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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.

[2]

Calculate the magnitude of the initial acceleration of the electron.

[2]

b(i).

Describe the subsequent motion of the electron.

[3]

b(ii).

$E = \frac{k \times q}{r^{2}}$ ✔

$E = \frac{8.99 \times 10^{9} \times 6.0 \times 10^{- 3}}{0 . 4^{2}}$ OR $E = 3.37 \times 10^{8} « N C^{- 1} »$ ✔

NOTE: Ignore any negative sign.

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

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

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

b(i).

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

b(ii).

[N/A]

b(i).

[N/A]

b(ii).