Date | November 2019 | Marks available | 2 | Reference code | 19N.2.SL.TZ0.5 |
Level | Standard level | Paper | Paper 2 | Time zone | 0 - no time zone |
Command term | Calculate | Question number | 5 | Adapted from | N/A |
Question
An electron is placed at a distance of 0.40 m from a fixed point charge of –6.0 mC.
Show that the electric field strength due to the point charge at the position of the electron is 3.4 × 108 N C–1.
Calculate the magnitude of the initial acceleration of the electron.
Describe the subsequent motion of the electron.
Markscheme
E=k×qr2E=k×qr2 ✔
E=8.99×109×6.0×10-30.42E=8.99×109×6.0×10−30.42 OR E=3.37×108 «N C-1» ✔
NOTE: Ignore any negative sign.
F=q×E OR F=1.6×10-19×3.4×108=5.4×10-11 «N» ✔
a=«5.4×10-119.1×10-31=» 5.9×1019« m s-2»✔
NOTE: Ignore any negative sign.
Award [1] for a calculation leading to a=« m s-2»
Award [2] for bald correct answer
the electron moves away from the point charge/to the right «along the line joining them» ✔
decreasing acceleration ✔
increasing speed ✔
NOTE: Allow ECF from MP1 if a candidate mistakenly evaluates the force as attractive so concludes that the acceleration will increase