A spherical oil drop has a radius of 1.2 × 10⁻⁶ m. The density of the oil is 940 kg m⁻³.
- (a)(i)Show that the mass of the oil drop is 6.8 × 10⁻¹⁵ kg.[2]
- (a)(ii)The oil drop is charged. Explain why it is impossible for the magnitude of the charge to be 8.0 × 10⁻²⁰ C.[1]
- (b)The charged oil drop in (a) is in a vacuum between two horizontal metal plates, as illustrated in Fig. 4.1. [Figure 4.1] The plates are separated by a distance of 8.0 mm. The electric field between the plates is uniform and has a field strength of 2.1 × 10⁵ V m⁻¹. The oil drop moves vertically downwards with a constant speed.
- (b)(i)Calculate the potential difference V between the plates.[2]
- (b)(ii)Explain how the motion of the oil drop shows that it is in equilibrium.[1]
- (b)(iii)Determine the charge on the oil drop. sign of charge[3]
- (c)The magnitude of the potential difference between the plates in (b) is decreased.
- (c)(i)Explain why the oil drop accelerates downwards.[2]
- (c)(ii)Describe the change to the pattern of the field lines (lines of force) representing the uniform electric field as the potential difference decreases.[1]
- (d)Two types of force, X and Y, can act on an oil drop when it is in air, but cannot act on an oil drop when it is in a vacuum. Force X can act on an oil drop when it is stationary or when it is moving. Force Y can only act on an oil drop when it is moving. State the name of:
- (d)(i)force X[1]
- (d)(ii)force Y.[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 (May/June 2019). The mark or mark type is shown in brackets.
- [1] m = ρ × V
- [1] = (4 / 3) × π × (1.2 × 10⁻⁶)³ × 940 = 6.8 × 10⁻¹⁵ (kg)
- [1] minimum charge (on drop) is 1.6 × 10⁻¹⁹ C
- [1] V = Ed
- [1] V = 2.1 × 10⁵ × 8.0 × 10⁻³ = 1.7 x 10³ V
- [1] constant velocity so no resultant force (so in equilibrium)
- [1] mg = Eq or mg = (V/d)q or F = mg and F = Eq
- [1] q = (6.8 x 10⁻¹⁵ × 9.81)/2.1 × 10⁵ = 3.2 x 10⁻¹⁹ C
- [1] sign of charge is negative
- [1] electric force decreases
- [1] weight > electric force or resultant force acts downwards
- [1] (field line) separation increases
- [1] upthrust (force)
- [1] air resistance/drag/viscous (force)
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →