Capacitance and Electrostatic Energy
Put two conductors close together and they become a reservoir for charge and energy. Capacitance measures that ability — and it is set entirely by geometry and the dielectric between, never by how much charge you happen to store.
- The definition of capacitance \(C = Q/V\) and why it depends only on geometry and dielectric.
- The four-step method for finding capacitance from the field.
- The parallel-plate, coaxial and spherical capacitor formulas.
- How a dielectric raises capacitance by a factor \(\varepsilon_r\).
- Combining capacitors in series and parallel.
- The energy stored in a capacitor and its connection to the field energy density.
What Capacitance Is
A capacitor is any pair of conductors carrying equal and opposite charges \(\pm Q\), separated by an insulator. The charge is proportional to the potential difference \(V\) between them, and the constant of proportionality is the capacitance \(C\), measured in farads (C/V):
The crucial point: \(C\) does not depend on \(Q\) or \(V\). Double the charge and the voltage doubles too, leaving the ratio fixed. Capacitance is a property of the geometry of the conductors and the permittivity of the dielectric between them — nothing else.
Every capacitance in this chapter comes from these same five moves. Assume a charge, push it through Gauss's law and the field-potential relation, and the assumed \(Q\) cancels at the end — exactly as it must, since \(C\) is charge-independent.
The Parallel-Plate Capacitor
Two plates of area \(A\), a distance \(d\) apart, filled with a dielectric \(\varepsilon\). From Chapter 4, the field between the plates is uniform, \(E = \rho_S/\varepsilon = Q/\varepsilon A\). Integrating across the gap gives \(V = Ed\), and the assumed \(Q\) cancels:
The message is intuitive: more plate area or a stronger dielectric stores more charge per volt (larger \(C\)); a wider gap stores less. This is the most common capacitor, and the template for the rest.
Coaxial & Spherical Capacitors
The same four-step method, applied to the cylindrical and spherical fields of Chapter 5, gives the other two standard geometries. The coaxial capacitor — two concentric cylinders of radii \(a
The spherical capacitor — two concentric spheres of radii \(a
Letting \(b\to\infty\) for the sphere gives the capacitance of an isolated sphere, \(C = 4\pi\varepsilon a\) — even a single conductor has capacitance, measured against infinity.
The Effect of a Dielectric
Every formula above carries the same factor \(\varepsilon = \varepsilon_0\varepsilon_r\). Replacing the air gap of a capacitor with a dielectric of constant \(\varepsilon_r\) therefore multiplies its capacitance by \(\varepsilon_r\) — storing several times more charge at the same voltage. Combined with the dielectric's higher breakdown strength (Chapter 7), this is why real capacitors are filled rather than empty.
Series & Parallel
Capacitors combine by rules that are the mirror image of resistors. In parallel, the plates share the same voltage and the charges add, so capacitances add. In series, the same charge sits on each and the voltages add, so the reciprocals add:
| Geometry | Capacitance | Field used |
|---|---|---|
| Parallel plate | \(\dfrac{\varepsilon A}{d}\) | Uniform (sheet) |
| Coaxial (length \(L\)) | \(\dfrac{2\pi\varepsilon L}{\ln(b/a)}\) | Line \(1/\rho\) |
| Spherical | \(\dfrac{4\pi\varepsilon ab}{b-a}\) | Point \(1/r^2\) |
| Isolated sphere | \(4\pi\varepsilon a\) | Point \(1/r^2\) |
Energy Stored
Charging a capacitor takes work, stored as electrostatic energy. Building up the charge bit by bit against the rising voltage gives the familiar trio of forms:
This must equal the field energy of Chapter 6. For a parallel-plate capacitor, substituting \(C=\varepsilon A/d\), \(V=Ed\) into \(\tfrac12 CV^2\) gives \(\tfrac12\varepsilon E^2\) times the volume \(Ad\) — recovering the energy density \(w_E=\tfrac12\varepsilon E^2\) exactly. The "energy in the capacitor" and the "energy in the field" are the same energy, counted two ways.
The same idea drives pulsed-power systems: energy charged slowly into a capacitor bank is the field energy released in a fast, high-voltage discharge — the operating principle of Marx generators and PEF chambers.
