Question 19M.2.HL.TZ1.a.ii
| Date | May 2019 | Marks available | [Maximum mark: 2] | Reference code | 19M.2.HL.TZ1.a.ii |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate | Question number | a.ii | Adapted from | N/A |
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 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]
1.5 × 106 × 55 × 10-6 = 83 «V» ✔
q «= CV»= 5.6 × 10-6 «C»✔
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.
