DP Physics (last assessment 2024)
Question 22N.2.HL.TZ0.9b.ii
| Date | November 2022 | Marks available | [Maximum mark: 2] | Reference code | 22N.2.HL.TZ0.9b.ii |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Determine | Question number | b.ii | Adapted from | N/A |
b.ii.
[Maximum mark: 2]
22N.2.HL.TZ0.9b.ii
A parallel-plate capacitor of capacitance 1.5 × 10−10 F is made from two metal plates separated by an air gap of 1.0 mm. The capacitor is initially charged to a potential difference of 24 V.
The charged capacitor is disconnected from the voltage supply and the separation between the plates is increased to 4.0 mm.
(b.ii)
Determine the work required to increase the separation of the plates.
[2]
Markscheme
work = (energy in the new arrangement) − (initial energy) ✓
«energy increased by factor 4 hence» work = 3 × 4.3 × 10−8 =1.3 × 10−7 «J» ✓
Allow ECF from 9(a).
Award [2] for a BCA.
Examiners report
Different strategies were observed here, some successful, some including a choice of wrong formulae.
