AP Physics C: Electricity and Magnetism Practice Test 7 – 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 5 μC and 12 μC and are held 0.25 m apart in a vacuum. What is the magnitude of the electrostatic force between them?
- (A) 17.261 N
- (B) 4.315 N
- (C) 12.946 N
- (D) 8.63 N
Show Answer
Answer: D – Use Coulomb’s law F = kq1q2/r2 = (8.99 × 109)(5 × 10-6)(12 × 10-6)/(0.25)2 = 8.63 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 0.3 m away?
- (A) 1.2 × 106 N/C
- (B) 2.4 × 106 N/C
- (C) 5.99 × 105 N/C
- (D) 1.8 × 106 N/C
Show Answer
Answer: A – E = kQ/r2 = (8.99 × 109)(12 × 10-6)/(0.3)2 = 1.2 × 106 N/C.
Q3. A closed Gaussian surface encloses a net charge of 6 μC. What is the total electric flux through the surface?
- (A) 1.36 × 106 N·m2/C
- (B) 6.78 × 105 N·m2/C
- (C) 3.39 × 105 N·m2/C
- (D) 1.02 × 106 N·m2/C
Show Answer
Answer: B – By Gauss’s law, Φ = Qenc/ε0 = (6 × 10-6)/(8.85 × 10-12) = 6.78 × 105 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 0.4 m from its centre?
- (A) 2.25 × 106 N/C
- (B) 5.62 × 105 N/C
- (C) 1.12 × 106 N/C
- (D) 1.69 × 106 N/C
Show Answer
Answer: C – Outside a spherical charge distribution, E = kQ/r2 = (8.99 × 109)(20 × 10-6)/(0.4)2 = 1.12 × 106 N/C.
Q5. A nonconducting sphere of radius 0.4 m carries a total charge of 9 μC uniformly distributed through its volume. What is the electric field magnitude at a point 0.2 m from its centre?
- (A) 5.06 × 105 N/C
- (B) 1.26 × 105 N/C
- (C) 3.79 × 105 N/C
- (D) 2.53 × 105 N/C
Show Answer
Answer: D – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R3 = 2.53 × 105 N/C.
Q6. An infinitely long straight wire carries a uniform linear charge density of 6 μC/m. What is the electric field magnitude at a perpendicular distance of 0.2 m from the wire?
- (A) 5.39 × 105 N/C
- (B) 1.08 × 106 N/C
- (C) 2.7 × 105 N/C
- (D) 8.09 × 105 N/C
Show Answer
Answer: A – For an infinite line, Gauss’s law gives E = λ/(2πε0r) = 2kλ/r = 5.39 × 105 N/C.
Q7. An infinite nonconducting sheet carries a uniform surface charge density of 5 μC/m2. What is the electric field magnitude near the sheet, on either side?
- (A) 5.65 × 105 N/C
- (B) 2.82 × 105 N/C
- (C) 1.41 × 105 N/C
- (D) 4.24 × 105 N/C
Show Answer
Answer: B – For an infinite sheet, E = σ/(2ε0) = (5 × 10-6)/(2 × 8.85 × 10-12) = 2.82 × 105 N/C.
Q8. A thin spherical shell of radius 0.2 m carries a total charge of 4 μC on its surface. What is the electric field magnitude at a point inside the shell, 0.1 m from its centre?
- (A) 3.6 × 106 N/C
- (B) 8.99 × 105 N/C
- (C) 0 N/C
- (D) 1.8 × 106 N/C
Show Answer
Answer: C – 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 4 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) 2.57 × 105 N/C
- (B) 6.42 × 104 N/C
- (C) 1.93 × 105 N/C
- (D) 1.28 × 105 N/C
Show Answer
Answer: D – Integrating dE = kλdx/x2 from x = 0.2 m to x = 0.7 m gives E = kλL/[a(a + L)] = 1.28 × 105 N/C.
Q10. Two point charges, 4 μC and 5 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) 4 × 105 N/C
- (B) 7.99 × 105 N/C
- (C) 2 × 105 N/C
- (D) 5.99 × 105 N/C
Show Answer
Answer: A – Each field at the midpoint is kq/(d/2)2; the fields point in opposite directions, so the net magnitude is |E2 – E1| = 4 × 105 N/C.
Q11. Two large parallel sheets carry surface charge densities of +6 μC/m2 and -2 μC/m2. What is the electric field magnitude in the region between the sheets?
