Capacitance and Energy Storage — AQA A-level Physics

SPEC 3.7.4 FIELDS AND THEIR CONSEQUENCES

Capacitance

Capacitors, energy storage, and exponential charge/discharge through resistors

What is Capacitance?

Definition

Capacitance is the charge stored per unit potential difference across a capacitor.

C = \frac{Q}{V}

SymbolQuantityUnit

C

CapacitanceFarads (F)

Q

Charge storedCoulombs (C)

V

Potential differenceVolts (V)

Key Concept

1 farad is an extremely large capacitance. In practice, capacitors are measured in microfarads (

μ F = 1 0^{- 6}

F), nanofarads (nF

= 1 0^{- 9}

F), or picofarads (pF

= 1 0^{- 12}

F).

Worked Example01

A capacitor stores 4.5 mC of charge when a potential difference of 9.0 V is applied across it. Calculate the capacitance of the capacitor.

1

Identify the values:

Q = 4.5 mC = 4.5 \times 1 0^{- 3} C

,

V = 9.0 V

2

Apply

C = \frac{Q}{V} = \frac{4.5 \times 1 0 ^{- 3}}{9.0}

3

C = 5.0 \times 1 0^{- 4} F = 500 μ F

Answer: C = 500 μF

Parallel Plate Capacitor

The capacitance of a parallel plate capacitor depends on the plate area, the separation, and the material between the plates:

Formula

C = \frac{A ε _{0} ε _{r}}{d}

FactorEffect on CRelationship

Plate area

A

Larger area → more charge stored

C \propto A

Separation

d

Smaller gap → stronger field

C \propto \frac{1}{d}

Dielectric

ε_{r}

Higher ᵣ → more capacitance

C \propto ε_{r}

Common Mistake

Always convert plate area to m² and separation to m before substituting into the formula. For example,

25 cm^{2} = 25 \times 1 0^{- 4} m^{2}

and

0.5 mm = 5.0 \times 1 0^{- 4} m

.

Dielectrics

A dielectric is an insulating material placed between the plates of a capacitor. It increases the capacitance by a factor of ᵣ (the relative permittivity or dielectric constant).

Key Relationship

The permittivity of the dielectric material:

ε = ε_{0} ε_{r}

The relative permittivity (dielectric constant) is the ratio of the permittivity of the material to the permittivity of free space:

ε_{r} = \frac{ε}{ε _{0}}

where

ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}

is the permittivity of free space.

Exam Tip

Relative permittivity has no units — it is a dimensionless ratio. Typical values: air ≈ 1.0, paraffin ≈ 2, glass ≈ 5–7, water ≈ 80.

Polar Molecule Action

When an electric field is applied across the dielectric, polar molecules rotate to align with the field. This creates a smaller internal electric field that opposes the applied field, reducing the overall electric field between the plates. This allows more charge to be stored for the same p.d., increasing capacitance.

02Worked Example

A parallel plate capacitor has plates of area 0.060 m² separated by a dielectric of thickness 0.50 mm and relative permittivity 2.4. Calculate the capacitance.

( $ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ )

1

Identify:

A = 0.060 m^{2}

,

d = 0.50 \times 1 0^{- 3} m

,

ε_{r} = 2.4

2

Apply

C = \frac{A ε _{0} ε _{r}}{d} = \frac{0.060 \times 8.85 \times 1 0 ^{- 12} \times 2.4}{0.50 \times 1 0 ^{- 3}}

3

Numerator:

0.060 \times 8.85 \times 1 0^{- 12} \times 2.4 = 1.274 \times 1 0^{- 12}

4

C = \frac{1.274 \times 1 0 ^{- 12}}{5.0 \times 1 0 ^{- 4}} = 2.5 \times 1 0^{- 9} F

Answer: C = 2.5 nF

Energy Stored

The energy stored in a capacitor equals the area under a charge–p.d. graph. Since

Q = C V

, the graph is a straight line through the origin and the area is a triangle:

Energy Equations

E = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C}

Exam Tip

Use whichever form of the energy equation is most convenient based on which two of Q, C, and V you know. All three are equivalent and given on the data sheet.

