AP Physics C: Electricity and Magnetism Practice Test 6 – 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 6 μC and are held 0.4 m apart in a vacuum. What is the magnitude of the electrostatic force between them?
- (A) 4.045 N
- (B) 1.011 N
- (C) 2.023 N
- (D) 3.034 N
Show Answer
Answer: C – Use Coulomb’s law F = kq1q2/r2 = (8.99 × 109)(6 × 10-6)(6 × 10-6)/(0.4)2 = 2.023 N.
Q2. A point charge of 12 μC is fixed in space. What is the magnitude of the electric field it produces at a point 0.1 m away?
- (A) 2.16 × 107 N/C
- (B) 5.39 × 106 N/C
- (C) 1.62 × 107 N/C
- (D) 1.08 × 107 N/C
Show Answer
Answer: D – E = kQ/r2 = (8.99 × 109)(12 × 10-6)/(0.1)2 = 1.08 × 107 N/C.
Q3. A closed Gaussian surface encloses a net charge of 12 μC. What is the total electric flux through the surface?
- (A) 1.36 × 106 N·m2/C
- (B) 2.71 × 106 N·m2/C
- (C) 6.78 × 105 N·m2/C
- (D) 2.03 × 106 N·m2/C
Show Answer
Answer: A – By Gauss’s law, Φ = Qenc/ε0 = (12 × 10-6)/(8.85 × 10-12) = 1.36 × 106 N·m2/C.
Q4. A solid conducting sphere of radius 0.2 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) 9.99 × 105 N/C
- (B) 4.99 × 105 N/C
- (C) 2.5 × 105 N/C
- (D) 7.49 × 105 N/C
Show Answer
Answer: B – 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 0.4 m carries a total charge of 12 μC uniformly distributed through its volume. What is the electric field magnitude at a point 0.1 m from its centre?
- (A) 3.37 × 105 N/C
- (B) 8.43 × 104 N/C
- (C) 1.69 × 105 N/C
- (D) 2.53 × 105 N/C
Show Answer
Answer: C – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R3 = 1.69 × 105 N/C.
Q6. An infinitely long straight wire carries a uniform linear charge density of 5 μC/m. What is the electric field magnitude at a perpendicular distance of 0.2 m from the wire?
- (A) 8.99 × 105 N/C
- (B) 2.25 × 105 N/C
- (C) 6.74 × 105 N/C
- (D) 4.49 × 105 N/C
Show Answer
Answer: D – For an infinite line, Gauss’s law gives E = λ/(2πε0r) = 2kλ/r = 4.49 × 105 N/C.
Q7. An infinite nonconducting sheet carries a uniform surface charge density of 4 μC/m2. What is the electric field magnitude near the sheet, on either side?
- (A) 2.26 × 105 N/C
- (B) 4.52 × 105 N/C
- (C) 1.13 × 105 N/C
- (D) 3.39 × 105 N/C
Show Answer
Answer: A – For an infinite sheet, E = σ/(2ε0) = (4 × 10-6)/(2 × 8.85 × 10-12) = 2.26 × 105 N/C.
Q8. A thin spherical shell of radius 0.1 m carries a total charge of 6 μC on its surface. What is the electric field magnitude at a point inside the shell, 0.05 m from its centre?
- (A) 2.16 × 107 N/C
- (B) 0 N/C
- (C) 5.39 × 106 N/C
- (D) 1.08 × 107 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, 0.2 m beyond one end of the rod?
- (A) 1.93 × 105 N/C
- (B) 4.82 × 104 N/C
- (C) 9.63 × 104 N/C
- (D) 1.44 × 105 N/C
Show Answer
Answer: C – Integrating dE = kλdx/x2 from x = 0.2 m to x = 0.7 m gives E = kλL/[a(a + L)] = 9.63 × 104 N/C.
Q10. Two point charges, 6 μC and 3 μC, are fixed 0.3 m apart on the x-axis. What is the magnitude of the net electric field at the midpoint between them?