- (A) 9.04 × 105 N/C
- (B) 4.52 × 105 N/C
- (C) 2.26 × 105 N/C
- (D) 6.78 × 105 N/C
Show Answer
Answer: B – Each sheet contributes σ/(2ε0) and the two contributions add between the sheets, giving E = (σ1 + σ2)/(2ε0) = 4.52 × 105 N/C.
Q12. An electric dipole has dipole moment 2 × 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) 8990 N/C
- (B) 2247.5 N/C
- (C) 4495 N/C
- (D) 6742.5 N/C
Show Answer
Answer: C – On the axis, far from the dipole, E ≈ 2kp/r3 = 4495 N/C.
Unit 9: Electric Potential
Q13. What is the electric potential at a point 0.4 m from a point charge of 10 μC, taking the potential to be zero at infinity?
- (A) 4.49 × 105 V
- (B) 1.12 × 105 V
- (C) 3.37 × 105 V
- (D) 2.25 × 105 V
Show Answer
Answer: D – V = kQ/r = (8.99 × 109)(10 × 10-6)/0.4 = 2.25 × 105 V.
Q14. A charge of 3 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) 1.2 × 10-3 J
- (B) 2.4 × 10-3 J
- (C) 6 × 10-4 J
- (D) -1.2 × 10-3 J
Show Answer
Answer: A – Work by the field is W = qΔV = (3 × 10-6)(400) = 1.2 × 10-3 J.
Q15. A conducting sphere of radius 0.1 m carries a charge of 12 μC. What is the electric potential at its surface, with zero potential at infinity?
- (A) 2.16 × 106 V
- (B) 1.08 × 106 V
- (C) 5.39 × 105 V
- (D) 1.62 × 106 V
Show Answer
Answer: B – Outside and at the surface, V = kQ/R = 1.08 × 106 V.
Q16. A thin rod of length 0.6 m carries a uniform linear charge density of 5 μC/m. What is the electric potential at a point on the rod’s axis, 0.2 m beyond one end?
- (A) 1.25 × 105 V
- (B) 3.12 × 104 V
- (C) 6.23 × 104 V
- (D) 9.35 × 104 V
Show Answer
Answer: C – Integrating dV = kλdx/x from x = 0.2 m to x = 0.8 m gives V = kλ ln[(L + a)/a] = 6.23 × 104 V.
Q17. A ring of radius 0.2 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) 5.98 × 105 V
- (B) 1.5 × 105 V
- (C) 4.49 × 105 V
- (D) 2.99 × 105 V
Show Answer
Answer: D – Every charge element is at distance √(R2 + x2) from the point, so V = kQ/√(R2 + x2) = 2.99 × 105 V.
Q18. Two point charges of 4 μC and 3 μC are held 0.4 m apart. What is the electric potential energy of this pair?
- (A) 0.27 J
- (B) 0.539 J
- (C) 0.135 J
- (D) 0.405 J
Show Answer
Answer: A – 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 = 1.5 m?
- (A) 150 V/m
- (B) -150 V/m
- (C) -300 V/m
- (D) -75 V/m
Show Answer
Answer: B – Ex = -dV/dx = -100x, so at x = 1.5 m, Ex = -150 V/m.
Q20. A thin conducting spherical shell of radius 0.15 m carries a charge of 15 μC. What is the electric potential at a point inside the shell, halfway to its centre?
- (A) 4.5 × 105 V
- (B) 1.8 × 106 V
- (C) 8.99 × 105 V
- (D) 1.35 × 106 V
Show Answer
Answer: C – Inside a conducting shell the potential is constant and equal to the surface value, V = kQ/R = 8.99 × 105 V.
Q21. A uniformly charged disk of radius 0.3 m carries a surface charge density of 3 μC/m2. What is the electric potential on the disk’s axis at a distance of 0.2 m from its centre?
- (A) 5.44 × 104 V
- (B) 1.36 × 104 V
- (C) 4.08 × 104 V
- (D) 2.72 × 104 V
Show Answer
Answer: D – Integrating rings gives V = 2πkσ( √(R2 + x2) – x ) = 2.72 × 104 V.
Unit 10: Conductors and Capacitors
Q22. A parallel-plate capacitor has plates of area 0.02 m2 separated by 4 mm in air. What is its capacitance?
- (A) 4.43 × 10-11 F
- (B) 8.85 × 10-11 F
- (C) 2.21 × 10-11 F
- (D) 6.64 × 10-11 F
Show Answer
Answer: A – C = ε0A/d = (8.85 × 10-12)(0.02)/(4 × 10-3) = 4.43 × 10-11 F.
