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Log-linear graphs for exponential changes

Edexcel IAL Physics · WPH16 Practical Skills in Physics II

Capacitor discharge, gamma absorption, radioactive decay and damped oscillations all follow exponential laws. A graph of ln y against x turns each into a straight line.

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Why it works

For y = y_0 e^-kx, take natural logs:

ln y = ln y_0 - kx, or ln y = -k x + ln y_0

Plot ln y against x: a straight line with gradient -k and intercept ln y_0. A straight line confirms the exponential relationship.

Practical	Equation	Plot	Gradient
CP11 capacitor discharge	V = V_0 e^-t/RC	ln V against t	-1/RC
CP15 gamma absorption	C = C_0 e^-μx	ln C against x	-μ
Radioactive decay	A = A_0 e^-λt	ln A against t	-λ
Damped oscillations	A = A_0 e^-λnT	ln A against n	-λT

Half values

The time (or thickness) for y to halve is ln 2 / k: for a capacitor, t_1/2 = RC ln 2; for absorption, x_1/2 = ln 2 / μ. From the graph, read where ln y has fallen by ln 2 = 0.693.

Base matters

These equations use e, so plot ln. If you plot log (base 10) instead, the gradient is -k / ln 10 = -0.434k, and you must convert. Examiners did not accept log graphs where ln was required by the question. y_0 = e^c, not 10^c, for an ln graph.

Worked example

A graph of ln (V / V) against t / s for a discharging capacitor passes through (0, 2.48) and (20.0, 1.48). Find RC, V_0, and C if R = 100 kΩ.

Gradient = (1.48 - 2.48) / (20.0 - 0.0) = -0.0500 s^-1
Gradient = -1/RC, so RC = 20.0 s
V_0 = e².48 = 11.9 V
C = RC / R = 20.0 / 1.00 x 10⁵ = 2.00 x 10^-4 F

Answer: RC = 20 s, V_0 = 11.9 V, C = 200 μF

Key idea: y = y_0 e^-kx gives ln y = -kx + ln y_0: gradient -k, intercept ln y_0 (y_0 = e^c). Half value = ln 2 / k. Use ln, not log, for e-based equations.

Check yourself

A graph of ln V against t for a discharging capacitor has gradient -0.0420 s^-1. Find RC in s, to 3 s.f.
Show answer
23.8 s. RC = 1 / 0.0420 = 23.8 s.
RC = 23.8 s and R = 47 kΩ. Find C in μF, to 3 s.f.
Show answer
506 μF. C = 23.8 / 47000 = 5.06 x 10^-4 F = 506 μF.
The intercept of a graph of ln (V / V) against t is 2.20. Find V_0 in V, to 3 s.f.
Show answer
9.03 V. V_0 = e².20 = 9.03 V.
Find the time for the p.d. to halve when RC = 20.0 s, in s, to 3 s.f.
Show answer
13.9 s. t = RC ln 2 = 20.0 x 0.693 = 13.9 s.
In core practical 15, a graph of ln C against x has gradient -0.052 mm^-1. Find the half-thickness of lead in mm, to 3 s.f.
Show answer
13.3 mm. μ = 0.052 mm^-1; x_1/2 = ln 2 / 0.052 = 13.3 mm.
A student plots log10 V instead of ln V against t for a capacitor with RC = 20.0 s. What gradient would the log10 graph have, in s^-1, to 3 s.f.?
Show answer
-0.0217 s^-1. Gradient = -(1/RC) / ln 10 = -0.0500 / 2.303 = -0.0217 s^-1.
Which graph gives a straight line for a capacitor discharging through a resistor?
- V against t
- V against 1/t
- log V against log t
- ln V against t
Show answer
ln V against t. V = V_0 e^-t/RC becomes ln V = -t/RC + ln V_0.
For a graph of ln y against x with intercept c, how is y_0 found?
- y_0 = 10^c
- y_0 = c
- y_0 = e^c
- y_0 = ln c
Show answer
y_0 = e^c. ln y_0 = c, so y_0 = e^c.

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Last updated 8 November 2026. Spotted a mistake? Tell us.