Question 22M.2.HL.TZ2.7
| Date | May 2022 | Marks available | [Maximum mark: 9] | Reference code | 22M.2.HL.TZ2.7 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Comment, Determine, Outline, Show that | Question number | 7 | Adapted from | N/A |
A metal sphere is charged positively and placed far away from other charged objects. The electric potential at a point on the surface of the sphere is 53.9 kV.
Outline what is meant by electric potential at a point.
[2]
the work done per unit charge ✓
In bringing a small/point/positive/test «charge» from infinity to the point ✓
Allow use of energy per unit charge for MP1
a) Well answered.
b) Generally, well answered, but there were quite a few using r + 2.8.
ci) Very few had problems to recognize the perpendicular angle
cii) Good simple calculation
ciii) Many had a good go at this, but a significant number tried to answer it based on forces.
The electric potential at a point a distance 2.8 m from the centre of the sphere is 7.71 kV. Determine the radius of the sphere.
[2]
use of Vr = constant ✓
0.40 m ✓
Allow [1] max if r + 2.8 used to get 0.47 m.
Allow [2] marks if they calculate Q at one potential and use it to get the distance at the other potential.
A small positively charged object moves towards the centre of the metal sphere. When the object is 2.8 m from the centre of the sphere, its speed is 3.1 m s−1. The mass of the object is 0.14 g and its charge is 2.4 × 10−8 C.
Comment on the angle at which the object meets equipotential surfaces around the sphere.
[1]
90° / perpendicular ✓
Show that the kinetic energy of the object is about 0.7 mJ.
[1]
OR 0.67 «mJ» seen ✓
Determine whether the object will reach the surface of the sphere.
[3]
«p.d. between point and sphere surface = » (53.9 kV – 7.71) «kV» OR 46.2 «kV» seen ✓
«energy required =» VQ « = 46 200 × 2.4 × 10-8» = 1.11 mJ ✓
this is greater than kinetic energy so will not reach sphere ✓
MP3 is for a conclusion consistent with the calculations shown.
Allow ECF from MP1