Q23. A capacitor of capacitance 50 pF is filled completely with a dielectric of dielectric constant 5, with no change in plate separation. What is the new capacitance?
- (A) 500 pF
- (B) 250 pF
- (C) 125 pF
- (D) 260 pF
Show Answer
Answer: B – C = κC0 = 5 × 50 pF = 250 pF.
Q24. A 20 pF capacitor is charged to 24 V. How much energy is stored in the capacitor?
- (A) 1.15 × 10-8 J
- (B) 2.88 × 10-9 J
- (C) 5.76 × 10-9 J
- (D) 8.64 × 10-9 J
Show Answer
Answer: C – U = (1/2)CV2 = 0.5 × (20 × 10-12) × (24)2 = 5.76 × 10-9 J.
Q25. Capacitors of 20 μF and 4 μF are connected in series. What is their equivalent capacitance?
- (A) 6.667 μF
- (B) 1.667 μF
- (C) 5 μF
- (D) 3.333 μF
Show Answer
Answer: D – 1/C = 1/20 + 1/4, so C = 3.333 μF.
Q26. Capacitors of 4 μF and 14 μF are connected in parallel across a 24 V battery. What total charge is drawn from the battery?
- (A) 4.32 × 10-4 C
- (B) 8.64 × 10-4 C
- (C) 2.16 × 10-4 C
- (D) 6.48 × 10-4 C
Show Answer
Answer: A – Ceq = 18 μF, so Q = CeqV = 4.32 × 10-4 C.
Q27. The plates of a parallel-plate capacitor carry surface charge densities of magnitude 4 μC/m2. What is the electric field magnitude between the plates?
- (A) 9.04 × 105 N/C
- (B) 4.52 × 105 N/C
- (C) 2.26 × 105 N/C
- (D) 6.78 × 105 N/C
Show Answer
Answer: B – Between the plates, E = σ/ε0 = 4.52 × 105 N/C.
Q28. A spherical capacitor consists of concentric conducting spheres of radii 0.12 m and 0.3 m. What is its capacitance?
- (A) 1.11 × 10-11 F
- (B) 4.45 × 10-11 F
- (C) 2.22 × 10-11 F
- (D) 3.34 × 10-11 F
Show Answer
Answer: C – C = 4πε0ab/(b – a) = 2.22 × 10-11 F.
Q29. The electric field in a region of space has magnitude 10000 N/C. What is the electric energy density in that region?
- (A) 8.85 × 10-4 J/m3
- (B) 2.21 × 10-4 J/m3
- (C) 6.64 × 10-4 J/m3
- (D) 4.42 × 10-4 J/m3
Show Answer
Answer: D – u = (1/2)ε0E2 = 0.5 × (8.85 × 10-12) × (10000)2 = 4.42 × 10-4 J/m3.
Unit 11: Electric Circuits
Q30. A 12 V battery is connected across a 6 Ω resistor. What current flows through the resistor?
- (A) 2 A
- (B) 4 A
- (C) 1 A
- (D) 3 A
Show Answer
Answer: A – I = V/R = 12/6 = 2 A.
Q31. Resistors of 30 Ω and 15 Ω are connected in parallel. What is their equivalent resistance?
- (A) 20 Ω
- (B) 10 Ω
- (C) 5 Ω
- (D) 15 Ω
Show Answer
Answer: B – 1/R = 1/30 + 1/15, so R = 10 Ω.
Q32. A resistor of 60 Ω is connected across a 12 V source. What power is dissipated in the resistor?
- (A) 4.8 W
- (B) 1.2 W
- (C) 2.4 W
- (D) 3.6 W
Show Answer
Answer: C – P = V2/R = (12)2/60 = 2.4 W.
Q33. A battery of emf 18 V and internal resistance 1 Ω is connected to an external resistor of 4 Ω. What is the terminal voltage of the battery?
- (A) 28.8 V
- (B) 7.2 V
- (C) 21.6 V
- (D) 14.4 V
Show Answer
Answer: D – Current I = emf/(R + r) = 3.6 A; terminal voltage V = IR = 14.4 V.
Q34. An RC circuit has resistance 2000 Ω and capacitance 20 μF. What is its time constant?
- (A) 0.04 s
- (B) 0.08 s
- (C) 0.02 s
- (D) 0.06 s
Show Answer
Answer: A – τ = RC = (2000)(20 × 10-6) = 0.04 s.
Q35. A capacitor of 10 μF, initially uncharged, is connected in series with a 5000 Ω resistor across a 12 V battery. What is the voltage across the capacitor after one time constant has elapsed?
