Question 19N.2.SL.TZ0.5

$E=\frac{k\times q}{r^2}$ ✔

$E=\frac{8.99\times {10}^9\times 6.0\times {10}^{-3}}{0.4^2}$  OR  $E=3.37\times {10}^8 «\mathrm{N} {\mathrm{C}}^{-1}»$ ✔

NOTE: Ignore any negative sign.

$F=q\times E$ OR  $F=1.6\times {10}^{-19}\times 3.4\times {10}^8=5.4\times {10}^{-11} «\mathrm{N}»$ ✔

$a=«\frac{5.4\times {10}^{-11}}{9.1\times {10}^{-31}}=» 5.9\times {10}^{19}« \mathrm{m} {\mathrm{s}}^{-2}»$✔

NOTE: Ignore any negative sign.
Award [1] for a calculation leading to $a=« \mathrm{m} {\mathrm{s}}^{-2}»$
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