Examstyle Questions: Electric Fields and Forces

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 μC of the charge on each plate
charge = ....................... μC
(3 Marks)
c) the energy stored in μJ stored in the capacitor.
energy = ................ μJ
(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 outline and marking scheme for question:

a) E = v/d = 2000/4 x 10^{3} = 5 x 10^{5}; N C^{1}/ V m^{1}
(2 Marks)
b) Q = CV; = 200 x 10^{12} x 2000l = 4 x 10^{7} = 0.40 (µC)
taking p as 1^{09} 1 mark / corect conversion of any answer to µC gets final mark
(3 Marks)
c) W = ½ CV^{2} ecf possible = ½ QV;
= 0.5 x 200 x 10^{12} x 4 x 10^{6} ; = 400 (µJ)
(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^{19}x 400 x 10^{3}
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^{15}J(1 Mark)
(2 Marks)
(Marks available: 6)