A-LevelPhysicsElectric fields9702/41 · Oct/Nov 2018 · Q69 Marks

Two parallel metal plates A and B are situated a distance 1.2 cm apart in a vacuum, as shown in [Figure 6.1]. Plate A is earthed and plate B is at a potential of –75 V. A helium nucleus is situated between the plates, a distance x from plate A. Initially, the helium nucleus is at rest on plate A where x = 0.

(a)(i)Define electric potential at a point.[2]
(a)(ii)State the relationship between electric potential and electric field strength at a point.[2]
(b)(i)The helium nucleus is free to move between the plates. By considering energy changes of the helium nucleus, explain why the speed at which it reaches plate B is independent of the separation of the plates.[2]
(b)(ii)As the helium nucleus (⁴₂He) moves from plate A towards plate B, its distance x from plate A increases. Calculate the speed of the nucleus after it has moved a distance x = 0.40 cm from plate A.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[B1] work done per unit charge
[B1] work done moving positive charge from infinity (to the point)
[M1] field strength = potential gradient
[A1] '-' sign included or directions discussed
[M1] gain in kinetic energy (= loss in potential energy) = charge × p.d. or qV = ½mv²
[A1] so v is independent of separation (because separation not in expressions)
[C1] (at x = 0.40 cm), potential = (–) 75 × 0.40/1.2 (= (–) 25 V)
[C1] ½mv² = qV OR v² = 2as with a = Vq/dm
[A1] ½ x 4 x 1.66 × 10⁻²⁷ x v² = 2 × 1.60 × 10⁻¹⁹ x 25 OR v² = (2 × 75 × 2 × 1.60 × 10⁻¹⁹ × 0.40 × 10⁻²) / (1.2 × 10⁻² × 4 × 1.66 × 10⁻²⁷), v = 4.9 x 10⁴ ms⁻¹

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