- (A) 2.4 × 106 N/C
- (B) 5.99 × 105 N/C
- (C) 1.8 × 106 N/C
- (D) 1.2 × 106 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| = 1.2 × 106 N/C.
Q11. Two large parallel sheets carry surface charge densities of +6 μC/m2 and -3 μC/m2. What is the electric field magnitude in the region between the sheets?
- (A) 5.08 × 105 N/C
- (B) 1.02 × 106 N/C
- (C) 2.54 × 105 N/C
- (D) 7.63 × 105 N/C
Show Answer
Answer: A – Each sheet contributes σ/(2ε0) and the two contributions add between the sheets, giving E = (σ1 + σ2)/(2ε0) = 5.08 × 105 N/C.
Q12. An electric dipole has dipole moment 4 × 10-9 C·m. What is the approximate electric field magnitude on the dipole axis at a distance of 0.2 m from its centre, far from the dipole?
- (A) 1.8 × 104 N/C
- (B) 8990 N/C
- (C) 4495 N/C
- (D) 1.35 × 104 N/C
Show Answer
Answer: B – On the axis, far from the dipole, E ≈ 2kp/r3 = 8990 N/C.
Unit 9: Electric Potential
Q13. What is the electric potential at a point 0.2 m from a point charge of 8 μC, taking the potential to be zero at infinity?
- (A) 7.19 × 105 V
- (B) 1.8 × 105 V
- (C) 3.6 × 105 V
- (D) 5.39 × 105 V
Show Answer
Answer: C – V = kQ/r = (8.99 × 109)(8 × 10-6)/0.2 = 3.6 × 105 V.
Q14. A charge of 2 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) 4 × 10-4 J
- (B) 1 × 10-4 J
- (C) -2 × 10-4 J
- (D) 2 × 10-4 J
Show Answer
Answer: D – Work by the field is W = qΔV = (2 × 10-6)(100) = 2 × 10-4 J.
Q15. A conducting sphere of radius 0.2 m carries a charge of 8 μC. What is the electric potential at its surface, with zero potential at infinity?
- (A) 3.6 × 105 V
- (B) 7.19 × 105 V
- (C) 1.8 × 105 V
- (D) 5.39 × 105 V
Show Answer
Answer: A – Outside and at the surface, V = kQ/R = 3.6 × 105 V.
Q16. A thin rod of length 0.5 m carries a uniform linear charge density of 3 μC/m. What is the electric potential at a point on the rod’s axis, 0.1 m beyond one end?
- (A) 9.66 × 104 V
- (B) 4.83 × 104 V
- (C) 2.42 × 104 V
- (D) 7.25 × 104 V
Show Answer
Answer: B – Integrating dV = kλdx/x from x = 0.1 m to x = 0.6 m gives V = kλ ln[(L + a)/a] = 4.83 × 104 V.
Q17. A ring of radius 0.3 m carries a total charge of 5 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) 1.8 × 105 V
- (B) 4.49 × 104 V
- (C) 8.99 × 104 V
- (D) 1.35 × 105 V
Show Answer
Answer: C – Every charge element is at distance √(R2 + x2) from the point, so V = kQ/√(R2 + x2) = 8.99 × 104 V.
Q18. Two point charges of 3 μC and 3 μC are held 0.3 m apart. What is the electric potential energy of this pair?
- (A) 0.539 J
- (B) 0.135 J
- (C) 0.405 J
- (D) 0.27 J
Show Answer
Answer: D – U = kq1q2/r = 0.27 J.
Q19. In a region of space the electric potential varies as V(x) = 50x2 + 10, where V is in volts and x is in metres. What is the x-component of the electric field at x = 1 m?
- (A) -100 V/m
- (B) 100 V/m
- (C) -200 V/m
- (D) -50 V/m
Show Answer
Answer: A – Ex = -dV/dx = -100x, so at x = 1 m, Ex = -100 V/m.
