Part 3 · Chapter 25

Capacitance

A capacitor is two conductors holding equal and opposite charge — the simplest device for storing electrical energy in a field. A camera's flash, the tuning of a radio, the smoothing of a power supply, and the memory in your phone all rest on one clean idea: how much charge a pair of conductors will hold per volt across them. That single number is the capacitance, and this chapter shows how to compute it, combine it, fill it with energy, and boost it with a dielectric.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min

i What you'll learn

Capacitance from its defining relation \(q = CV\), with the SI unit the farad (1 F = 1 C/V).
A four-step recipe for computing \(C\) from geometry, giving the parallel-plate result \(C = \dfrac{\varepsilon_{0}A}{d}\) plus the cylindrical, spherical, and isolated-sphere cases.
Combining capacitors: parallel add directly (\(C_{\text{eq}} = \sum C_{j}\)); series add as reciprocals (\(\dfrac{1}{C_{\text{eq}}} = \sum \dfrac{1}{C_{j}}\)).
The energy stored in a charged capacitor, \(U = \dfrac{q^{2}}{2C} = \tfrac{1}{2}CV^{2}\), and the energy density of any electric field, \(u = \tfrac{1}{2}\varepsilon_{0}E^{2}\).
How a dielectric multiplies capacitance by \(\kappa\), weakening the internal field, and how Gauss' law generalises to \(\varepsilon_{0}\oint \kappa\vec{E}\cdot d\vec{A} = q\).

Section 25-1

What Is Physics?

One goal of physics is to supply the basic science behind the devices engineers build. This chapter focuses on one of the most common of all: the capacitor, a device that stores electrical energy. The batteries in a camera, for instance, trickle energy slowly into a capacitor in the flash unit; the capacitor then dumps it all at once, fast enough to fire a brilliant burst of light. A battery alone could never deliver energy that quickly.

The same physics reaches far beyond gadgets. Meteorologists model Earth's atmosphere as one enormous spherical capacitor that periodically discharges as lightning; the static a skier builds gliding over dry snow is charge stored in a body-sized capacitor that sparks to ground. Wherever charge accumulates on conductors and energy hides in an electric field, the language of capacitance applies. Our first question is the most basic one — how much charge can a given pair of conductors hold?

Section 25-2

Capacitance

Every capacitor, whatever its shape, is two conductors — called plates — isolated from each other and their surroundings. When charged, one plate carries \(+q\) and the other \(-q\); we call the magnitude \(q\) "the charge on the capacitor," even though its net charge is zero. Because the plates are conductors, each is an equipotential, and there is a potential difference \(V\) between them. Charge and voltage rise together in strict proportion:

Definition of capacitance

\[ q = CV \]

The constant \(C\) is the capacitance — a measure of how much charge the plates hold per volt between them. It depends only on geometry, not on \(q\) or \(V\). The SI unit is the farad: \(1\,\mathrm{F} = 1\,\mathrm{C/V}\). The farad is enormous, so practical capacitors are rated in microfarads (\(1\,\mu\mathrm{F} = 10^{-6}\,\mathrm{F}\)) and picofarads (\(1\,\mathrm{pF} = 10^{-12}\,\mathrm{F}\)).

How a battery charges a capacitor. Close a switch and the battery's field drives electrons off one plate and onto the other, until the plate-to-plate voltage matches the battery's terminal voltage. Then the field in the connecting wires vanishes, the flow stops, and the capacitor sits fully charged at \(q = CV\) — holding that charge until a circuit lets it discharge.

Section 25-3

Calculating the Capacitance

To find \(C\) for a given geometry we follow a reliable four-step plan: (1) assume a charge \(q\) on the plates; (2) find the field \(\vec{E}\) between them with Gauss' law; (3) integrate the field to get the potential difference \(V\); (4) divide to obtain \(C = q/V\). In every case we choose a Gaussian surface enclosing the charge on the positive plate, where \(\vec{E}\) is uniform and parallel to \(d\vec{A}\), so Gauss' law collapses to a simple form.

The two working equations

\[ q = \varepsilon_{0} E A \qquad\text{and}\qquad V = \int_{-}^{+} E\,ds \]

The first is Gauss' law for the convenient surface (\(A\) = area pierced by the field). The second integrates the field along a path from the negative to the positive plate, chosen to run along a field line so the dot product is just \(E\,ds\). Combining them with \(C = q/V\) yields the capacitance.

