Define the coulomb.
- (a)Define the coulomb.[1]
- (b)(i)Two vertical metal plates in a vacuum have a separation of 4.0cm. A potential difference of 2.0 × 10² V is applied between the plates. Fig. 5.1 shows a side view of this arrangement. [Figure 5.1] A smoke particle is in the uniform electric field between the plates. The particle has weight 3.9 × 10⁻¹⁵N and charge -8.0 × 10⁻¹⁹C. Show that the electric force acting on the particle is 4.0 × 10⁻¹⁵ N.[2]
- (b)(ii)On Fig. 5.1, draw labelled arrows to show the directions of the two forces acting on the smoke particle.[1]
- (b)(iii)The resultant force acting on the particle is F. Determine 1. the magnitude of F, 2. the angle of F to the horizontal.[3]
- (c)The electric field in (b) is switched on at time t = 0 when the particle is at a horizontal displacement s = 2.0 cm from the left-hand plate. At time t = 0 the horizontal velocity of the particle is zero. The particle is then moved by the electric field until it hits a plate at time t = T. On Fig. 5.2, sketch the variation with time t of the horizontal displacement s of the particle from the left-hand plate. [Figure 5.2][2]
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] (coulomb is) ampere second
- [C1] E = V/d or E = F/Q
- [A1] F = VQ/d = (2.0 × 10² × 8.0 × 10⁻¹⁹)/4.0 × 10⁻² = 4.0 × 10⁻¹⁵ N
- [B1] arrow pointing to the left labelled 'electric force' and arrow pointing downwards labelled 'weight'
- [C1] resultant force = √ [(3.9 × 10⁻¹⁵)² + (4.0 × 10⁻¹⁵)²]
- [A1] magnitude of F = 5.6 x 10⁻¹⁵ N
- [A1] angle = tan⁻¹ (3.9 × 10⁻¹⁵/4.0 × 10⁻¹⁵) = 44°
- [M1] downward sloping line from (0, 2.0)
- [A1] magnitude of gradient of line increases with time and line ends at (T, 0)
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