Module 4: Electricity and Magnetism — Worked Solutions (Preliminary Physics)
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Worked examples for Preliminary Physics Module 4: Electricity and Magnetism. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Physics — Module 4: Electricity and Magnetism. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
For circuits, combine resistances first, then apply Ohm's law $V = IR$ and the power relationship $P = VI$. For electrostatics, use Coulomb's law with $k = 8.99 \times 10^9\ \text{N m}^2\,\text{C}^{-2}$.
Example 1 — Ohm's law in a series circuit
Question
A $12\ \text{V}$ battery is connected to two resistors, $4.0\ \Omega$ and $8.0\ \Omega$, joined in series. Find the current drawn from the battery, the voltage across the $8.0\ \Omega$ resistor, and the total power dissipated in the circuit.
Solution
In series, resistances add: $R_{\text{total}} = 4.0 + 8.0 = 12\ \Omega$.
Current from the battery (same everywhere in series): $I = \dfrac{V}{R} = \dfrac{12}{12} = 1.0\ \text{A}$.
Voltage across the $8.0\ \Omega$ resistor: $V = IR = 1.0 \times 8.0 = 8.0\ \text{V}$.
Total power: $P = VI = 12 \times 1.0 = 12\ \text{W}$.
Current is the same through series components — that's the fact that unlocks everything else here.
In a series circuit there's only one path, so the resistances simply add up: $R_{\text{total}} = 4.0 + 8.0 = 12\ \Omega$.
Because the whole battery voltage drives current through that combined resistance, Ohm's law gives the current: $I = \dfrac{V}{R} = \dfrac{12}{12} = 1.0\ \text{A}$. And since there's only one loop, this same $1.0\ \text{A}$ flows through both resistors.
The voltage across the $8.0\ \Omega$ resistor is then $V = IR = 1.0 \times 8.0 = 8.0\ \text{V}$ — the larger resistor takes the larger share of the $12\ \text{V}$, which makes sense.
For the power the source delivers, $P = VI = 12 \times 1.0 = 12\ \text{W}$. The reason current is constant in series is that charge has nowhere else to go — it must pass through every component in turn.
Series: resistances add.
- $R_{\text{total}} = 4.0 + 8.0 = 12\ \Omega$
- $I = V/R = 12/12 = 1.0\ \text{A}$
- $V_{8\Omega} = IR = 1.0 \times 8.0 = 8.0\ \text{V}$
- $P = VI = 12 \times 1.0 = 12\ \text{W}$
$I = 1.0\ \text{A}$; $V_{8\Omega} = 8.0\ \text{V}$; $P = 12\ \text{W}$.
Where the marks go
- 1 mark: Adds series resistances: $R_{\text{total}} = 12\ \Omega$
- 1 mark: Applies $I = V/R$ to get $I = 1.0\ \text{A}$
- 1 mark: Correct voltage across $8.0\ \Omega$: $V = IR = 8.0\ \text{V}$
- 1 mark: Correct total power $P = VI = 12\ \text{W}$
Key idea
In series the current is the same throughout and resistances add; then $V = IR$ and $P = VI$ give the rest.
Example 2 — Coulomb's law for two point charges
Question
Two small charges, $+3.0\ \mu\text{C}$ and $-2.0\ \mu\text{C}$, are held $0.20\ \text{m}$ apart in air. Taking $k = 8.99 \times 10^9\ \text{N m}^2\,\text{C}^{-2}$, find the magnitude of the electrostatic force between them and state whether it is attractive or repulsive.
Solution
Use Coulomb's law: $F = \dfrac{k q_1 q_2}{r^2}$.
Convert: $q_1 = 3.0 \times 10^{-6}\ \text{C}$, $q_2 = 2.0 \times 10^{-6}\ \text{C}$, $r = 0.20\ \text{m}$.
$F = \dfrac{(8.99 \times 10^9)(3.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.20)^2} = \dfrac{8.99 \times 10^9 \times 6.0 \times 10^{-12}}{0.040}$.
Numerator $= 5.394 \times 10^{-2}$; divide by $0.040$: $F = 1.35\ \text{N}$.
Opposite signs → attractive. Don't forget to square $r$ and convert microcoulombs.
Coulomb's law tells us the force between two point charges: $F = \dfrac{k q_1 q_2}{r^2}$, using the magnitudes of the charges.
First convert the microcoulombs to coulombs: $q_1 = 3.0 \times 10^{-6}\ \text{C}$ and $q_2 = 2.0 \times 10^{-6}\ \text{C}$, with $r = 0.20\ \text{m}$.
Substituting: $F = \dfrac{(8.99 \times 10^9)(3.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.20)^2}$. The product of the charges is $6.0 \times 10^{-12}$, and $r^2 = 0.040\ \text{m}^2$, so $F = \dfrac{8.99 \times 10^9 \times 6.0 \times 10^{-12}}{0.040} = 1.35\ \text{N}$.
Because the charges have opposite signs, the force pulls them together — it's attractive. The most common slips here are forgetting to square the distance and leaving the charges in microcoulombs, so always check those two things.
Coulomb's law: $F = \dfrac{k q_1 q_2}{r^2}$.
- $q_1 = 3.0 \times 10^{-6}\ \text{C}$, $q_2 = 2.0 \times 10^{-6}\ \text{C}$, $r = 0.20\ \text{m}$
- $F = \dfrac{(8.99 \times 10^9)(6.0 \times 10^{-12})}{0.040}$
- $F = 1.35\ \text{N}$
Opposite signs → attractive; $F = 1.35\ \text{N}$.
Where the marks go
- 1 mark: Selects Coulomb's law and converts charges to coulombs
- 1 mark: Correct substitution with $r^2 = 0.040\ \text{m}^2$
- 1 mark: Correct magnitude $F = 1.35\ \text{N}$ and identifies it as attractive
Key idea
Coulomb's law $F = \dfrac{k q_1 q_2}{r^2}$ uses charge magnitudes; opposite signs attract, like signs repel, and force falls off with $r^2$.
Frequently asked questions
Step-by-step solutions to Electricity and Magnetism questions in Preliminary Physics, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Physics, using the methods and notation expected in exams and assessments.