Q20. A thin conducting spherical shell of radius 0.1 m carries a charge of 12 μC. What is the electric potential at a point inside the shell, halfway to its centre?
- (A) 2.16 × 106 V
- (B) 1.08 × 106 V
- (C) 5.39 × 105 V
- (D) 1.62 × 106 V
Show Answer
Answer: B – Inside a conducting shell the potential is constant and equal to the surface value, V = kQ/R = 1.08 × 106 V.
Q21. A uniformly charged disk of radius 0.3 m carries a surface charge density of 4 μC/m2. What is the electric potential on the disk’s axis at a distance of 0.1 m from its centre?
- (A) 9.77 × 104 V
- (B) 2.44 × 104 V
- (C) 4.89 × 104 V
- (D) 7.33 × 104 V
Show Answer
Answer: C – Integrating rings gives V = 2πkσ( √(R2 + x2) – x ) = 4.89 × 104 V.
Unit 10: Conductors and Capacitors
Q22. A parallel-plate capacitor has plates of area 0.05 m2 separated by 2 mm in air. What is its capacitance?
- (A) 4.42 × 10-10 F
- (B) 1.11 × 10-10 F
- (C) 3.32 × 10-10 F
- (D) 2.21 × 10-10 F
Show Answer
Answer: D – C = ε0A/d = (8.85 × 10-12)(0.05)/(2 × 10-3) = 2.21 × 10-10 F.
Q23. A capacitor of capacitance 50 pF is filled completely with a dielectric of dielectric constant 3, with no change in plate separation. What is the new capacitance?
- (A) 150 pF
- (B) 300 pF
- (C) 75 pF
- (D) 160 pF
Show Answer
Answer: A – C = κC0 = 3 × 50 pF = 150 pF.
Q24. A 20 pF capacitor is charged to 50 V. How much energy is stored in the capacitor?
- (A) 5 × 10-8 J
- (B) 2.5 × 10-8 J
- (C) 1.25 × 10-8 J
- (D) 3.75 × 10-8 J
Show Answer
Answer: B – U = (1/2)CV2 = 0.5 × (20 × 10-12) × (50)2 = 2.5 × 10-8 J.
Q25. Capacitors of 6 μF and 3 μF are connected in series. What is their equivalent capacitance?
- (A) 4 μF
- (B) 1 μF
- (C) 2 μF
- (D) 3 μF
Show Answer
Answer: C – 1/C = 1/6 + 1/3, so C = 2 μF.
Q26. Capacitors of 10 μF and 8 μF are connected in parallel across a 12 V battery. What total charge is drawn from the battery?
- (A) 4.32 × 10-4 C
- (B) 1.08 × 10-4 C
- (C) 3.24 × 10-4 C
- (D) 2.16 × 10-4 C
Show Answer
Answer: D – Ceq = 18 μF, so Q = CeqV = 2.16 × 10-4 C.
Q27. The plates of a parallel-plate capacitor carry surface charge densities of magnitude 2 μC/m2. What is the electric field magnitude between the plates?
- (A) 2.26 × 105 N/C
- (B) 4.52 × 105 N/C
- (C) 1.13 × 105 N/C
- (D) 3.39 × 105 N/C
Show Answer
Answer: A – Between the plates, E = σ/ε0 = 2.26 × 105 N/C.
Q28. A spherical capacitor consists of concentric conducting spheres of radii 0.1 m and 0.2 m. What is its capacitance?
- (A) 4.45 × 10-11 F
- (B) 2.22 × 10-11 F
- (C) 1.11 × 10-11 F
- (D) 3.34 × 10-11 F
Show Answer
Answer: B – C = 4πε0ab/(b – a) = 2.22 × 10-11 F.
Q29. The electric field in a region of space has magnitude 2000 N/C. What is the electric energy density in that region?
