Cambridge A Level Physics 9702 — 2022 May/June Paper 5 · Variant 1

1 Two parallel metal plates, each of area A, are separated by a small distance d, as shown in Fig. 1.1. area A metal plates d Fig. 1.1 (not to scale) The plates are initially charged using a power supply. The plates are then connected to an uncharged capacitor. The potential difference V across the capacitor is measured. It is suggested that V is related to d by the relationship W Cd = 1 + V KA where C is the capacitance of the capacitor, and K and W are constants. Plan a laboratory experiment to test the relationship between V and d. Draw a diagram showing the arrangement of your equipment. Explain how the results could be used to determine values for K and W. In your plan you should include: ● the procedure to be followed ● the measurements to be taken ● the control of variables ● the analysis of the data ● any safety precautions to be taken. 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[15]

Mark scheme: 1 Defining the problem d is the independent variable and V is the dependent variable or vary d and measure V 1 keep A or area (of overlap) of plates constant 1 Methods of data collection labelled diagram of workable experiment including:  circuit diagram with voltmeter connected in parallel with the capacitor  capacitor and voltmeter connected to the metal plates with no power supply in discharge part of the circuit  correct symbols for capacitor and voltmeter 1 method to charge parallel plates, e.g. separate circuit diagram showing plates connected to a d.c. power supply or combined circuit with switches and d.c. power supply 1 use calipers to measure d or use micrometer/calipers to measure thickness of spacers 1 use rule(r) to measure lengths to determine A and A = length  breadth 1 Method of analysis plot a graph of 1 V against d or equivalent (e.g. d against 1 V ) (Do not accept log graphs.) 1 -intercept or gradient gradient y C C K K A AW      (for d against 1 V : -intercept y C K A   ) 1 Question Answer Marks 1 1 -intercept W y  (for d against 1 V : gradient or -intercept W y  gradient C W AK   ) 1 Additional detail including safety considerations 6 D1 use gloves to prevent electric shock or do not touch metal plates to avoid shocks D2 keep the initial p.d. across plates or initial charge constant D3 method to determine the value of C, e.g. description of an experiment to measure p.d. or current against time during discharge through a resistor D4 method of operation of circuit(s) using switch(es) D5 description of method to fully discharge capacitor, e.g. between experiments, short-circuit the capacitor or use of switch in parallel with capacitor D6 repeat measurements of d at different points across plates and average D7 repeat measurements of V for same d and average V D8 bottom plate resting on insulating material or top plate supported by strings D9 use high voltage power supply to increase charge on plates or use a very small value of capacitance to increase voltmeter reading D10 relationship valid if a straight line is produced (not passing through the origin)

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2 A student investigates the relationship between the luminosity L of a star and its mass M for a set of stars known as main-sequence stars. It is suggested that L and M are related by the equation L = SZM n where S is the luminosity of the Sun, and Z and n are constants. (a) A graph is plotted of lg L on the y-axis against lg M on the x-axis. Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of M and L are given in Table 2.1. Table 2.1 M / 1030 kg L / 1028 W lg (M / 1030 kg) lg (L / 1028 W) 4.8 ± 0.4 1.4 6.4 ± 0.4 3.1 12 ± 2 32 23 ± 2 350 43 ± 4 3600 91 ± 4 66 000 Calculate and record values of lg (M / 1030 kg) and lg (L / 1028 W) in Table 2.1. Include the absolute uncertainties in lg (M / 1030 kg). [2] (c) (i) Plot a graph of lg (L / 1028 W) against lg (M / 1030 kg). Include error bars for lg (M / 1030 kg). [2] (ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines. [2] (iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer. gradient = ......................................................... [2]

Mark scheme: 2(a) gradient = n y-intercept = lgSZ 2(b) lg (M / 1030kg) lg (L / 1028W) 0.68 or 0.681  0.03 or 0.04 0.15 or 0.146 0.81 or 0.806  0.02 or 0.03 0.49 or 0.491 1.08 or 1.079  0.07 or 0.08 1.51 or 1.505 1.36 or 1.362  0.04 2.54 or 2.544 1.63 or 1.633  0.04 3.56 or 3.556 1.96 or 1.959  0.02 4.82 or 4.820 Values of lg (M / 1030kg) and lg (L / 1028W) correct as shown above. 1 Absolute uncertainties in lg (M / 1030kg) correct as shown above. 1 2(c)(i) Six points from (b) plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square. 1 Error bars in lg (M / 1030 kg) plotted correctly. All error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical. 1 2(c)(ii) Straight line of best fit drawn. Points must be balanced. Do not accept line from top point to bottom point. Line must pass between (0.92, 1.0) and (0.96, 1.0) and between (1.86, 4.5) and (1.90, 4.5) 1 Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted. 1 Question Answer Marks 2(c)(iii) Gradient determined with clear substitution of data points into y / x. Distance between data points must be greater than half the length of the drawn line. 1 Gradient of worst acceptable line determined. uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 1 2(c)(iv) y-intercept determined by substitution into y = mx + c. 1 y-intercept of worst acceptable line determined by substitution into y = mx + c. uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line or uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) Do not allow methods using a false origin. 1 Question Answer Marks 2(d) gradient n  (c)(iii) and n and Z both given to two or three significant figures. 1 Value of Z determined using y-intercept. Correct method must be seen. -intercept 28 28 26 10 10 10 10 3.85 10 y Z S      (c)(iv) or -intercept lg 28 10 10 y S Z    or 26 lg 3.85 10 28 10 10 Z     (c)(iv) 1 Absolute uncertainty in n = absolute uncertainty in gradient and   -intercept WAL -intercept 28 10 10 10 y y Z S     Correct substitution of numbers must be seen. 1 2(e) L determined from (d) or (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s). Do not accept incorrect POT for n or Z. L = 3.85  1026  (d)  3.0(c)(iii) or lg lg3.0 -intercept L y    (c)(iii) 1

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