Q5 · 2021 · Paper 2Simple familiar4 marks
QUESTION 5 (4 marks)
An alpha particle with a charge of +3.2 × 10–19 C moves through an electric field, accelerating from rest through a potential difference of 240 V. Determine the velocity of the particle at the end of its acceleration, expressing your answer in scientific notation.
WORKED SOLUTION
4 marksQCAA guide · typeset solution
ANSWER
See worked solution below.
Worked solution
The change in potential energy of an electric charge
moving through an electric field is equivalent to the work
done on the charge.
𝑉= Δ𝑈
𝑞
Δ𝑈= 𝑉𝑞
= 240 × 3.2 × 10−19
= 7.68 × 10−17 𝐽= 𝑊
The work done on an object is equal to the change in
kinetic energy.
𝐸𝑘= 1
2 𝑚𝑣2
7.68 × 10−17 = 1
2 × 6.64 × 10−27 × 𝑣2
𝑣2 =
7.68 × 10−17
1
2 × 6.64 × 10−27
𝑣= √
7.68 × 10−17
1
2 × 6.64 × 10−27
Velocity = 1.5 × 105 m s−1 (to 2 significant figures)
• recognises the scenario relates to work done on a
moving charge in an electric field [1 mark]
• identifies that work done on the charge equates to
its kinetic energy [1 mark]
• provides appropriate mathematical reasoning
[1 mark]
• determines the velocity [1 mark]
One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.
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