- (A) 3.54 × 10-5 J/m3
- (B) 8.85 × 10-6 J/m3
- (C) 1.77 × 10-5 J/m3
- (D) 2.65 × 10-5 J/m3
Show Answer
Answer: C – u = (1/2)ε0E2 = 0.5 × (8.85 × 10-12) × (2000)2 = 1.77 × 10-5 J/m3.
Unit 11: Electric Circuits
Q30. A 36 V battery is connected across a 4 Ω resistor. What current flows through the resistor?
- (A) 18 A
- (B) 4.5 A
- (C) 13.5 A
- (D) 9 A
Show Answer
Answer: D – I = V/R = 36/4 = 9 A.
Q31. Resistors of 20 Ω and 20 Ω are connected in parallel. What is their equivalent resistance?
- (A) 10 Ω
- (B) 20 Ω
- (C) 5 Ω
- (D) 15 Ω
Show Answer
Answer: A – 1/R = 1/20 + 1/20, so R = 10 Ω.
Q32. A resistor of 60 Ω is connected across a 120 V source. What power is dissipated in the resistor?
- (A) 480 W
- (B) 240 W
- (C) 120 W
- (D) 360 W
Show Answer
Answer: B – P = V2/R = (120)2/60 = 240 W.
Q33. A battery of emf 12 V and internal resistance 2 Ω is connected to an external resistor of 6 Ω. What is the terminal voltage of the battery?
- (A) 18 V
- (B) 4.5 V
- (C) 9 V
- (D) 13.5 V
Show Answer
Answer: C – Current I = emf/(R + r) = 1.5 A; terminal voltage V = IR = 9 V.
Q34. An RC circuit has resistance 10000 Ω and capacitance 20 μF. What is its time constant?
- (A) 0.4 s
- (B) 0.1 s
- (C) 0.3 s
- (D) 0.2 s
Show Answer
Answer: D – τ = RC = (10000)(20 × 10-6) = 0.2 s.
Q35. A capacitor of 20 μF, initially uncharged, is connected in series with a 2000 Ω resistor across a 24 V battery. What is the voltage across the capacitor after one time constant has elapsed?
- (A) 15.171 V
- (B) 30.342 V
- (C) 7.585 V
- (D) 22.756 V
Show Answer
Answer: A – After one time constant, V = V0(1 – e-1) = 0.632 × 24 = 15.171 V.
Q36. A capacitor charged to 10 V discharges through a 2000 Ω resistor. Its capacitance is 40 μF. What is the voltage across the capacitor after two time constants?
- (A) 2.707 V
- (B) 1.353 V
- (C) 0.677 V
- (D) 2.03 V
Show Answer
Answer: B – τ = RC = 0.08 s; V = V0e-t/τ = 10e-2 = 1.353 V.
Q37. A capacitor discharges through a 500 Ω resistor. Its capacitance is 50 μF. How long does it take for the charge on the capacitor to fall to 0.5 of its initial value?
- (A) 0.035 s
- (B) 8.66 × 10-3 s
- (C) 0.017 s
- (D) 0.026 s
Show Answer
Answer: C – q = Q0e-t/RC; setting q/Q0 = 0.5 gives t = -RC ln(0.5) = 0.017 s.
Q38. A battery of emf 12 V is connected in series with resistors of 12 Ω and 12 Ω. What is the voltage across the 12 Ω resistor?
- (A) 12 V
- (B) 3 V
- (C) 9 V
- (D) 6 V
Show Answer
Answer: D – Series current I = 12/(12 + 12) = 0.5 A; V = IR2 = 6 V.
Q39. A current of 3 A flows through a resistor of 15 Ω. What power is dissipated as heat in the resistor?
- (A) 135 W
- (B) 270 W
- (C) 67.5 W
- (D) 202.5 W
Show Answer
Answer: A – P = I2R = (3)2(15) = 135 W.
