A charged oil drop is in a vacuum between two horizontal metal plates. A uniform electric field is produced between the plates by applying a potential difference of 1340 V across them, as shown in [Figure 2.1]. The separation of the plates is 1.4 × 10⁻² m. The oil drop of weight 4.6 × 10⁻¹⁴ N remains stationary at a point mid-way between the plates.

1. (a)(i)Calculate the magnitude of the electric field strength. electric field strength = ...................................................... NC⁻¹[2]
2. (a)(ii)Determine the magnitude and the sign of the charge on the oil drop. magnitude of charge = ...................................................... C sign of charge ....................................................[3]
3. (b)(i)Compare the new pattern of the electric field lines between the plates with the original pattern.[2]
4. (b)(ii)Determine the magnitude of the resultant force acting on the oil drop. resultant force = ............................................................ N[1]
5. (b)(iii)Show that the magnitude of the acceleration of the oil drop is 20 m s⁻².[2]
6. (b)(iv)Assume that the radius of the oil drop is negligible. Use the information in (b)(iii) to calculate the time taken for the oil drop to move to the bottom metal plate from its initial position mid-way between the plates. time = ............................................................ s[2]
7. (c)The oil drop in (b) starts to move at time t = 0. The distance of the oil drop from the bottom plate is x. On [Figure 2.2], sketch the variation with time t of distance x for the movement of the drop from its initial position until it hits the surface of the bottom plate. Numerical values of t are not required.[2]

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