Capacitance: Definition
A capacitor is one of the most fundamental components in electronics and physics — a device specifically designed to store electrical charge. Think of it like a rechargeable bucket: you can pour charge into it (charging), store that charge for a while, and then let it flow out (discharging) whenever it is needed. The key question is: how good is a particular capacitor at storing charge? That is precisely what capacitance measures. Capacitance is defined as the amount of electric charge stored on a conductor per unit of potential difference (voltage) across it. In other words, it tells you how much charge you get for each volt you apply. A capacitor with a large capacitance is like a wide bucket — it holds a lot of charge even at a modest voltage. A small capacitance is like a narrow tube — the voltage builds up quickly even for a small amount of stored charge. The concept applies to any pair of conductors separated by an insulator (called a dielectric), but the most common example is the parallel-plate capacitor: two flat metal plates facing each other with a small gap between them. Understanding capacitance is essential for analysing circuits, energy storage, camera flashes, computer memory, touchscreens, and countless other technologies that shape modern life.
- Charge Accumulates on Conductors
- Potential Difference Opposes Further Charging
- Defining Capacitance
- What Determines Capacitance?
- Rearranging the Definition
Charge Accumulates on Conductors
When a battery is connected across two conducting plates, positive charge accumulates on one plate and an equal negative charge accumulates on the other. The plates are separated by an insulator, so no charge physically crosses the gap — but an electric field builds up between the plates.
Potential Difference Opposes Further Charging
As charge builds up, a potential difference V (voltage) develops between the plates. This voltage opposes further charge flow — the more charge already stored, the harder it is to push more on. Experiments show that the charge Q stored is directly proportional to the voltage V across the capacitor: Q ∝ V. The constant of proportionality is the capacitance C.
Defining Capacitance
The capacitance C is defined by the equation C = Q / V. Here Q is the magnitude of charge on either plate (in coulombs, C) and V is the potential difference across the capacitor (in volts, V). The SI unit of capacitance is the farad (F), named after Michael Faraday: 1 F = 1 C V⁻¹. One farad is an enormous capacitance in practice; most everyday capacitors are measured in microfarads (μF, 10⁻⁶ F), nanofarads (nF, 10⁻⁹ F), or picofarads (pF, 10⁻¹² F).
What Determines Capacitance?
For a parallel-plate capacitor, the capacitance depends on three physical factors: the area A of the plates (larger area → more room for charge → higher C), the separation d between the plates (smaller gap → stronger field → lower V for same Q → higher C), and the permittivity ε of the material between the plates (insulators with higher permittivity allow more charge storage). This gives C = ε₀εᵣA / d, where ε₀ = 8.85 × 10⁻¹² F m⁻¹ is the permittivity of free space and εᵣ is the relative permittivity of the dielectric material. However, the defining relationship C = Q/V always holds regardless of the geometry.
Rearranging the Definition
The defining equation C = Q/V can be rearranged in two useful ways: Q = CV (to find the charge stored when you know the capacitance and voltage), and V = Q/C (to find the voltage when you know the charge stored). These rearrangements are used constantly when solving circuit problems.
The Key Equation
- C — Capacitance (farad (F)): The ability of a capacitor to store charge per unit voltage. 1 F = 1 C V⁻¹.
- Q — Electric charge stored (coulomb (C)): The magnitude of charge on either plate of the capacitor. Both plates carry equal and opposite charges.
- V — Potential difference (volt (V)): The voltage across the two plates of the capacitor, equal to the work done per unit charge moving between them.
Worked Example
Problem. A capacitor is connected to a 12 V battery and stores a charge of 4.8 × 10⁻⁴ C on its plates. (a) Calculate the capacitance of the capacitor. (b) How much charge would the same capacitor store if the voltage were increased to 20 V?
Answer. The capacitance is 40 μF. At 20 V, the capacitor stores 8.0 × 10⁻⁴ C (doubling the voltage doubles the stored charge).
Origins · 1 minMichael Faraday and how this was discovered
Michael Faraday was a self-educated experimental genius who transformed our understanding of electricity and magnetism. He discovered electromagnetic induction, established the concept of the electric field, and conducted pioneering experiments on capacitors — including careful studies of how different dielectric materials affect the ability of a capacitor to store charge. The SI unit of capacitance, the farad (F), was named in his honour. Faraday came from a very poor family and had almost no formal schooling, yet became one of the most influential scientists in history. He started as a bookbinder's apprentice and taught himself chemistry and physics by reading the books he was sent to bind.
Related Concepts
- Electric Potential and Potential Difference — Capacitance is defined as charge per unit potential difference, so understanding voltage (potential difference) is essential before working with capacitance.
- Electric Charge and Coulomb's Law — The quantity Q in C = Q/V is electric charge. Knowing what charge is and how it behaves underpins the whole concept of capacitance.
- Energy Stored in a Capacitor — Once capacitance is defined, the natural next step is calculating how much electrical energy is stored, given by E = ½CV².
- The Parallel-Plate Capacitor — The parallel-plate geometry gives a concrete formula C = ε₀εᵣA/d that links capacitance to physical dimensions.