Worked Solutions
Module 4: Electricity and Magnetism — Worked Solutions (Preliminary Physics)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our Preliminary Physics course →
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.
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 $r^2$.