Capacitors and capacitance

1. Overview

Capacitance is the measure of a system's ability to store electric charge per unit potential difference. In any capacitor, work must be done to separate opposite charges; this work is then stored as electrical potential energy within the electric field between the conductors. The fundamental principle of a capacitor relies on the fact that the charge $Q$ residing on the conductors is directly proportional to the potential difference $V$ established between them. This relationship holds true for various geometries, including the standard parallel plate arrangement and isolated spherical conductors. Understanding capacitance is essential for analyzing how energy is managed in electronic timing circuits, smoothing circuits, and high-power pulse applications.

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Key Definitions

• Capacitance ($C$): The charge stored per unit potential difference across a conductor or between two conductors.
• Farad ($\text{F}$): The SI unit of capacitance. One Farad is defined as one coulomb of charge stored per volt of potential difference ($1\text{ F} = 1\text{ C V}^{-1}$).
• Dielectric: An electrically insulating material (such as mica, paper, or ceramic) placed between capacitor plates to increase the capacitance by reducing the internal electric field for a given charge.
• Parallel Plate Capacitor: A device consisting of two parallel conducting plates of a certain area, separated by a small distance containing an insulator.
• Isolated Spherical Conductor: A single conducting sphere that stores charge. Its capacitance is defined relative to a theoretical second "plate" at infinity where the potential is zero.

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Content

3.1 The Fundamental Equation of Capacitance

When a capacitor is connected to a direct current (DC) source, such as a battery, electrons are displaced from one conducting plate and deposited onto the other. This process continues until the potential difference across the capacitor plates equals the electromotive force (e.m.f.) of the source.

The relationship is defined by the equation:

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

• $Q$: The magnitude of the charge on one of the plates (measured in Coulombs, $\text{C}$). Note: The net charge of a capacitor is zero, but $Q$ refers to the charge separated.
• $V$: The potential difference (p.d.) across the plates (measured in Volts, $\text{V}$).
• $C$: The capacitance (measured in Farads, $\text{F}$).

Physical Interpretation: A capacitor with a high capacitance can store a large amount of charge for a relatively small potential difference. In a circuit, the capacitance is a constant determined by the physical geometry (area and separation of plates) and the material between them.

3.2 Capacitance of an Isolated Spherical Conductor

While we usually think of capacitors as two plates, a single isolated sphere can also hold charge and thus possesses capacitance. To calculate this, we consider the sphere's potential relative to a point at infinity (where $V = 0$).

Derivation of $C = 4\pi\epsilon_0 r$:

1. Consider an isolated sphere of radius $r$ carrying a charge $+Q$.
2. From the study of Electric Fields, the electric potential $V$ at the surface of a sphere is given by: $V = \frac{Q}{4\pi\epsilon_0 r}$ (where $\epsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12} \text{ F m}^{-1}$)
3. The definition of capacitance is $C = \frac{Q}{V}$.
4. Rearranging the potential formula to find $\frac{Q}{V}$: $V \cdot (4\pi\epsilon_0 r) = Q$ $\frac{Q}{V} = 4\pi\epsilon_0 r$
5. Therefore: $C = 4\pi\epsilon_0 r$

Key Insight: The capacitance of a sphere depends solely on its radius. The larger the sphere, the more charge it can hold for a given potential. This is why large Van de Graaff generator domes can reach much higher voltages before discharging than smaller ones.

3.3 Capacitors in Parallel

When capacitors are connected in parallel, they all experience the same potential difference $V$ provided by the source.

Derivation of the Parallel Formula:

1. Conservation of Charge: The total charge $Q_{total}$ supplied by the source is distributed among the capacitors: $Q_{total} = Q_1 + Q_2 + Q_3 + \dots$
2. Substitution: Using $Q = CV$, we substitute for each capacitor: $C_{total}V = C_1V + C_2V + C_3V + \dots$
3. Simplification: Since $V$ is the same for all components in parallel, we divide the entire equation by $V$: $C_{total} = C_1 + C_2 + C_3 + \dots$

Physical Reasoning: Connecting capacitors in parallel is equivalent to increasing the total surface area of the plates. Since $C \propto \text{Area}$, the total capacitance increases.

