Capacitance

Capacitance

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Explore the simulation above to develop intuition for this topic.

Intuition

A capacitor is like a rechargeable battery that charges instantly: Imagine two metal plates separated by a small gap. When you connect a battery, positive charge accumulates on one plate and negative charge on the other. The gap prevents current from flowing through, but the electric field between the plates stores energy. When you disconnect the battery, the capacitor holds its charge — like a tiny, fast-charging battery.

Why it matters: Capacitors store energy in electric fields and are used in camera flashes (instant high-power discharge), computer memory (storing bits as charged/uncharged), power supplies (smoothing voltage ripple), and tuning circuits (selecting radio frequencies). They’re one of the most common electronic components.

The key insight: Capacitors charge and discharge exponentially — they reach ~63% of full charge in one time constant ($\tau = RC$) and ~99% in 5 time constants. This exponential behaviour is why capacitor circuits have characteristic “charging curves” that look the same regardless of the specific values of R and C.

1. Capacitance

Definition. The capacitance $C$ of a conductor is the charge stored per unit potential Difference:

$\boxed{C = \frac{Q}{V}}$

SI unit: farad (F), where 1 F = 1 C V$^{-1}$.

In practice, capacitances range from picofarads (pF) to millifarads (mF).

Intuition. A capacitor is a device that stores charge. When a p.d. Is applied across two Parallel conducting plates, one plate gains $+Q$ and the other gains $-Q$. The larger the plates and The closer they are, the more charge can be stored for a given p.d. — hence the larger the Capacitance.

2. Parallel Plate Capacitor

Derivation of $C = \varepsilon_0 A/d$

Consider two parallel plates of area $A$ separated by distance $d$With a vacuum between them. A Charge $+Q$ is placed on one plate and $-Q$ on the other.

The electric field between the plates is uniform:

$E = \frac{Q}{\varepsilon_0 A}$

(This comes from Gauss’s law: $\oint \mathbf{E} \cdot d\mathbf{A} = Q_{\mathrm{enclosed}}/\varepsilon_0$Applied to a Gaussian Surface enclosing one plate.)

The p.d. Between the plates is:

$V = Ed = \frac{Qd}{\varepsilon_0 A}$

Therefore:

$\boxed{C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}}$

With a dielectric material of relative permittivity $\varepsilon_r$ between the plates:

$\boxed{C = \frac{\varepsilon_0 \varepsilon_r A}{d}}$

Intuition. Larger plate area allows more charge to be stored. Smaller separation increases the Electric field (and hence the p.d.) for a given charge, but since $C = Q/V$ and $V$ increases more Slowly than $Q$ with decreasing $d$The net effect is that $C$ increases. A dielectric material Polarises in the field, partially cancelling it and allowing more charge to be stored.

3. Energy Stored in a Capacitor

Derivation of $E = \frac{1}{2}CV^2$ from $dE = V\,dQ$

As charge builds up on the capacitor, the p.d. Increases. The work done to transfer a small charge $dQ$ at p.d. $V$ is:

$dE = V\,dQ$

Since $V = Q/C$:

$E = \int_0^Q V\,dQ = \int_0^Q \frac{Q}{C}\,dQ = \frac{1}{C}\left[\frac{Q^2}{2}\right]_0^Q = \frac{Q^2}{2C}$

Using $Q = CV$:

$E = \frac{Q^2}{2C} = \frac{(CV)^2}{2C} = \boxed{\frac{1}{2}CV^2}$

Alternative forms using $Q = CV$ and $V = Q/C$:

$E = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$

Graphical interpretation. The energy stored equals the area under the $V$-$Q$ graph (a straight Line through the origin). This area is a triangle of base $Q$ and height $V$: $E = \frac{1}{2}QV$.

