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
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
Rearrange Coulomb’s law for r.
r = √(k·|q₁·q₂|/F).
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
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
Reduce the network step by step. In series, resistances add. Two resistors in parallel combine as R₁R₂/(R₁ + R₂).
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.
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.
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.