Section 25-4

The Parallel-Plate Capacitor

Take two large flat plates of area \(A\) separated by a small gap \(d\), so close that we may ignore the fringing of the field at the edges and treat \(\vec{E}\) as uniform between them. Gauss' law gives \(q = \varepsilon_{0}EA\), and because the field is constant the potential difference is simply \(V = Ed\). Dividing one by the other clears the charge and field entirely:

Capacitance of a parallel-plate capacitor

\[ C = \frac{q}{V} = \frac{\varepsilon_{0} E A}{E d} = \frac{\varepsilon_{0} A}{d} \]

Pure geometry, as promised — bigger plates or a smaller gap give more capacitance. This relation also lets us recast the permittivity constant in a unit handy for circuit work: \(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{F/m} = 8.85\,\mathrm{pF/m}\).

🔌

Why the farad is so large

A 1 F parallel-plate capacitor with a 1 mm gap would need plates of about 100 km².

From \(C = \varepsilon_{0}A/d\), one farad demands an absurd area, which is why real capacitors live in the µF and pF range — and why squeezing the gap, enlarging the area by rolling up foil, or inserting a dielectric are the engineer's three levers for raising \(C\).

Section 25-5

Cylindrical and Spherical Capacitors

The same four steps handle curved geometries; only the Gaussian surface changes. For a cylindrical capacitor — two coaxial cylinders of radii \(a\) and \(b\) and length \(L \gg b\) — the field is radial, \(E = q/(2\pi\varepsilon_{0}Lr)\), and integrating from \(a\) to \(b\) introduces a logarithm. For a spherical capacitor — concentric shells of radii \(a\) and \(b\) — the field is that of a point charge and the integral gives a difference of reciprocals.

Table 25-1 · Capacitance by geometry (each is \(\varepsilon_{0}\) times a length)
Geometry	Capacitance	Eq.
Parallel plates, area \(A\), gap \(d\)	\(C = \dfrac{\varepsilon_{0}A}{d}\)	25-9
Coaxial cylinders, radii \(a,b\), length \(L\)	\(C = \dfrac{2\pi\varepsilon_{0}L}{\ln(b/a)}\)	25-14
Concentric spheres, radii \(a,b\)	\(C = 4\pi\varepsilon_{0}\dfrac{ab}{b-a}\)	25-17
Isolated sphere, radius \(R\)	\(C = 4\pi\varepsilon_{0}R\)	25-18

The isolated sphere is a limit. Let the outer shell of a spherical capacitor expand to infinity (\(b\to\infty\)) in \(C = 4\pi\varepsilon_{0}\,ab/(b-a)\) and you are left with \(C = 4\pi\varepsilon_{0}R\) for a lone conductor of radius \(R\) — its "missing plate" being the distant walls of the room. Every capacitance formula here is \(\varepsilon_{0}\) multiplied by a length, a useful sanity check on any derivation.

Section 25-6

Capacitors in Parallel and in Series

A network of capacitors can often be replaced by a single equivalent capacitor. Two wiring patterns cover most cases. In parallel, the plates are joined so every capacitor feels the same voltage \(V\) ("par-V"); their charges add. In series, they are wired one after another so every capacitor carries the same charge \(q\) ("seri-q"); their voltages add.

Equivalent capacitance

\[ \text{Parallel:}\quad C_{\text{eq}} = \sum_{j=1}^{n} C_{j} \qquad\qquad \text{Series:}\quad \frac{1}{C_{\text{eq}}} = \sum_{j=1}^{n} \frac{1}{C_{j}} \]

Parallel capacitances simply add — the combination behaves like one big capacitor with more plate area. Series capacitances add as reciprocals, so the result is always smaller than the least capacitor in the chain — like stacking gaps to make one wider gap.

🧩

The reduce-and-rebuild strategy

Collapse the network to one capacitor, then work backwards using "par-V" and "seri-q".

Reduce step by step to a single \(C_{\text{eq}}\), find its charge from \(q = C_{\text{eq}}V\), then expand back out: series members inherit that same charge, parallel members inherit that same voltage. Two simple rules unwind even a tangled combination.

Section 25-7

Energy Stored in an Electric Field

Charging a capacitor takes work: each additional increment of charge \(dq\) must be pushed against the field already built up, at cost \(dW = (q/C)\,dq\). Integrating from \(0\) to the final charge \(q\) gives the total work, which is stored as electric potential energy in the field between the plates.

Energy stored in a charged capacitor

\[ U = \int_{0}^{q} \frac{q'}{C}\,dq' = \frac{q^{2}}{2C} = \tfrac{1}{2}CV^{2} \]

Both forms hold for any geometry. Use \(q^{2}/2C\) when the charge is fixed (battery disconnected) and \(\tfrac{1}{2}CV^{2}\) when the voltage is fixed (battery still connected). This is the energy a camera flash or a defibrillator releases in a single fast pulse.

