Free Pearson Edexcel International A Level Physics lesson
Capacitors and exponential change
Store separated charge, then watch a circuit release its energy.
What you’ll learn
- Use capacitance and derive stored energy.
- Interpret charge and discharge curves.
- Extract RC from logged data and evaluate instrument loading.
Charge storage needs energy
A capacitor stores energy by separating charge. If one plate carries +Q, the other carries −Q, so Q in C = Q/V is the magnitude on one plate, not the algebraic sum of both. Capacitance is measured in farads, where 1 F = 1 C V⁻¹. For a fixed ideal capacitor, Q is proportional to V.
Transferring a small extra charge requires work VΔQ. Since the potential difference rises during charging, the total energy is the area under V plotted against Q. A straight line from the origin gives a triangle with area ½QV. Substituting Q = CV gives W = ½CV² = Q²/(2C), three equivalent expressions for the same stored energy.
Doubling voltage on an unchanged capacitor doubles charge but quadruples stored energy. Distinguish the energy stored from all the energy supplied by a charging source; a resistor can dissipate energy during charging. Respect capacitor voltage ratings and polarity in real circuits, and discharge safely through an appropriate resistor before handling a circuit.
C = Q/V; W = area under V–Q = ½QV = ½CV² = Q²/(2C)
Symbols and units
- C is capacitance in F = C V⁻¹
- Q is stored charge magnitude in C
- V is potential difference in V
- W is stored energy in J.
How the equation explains the idea
- For a constant capacitance, V = Q/C. Charging from zero requires work equal to the triangular area under V against Q: W = QV/2.
- Substitute Q = CV to get W = CV²/2, or V = Q/C to get Q²/(2C). Doubling V at fixed C quadruples energy; doubling C at fixed Q halves energy.
Conditions. Ideal linear capacitor with C = 1.0 mF, charged quasistatically from zero. Charge is plotted in milli-coulombs, so each V mC of area corresponds to 1 mJ.
Stored charge / mC; Capacitor voltage / V.
At 4 mC the voltage is 4 V. The triangular area is ½ × 4 × 4 = 8 V mC = 8 mJ. Using final voltage times all charge would overestimate energy by a factor of two.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A small backup capacitor
- C = 470 μF and V = 12 V.
- Q = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C.
- W = ½ × 470 × 10⁻⁶ × 12² = 0.03384 J.
The plate charge magnitude is 5.64 mC and stored energy is 33.8 mJ.
Keep this idea. Charge scales with voltage, but capacitor energy scales with voltage squared.
A constant fraction leaves in each time interval
During discharge through a resistor, the capacitor drives a current of magnitude I = V/R. As charge falls, V = Q/C falls, so the current and rate of charge loss fall too. This feedback gives exponential decay rather than a straight decline. Each equal time interval removes the same fraction of the remaining charge, not the same amount.
The charge equation is Q = Q₀e^(−t/RC). Since voltage is Q/C and current magnitude is V/R, V and I have the same exponential time factor. The time constant τ = RC is the time to reach 1/e, about 36.8%, of the initial discharge value. During charging from an ideal fixed source, capacitor voltage rises towards the source voltage as V = Vs(1 − e^(−t/RC)).
Taking natural logarithms of a dimensionless voltage ratio gives ln(V/V₀) = −t/RC. A graph against time is straight for the ideal model. A nonzero meter offset produces curvature at low voltages, where the offset becomes a large fraction of the signal. A half-life is τ ln 2, not τ, and stored energy falls twice as fast in its exponent because it depends on V².
Q/Q₀ = V/V₀ = I/I₀ = e^(−t/RC); τ = RC; ln(V/V₀) = −t/RC; Vcharge = Vs(1 − e^(−t/RC))
Symbols and units
- R is effective discharge resistance in Ω
- C is capacitance in F
- τ = RC is time constant in s
- t is time in s
- Q₀ is initial charge magnitude in C, V₀ is initial voltage magnitude in V and I₀ is initial current magnitude in A.
