Physics 2: Electricity and Magnetism: practice problems and exam questions

Electric charges, forces and fields, DC circuits, capacitors and RC circuits, and alternating current: the second semester of university physics.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Coulomb’s law & electric fieldsforce between two charges · field between two charges · forces at a right anglecalculations
  2. Series & parallel resistors3 resistors · 4 resistors · ladder of 6 resistorscalculations
  3. RC & RL transientsRC charging · RL circuit · capacitor with initial chargecalculations
  4. Magnetic fields & inductionfields of currents and magnetic forces · charged particles, Faraday and Lenz · generators, inductance and particle beamscalculations
  5. AC impedance & phasorsseries RL · series RLC · capacitor voltage and resonancecalculations

Exam tips for Physics 2: Electricity and Magnetism

  • Forces and fields from several charges add as vectors: draw each one before you add.
  • Convert µC to C and cm to m before using Coulomb’s law.
  • A capacitor blocks DC once it is charged; the time constant τ = RC sets how fast that happens.
  • For magnetic forces and induced currents, sketch ⊗ and ⊙ and use the right-hand rule every time: most lost points are wrong directions.

Worked examples

Coulomb’s law & electric fieldsEasy
++4 µC++1 µCr = ?

Point charges of +4 µC and +1 µC exert a force of 20 N on each other. How far apart r are they? Give the answer in cm. Use k = 8.99 × 10⁹ N·m²/C².

q1 = +4 µCq2 = +1 µCF = 20 N

Find r (cm)

Show hints
  1. Rearrange Coulomb’s law for r.
  2. r = √(k·|q₁·q₂|/F).
  3. r² = 0.001798 m².
Show the answer and the step-by-step solution

Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.

Solve for the distance

r = √(8.99 × 10⁹·4 × 10⁻⁶·1 × 10⁻⁶/20) = 0.0424 m
r = 4.24 cm

Answer

r = 4.24 cm
Series & parallel resistorsMedium
R1​ = 6 ΩIR1​R3​ = 3 Ω+−V1​18 VR2​3 ΩAR4​6 ΩB

Find the current IR1 delivered by the source, which flows through R1.

V1 = 18 VR1 = 6 ΩR2 = 3 ΩR3 = 3 ΩR4 = 6 Ω

Find IR1 (A)

Show hints
  1. Reduce the network step by step. In series, resistances add. Two resistors in parallel combine as R₁R₂/(R₁ + R₂).
  2. R₃ and R₄ carry the same current, so they are in series. That branch is in parallel with R₂, and the result is in series with R₁.
  3. Req = 8.25 Ω.
Show the answer and the step-by-step solution

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 6 ΩR3​ = 3 Ω+−V1​18 VR2​3 ΩAR4​6 ΩB

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R3 + R4 = 3 + 6 = 9 Ω
R2 ∥ 9 = 3·9/(3+9) = 2.25 Ω
Req = R1 + 2.25 = 6 + 2.25 = 8.25 Ω

Current from the source

The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.

IR1 = V1/Req = 18/8.25 = 2.182 A

Answer

IR1 = 2.182 A

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

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