- (A) 15.171 V
- (B) 7.585 V
- (C) 3.793 V
- (D) 11.378 V
Show Answer
Answer: B – After one time constant, V = V0(1 – e-1) = 0.632 × 12 = 7.585 V.
Q36. A capacitor charged to 20 V discharges through a 1000 Ω resistor. Its capacitance is 40 μF. What is the voltage across the capacitor after two time constants?
- (A) 5.413 V
- (B) 1.353 V
- (C) 2.707 V
- (D) 4.06 V
Show Answer
Answer: C – τ = RC = 0.04 s; V = V0e-t/τ = 20e-2 = 2.707 V.
Q37. A capacitor discharges through a 2000 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) 0.139 s
- (B) 0.035 s
- (C) 0.104 s
- (D) 0.069 s
Show Answer
Answer: D – q = Q0e-t/RC; setting q/Q0 = 0.25 gives t = -RC ln(0.25) = 0.069 s.
Q38. A battery of emf 12 V is connected in series with resistors of 6 Ω and 24 Ω. What is the voltage across the 24 Ω resistor?
- (A) 9.6 V
- (B) 19.2 V
- (C) 4.8 V
- (D) 14.4 V
Show Answer
Answer: A – Series current I = 12/(6 + 24) = 0.4 A; V = IR2 = 9.6 V.
Q39. A current of 3 A flows through a resistor of 10 Ω. What power is dissipated as heat in the resistor?
- (A) 180 W
- (B) 90 W
- (C) 45 W
- (D) 135 W
Show Answer
Answer: B – P = I2R = (3)2(10) = 90 W.
Q40. A total current of 6 A splits between two parallel resistors of 4 Ω and 12 Ω. What current flows through the 4 Ω resistor?
- (A) 9 A
- (B) 2.25 A
- (C) 4.5 A
- (D) 6.75 A
Show Answer
Answer: C – By current division, I1 = ItotalR2/(R1 + R2) = 4.5 A.
Q41. A series RC circuit with resistance 12 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) 6 V
- (C) 48 V
- (D) 24 V
Show Answer
Answer: D – 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 3 μC moves at 4000 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the particle?
- (A) 6 × 10-3 N
- (B) 0.012 N
- (C) 3 × 10-3 N
- (D) 9 × 10-3 N
Show Answer
Answer: A – F = qvB = (3 × 10-6)(4000)(0.5) = 6 × 10-3 N.
Q43. A straight wire of length 0.5 m carries a current of 10 A perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the wire?
- (A) 5 N
- (B) 2.5 N
- (C) 1.25 N
- (D) 3.75 N
Show Answer
Answer: B – F = ILB = (10)(0.5)(0.5) = 2.5 N.
Q44. A long straight wire carries a current of 10 A. What is the magnetic field magnitude at a distance of 0.05 m from the wire?
- (A) 8 × 10-5 T
- (B) 2 × 10-5 T
- (C) 4 × 10-5 T
- (D) 6 × 10-5 T
Show Answer
Answer: C – B = μ0I/(2πr) = 4 × 10-5 T.
Q45. A circular loop of radius 0.1 m carries a current of 4 A. What is the magnetic field magnitude at the centre of the loop?
- (A) 5.03 × 10-5 T
- (B) 1.26 × 10-5 T
- (C) 3.77 × 10-5 T
- (D) 2.51 × 10-5 T
Show Answer
Answer: D – B = μ0I/(2R) = 2.51 × 10-5 T.
Q46. A solenoid has 650 turns per metre and carries a current of 3 A. What is the magnetic field magnitude inside the solenoid, far from its ends?
- (A) 2.45 × 10-3 T
- (B) 1.23 × 10-3 T
- (C) 4.9 × 10-3 T
- (D) 3.68 × 10-3 T
Show Answer
Answer: A – B = μ0nI = (4π × 10-7)(650)(3) = 2.45 × 10-3 T.
Q47. An electron (mass 9.11 × 10-31 kg, charge magnitude 1.60 × 10-19 C) moves at 3000000 m/s perpendicular to a uniform magnetic field of 0.02 T. What is the radius of its circular path?
- (A) 4.27 × 10-4 m
- (B) 8.54 × 10-4 m
- (C) 1.71 × 10-3 m
- (D) 1.28 × 10-3 m
Show Answer
Answer: B – r = mv/(qB) = (9.11 × 10-31)(3000000)/[(1.60 × 10-19)(0.02)] = 8.54 × 10-4 m.