Worked Example03

A 470 μF capacitor is charged to a p.d. of 12 V. Calculate the energy stored, and the charge on the capacitor.

1

C = 470 \times 1 0^{- 6} F

,

V = 12 V

2

Energy:

E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 470 \times 1 0^{- 6} \times 1 2^{2}

3

E = \frac{1}{2} \times 470 \times 1 0^{- 6} \times 144 = 0.034 J = 34 mJ

4

Charge:

Q = C V = 470 \times 1 0^{- 6} \times 12 = 5.64 \times 1 0^{- 3} C = 5.6 mC

Answer: E = 34 mJ, Q = 5.6 mC

Charging Through a Resistor

When a capacitor charges through a resistor, charge builds up exponentially towards a maximum value

Q_{0}

:

Charging Equations

Q = Q_{0} (1 - e^{- t / RC})

V = V_{0} (1 - e^{- t / RC})

I = I_{0} e^{- t / RC}

Important

During charging, the current always decays exponentially (same as discharge). Only Q and V follow the

(1 - e^{- t / RC})

pattern. The initial current is

I_{0} = \frac{V _{0}}{R}

, where

V_{0}

is the supply EMF and

R

is the resistance in the circuit.

Graph Interpretation

The area under the current–time graph gives the total charge stored on the capacitor. The gradient of the charge–time graph at any point gives the current at that instant.

Discharging Through a Resistor

When a charged capacitor discharges through a resistor, all three quantities (Q, V, I) decay exponentially:

Discharge Equations

Q = Q_{0} e^{- t / RC}

V = V_{0} e^{- t / RC}

I = I_{0} e^{- t / RC}

Key Concept

For discharge, all three quantities (Q, V, and I) follow exactly the same exponential decay shape. After one time constant ( = RC), each has fallen to 37% of its initial value.

The initial discharge current is found from:

I_{0} = \frac{V _{0}}{R}

where

R

is the resistance in the discharge circuit.

The Time Constant

Definition

The time constant is the time for the charge (or p.d. or current) to fall to

\frac{1}{e}

(≈ 37%) of its initial value during discharge, or rise to

1 - \frac{1}{e}

(≈ 63%) of its final value during charging.

τ = RC

The time constant is measured in seconds (s). The bigger the value of RC, the slower the rate at which the capacitor charges and discharges.

Half-Life

The time for charge to halve:

T_{1/2} = 0.69 RC = ln 2 \times RC

Time ElapsedDischarge: Fraction RemainingCharging: Fraction of Final Value

t = RC

0.37 (37%)0.63 (63%)

t = 2 RC

0.14 (14%)0.86 (86%)

t = 3 RC

0.050 (5.0%)0.95 (95%)

t = 5 RC

0.0067 (0.67%)0.993 (99.3%)

Exam Tip

A capacitor is considered "fully charged" or "fully discharged" after about 5 time constants (5RC).

Worked Example04

A 220 μF capacitor is fully charged to 9.0 V and then discharged through a 47 kΩ resistor. Calculate: (a) the initial charge, (b) the time constant, (c) the charge remaining after 15 s.

1

(a)

Q_{0} = C V = 220 \times 1 0^{- 6} \times 9.0 = 1.98 \times 1 0^{- 3} C = 1.98 mC

2

(b)

τ = RC = 47 \times 1 0^{3} \times 220 \times 1 0^{- 6} = 10.34 s \approx 10.3 s

3

(c)

Q = Q_{0} e^{- t / RC} = 1.98 \times 1 0^{- 3} \times e^{- 15/10.34}

4

Q = 1.98 \times 1 0^{- 3} \times e^{- 1.451} = 1.98 \times 1 0^{- 3} \times 0.2343

5

Q = 4.64 \times 1 0^{- 4} C \approx 0.46 mC

Answer: (a) 1.98 mC, (b) 10.3 s, (c) 0.46 mC

Worked Example05

A capacitor discharges through a resistor. The voltage across the capacitor is 8.0 V at t = 0 and 2.9 V at t = 4.0 s. Calculate the time constant of the circuit.