3.4 Capacitors in Series

In a series circuit, the capacitors are connected end-to-end. This creates a unique situation regarding charge distribution.

The Principle of Charge in Series: When a battery is connected, it pulls electrons from the left plate of $C_1$ and pushes them onto the right plate of $C_n$. The intermediate plates become charged through induction. For every electron that arrives at one plate, one electron is repelled from the opposite plate. Consequently, the magnitude of charge $Q$ on every capacitor in series is identical.

Derivation of the Series Formula:

1. Kirchhoff’s Second Law: The total potential difference $V_{total}$ across the combination is the sum of the individual potential differences: $V_{total} = V_1 + V_2 + V_3 + \dots$
2. Substitution: Using $V = \frac{Q}{C}$, we substitute for each potential difference: $\frac{Q}{C_{total}} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \dots$
3. Simplification: Since the charge $Q$ is common to all capacitors in series, we divide the entire equation by $Q$: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots$

Physical Reasoning: Connecting capacitors in series effectively increases the total separation distance between the "outermost" plates. Since capacitance is inversely proportional to the separation of plates ($C \propto 1/d$), the total capacitance decreases. The total capacitance in series is always less than the smallest individual capacitor in the string.

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3.5 Worked Examples

Worked example 1 — Basic Definition and Units

A capacitor stores $4.5 \times 10^{-4} \text{ C}$ of charge when the potential difference across its plates is $15 \text{ V}$. Calculate the capacitance in microfarads ($\mu\text{F}$).

• Step 1: State the formula. $C = \frac{Q}{V}$
• Step 2: Substitute the values. $C = \frac{4.5 \times 10^{-4}}{15}$
• Step 3: Calculate the value in Farads. $C = 3.0 \times 10^{-5} \text{ F}$
• Step 4: Convert to microfarads. $3.0 \times 10^{-5} \text{ F} = 30 \times 10^{-6} \text{ F} = 30 \mu\text{F}$
• Answer: $30 \mu\text{F}$

Worked example 2 — Isolated Sphere

A metal sphere used in a high-voltage experiment has a diameter of $40 \text{ cm}$. Calculate the charge required to raise the potential of the sphere to $50 \text{ kV}$.

• Step 1: Identify the radius. $r = \frac{\text{diameter}}{2} = \frac{0.40}{2} = 0.20 \text{ m}$
• Step 2: Calculate the capacitance of the sphere. $C = 4\pi\epsilon_0 r$ $C = 4 \times \pi \times (8.85 \times 10^{-12}) \times 0.20$ $C \approx 2.22 \times 10^{-11} \text{ F}$
• Step 3: Use $Q = CV$ to find the charge. $V = 50 \text{ kV} = 50,000 \text{ V}$ $Q = (2.22 \times 10^{-11}) \times 50,000$ $Q = 1.11 \times 10^{-6} \text{ C}$
• Answer: $1.1 \mu\text{C}$ (to 2 s.f.)

Worked example 3 — Complex Capacitor Network

Three capacitors are arranged as follows: a $20 \mu\text{F}$ and a $30 \mu\text{F}$ capacitor are connected in parallel. This parallel combination is then connected in series with a $10 \mu\text{F}$ capacitor. Calculate the total capacitance of the circuit.

• Step 1: Calculate the parallel branch capacitance ($C_p$). $C_p = C_1 + C_2$ $C_p = 20 \mu\text{F} + 30 \mu\text{F} = 50 \mu\text{F}$
• Step 2: Calculate the total series capacitance ($C_{total}$). The circuit is now effectively a $50 \mu\text{F}$ capacitor in series with a $10 \mu\text{F}$ capacitor. $\frac{1}{C_{total}} = \frac{1}{50} + \frac{1}{10}$ $\frac{1}{C_{total}} = 0.02 + 0.10 = 0.12$
• Step 3: Take the reciprocal. $C_{total} = \frac{1}{0.12} \approx 8.333 \mu\text{F}$
• Answer: $8.3 \mu\text{F}$ (to 2 s.f.)

