Two parallel metal plates in a vacuum are separated by a distance of 15mm, as shown in Fig. 6.1. A uniform electric field is produced between the plates by applying a potential difference between them. A particle of mass 1.7 × 10⁻²⁷ kg and charge +1.6 × 10⁻¹⁹C is initially at rest at point A on one plate. The particle is moved by the electric field to point B on the other plate. The particle reaches point B with kinetic energy 2.4 × 10⁻¹⁶ J. [Figure 6.1]
- (a)Define electric field strength.[1]
- (b)Two parallel metal plates in a vacuum are separated by a distance of 15mm, as shown in Fig. 6.1. A uniform electric field is produced between the plates by applying a potential difference between them. A particle of mass 1.7 × 10⁻²⁷ kg and charge +1.6 × 10⁻¹⁹C is initially at rest at point A on one plate. The particle is moved by the electric field to point B on the other plate. The particle reaches point B with kinetic energy 2.4 × 10⁻¹⁶ J.
- (b)(i)Calculate the speed of the particle at point B.[2]
- (b)(ii)State the work done by the electric field to move the particle from A to B.[1]
- (b)(iii)Use your answer in (ii) to determine the force on the particle.[2]
- (b)(iv)Determine the potential difference between the plates.[3]
- (b)(v)On Fig. 6.2, sketch a graph to show the variation of the kinetic energy of the particle with the distance x from point A along the line AB. Numerical values for the kinetic energy are not required. [Figure 6.2][1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] force per unit positive charge
- [C1] EK = ½mv²
- [A1] 2.4 x 10⁻¹⁶ = ½ x 1.7 × 10⁻²⁷ x v² v = 5.3 x 10⁵ms⁻¹
- [A1] work done = 2.4 × 10⁻¹⁶ J
- [C1] W = Fs
- [A1] F = 2.4 × 10⁻¹⁶ / 15 × 10⁻³ = 1.6 x 10⁻¹⁴ N
- [C1] V = Fd/Q or V = W/Q or E = V/d and E = F/Q
- [C1] V = (1.6 × 10⁻¹⁴ × 15 × 10⁻³)/1.6 × 10⁻¹⁹ or 2.4 × 10⁻¹⁶/1.6 × 10⁻¹⁹
- [A1] = 1500 V
- [B1] straight line with positive gradient starting at the origin and going as far as x = 15mm
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