Q48. A rectangular coil of 150 turns and area 0.03 m2 carries a current of 3 A in a magnetic field of 0.4 T. What is the maximum torque on the coil?
- (A) 2.7 N·m
- (B) 10.8 N·m
- (C) 5.4 N·m
- (D) 8.1 N·m
Show Answer
Answer: C – τmax = NIAB = (150)(3)(0.03)(0.4) = 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 0.4 T. What is the speed of the particle?
- (A) 2000 m/s
- (B) 500 m/s
- (C) 1500 m/s
- (D) 1000 m/s
Show Answer
Answer: D – Undeflected motion requires qE = qvB, so v = E/B = 400/0.4 = 1000 m/s.
Q50. Two long parallel wires 0.2 m apart carry currents of 10 A and 10 A in opposite directions. What is the magnetic field magnitude at the midpoint between the wires?
- (A) 4 × 10-5 T
- (B) 8 × 10-5 T
- (C) 2 × 10-5 T
- (D) 6 × 10-5 T
Show Answer
Answer: A – Each wire contributes μ0I/(2πr) with r = 0.1 m, and the contributions add at the midpoint, giving B = 4 × 10-5 T.
Unit 13: Electromagnetic Induction
Q51. A flat loop of area 0.04 m2 lies perpendicular to a uniform magnetic field of 0.4 T. What is the magnetic flux through the loop?
- (A) 0.032 Wb
- (B) 0.016 Wb
- (C) 8 × 10-3 Wb
- (D) 0.024 Wb
Show Answer
Answer: B – Φ = BA cosθ = (0.4)(0.04) × 1 = 0.016 Wb.
Q52. A coil of 200 turns and area 0.02 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) 16 V
- (B) 4 V
- (C) 8 V
- (D) 12 V
Show Answer
Answer: C – |ε| = N A ΔB/Δt = (200)(0.02)(0.4)/0.2 = 8 V.
Q53. A conducting rod of length 0.4 m moves at 3 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) 0.72 V
- (B) 0.18 V
- (C) 0.54 V
- (D) 0.36 V
Show Answer
Answer: D – ε = BLv = (0.3)(0.4)(3) = 0.36 V.
Q54. A loop of resistance 12 Ω experiences an induced emf of 12 V. What is the magnitude of the induced current in the loop?
- (A) 1 A
- (B) 2 A
- (C) 0.5 A
- (D) 1.5 A
Show Answer
Answer: A – I = ε/R = 12/12 = 1 A.
Q55. A solenoid has 500 turns per metre, cross-sectional area 0.02 m2, and length 0.5 m. What is its self-inductance?
- (A) 6.28 × 10-3 H
- (B) 3.14 × 10-3 H
- (C) 1.57 × 10-3 H
- (D) 4.71 × 10-3 H
Show Answer
Answer: B – L = μ0n2Al = 3.14 × 10-3 H.
Q56. An LR circuit has inductance 10 H and resistance 10 Ω. What is its inductive time constant?
- (A) 2 s
- (B) 0.5 s
- (C) 1 s
- (D) 1.5 s
Show Answer
Answer: C – τ = L/R = 10/10 = 1 s.
Q57. An inductor of 5 H carries a current of 3 A. How much energy is stored in its magnetic field?
- (A) 11.2 J
- (B) 45 J
- (C) 33.8 J
- (D) 22.5 J
Show Answer
Answer: D – U = (1/2)LI2 = 0.5 × 5 × (3)2 = 22.5 J.
Q58. An LC circuit has inductance 10 mH and capacitance 25 μF. What is its natural frequency of oscillation?
- (A) 318.31 Hz
- (B) 636.62 Hz
- (C) 159.155 Hz
- (D) 477.465 Hz
Show Answer
Answer: A – f = 1/[2π√(LC)] = 1/[2π√((10 × 10-3)(25 × 10-6))] = 318.31 Hz.
Q59. A coil of 50 turns and area 0.04 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) 80 V
- (B) 40 V
- (C) 20 V
- (D) 60 V
Show Answer
Answer: B – εmax = NBAω = (50)(0.4)(0.04)(50) = 40 V.
Q60. In an LC circuit, the maximum current is 3 A. The inductance is 20 mH and the capacitance is 10 μF. What is the maximum voltage across the capacitor?
- (A) 268.328 V
- (B) 67.082 V
- (C) 134.164 V
- (D) 201.246 V
Show Answer
Answer: C – Energy conservation gives (1/2)LI2max = (1/2)CV2max, so Vmax = Imax√(L/C) = 134.164 V.
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