NCEA Level 2 Physics 91173 Electricity and electromagnetism: past papers, practice and marking
- NCEA Level 2
- Physics
- AS91173
- 6 credits
- External exam
About 91173
AS91173, Demonstrate understanding of electricity and electromagnetism. Electric fields between plates, charges and energy, DC circuits and what changes when you add or remove a part, forces on wires and moving charges, and induced voltage: the 6-credit external, in the seven question types it keeps reusing.
What 91173 covers
- Electric fieldsField lines between parallel plates, E = V/d in V m⁻¹ or N C⁻¹, F = Eq, number of charges n = q/e, the effect of changing the voltage or the plate separation
- Electric potential energyΔEp = Eqd with d moved parallel to the field, no work across the field lines, ΔEp = Vq, Ep to Ek for the speed of an accelerated charge
- Charges in fieldsA suspended droplet with mg balanced by Eq, the sign and number of charges, the parabolic path of a charge crossing a uniform field
- DC circuitsV = IR, P = IV = I²R = V²/R, P = ΔE/t, I = q/t, V = ΔE/q, series and parallel resistance, current and voltage for each component, minutes to seconds
- Circuit changesAdding or removing a component at a fixed supply voltage: total resistance, then current, then the voltage and power of each component, and badly wired circuits
- Magnetic forcesF = BIL with the length in the field, direction by a right-hand rule, zero force parallel to the field, coils of many turns, F = Bqv and why a moving charge feels a force
- Induced voltageV = BvL for a conductor moving across a field, which end is positive and why, turns and both sides of a coil, maximum at 90° and none parallel, the opposing force
91173 past papers
Open a 91173 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 2 Physics past papers
Practice questions
Question 1: Electric fields
In an air purifier, two plates 8.0 mm apart have 5.0 kV across them. Smoke particles pick up a charge of +3.2 × 10⁻¹⁸ C and are pulled onto a plate.
- Calculate the electric field strength between the plates.(up to Achieved)
- Calculate the electric force on a charged smoke particle.(up to Achieved)
Check your answer
Part a: what gets the marks
- Convert kV and mm to SI units: E = V/d = 5000/0.0080 = 6.3 × 10⁵ V m⁻¹
A full answer: E = 6.3 × 10⁵ V m⁻¹ (6.25 × 10⁵ V m⁻¹ before rounding)
Part b: what gets the marks
- F = Eq, using the unrounded E: F = Eq = 6.25 × 10⁵ × 3.2 × 10⁻¹⁸ = 2.0 × 10⁻¹² N
A full answer: F = 2.0 × 10⁻¹² N
What gets each grade
Question 2: Electric fields
In an air purifier, two plates 8.0 mm apart have 5.0 kV across them. Smoke particles pick up a charge of +3.2 × 10⁻¹⁸ C and are pulled onto a plate by a force of 2.0 × 10⁻¹² N.
- Calculate how many electrons each particle has lost, and explain how you know they were lost rather than gained.(up to Merit)
- The engineer wants a stronger pull on the particles. Discuss the effect of (i) doubling the voltage to 10 kV, (ii) halving the separation to 4.0 mm, and (iii) doing both.(up to Excellence)
Check your answer
Part a: what gets the marks
- Number of elementary charges: n = q/e = 3.2 × 10⁻¹⁸/1.6 × 10⁻¹⁹ = 20
- The charge is positive, so the particle has fewer electrons than protons: it has lost 20 electrons rather than gained them.
A full answer: Each particle has lost 20 electrons. Its charge is positive, so it has fewer electrons than protons; gaining electrons would have made it negative.
Part b: what gets the marks
- E = V/d, so doubling V to 10 kV doubles E, and halving d to 4.0 mm also doubles E.
- Doing both gives E = 10 000/0.0040 = 2.5 × 10⁶ V m⁻¹, four times the original field.
- F = Eq with q fixed, so the force follows E: twice as strong for (i) or (ii), and four times as strong (8.0 × 10⁻¹² N) for both.
A full answer: (i) Doubling V doubles E to 1.25 × 10⁶ V m⁻¹, so F doubles to 4.0 × 10⁻¹² N. (ii) Halving d also doubles E, with the same result. (iii) Doing both quadruples E to 2.5 × 10⁶ V m⁻¹, so the force is four times bigger: 8.0 × 10⁻¹² N. Since F = Eq and the charge on each particle is fixed, the force on the particles changes by exactly the same factor as E = V/d.