Energy Density

For a parallel plate capacitor, the energy per unit volume between the plates:

$u = \frac{E}{\mathrm{volume}} = \frac{\frac{1}{2}CV^2}{Ad} = \frac{\frac{1}{2}\frac{\varepsilon_0 A}{d} \cdot V^2}{Ad} = \frac{\varepsilon_0 V^2}{2d^2} = \frac{1}{2}\varepsilon_0 E_{\mathrm{field}}^2$

4. Capacitors in Series and Parallel

Parallel Combination

$C_{\mathrm{total}} = C_1 + C_2 + \cdots + C_n$

Proof. All capacitors have the same p.d. $V$. Total charge: $Q = Q_1 + Q_2 + \cdots = C_1 V + C_2 V + \cdots = (C_1 + C_2 + \cdots)V$. Since $Q = C_{\mathrm{total}} V$: $C_{\mathrm{total}} = C_1 + C_2 + \cdots$ $\square$

Series Combination

$\frac{1}{C_{\mathrm{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}$

Proof. All capacitors store the same charge $Q$. Total p.d.: $V = V_1 + V_2 + \cdots = Q/C_1 + Q/C_2 + \cdots = Q(1/C_1 + 1/C_2 + \cdots)$. Since $V = Q/C_{\mathrm{total}}$: $1/C_{\mathrm{total}} = 1/C_1 + 1/C_2 + \cdots$ $\square$

5. RC Circuits: Charging

Consider a capacitor $C$ charged through a resistor $R$ from a supply of e.m.f. $\mathcal{E}$.

Derivation of $Q = Q_0(1 - e^{-t/RC})$

By Kirchhoff’s second law at any instant:

$\mathcal{E} = V_R + V_C = IR + \frac{Q}{C}$

Since $I = dQ/dt$:

$\mathcal{E} = R\frac{dQ}{dt} + \frac{Q}{C}$

Rearranging:

$\frac{dQ}{dt} = \frac{\mathcal{E}}{R} - \frac{Q}{RC}$

Let $Q_0 = C\mathcal{E}$ (the maximum charge when fully charged):

$\frac{dQ}{dt} = \frac{Q_0 - Q}{RC}$

Separating variables and integrating:

$\int_0^Q \frac{dQ}{Q_0 - Q} = \int_0^t \frac{dt}{RC}$

$\left[-\ln(Q_0 - Q)\right]_0^Q = \frac{t}{RC}$

$-\ln(Q_0 - Q) + \ln Q_0 = \frac{t}{RC}$

$\ln\left(\frac{Q_0}{Q_0 - Q}\right) = \frac{t}{RC}$

$\frac{Q_0}{Q_0 - Q} = e^{t/RC}$

$\boxed{Q = Q_0\left(1 - e^{-t/RC}\right)}$

The current is:

$I = \frac{dQ}{dt} = \frac{Q_0}{RC}e^{-t/RC} = \frac{\mathcal{E}}{R}e^{-t/RC} = I_0 e^{-t/RC}$

The p.d. Across the capacitor is:

$V_C = \frac{Q}{C} = \mathcal{E}\left(1 - e^{-t/RC}\right)$

6. RC Circuits: Discharging

When a charged capacitor discharges through a resistor:

$\mathcal{E} = 0 \implies 0 = IR + \frac{Q}{C}$

$R\frac{dQ}{dt} = -\frac{Q}{C}$

$\frac{dQ}{dt} = -\frac{Q}{RC}$

Separating and integrating:

$\int_{Q_0}^Q \frac{dQ}{Q} = -\int_0^t \frac{dt}{RC}$

$\ln Q - \ln Q_0 = -\frac{t}{RC}$

$\boxed{Q = Q_0 e^{-t/RC}}$

$I = \frac{dQ}{dt} = -\frac{Q_0}{RC}e^{-t/RC} = -I_0 e^{-t/RC}$

$V_C = \frac{Q}{C} = V_0 e^{-t/RC}$

7. The Time Constant

Definition. The time constant $\tau = RC$ has units of seconds and characterises the rate of Charging/discharging.

Proof that 63.2% of charge is stored in one time constant

At $t = \tau = RC$:

$\frac{Q}{Q_0} = 1 - e^{-1} = 1 - \frac{1}{e} = 1 - 0.368 = 0.632$

So 63.2% of the maximum charge is stored.