Worked Examples
Problem. Plates of area \(100\ \text{cm}^2\), gap \(1\ \text{mm}\), dielectric \(\varepsilon_r=4\). Find \(C\).
Solution. Apply \(C=\varepsilon_0\varepsilon_r A/d\):
Problem. A coaxial cable has \(a=1\ \text{mm}\), \(b=3\ \text{mm}\), \(\varepsilon_r=2.3\), length \(1\ \text{m}\). Find \(C\).
Solution. Apply \(C=2\pi\varepsilon L/\ln(b/a)\):
Problem. Find the capacitance of an isolated sphere of radius \(10\ \text{cm}\) in air.
Solution. Apply \(C=4\pi\varepsilon_0 a\):
Problem. \(C_1=2\ \mu\text{F}\) and \(C_2=4\ \mu\text{F}\). Find their series and parallel combinations.
Solution. Add directly for parallel; add reciprocals for series:
Problem. A \(100\ \mu\text{F}\) capacitor is charged to \(400\ \text{V}\). Find the stored energy.
Solution. Apply \(W_E=\tfrac12 CV^2\):
Released in microseconds, this 8 J becomes the pulse energy of a small pulsed-power stage.
Problem. An isolated capacitor holds charge \(Q\) at \(V_0\). A dielectric \(\varepsilon_r=3\) is inserted. Find the new voltage and energy.
Solution. With \(Q\) fixed, \(C\) triples so \(V\) falls; energy \(Q^2/2C\) drops:
The "missing" energy does work pulling the dielectric into the gap.
Chapter Summary
\(C=Q/V\) in farads; depends only on geometry and dielectric, not on \(Q\) or \(V\).
Assume \(Q\), find \(\vec{D}\), then \(\vec{E}\), integrate for \(V\), divide; \(Q\) cancels.
Plate \(\varepsilon A/d\); coaxial \(2\pi\varepsilon L/\ln(b/a)\); spherical \(4\pi\varepsilon ab/(b-a)\).
Raises \(C\) by \(\varepsilon_r\); behaviour differs for fixed \(V\) vs. fixed \(Q\).
Parallel: \(C\) adds. Series: \(1/C\) adds. Opposite of resistors.
\(W_E=\tfrac12 CV^2=\tfrac12 QV=Q^2/2C\); equals the field energy \(\int\tfrac12\varepsilon E^2\,dv\).
Problems
For each item, identify the geometry, apply the four-step method or a standard formula, then handle any combination or energy step. Difficulty rises down the list.
- Plates of area \(50\ \text{cm}^2\) are \(2\ \text{mm}\) apart in air. Find \(C\).
- Find the gap needed for a \(1\ \text{nF}\) air capacitor with \(200\ \text{cm}^2\) plates.
- A coaxial cable has \(b/a=2.5\), \(\varepsilon_r=2\). Find its capacitance per metre.
- A spherical capacitor has \(a=2\ \text{cm}\), \(b=5\ \text{cm}\), air-filled. Find \(C\).
- Three capacitors \(1,2,3\ \mu\text{F}\) are connected in series, then in parallel. Find both totals.
- A \(220\ \mu\text{F}\) capacitor is charged to \(50\ \text{V}\). Find the charge and stored energy.
- Derive the parallel-plate formula from the four-step method, showing the assumed \(Q\) cancels.
- A parallel-plate capacitor is half-filled (in series, by thickness) with two dielectrics \(\varepsilon_{r1},\varepsilon_{r2}\). Find the equivalent capacitance.
- The same capacitor is half-filled side-by-side (in parallel, by area). Find the equivalent capacitance.
- An air capacitor stores \(2\ \text{J}\) at \(1\ \text{kV}\). A dielectric \(\varepsilon_r=4\) is inserted at constant voltage. Find the new energy.
- Show that the energy of a charged isolated sphere equals \(\int\tfrac12\varepsilon_0 E^2\,dv\) integrated from its surface outward.
- A capacitor bank must deliver \(500\ \text{J}\) at \(20\ \text{kV}\). Find the required capacitance and the stored charge.