A-LevelPhysicsElectric fields9702/23 · Oct/Nov 2017 · Q510 Marks
(a) Define electric field strength. (b) Two parallel metal plates in a vacuum are separated by 0.045m. A potential difference V is applied between the plates, as shown in Fig. 5.1. [Figure 5.1] A proton is initially at rest on the surface of the positive plate. The proton in the uniform electric field takes a time of 1.5 × 10⁻⁷ s to reach the negative plate. (c) An α particle is now accelerated between the two metal plates in (b) by the electric field. Calculate the ratio acceleration of α particle / acceleration of proton
- (a)Define electric field strength.[1]
- (b)(i)Show that the acceleration of the proton is 4.0 × 10¹² m s⁻².[2]
- (b)(ii)Calculate the electric force on the proton.[1]
- (b)(iii)1the electric field strength,[2]
- (b)(iii)2the potential difference V between the plates.[2]
- (c)An α particle is now accelerated between the two metal plates in (b) by the electric field. Calculate the ratio acceleration of α particle acceleration of proton[2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- 5(a) force per unit positive charge B1
- 5(b)(i) s = ½at² C1 a = (2 × 0.045) / (1.5 × 10⁻⁷)² = 4.0(0) × 10¹² m s⁻² A1
- 5(b)(ii) F = 1.67 × 10⁻²⁷ × 4.0 × 10¹² = 6.7 (6.68) × 10⁻¹⁵ N A1
- 5(b)(iii)1. E = F/Q C1 = 6.68 x 10⁻¹⁵ / 1.6 × 10⁻¹⁹ A1 = 4.2 (4.18) × 10⁴ N C⁻¹
- 5(b)(iii)2. E = V/d C1 V = 4.18 x 10⁴ x 0.045 A1 = 1.9 x 10³ V
- 5(c) a = Eq / m C1 ratio = (2×1.6×10⁻¹⁹) × (1.67×10⁻²⁷) / (1.6×10⁻¹⁹)×(4×1.66×10⁻²⁷) or 2×1 / 1x4 A1 = 0.50
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