For discharging: $\frac{Q}{Q_0} = e^{-1} = 0.368$So 36.8% remains (63.2% is lost).

Other useful values:

Intuition. The time constant is the “characteristic time” of the circuit. A large resistance Limits the charging current, and a large capacitance requires more charge — both increase the time Needed. After $5\tau$The capacitor is effectively fully charged or discharged.

8. Dielectrics in Detail

Polarisation Mechanism

A dielectric is an insulating material placed between the plates of a capacitor. When an External electric field $E_0$ is applied (by charging the plates), the molecules within the Dielectric respond in one of two ways depending on the material:

1. Polar dielectrics (e.g. Water, HCl): Molecules possess a permanent electric dipole moment. In the absence of a field, these dipoles are randomly oriented. When $E_0$ is applied, the dipoles tend to align with the field. Thermal agitation prevents perfect alignment, but a net polarisation emerges.

2. Non-polar dielectrics (e.g. Polystyrene, glass): Molecules have no permanent dipole moment. The applied field distorts the electron cloud relative to the nucleus, inducing a dipole moment proportional to the applied field.

In both cases, the net effect is the same: the dielectric material becomes polarised — the Positive and negative charges within the molecules shift slightly in opposite directions along the Field lines.

Bound Surface Charges and Field Reduction

Consider a dielectric slab inserted between the plates of a parallel plate capacitor charged to $\pm Q$.

The polarisation creates a layer of bound charge on the surfaces of the dielectric adjacent to The plates:

• The surface of the dielectric facing the $+Q$ plate acquires a negative bound charge density $-\sigma_b$.
• The surface facing the $-Q$ plate acquires a positive bound charge density $+\sigma_b$.

These bound charges produce their own electric field $E_b$ that opposes the applied field $E_0$:

$E_b = \frac{\sigma_b}{\varepsilon_0}$

The net field inside the dielectric is therefore:

$E_{\mathrm{eff}} = E_0 - E_b = E_0 - \frac{\sigma_b}{\varepsilon_0}$

Derivation of $E_{\mathrm{eff}} = E_0/\varepsilon_r$

The polarisation $\mathbf{P}$ of the dielectric is defined as the dipole moment per unit volume. For a linear, isotropic dielectric, the polarisation is proportional to the effective field:

$P = \chi_e \varepsilon_0 E_{\mathrm{eff}}$

Where $\chi_e$ is the electric susceptibility of the material.

The bound surface charge density is related to the polarisation by $\sigma_b = P$So:

$E_{\mathrm{eff}} = E_0 - \frac{P}{\varepsilon_0} = E_0 - \chi_e E_{\mathrm{eff}}$

Rearranging:

$E_{\mathrm{eff}}(1 + \chi_e) = E_0$

$\boxed{E_{\mathrm{eff}} = \frac{E_0}{1 + \chi_e}}$

The relative permittivity (dielectric constant) is defined as:

$\varepsilon_r = 1 + \chi_e$

Therefore:

$\boxed{E_{\mathrm{eff}} = \frac{E_0}{\varepsilon_r}}$

This is the key result: the dielectric reduces the internal electric field by a factor of $\varepsilon_r$.

Capacitance Increase: $C = \varepsilon_r C_0$

The p.d. Between the plates with a dielectric inserted is:

$V = E_{\mathrm{eff}} \cdot d = \frac{E_0}{\varepsilon_r} \cdot d = \frac{V_0}{\varepsilon_r}$

Since the charge $Q$ on the plates is unchanged (assuming the capacitor was isolated before Insertion), the new capacitance is:

$C = \frac{Q}{V} = \frac{Q}{V_0/\varepsilon_r} = \varepsilon_r \frac{Q}{V_0} = \boxed{\varepsilon_r C_0}$

The capacitance increases by a factor of $\varepsilon_r$. Physically, the reduced internal field Means a smaller p.d. For the same charge — the capacitor can store more charge for the same p.d.

