ω. What is the terminal voltage of the battery?
- (A) 32 V
- (B) 8 V
- (C) 16 V
- (D) 24 V
Show Answer
Answer: C – Current I = emf/(R + r) = 4 A; terminal voltage V = IR = 16 V.
Q34. An RC circuit has resistance 3000 Ω and capacitance 40 μF. What is its time constant?
- (A) 0.06 s
- (B) 0.24 s
- (C) 0.18 s
- (D) 0.12 s
Show Answer
Answer: D – τ = RC = (3000)(40 × 10-6) = 0.12 s.
Q35. A capacitor of 20 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) 12.642 V
- (B) 25.285 V
- (C) 6.321 V
- (D) 18.964 V
Show Answer
Answer: A – After one time constant, V = V0(1 – e-1) = 0.632 × 20 = 12.642 V.
Q36. A capacitor charged to 15 V discharges through a 3000 Ω resistor. Its capacitance is 30 μF. What is the voltage across the capacitor after two time constants?
- (A) 1.02 V
- (B) 2.03 V
- (C) 4.06 V
- (D) 3.05 V
Show Answer
Answer: B – τ = RC = 0.09 s; V = V0e-t/τ = 15e-2 = 2.03 V.
Q37. A capacitor discharges through a 4000 Ω resistor. Its capacitance is 20 μF. How long does it take for the charge on the capacitor to fall to 0.25 of its initial value?
- (A) 0.0555 s
- (B) 0.222 s
- (C) 0.111 s
- (D) 0.166 s
Show Answer
Answer: C – q = Q0e-t/RC; setting q/Q0 = 0.25 gives t = -RC ln(0.25) = 0.111 s.
Q38. A battery of emf 12 V is connected in series with resistors of 6 Ω and 12 Ω. What is the voltage across the 12 Ω resistor?
- (A) 16 V
- (B) 4 V
- (C) 12 V
- (D) 8 V
Show Answer
Answer: D – Series current I = 12/(6 + 12) = 0.667 A; V = IR2 = 8 V.
Q39. A current of 2 A flows through a resistor of 15 Ω. What power is dissipated as heat in the resistor?
- (A) 60 W
- (B) 120 W
- (C) 30 W
- (D) 90 W
Show Answer
Answer: A – P = I2R = (2)2(15) = 60 W.
Q40. A total current of 6 A splits between two parallel resistors of 6 Ω and 18 Ω. What current flows through the 6 Ω resistor?
- (A) 9 A
- (B) 4.5 A
- (C) 2.25 A
- (D) 6.75 A
Show Answer
Answer: B – By current division, I1 = ItotalR2/(R1 + R2) = 4.5 A.
Q41. A series RC circuit with resistance 12 Ω and capacitance 20 μF is connected to a 18 V battery and left for a very long time. What is the final voltage across the capacitor?
- (A) 9 V
- (B) 36 V
- (C) 18 V
- (D) 27 V
Show Answer
Answer: C – After a long time the charging current falls to zero, so the full battery voltage appears across the capacitor: 18 V.
Unit 12: Magnetic Fields and Electromagnetism
Q42. A particle of charge 4 μC moves at 3000 m/s perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the particle?
- (A) 4.8 × 10-3 N
- (B) 1.2 × 10-3 N
- (C) 3.6 × 10-3 N
- (D) 2.4 × 10-3 N
Show Answer
Answer: D – F = qvB = (4 × 10-6)(3000)(0.2) = 2.4 × 10-3 N.
Q43. A straight wire of length 0.2 m carries a current of 2 A perpendicular to a uniform magnetic field of 0.8 T. What is the magnitude of the magnetic force on the wire?
- (A) 0.32 N
- (B) 0.64 N
- (C) 0.16 N
- (D) 0.48 N
Show Answer
Answer: A – F = ILB = (2)(0.2)(0.8) = 0.32 N.
Q44. A long straight wire carries a current of 6 A. What is the magnetic field magnitude at a distance of 0.04 m from the wire?
- (A) 1.5 × 10-5 T
- (B) 3 × 10-5 T
- (C) 6 × 10-5 T
- (D) 4.5 × 10-5 T
Show Answer
Answer: B – B = μ0I/(2πr) = 3 × 10-5 T.
Q45. A circular loop of radius 0.06 m carries a current of 5 A. What is the magnetic field magnitude at the centre of the loop?
- (A) 2.62 × 10-5 T
- (B) 1.05 × 10-4 T
- (C) 5.24 × 10-5 T
- (D) 7.85 × 10-5 T
Show Answer
Answer: C – At the centre of a loop, B = μ0I/(2R) = 5.24 × 10-5 T.
Q46. A solenoid has 1000 turns per metre and carries a current of 3 A. What is the magnetic field magnitude inside the solenoid, far from its ends?
- (A) 7.54 × 10-3 T
- (B) 1.88 × 10-3 T
- (C) 5.65 × 10-3 T
- (D) 3.77 × 10-3 T
Show Answer
Answer: D – B = μ0nI = (4π × 10-7)(1000)(3) = 3.77 × 10-3 T.
Q47. An electron (mass 9.11 × 10-31 kg, charge magnitude 1.60 × 10-19 C) moves at 1000000 m/s perpendicular to a uniform magnetic field of 0.01 T. What is the radius of its circular path?
