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Date	May 2019	Marks available	3	Reference code	19M.2.HL.TZ1.8
Level	Higher level	Paper	Paper 2	Time zone	1
Command term	Calculate	Question number	8	Adapted from	N/A

Question

A student makes a parallel-plate capacitor of capacitance 68 nF from aluminium foil and plastic film by inserting one sheet of plastic film between two sheets of aluminium foil.

The aluminium foil and the plastic film are 450 mm wide.

The plastic film has a thickness of 55 μm and a permittivity of 2.5 × 10⁻¹¹ C² N^–1 m^–2.

The student uses a switch to charge and discharge the capacitor using the circuit shown. The ammeter is ideal.

The emf of the battery is 12 V.

Calculate the total length of aluminium foil that the student will require.

[3]

a.i.

The plastic film begins to conduct when the electric field strength in it exceeds 1.5 MN C^–1. Calculate the maximum charge that can be stored on the capacitor.

[2]

a.ii.

The resistor R in the circuit has a resistance of 1.2 kΩ. Calculate the time taken for the charge on the capacitor to fall to 50 % of its fully charged value.

[3]

b.i.

The ammeter is replaced by a coil. Explain why there will be an induced emf in the coil while the capacitor is discharging.

[2]

b.ii.

Suggest one change to the discharge circuit, apart from changes to the coil, that will increase the maximum induced emf in the coil.

[2]

b.iii.

Markscheme

length = $\frac{{d \times C}}{{{\text{width}} \times \varepsilon }}$ ✔

= 0.33 «m» ✔

so 0.66/0.67 «m» «as two lengths required» ✔

a.i.

1.5 × 10⁶ × 55 × 10^-6 = 83 «V» ✔

q «= CV»= 5.6 × 10^-6 «C»✔

a.ii.

$0.5 = {{\text{e}}^{ - \frac{t}{{RC}}}} = {{\text{e}}^{ - \frac{t}{{1200 \times 6.8 \times {{10}^{ - 8}}}}}}$

t = «−» 1200 × 6.8 × 10⁻⁸ × ln0.5 ✔

5.7 × 10⁻⁵ «s» ✔

use of t $\frac{1}{2}$ = RC × ln2 ✔

1200 × 6.8 × 10⁻⁸ × 0.693 ✔

5.7 × 10⁻⁵ «s» ✔

b.i.

mention of Faraday’s law ✔

indicating that changing current in discharge circuit leads to change in flux in coil/change in magnetic field «and induced emf» ✔

b.ii.

decrease/reduce ✔

resistance (R) OR capacitance (C) ✔

b.iii.

Examiners report

Many candidates were able to use the proper equation to calculate the length of one piece of aluminum foil for the first two marks, but very few doubled the length for the final mark.

a.i.

This question was challenging for many candidates. While some candidates were able to use proper equations for capacitors to determine the charge some of the candidates attempted to use electrostatic equations for the electric field around a point charge to solve this problem.

a.ii.

This question was also challenging for many candidates, with not an insignificant number leaving it blank. The candidates who did attempt it generally set up a correct equation, but ran into some simple calculation and power of ten errors. Some candidates attempted to solve the equation using basic circuit equations, which did not receive any marks.

b.i.

This is an explain question, so there was an expectation for a fairly detailed response. Many candidates missed the fact that the discharging capacitor is causing the current in the coil to change in time, and that this is what is inducing the emf in the coil. Many simply stated that the current created a magnetic field with not complete explanation of induction.

b.ii.

Candidates who recognized that something about the discharge circuit (not the charging circuit) needed to be changed generally suggested that something had to change with the resistance or capacitance. It should be noted that even though this was the last question on the exam, it was attempted at a higher rate than many of the other questions on the exam.

b.iii.

Syllabus sections

Additional higher level (AHL) » Topic 11: Electromagnetic induction » 11.3 – Capacitance

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Additional higher level (AHL) » Topic 11: Electromagnetic induction

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