Absolute vs Relative Permittivity

The absolute permittivity $\varepsilon$ of a material is:

$\varepsilon = \varepsilon_0 \varepsilon_r$

Where $\varepsilon_0 = 8.854 \times 10^{-12}$ F m$^{-1}$ is the permittivity of free space (vacuum).

For a parallel plate capacitor with a dielectric:

$C = \frac{\varepsilon A}{d} = \frac{\varepsilon_0 \varepsilon_r A}{d}$

Dielectric Strength

The dielectric strength is the maximum electric field a material can withstand before electrical Breakdown occurs (the dielectric becomes conducting). When $E$ exceeds this threshold, electrons are Ripped from their atoms and a conducting path forms — this is what happens in a lightning strike, For instance.

The maximum p.d. A capacitor can tolerate is:

$V_{\mathrm{max}} = E_{\mathrm{breakdown}} \cdot d$

Common Dielectrics

Effect of Dielectric on Energy Stored

Two distinct cases must be considered:

Case 1: Capacitor isolated (charge $Q$ fixed). The dielectric is inserted after the capacitor Has been charged and disconnected from the supply. Since $Q$ is constant and $C$ increases by $\varepsilon_r$The energy decreases:

$E_{\mathrm{new}} = \frac{Q^2}{2C_{\mathrm{new}}} = \frac{Q^2}{2\varepsilon_r C_0} = \frac{E_0}{\varepsilon_r}$

The “missing” energy is extracted by the work done pulling the dielectric into the field (the Polarised dielectric is attracted into the gap).

Case 2: Capacitor connected to supply (p.d. $V$ fixed). The dielectric is inserted while the Capacitor remains connected to the battery. Since $V$ is constant and $C$ increases by $\varepsilon_r$The energy increases:

$E_{\mathrm{new}} = \frac{1}{2}C_{\mathrm{new}}V^2 = \frac{1}{2}\varepsilon_r C_0 V^2 = \varepsilon_r E_0$

The additional energy comes from the battery, which supplies extra charge $Q' = (\varepsilon_r - 1)CV$ to the plates.

9. Charge Sharing Between Capacitors

Setup

Consider a capacitor $C_1$ charged to p.d. $V_1$Carrying charge $Q_1 = C_1 V_1$. It is then Connected (via ideal conducting wires) to an initially uncharged capacitor $C_2$. Charge Redistributes between them until both capacitors reach the same potential difference.

Conservation of Charge

Charge is conserved: the total charge before and after connection must be equal:

$Q_1 = Q_1' + Q_2'$

Where primes denote the final state. The battery (if any) is disconnected, so no charge enters or Leaves the system.

Common Potential Difference

After connection, both capacitors are in parallel (they share the same two nodes), so they reach the Same p.d. $V'$:

$Q_1' = C_1 V', \quad Q_2' = C_2 V'$

Substituting into the charge conservation equation:

$C_1 V_1 = C_1 V' + C_2 V' = (C_1 + C_2)V'$

Solving for $V'$:

$\boxed{V' = \frac{C_1 V_1}{C_1 + C_2}}$

Final Charges

$Q_1' = C_1 V' = \frac{C_1^2 V_1}{C_1 + C_2}, \quad Q_2' = C_2 V' = \frac{C_1 C_2 V_1}{C_1 + C_2}$

Energy is NOT Conserved

Initial energy stored:

$E_i = \frac{1}{2}C_1 V_1^2$

Final energy stored:

$E_f = \frac{1}{2}C_1 V'^2 + \frac{1}{2}C_2 V'^2 = \frac{1}{2}(C_1 + C_2)\left(\frac{C_1 V_1}{C_1 + C_2}\right)^2 = \frac{1}{2}\frac{C_1^2 V_1^2}{C_1 + C_2}$

The energy lost:

