A potential difference of 1.2 × 10³ V is applied between a pair of horizontal metal plates in a vacuum, as shown in Fig. 3.1. [Figure 3.1] The separation of the plates is 3.6cm. The electric field between the plates is uniform. A particle of mass 5.9 × 10⁻⁶kg and charge –4.2 × 10⁻⁹C enters the field at point X with a horizontal velocity of 0.75ms⁻¹ along a line midway between the two plates. The particle is deflected by the field and hits the top plate at point Y.
- (a)(i)a gravitational field[1]
- (a)(ii)an electric field.[1]
- (b)(i)Calculate the magnitude of the electric force acting on the particle in the field.[3]
- (b)(ii)By considering the resultant vertical force acting on the particle, show that the acceleration of the particle in the electric and gravitational fields is 14ms⁻².[4]
- (b)(iii) 1.the time taken for the particle to move from X to Y[2]
- (b)(iii) 2.the distance p of point Y from the left-hand edge of the top plate.[1]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [1] mass
- [1] charge
- [1] E = V/d or E = F/q
- [1] F = (1.2 × 103 × 4.2 × 10-9) / 3.6 × 10-2
- [1] = 1.4 x 10-4 N
- [1] W = mg
- [1] = 5.9 x 10-6 × 9.81
- [1] resultant force = 1.4 × 10–4 – (5.9 × 10–6 × 9.81)
- [1] a = F/m
- [1] a = [1.4 × 10-4 – (5.9 × 10–6 × 9.81)] / [5.9 × 10-6] = 14 m s-2
- [1] s = ut + ½at2
- [1] 1.8 × 10-2 = ½ × 14 × t2
- [1] t = 0.051 s
- [1] p = 0.75 × 0.051
- [1] = 0.038 m
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