Q40. A total current of 9 A splits between two parallel resistors of 4 Ω and 18 Ω. What current flows through the 4 Ω resistor?
- (A) 14.727 A
- (B) 7.364 A
- (C) 3.682 A
- (D) 11.045 A
Show Answer
Answer: B – By current division, I1 = ItotalR2/(R1 + R2) = 7.364 A.
Q41. A series RC circuit with resistance 8 Ω and capacitance 10 μF is connected to a 12 V battery and left for a very long time. What is the final voltage across the capacitor?
- (A) 6 V
- (B) 3 V
- (C) 12 V
- (D) 24 V
Show Answer
Answer: C – After a long time the charging current falls to zero, so the full battery voltage appears across the capacitor: 12 V.
Unit 12: Magnetic Fields and Electromagnetism
Q42. A particle of charge 4 μC moves at 2000 m/s perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the magnetic force on the particle?
- (A) 6.4 × 10-3 N
- (B) 1.6 × 10-3 N
- (C) 4.8 × 10-3 N
- (D) 3.2 × 10-3 N
Show Answer
Answer: D – F = qvB = (4 × 10-6)(2000)(0.4) = 3.2 × 10-3 N.
Q43. A straight wire of length 0.4 m carries a current of 5 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the wire?
- (A) 1 N
- (B) 2 N
- (C) 0.5 N
- (D) 1.5 N
Show Answer
Answer: A – F = ILB = (5)(0.4)(0.5) = 1 N.
Q44. A long straight wire carries a current of 5 A. What is the magnetic field magnitude at a distance of 0.1 m from the wire?
- (A) 2 × 10-5 T
- (B) 1 × 10-5 T
- (C) 5 × 10-6 T
- (D) 1.5 × 10-5 T
Show Answer
Answer: B – B = μ0I/(2πr) = 1 × 10-5 T.
Q45. A circular loop of radius 0.05 m carries a current of 10 A. What is the magnetic field magnitude at the centre of the loop?
- (A) 2.51 × 10-4 T
- (B) 6.28 × 10-5 T
- (C) 1.26 × 10-4 T
- (D) 1.88 × 10-4 T
Show Answer
Answer: C – B = μ0I/(2R) = 1.26 × 10-4 T.
Q46. A solenoid has 500 turns per metre and carries a current of 2 A. What is the magnetic field magnitude inside the solenoid, far from its ends?
- (A) 2.51 × 10-3 T
- (B) 6.28 × 10-4 T
- (C) 1.88 × 10-3 T
- (D) 1.26 × 10-3 T
Show Answer
Answer: D – B = μ0nI = (4π × 10-7)(500)(2) = 1.26 × 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 0.01 T. What is the radius of its circular path?
- (A) 1.14 × 10-3 m
- (B) 2.28 × 10-3 m
- (C) 5.69 × 10-4 m
- (D) 1.71 × 10-3 m
Show Answer
Answer: A – r = mv/(qB) = (9.11 × 10-31)(2000000)/[(1.60 × 10-19)(0.01)] = 1.14 × 10-3 m.
Q48. A rectangular coil of 200 turns and area 0.02 m2 carries a current of 2 A in a magnetic field of 0.5 T. What is the maximum torque on the coil?
- (A) 8 N·m
- (B) 4 N·m
- (C) 2 N·m
- (D) 6 N·m
Show Answer
Answer: B – τmax = NIAB = (200)(2)(0.02)(0.5) = 4 N·m.
Q49. In a velocity selector, a charged particle passes undeflected through crossed electric and magnetic fields of 400 V/m and 0.5 T. What is the speed of the particle?
- (A) 1600 m/s
- (B) 400 m/s
- (C) 800 m/s
- (D) 1200 m/s
Show Answer
Answer: C – Undeflected motion requires qE = qvB, so v = E/B = 400/0.5 = 800 m/s.
Q50. Two long parallel wires 0.1 m apart carry currents of 10 A and 5 A in opposite directions. What is the magnetic field magnitude at the midpoint between the wires?