- (A) 5.69 × 10-4 m
- (B) 1.14 × 10-3 m
- (C) 2.85 × 10-4 m
- (D) 8.54 × 10-4 m
Show Answer
Answer: A – r = mv/(qB) = (9.11 × 10-31)(1000000)/[(1.60 × 10-19)(0.01)] = 5.69 × 10-4 m.
Q48. A rectangular coil of 200 turns and area 0.02 m2 carries a current of 2 A in a magnetic field of 0.4 T. What is the maximum torque on the coil?
- (A) 6.4 N·m
- (B) 3.2 N·m
- (C) 1.6 N·m
- (D) 4.8 N·m
Show Answer
Answer: B – τmax = NIAB = (200)(2)(0.02)(0.4) = 3.2 N·m.
Q49. In a velocity selector, a charged particle passes undeflected through crossed electric and magnetic fields of 200 V/m and 0.5 T. What is the speed of the particle?
- (A) 800 m/s
- (B) 200 m/s
- (C) 400 m/s
- (D) 600 m/s
Show Answer
Answer: C – Undeflected motion requires qE = qvB, so v = E/B = 200/0.5 = 400 m/s.
Q50. Two long parallel wires 0.3 m apart carry currents of 6 A and 12 A in opposite directions. What is the magnetic field magnitude at the midpoint between the wires?
- (A) 1.2 × 10-5 T
- (B) 4.8 × 10-5 T
- (C) 3.6 × 10-5 T
- (D) 2.4 × 10-5 T
Show Answer
Answer: D – Each wire contributes μ0I/(2πr) with r = 0.15 m, and the contributions add at the midpoint, giving B = 2.4 × 10-5 T.
Unit 13: Electromagnetic Induction
Q51. A flat loop of area 0.02 m2 lies perpendicular to a uniform magnetic field of 0.4 T. What is the magnetic flux through the loop?
- (A) 8 × 10-3 Wb
- (B) 0.016 Wb
- (C) 4 × 10-3 Wb
- (D) 0.012 Wb
Show Answer
Answer: A – Φ = BA cosθ = (0.4)(0.02) × 1 = 8 × 10-3 Wb.
Q52. A coil of 100 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) 4 V
- (B) 2 V
- (C) 1 V
- (D) 3 V
Show Answer
Answer: B – |ε| = N A ΔB/Δt = (100)(0.01)(0.4)/0.2 = 2 V.
Q53. A conducting rod of length 0.5 m moves at 3 m/s perpendicular to a uniform magnetic field of 1.0 T, sliding on conducting rails. What is the motional emf induced between the ends of the rod?
- (A) 0.75 V
- (B) 3 V
- (C) 1.5 V
- (D) 2.25 V
Show Answer
Answer: C – ε = BLv = (1.0)(0.5)(3) = 1.5 V.
Q54. A loop of resistance 8 Ω experiences an induced emf of 16 V. What is the magnitude of the induced current in the loop?
- (A) 1 A
- (B) 4 A
- (C) 3 A
- (D) 2 A
Show Answer
Answer: D – I = ε/R = 16/8 = 2 A.
Q55. A solenoid has 1000 turns per metre, cross-sectional area 0.04 m2, and length 0.2 m. What is its self-inductance?
- (A) 0.01 H
- (B) 0.02 H
- (C) 5.03 × 10-3 H
- (D) 0.015 H
Show Answer
Answer: A – L = μ0n2Al = 0.01 H.
Q56. An LR circuit has inductance 10 H and resistance 20 Ω. What is its inductive time constant?
- (A) 1 s
- (B) 0.5 s
- (C) 0.25 s
- (D) 0.75 s
Show Answer
Answer: B – τ = L/R = 10/20 = 0.5 s.
Q57. An inductor of 0.5 H carries a current of 2 A. How much energy is stored in its magnetic field?
- (A) 2 J
- (B) 0.5 J
- (C) 1 J
- (D) 1.5 J
Show Answer
Answer: C – U = (1/2)LI2 = 0.5 × 0.5 × (2)2 = 1 J.
Q58. An LC circuit has inductance 10 mH and capacitance 10 μF. What is its natural frequency of oscillation?
- (A) 1006.584 Hz
- (B) 251.646 Hz
- (C) 754.938 Hz
- (D) 503.292 Hz
Show Answer
Answer: D – f = 1/[2π√(LC)] = 1/[2π√((10 × 10-3)(10 × 10-6))] = 503.292 Hz.
Q59. A coil of 100 turns and area 0.02 m2 rotates at angular speed 100 rad/s in a uniform magnetic field of 0.2 T. What is the maximum induced emf in the coil?
- (A) 40 V
- (B) 80 V
- (C) 20 V
- (D) 60 V
Show Answer
Answer: A – εmax = NBAω = (100)(0.2)(0.02)(100) = 40 V.
Q60. In an LC circuit, the maximum current is 2 A. The inductance is 20 mH and the capacitance is 20 μF. What is the maximum voltage across the capacitor?
- (A) 126.491 V
- (B) 63.246 V
- (C) 31.623 V
- (D) 94.868 V
Show Answer
Answer: B – Energy conservation gives (1/2)LI2max = (1/2)CV2max, so Vmax = Imax√(L/C) = 63.246 V.
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