$\Delta E = E_i - E_f = \frac{1}{2}C_1 V_1^2 - \frac{1}{2}\frac{C_1^2 V_1^2}{C_1 + C_2} = \frac{1}{2}C_1 V_1^2\left(1 - \frac{C_1}{C_1 + C_2}\right)$

$\boxed{\Delta E = \frac{1}{2}\frac{C_1 C_2}{C_1 + C_2}V_1^2}$

Where does the energy go? The energy loss is real and irrecoverable. During the charge Redistribution, a transient current flows through the connecting wires. The resistance of the wires (however small) dissipates the energy as heat. Additionally, the changing currents produce Electromagnetic radiation. The “missing” energy is accounted for by Joule heating and radiative Losses.

Worked Example

A $10\,\mu$F capacitor is charged to $100$ V and then connected to an uncharged $30\,\mu$F Capacitor. Find the final p.d., the final charge on each capacitor, and the energy lost.

Step 1: Initial charge. $Q_1 = C_1 V_1 = 10 \times 10^{-6} \times 100 = 1.0 \times 10^{-3}$ C.

Step 2: Final p.d. $V' = \frac{C_1 V_1}{C_1 + C_2} = \frac{10 \times 100}{10 + 30} = \frac{1000}{40} = 25$ V.

Step 3: Final charges. $Q_1' = 10 \times 10^{-6} \times 25 = 250\,\mu$C. $Q_2' = 30 \times 10^{-6} \times 25 = 750\,\mu$C. Check: $250 + 750 = 1000\,\mu$C $= Q_1$.

Step 4: Energy. $E_i = \frac{1}{2} \times 10 \times 10^{-6} \times 10000 = 0.050$ J. $E_f = \frac{1}{2} \times 10 \times 10^{-6} \times 625 + \frac{1}{2} \times 30 \times 10^{-6} \times 625 = 3.125 \times 10^{-3} + 9.375 \times 10^{-3} = 12.5 \times 10^{-3}$ J. $\Delta E = 50.0 - 12.5 = 37.5$ mJ.

Using the formula: $\Delta E = \frac{1}{2}\frac{10 \times 30}{10 + 30} \times 10000 \times 10^{-6} = \frac{1}{2} \times 7.5 \times 10^{-6} \times 10000 = 37.5 \times 10^{-3}$ J $= 37.5$ mJ.

10. RC Circuit Applications

Integrator Circuit

Consider a series $RC$ circuit where the output is taken across the capacitor:

$V_{\mathrm{in}} = V_R + V_C = IR + \frac{Q}{C}$

If $RC \gg T$ (the time constant is much larger than the period of the input signal), then $V_R \gg V_C$ at all times during one cycle, so $V_{\mathrm{in}} \approx V_R = IR$. Since $I = dQ/dt$:

$V_{\mathrm{in}} \approx RC\frac{dV_C}{dt}$

Therefore:

$V_{\mathrm{out}} = V_C \approx \frac{1}{RC}\int_0^t V_{\mathrm{in}}\,dt$

The output is approximately the integral of the input. For a square wave input, the output Approximates a triangular wave.

Differentiator Circuit

Consider a series $RC$ circuit where the output is taken across the resistor:

$V_{\mathrm{in}} = \frac{Q}{C} + IR$

If $RC \ll T$ (the time constant is much smaller than the period), then $V_C \gg V_R$So $V_{\mathrm{in}} \approx V_C = Q/C$. The current is:

$I = \frac{dQ}{dt} = C\frac{dV_C}{dt} \approx C\frac{dV_{\mathrm{in}}}{dt}$

Therefore:

$V_{\mathrm{out}} = V_R = IR \approx RC\frac{dV_{\mathrm{in}}}{dt}$

The output is approximately the derivative of the input. For a square wave input, the output Produces sharp spikes at the transitions.

Timing Circuits

The exponential charging/discharging behaviour of RC circuits provides a natural timing mechanism. A Common application is the 555 timer IC, which uses an RC network to define the timing period. The 555 timer charges an external capacitor through a resistor; when the voltage across the Capacitor reaches $2/3$ of the supply voltage, the output switches. The timing period is:

$T \approx 0.693 \cdot RC$

This principle is used in:

• Monostable circuits: a single pulse of fixed duration is produced in response to a trigger input.
• Astable circuits: a continuous square wave output with a frequency determined by $R$ and $C$.