- (A) 1.2 × 10-4 T
- (B) 3 × 10-5 T
- (C) 9 × 10-5 T
- (D) 6 × 10-5 T
Show Answer
Answer: D – Each wire contributes μ0I/(2πr) with r = 0.05 m, and the contributions add at the midpoint, giving B = 6 × 10-5 T.
Unit 13: Electromagnetic Induction
Q51. A flat loop of area 0.05 m2 lies perpendicular to a uniform magnetic field of 0.4 T. What is the magnetic flux through the loop?
- (A) 0.02 Wb
- (B) 0.04 Wb
- (C) 0.01 Wb
- (D) 0.03 Wb
Show Answer
Answer: A – Φ = BA cosθ = (0.4)(0.05) × 1 = 0.02 Wb.
Q52. A coil of 500 turns and area 0.01 m2 experiences a magnetic field change of 0.2 T in 0.1 s, with the field perpendicular to the coil. What is the magnitude of the induced emf?
- (A) 20 V
- (B) 10 V
- (C) 5 V
- (D) 15 V
Show Answer
Answer: B – |ε| = N A ΔB/Δt = (500)(0.01)(0.2)/0.1 = 10 V.
Q53. A conducting rod of length 0.6 m moves at 5 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) 1.8 V
- (B) 0.45 V
- (C) 0.9 V
- (D) 1.35 V
Show Answer
Answer: C – ε = BLv = (0.3)(0.6)(5) = 0.9 V.
Q54. A loop of resistance 6 Ω experiences an induced emf of 6 V. What is the magnitude of the induced current in the loop?
- (A) 2 A
- (B) 0.5 A
- (C) 1.5 A
- (D) 1 A
Show Answer
Answer: D – I = ε/R = 6/6 = 1 A.
Q55. A solenoid has 500 turns per metre, cross-sectional area 0.01 m2, and length 0.5 m. What is its self-inductance?
- (A) 1.57 × 10-3 H
- (B) 3.14 × 10-3 H
- (C) 7.85 × 10-4 H
- (D) 2.36 × 10-3 H
Show Answer
Answer: A – L = μ0n2Al = 1.57 × 10-3 H.
Q56. An LR circuit has inductance 4 H and resistance 20 Ω. What is its inductive time constant?
- (A) 0.4 s
- (B) 0.2 s
- (C) 0.1 s
- (D) 0.3 s
Show Answer
Answer: B – τ = L/R = 4/20 = 0.2 s.
Q57. An inductor of 4 H carries a current of 4 A. How much energy is stored in its magnetic field?
- (A) 64 J
- (B) 16 J
- (C) 32 J
- (D) 48 J
Show Answer
Answer: C – U = (1/2)LI2 = 0.5 × 4 × (4)2 = 32 J.
Q58. An LC circuit has inductance 20 mH and capacitance 25 μF. What is its natural frequency of oscillation?
- (A) 450.158 Hz
- (B) 112.54 Hz
- (C) 337.619 Hz
- (D) 225.079 Hz
Show Answer
Answer: D – f = 1/[2π√(LC)] = 1/[2π√((20 × 10-3)(25 × 10-6))] = 225.079 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 0.4 T. What is the maximum induced emf in the coil?
- (A) 20 V
- (B) 40 V
- (C) 10 V
- (D) 30 V
Show Answer
Answer: A – εmax = NBAω = (50)(0.4)(0.02)(50) = 20 V.
Q60. In an LC circuit, the maximum current is 4 A. The inductance is 40 mH and the capacitance is 20 μF. What is the maximum voltage across the capacitor?
- (A) 357.771 V
- (B) 178.885 V
- (C) 89.443 V
- (D) 268.328 V
Show Answer
Answer: B – Energy conservation gives (1/2)LI2max = (1/2)CV2max, so Vmax = Imax√(L/C) = 178.885 V.
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