Worked example 4 — Potential Distribution in Series

Two capacitors, $C_1 = 100 \mu\text{F}$ and $C_2 = 470 \mu\text{F}$, are connected in series to a $12 \text{ V}$ battery. Calculate the potential difference across the $100 \mu\text{F}$ capacitor.

• Step 1: Find the total capacitance. $\frac{1}{C_{total}} = \frac{1}{100} + \frac{1}{470} \approx 0.012127$ $C_{total} \approx 82.46 \mu\text{F}$
• Step 2: Find the total charge $Q$ stored in the series chain. $Q = C_{total} \times V_{total}$ $Q = 82.46 \times 10^{-6} \times 12 \approx 9.895 \times 10^{-4} \text{ C}$
• Step 3: Calculate $V_1$ using the fact that $Q$ is the same for all series capacitors. $V_1 = \frac{Q}{C_1}$ $V_1 = \frac{9.895 \times 10^{-4}}{100 \times 10^{-6}} = 9.895 \text{ V}$
• Answer: $9.9 \text{ V}$ (to 2 s.f.)

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Key Equations

Equation	Description	Status
$C = \frac{Q}{V}$	Definition of Capacitance	Data Sheet
$C = 4\pi\epsilon_0 r$	Capacitance of an isolated sphere	Memorise
$C_{total} = C_1 + C_2 + \dots$	Capacitors in parallel	Memorise
$\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$	Capacitors in series	Memorise
$\epsilon_0 \approx 8.85 \times 10^{-12} \text{ F m}^{-1}$	Permittivity of free space	Data Sheet

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Common Mistakes to Avoid

• ❌ Wrong: Applying resistor combination rules to capacitors. ✓ Right: Remember they are the inverse of resistor rules. Parallel capacitors add directly ($C_1 + C_2$); series capacitors use the reciprocal sum.
• ❌ Wrong: Summing the charges on both plates to find $Q$. ✓ Right: $Q$ is the magnitude of charge on one plate only. If a capacitor is labeled "$10 \mu\text{C}$", it means $+10 \mu\text{C}$ on one plate and $-10 \mu\text{C}$ on the other.
• ❌ Wrong: Forgetting to invert the final value in series calculations. ✓ Right: After calculating $\frac{1}{C_1} + \frac{1}{C_2}$, you must perform the $x^{-1}$ operation to find $C_{total}$.
• ❌ Wrong: Ignoring unit prefixes like $\mu$ ($10^{-6}$), $n$ ($10^{-9}$), or $p$ ($10^{-12}$). ✓ Right: Always convert to Farads ($\text{F}$) before plugging numbers into $C = Q/V$ or $C = 4\pi\epsilon_0 r$.
• ❌ Wrong: Assuming the potential difference is the same for capacitors in series. ✓ Right: In series, the charge is the same; the potential difference is shared (inversely proportional to capacitance).

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Exam Tips

1. Derivation Precision: If asked to derive the series or parallel formula, start with a statement of physics laws. For parallel, state "The total charge is the sum of individual charges ($Q = Q_1 + Q_2$)." For series, state "The total p.d. is the sum of individual p.d.s ($V = V_1 + V_2$)."
2. The "Isolated Sphere" Context: If a question mentions a planet, a Van de Graaff dome, or a "charged drop of oil," immediately think of the $C = 4\pi\epsilon_0 r$ formula.
3. Significant Figures: Cambridge 9702 is strict on s.f. If the data provided is $2.0 \mu\text{F}$ and $12 \text{ V}$, give your answer to 2 s.f. ($24 \mu\text{C}$), not 1 or 4.
4. Ratio Questions: Many exam questions ask what happens if the radius of a sphere doubles. Use the formula to show $C \propto r$, so the capacitance also doubles.
5. Circuit Symbols: When drawing a capacitor, ensure the two lines are parallel and of equal length. If one is shorter or thicker, it is a battery/cell symbol and you will lose marks.
6. Unit Conversions: Practice converting between $\text{pF}$, $\text{nF}$, and $\mu\text{F}$ quickly.
   • $1 \mu\text{F} = 10^3 \text{ nF} = 10^6 \text{ pF}$
   • $1 \text{ pF} = 10^{-12} \text{ F}$