Fig. 4.1 shows two large parallel insulated capacitor plates, seperated by an air gap of 4.0 x 10-3 m. The capacitance of the arrangement is 200 pF. The plates are connected by a switch to a 2000 V d. d. power supply. The switch is closed and then opened. Calculate a) the magnitude of the electric field strength between the plates givinga suitable unit ofr your answer. electric field strength = .................... unit ...... (2 Marks) b) the magnitude in μ C of the charge on each plate charge = ....................... μ C (3 Marks) c) the energy stored in μ J stored in the capacitor. energy = ................ μ J (3 Marks) (Marks available: 8) The diagram shows a pair of flat, wide metal plates. They are parallel and connected to a constant 2000 V supply. a) What is the electric field (E) between the plates? (2 Marks) b) A drop of oil (D), between the plates carries a change of 10 electrons (each 1.6 x 10-19 C). What is the force on the drop? (2 Marks) c) If the drop moves a distance of 2.5 mm towards the positive plate, how much electrical energy is transferred? (2 Marks) (Marks available: 6) Answer Answer outline and marking scheme for question: a) E = v/d = 2000/4 x 10-3 = 5 x 105; N C-1/ V m-1 (2 Marks) b) Q = CV; = 200 x 10-12 x 2000l = 4 x 10-7 = 0.40 (µ C) taking p as 10-9 -1 mark / corect conversion of any answer to µ C gets final mark (3 Marks) c) W = ½ CV2 ecf possible = ½ QV; = 0.5 x 200 x 10-12 x 4 x 106 ; = 400 (µ J) (3 Marks) (Marks available: 8) a) E = V = 2000 (1 Mark) = 400 kV m-1 (or kJC-1) (1 Mark) d 5 x 10-3 (2 Marks) b) F = qE (1 Mark) = 10 x 1.6 x 10-19x 400 x 103 F = 6.4 x 10-13 N (1 Mark) (2 Marks) c) W = Fd (1 Mark) = 6.4 x 10-13 x 2.5 x 10-3 W = 1.6 x 10-15J(1 Mark) (2 Marks) (Marks available: 6)