1

Use

V = V_{0} e^{- t / RC}

:

2.9 = 8.0 e^{- 4.0/ RC}

2

Divide:

\frac{2.9}{8.0} = e^{- 4.0/ RC}

so

0.3625 = e^{- 4.0/ RC}

3

Take natural log:

ln (0.3625) = \frac{- 4.0}{RC}

4

- 1.015 = \frac{- 4.0}{RC}

so

RC = \frac{4.0}{1.015} = 3.94 s \approx 3.9 s

Answer: = RC = 3.9 s

Worked Example06

A 100 μF capacitor is charged through a 50 kΩ resistor from a 6.0 V supply. Calculate: (a) the time constant, (b) the charge on the capacitor after 8.0 s, (c) the current at t = 8.0 s.

1

(a)

τ = RC = 50 \times 1 0^{3} \times 100 \times 1 0^{- 6} = 5.0 s

2

(b)

Q_{0} = C V_{0} = 100 \times 1 0^{- 6} \times 6.0 = 6.0 \times 1 0^{- 4} C

3

Q = Q_{0} (1 - e^{- t / RC}) = 6.0 \times 1 0^{- 4} (1 - e^{- 8.0/5.0})

4

Q = 6.0 \times 1 0^{- 4} (1 - e^{- 1.6}) = 6.0 \times 1 0^{- 4} (1 - 0.2019) = 6.0 \times 1 0^{- 4} \times 0.7981

5

Q = 4.79 \times 1 0^{- 4} C \approx 0.48 mC

6

(c)

I_{0} = \frac{V _{0}}{R} = \frac{6.0}{50 \times 1 0 ^{3}} = 1.2 \times 1 0^{- 4} A

7

I = I_{0} e^{- t / RC} = 1.2 \times 1 0^{- 4} \times e^{- 1.6} = 1.2 \times 1 0^{- 4} \times 0.2019 = 2.4 \times 1 0^{- 5} A

Answer: (a) 5.0 s, (b) 0.48 mC, (c) 24 μA

Worked Example07

A 330 μF capacitor is charged to 12 V and then discharged through a resistor. The energy stored falls to 25% of its initial value. Calculate the p.d. across the capacitor at this point, and the time taken if R = 22 kΩ.

1

Since

E = \frac{1}{2} C V^{2}

, if energy is 25% of initial:

\frac{1}{2} C V^{2} = 0.25 \times \frac{1}{2} C V_{0}^{2}

2

V^{2} = 0.25 V_{0}^{2}

so

V = 0.50 \times V_{0} = 0.50 \times 12 = 6.0 V

3

Using

V = V_{0} e^{- t / RC}

:

6.0 = 12 e^{- t / RC}

so

0.5 = e^{- t / RC}

4

ln (0.5) = \frac{- t}{RC}

so

t = - RC ln (0.5) = RC ln 2

5

RC = 22 \times 1 0^{3} \times 330 \times 1 0^{- 6} = 7.26 s

6

t = 7.26 \times 0.693 = 5.03 s \approx 5.0 s

Answer: V = 6.0 V, t = 5.0 s

Log-Linear Plotting

Taking natural logarithms of the discharge equation gives a linear relationship:

Derivation

Q = Q_{0} e^{- t / RC}

ln Q = ln Q_{0} - \frac{t}{RC}

This is in the form

y = m x + c

, where plotting ln Q against t gives a straight line with gradient

= - \frac{1}{RC}

and y-intercept

= ln Q_{0}

.

The same applies for voltage:

ln V = ln V_{0} - \frac{t}{RC}

, giving the same gradient.

Therefore:

RC = \frac{- 1}{gradient}

Required Practical 9

In the investigation of capacitor charge and discharge, you use log-linear plotting (ln Q or ln V against t) to determine the time constant RC from the gradient of the best-fit line.