Smoothing Circuits

In AC-to-DC power supplies, a rectifier converts AC to pulsating DC. A capacitor placed in parallel With the load acts as a smoothing filter:

• During the peak of the rectified waveform, the capacitor charges to the peak voltage.
• Between peaks, the capacitor discharges through the load resistor $R_L$Maintaining the output voltage.

The ripple voltage is approximately:

$\Delta V \approx \frac{I_{\mathrm{load}}}{fC}$

Where $I_{\mathrm{load}}$ is the load current and $f$ is the frequency of the rectified AC. A larger $C$ produces smaller ripple, hence smoother DC output.

Flash Photography

A camera flash uses a capacitor to store energy and release it rapidly through a xenon gas tube:

1. A battery slowly charges a capacitor ($C \approx 100$ — $1000\,\mu$F) to a high voltage ($V \approx 300$ V) through a step-up converter circuit.
2. When the shutter is triggered, the capacitor discharges through the flash tube in a time of order $\tau = RC$Where $R$ is the resistance of the ionised gas (very small, $\lt$ 1 $\Omega$).
3. The rapid discharge ($\tau \approx 1$ ms) delivers a large pulse of power: $P = E/\tau \approx \frac{1}{2}CV^2 / \tau$.

Example. A $330\,\mu$F capacitor charged to $300$ V stores $E = \frac{1}{2} \times 330 \times 10^{-6} \times 90000 = 14.85$ J. If discharged in $1$ ms, the Average power is $14.85$ kW.

11. Measuring Capacitance Experimentally

Method 1: RC Discharge Curve

This is the most common method at A-level.

Procedure.

1. Charge the capacitor to a known p.d. $V_0$.
2. Disconnect the supply and immediately start recording the p.d. $V$ across the capacitor at regular time intervals as it discharges through a known resistor $R$.
3. Plot $\ln(V)$ against $t$.

Derivation of the method. From the discharge equation:

$V = V_0 e^{-t/RC}$

Taking natural logarithms:

$\ln V = \ln V_0 - \frac{t}{RC}$

This is a straight line of the form $y = mx + c$ where:

• Gradient $m = -1/(RC)$
• y-intercept $c = \ln V_0$

From the gradient, since $R$ is known:

$C = -\frac{1}{mR}$

Uncertainty analysis. The uncertainty in $C$ depends on:

• Uncertainty in the resistor $R$ ( $\pm 1$% for a standard resistor).
• Uncertainty in the gradient of the $\ln(V)$ vs $t$ graph (from the line of best fit).
• Systematic errors: the voltmeter has its own resistance $R_V$ in parallel with the capacitor. If $R_V$ is not much larger than $R$The effective discharge resistance is $R_{\mathrm{eff}} = \frac{R \cdot R_V}{R + R_V}$Leading to a systematic underestimate of $C$.

Method 2: Repeating Capacitor (Ballistic Galvanometer)

A ballistic galvanometer measures total charge passing through it. If a capacitor $C$ is charged To p.d. $V$ and then fully discharged through the galvanometer, the total charge $Q = CV$ flows, and The galvanometer deflection $\theta$ is proportional to $Q$:

$\theta = kQ = kCV$

Where $k$ is the galvanometer constant. By measuring $\theta$ for a known $V$, $C$ can be determined If $k$ is calibrated using a standard capacitor.

Method 3: Capacitance Bridge

A capacitance bridge (analogous to a Wheatstone bridge for resistors) compares an unknown Capacitance $C_x$ with a known standard $C_s$. At balance:

$\frac{C_x}{C_s} = \frac{R_3}{R_4}$

Where $R_3$ and $R_4$ are known resistances. This method is capable of high precision but is less Common at A-level.

Problem Set