A student investigates the discharge of a capacitor through a resistor using the circuit shown in Fig. 2.1. [Figure 2.1] The student initially closes the switch and charges the capacitor. The switch is then opened and a stop-watch is started. The capacitor discharges through the resistor. At time t the potential difference V across the capacitor is measured. It is suggested that V and t are related by the equation V = (Q₀/C)e^(-t/RC) where Q₀ is the charge of the fully charged capacitor, C is the capacitance of the capacitor and R is the resistance of the resistor.

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

1. [1] gradient = -1/CR
2. [1] y-intercept = In (Qo/C)
3. [1] In (V/V) values: 1.82 or 1.825, 1.53 or 1.526, 1.22 or 1.224, 0.96 or 0.956, 0.69 or 0.693, 0.34 or 0.336
4. [1] Absolute uncertainties in In V from ± 0.03 or ± 0.04 to ± 0.13 or ± 0.14 or ± 0.15.
5. [1] Six points plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square.
6. [1] Error bars in In V plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
7. [1] Line of best fit drawn. Points must be balanced. Do not accept top point to bottom point. Line must pass between (4.0, 1.6) and (5.0, 1.6) and between (26.5, 0.5) and (28.0, 0.5)
8. [1] Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted.
9. [1] Gradient determined with clear substitution of data points into Δy / Δx. Distance between data points must be at least half the length of the drawn line. Gradient must be negative.
10. [1] uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = 1/2 (steepest worst line gradient – shallowest worst line gradient)
11. [1] y-intercept read from y-axis to less than half a small square or y-intercept determined from substitution into y = mx + c.
12. [1] C determined using gradient and C and Q₀ given to two or three significant figures. Correct substitution of numbers required. -1 / (39×10³ × gradient (c)(iii))
13. [1] Q₀ determined using y-intercept. Q₀ = C × ey-intercept = C × e(c)(iv)
14. [1] C determined using gradient and Q₀ determined using y-intercept and dimensionally correct units for C (F or sΩ⁻¹) and Q₀ (C or Vs⁻¹ or As).
15. [1] ΔC = (0.05 + Δgradient/gradient) × C
16. [1] V determined from (d)(i) (or (c)(iii) and (c)(iv)) with correct substitution shown and correct power of ten. V = Q₀/C × e^(-60/CR) = e^(y-intercept) × e^(gradient×60) or ln V = - (t / RC) + ln (Q₀/C) = - (60 / 39000) × (d)(i) + ln (Q₀/C) or ln V = 60 × gradient + y-intercept or ln V = 60 x (c)(iii) + (c)(iv)