Worked Example08

A student discharges a capacitor through a 47 kΩ resistor and plots a graph of ln V against t. The best-fit line has a gradient of −0.025 s⁻¹ and a y-intercept of 2.48. Calculate: (a) the time constant, (b) the capacitance of the capacitor, (c) the initial voltage across the capacitor.

1

The gradient of ln V against t is

- \frac{1}{RC}

, so:

RC = \frac{- 1}{gradient} = \frac{- 1}{- 0.025} = 40 s

2

C = \frac{RC}{R} = \frac{40}{47 \times 1 0 ^{3}} = 8.5 \times 1 0^{- 4} F = 850 μ F

3

The y-intercept is

ln V_{0}

, so:

V_{0} = e^{2.48} = 11.9 V \approx 12 V

Answer: (a) = 40 s, (b) C = 850 μF, (c) V₀ = 12 V

Practice

Question 1: A capacitor of capacitance 680 μF is charged to a p.d. of 15 V. Calculate: (a) the charge stored, and (b) the energy stored in the capacitor.

1

(a)

Q = C V = 680 \times 1 0^{- 6} \times 15 = 0.0102 C = 10.2 mC

2

(b)

E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 680 \times 1 0^{- 6} \times 1 5^{2} = \frac{1}{2} \times 680 \times 1 0^{- 6} \times 225 = 0.0765 J \approx 77 mJ

Answer: (a) 10.2 mC, (b) 77 mJ

Practice

Question 2: A 47 μF capacitor is fully charged to 5.0 V and then discharged through a 100 kΩ resistor. Calculate the voltage across the capacitor after 6.0 s.

1

RC = 100 \times 1 0^{3} \times 47 \times 1 0^{- 6} = 4.7 s

2

V = V_{0} e^{- t / RC} = 5.0 \times e^{- 6.0/4.7} = 5.0 \times e^{- 1.277}

3

V = 5.0 \times 0.2788 = 1.39 V \approx 1.4 V

Answer: V = 1.4 V

Practice

Question 3: A capacitor is charged from a 10 V supply through a resistor. The time constant of the circuit is 3.0 s. Calculate: (a) the p.d. across the capacitor after 5.0 s, and (b) the time taken for the capacitor to reach 9.0 V.

1

(a)

V = V_{0} (1 - e^{- t / RC}) = 10 (1 - e^{- 5.0/3.0}) = 10 (1 - e^{- 1.667})

2

V = 10 (1 - 0.1889) = 10 \times 0.8111 = 8.1 V

3

(b)

9.0 = 10 (1 - e^{- t /3.0})

so

0.9 = 1 - e^{- t /3.0}

so

e^{- t /3.0} = 0.1

4

\frac{- t}{3.0} = ln (0.1) = - 2.303

so

t = 3.0 \times 2.303 = 6.9 s

Answer: (a) 8.1 V, (b) 6.9 s

Summary

Key Formulas

ConceptFormula

Capacitance

C = \frac{Q}{V}

Parallel plate capacitor

C = \frac{A ε _{0} ε _{r}}{d}

Relative permittivity

ε_{r} = \frac{ε}{ε _{0}}

Energy stored

E = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C}

Time constant

τ = RC

Half-life

T_{1/2} = 0.69 RC

Initial current

I_{0} = \frac{V _{0}}{R}

Discharge

Q = Q_{0} e^{- t / RC}

Charging

Q = Q_{0} (1 - e^{- t / RC})

Log-linear analysis

ln Q = ln Q_{0} - \frac{t}{RC}

, gradient

= \frac{- 1}{RC}

Key Exam Skills

Rearrange exponential equations using natural logarithms to find t or RC

Use log-linear graphs to determine time constants from experimental data

Remember that current always decays exponentially — even during charging

Link energy stored to the area under a Q–V graph

The area under an I–t graph gives total charge; the gradient of a Q–t graph gives the current

Always convert to SI units (m² and m) before using

C = A ε_{0} ε_{r} / d