How the equation explains the idea
- During ideal discharge, current magnitude is V/R = Q/(RC). Because this leaves the capacitor, dQ/dt = −Q/(RC), producing Q/Q₀ = exp(−t/RC).
- After one time constant, charge and voltage are e⁻¹ ≈ 0.368 of their initial values; after two, e⁻² ≈ 0.135. Equal time intervals remove an equal fraction, not an equal amount.
- Charging towards a constant supply gives V = Vs(1 − exp(−t/RC)), reaching about 0.632Vs after one time constant.
Conditions. Ideal R = 20 kΩ and C = 100 μF, giving τ = RC = 2.0 s. One experiment discharges from 5 V; the other charges from zero towards a 5 V ideal supply.
Time / s; Capacitor voltage / V.
At t = 2 s, discharge voltage is 5/e ≈ 1.84 V and charging voltage is 5(1 − 1/e) ≈ 3.16 V. Discharge loses the same fraction in equal times, whereas charging approaches the supply asymptotically.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A two-second circuit
- R = 100 kΩ and C = 20 μF give RC = 2.0 s.
- After 2.0 s, a 9.0 V initial voltage becomes 9.0/e = 3.31 V.
- At 4.0 s it becomes 9.0e⁻² = 1.22 V.
The second two-second interval removes the same fraction of a smaller remaining voltage.
Keep this idea. Use RC for the fractional decay timescale and logarithms to test an exponential model.
Measure a decay without changing it too much
Core practical 11 records potential difference across a capacitor as it charges or discharges through a resistor. Select R and C so that the time constant is comfortably longer than the sampling interval but short enough to observe several time constants. A data logger or oscilloscope must have a suitable voltage range and known timebase; check its zero and capacitor polarity.
Connect the voltage input across the capacitor and use a switch arrangement that cleanly selects charge or discharge. Log before switching so the start and initial voltage are visible. Keep the supply, resistor and temperature stable, then obtain several discharge runs. An instrument input resistance comparable with R provides an extra discharge route: use the parallel resistance in the model or choose a much higher-resistance input.
Analyse the curve using the V₀/e time or a log plot across the well-resolved voltage range. Include resistor tolerance and gradient uncertainty when estimating C. Inspect whether successive time intervals show a constant voltage ratio; a curved log plot can indicate leakage, loading or an offset. Repeating an unchanged loaded circuit improves scatter estimates but does not remove loading bias.
Gradient of ln(V/Vref) against t = −1/(Reffective C); C = −1/(R gradient)
Symbols and units
- V and Vref are measured voltage and a fixed positive reference voltage in V
- t is time in s
- R is effective resistance in Ω
- C is capacitance in F.
How the equation explains the idea
- Taking the natural log of V/Vref makes the argument dimensionless and gives a straight line with gradient −1/(RC). The reference changes the intercept, not the slope.
- Solve C = −1/(R × gradient). A meter connected in parallel can reduce effective R, so using the resistor label alone can bias the inferred capacitance.
Conditions. Computed discharge V = 5 exp(−t/2) V with effective R = 20 kΩ, C = 100 μF. Natural logarithms use the fixed 1 V reference; no measurement noise or offset is included.
Time / s; ln(V / 1 V) / dimensionless.
The line has slope −0.5 s⁻¹ and intercept ln 5 ≈ 1.609. Thus C = −1/(20000 × −0.5) = 0.0001 F. A different voltage reference shifts the intercept but leaves the slope unchanged.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Extract capacitance from a slope
- A log-voltage plot has gradient −0.250 s⁻¹.
- The time constant is −1/(−0.250) = 4.00 s.
- With R = 200 kΩ and negligible loading, C = 4.00/200000 = 20.0 μF.
The capacitance follows from the reciprocal slope and known resistance.
Keep this idea. Choose sampling and input resistance that resolve the decay without materially altering it.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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