Section 25-8

Energy Density

Where does the stored energy actually reside? In the field. Dividing the energy of a parallel-plate capacitor by the volume \(Ad\) between its plates, and using \(E = V/d\), gives an energy per unit volume that depends only on the field strength:

Energy density of an electric field

\[ u = \frac{U}{Ad} = \tfrac{1}{2}\varepsilon_{0}E^{2} \]

Although derived for a capacitor, this result is completely general: any point where an electric field exists holds energy with density \(\tfrac{1}{2}\varepsilon_{0}E^{2}\), whatever the source of the field. The field is not just a bookkeeping device — it is a genuine reservoir of energy spread through space.

The field is the energy. Two identical capacitors with the same charge but one with twice the gap (hence twice the volume and half the capacitance) store twice the energy — even though their fields are equal in strength. The extra energy lives in the extra volume of field, exactly as \(u = \tfrac{1}{2}\varepsilon_{0}E^{2}\) demands.

Section 25-9

Capacitor with a Dielectric

Faraday discovered in 1837 that filling the gap with an insulating material — a dielectric such as mica, oil, or plastic — multiplies the capacitance by a factor \(\kappa\), the dielectric constant of the material. A vacuum has \(\kappa = 1\) by definition; paper is about 3.5, water about 80, and some ceramics exceed 100.

Effect of a dielectric

\[ C = \kappa\, C_{\text{air}} \qquad\text{and, in any field-containing equation,}\qquad \varepsilon_{0} \to \kappa\varepsilon_{0} \]

A dielectric does two useful things at once: it raises \(C\) and, by physically separating the plates, lets them hold a larger voltage before the material breaks down. Each dielectric has a dielectric strength, the maximum field \(E\) it can withstand before it punctures and conducts.

The microscopic picture explains why. In an external field the molecules of the dielectric — whether they carry permanent dipole moments (polar, like water) or acquire induced ones (nonpolar) — line up so as to produce their own field opposing the applied one. The net field inside the slab is therefore weaker, the voltage for a given charge is smaller, and so \(C = q/V\) rises.

🧱

Battery connected, or not?

Insert a slab with the battery on and the charge grows; with the battery off the voltage drops. Either way C rises by κ.

If \(V\) is held fixed (battery connected), the extra capacitance pulls in more charge, \(q = \kappa C_{\text{air}}V\). If \(q\) is held fixed (battery removed), the voltage falls to \(V/\kappa\) and the stored energy decreases — the capacitor does work pulling the slab in. Both outcomes are consistent with \(q = CV\) and a larger \(C\).

Section 25-10

Dielectrics and Gauss' Law

Gauss' law in Chapter 23 assumed a vacuum. With a dielectric present, the Gaussian surface encloses two kinds of charge: the free charge \(q\) on the conducting plate (which can move) and the induced charge \(q'\) on the dielectric face (which cannot). Accounting for the induced charge by folding \(\kappa\) into the integral gives the most general form of the law.

Gauss' law with a dielectric

\[ \varepsilon_{0}\oint \kappa\vec{E}\cdot d\vec{A} = q \]

Here \(q\) is the free charge only; the induced surface charge is handled entirely by the factor \(\kappa\) on the left. Keeping \(\kappa\) inside the integral allows for surfaces where it is not constant. Set \(\kappa = 1\) and this reduces to the vacuum Gauss' law of Chapter 23.

Why the induced charge is smaller than the free charge. Working through the law gives \(q' = q\left(1 - 1/\kappa\right)\): the induced charge is always less than the free charge, and vanishes when \(\kappa = 1\). Because \(\kappa > 1\) always, a dielectric can only weaken the field a given free charge would otherwise produce — never reverse or overpower it.

Worked Examples

Putting It to Work

1 A parallel-plate capacitor

Problem. Two square plates of side 0.12 m are separated by an air gap of 1.0 mm. Find the capacitance, and the charge stored at 50 V.

Solution. Use \(C = \varepsilon_{0}A/d\), then \(q = CV\).

C = ε₀A/d, then q = CV

\[ C = \frac{(8.85\times10^{-12})(0.12)^{2}}{1.0\times10^{-3}} \approx 1.3\times10^{-10}\,\mathrm{F} = 130\,\mathrm{pF}, \qquad q = CV \approx 6.4\,\mathrm{nC} \]

A modest 130 pF — confirming how large the farad really is, and why practical values sit in the picofarad-to-microfarad range.

2 Parallel versus series

Problem. Capacitors of 12.0 µF and 5.30 µF are combined. Find the equivalent capacitance (a) in parallel and (b) in series.

Solution. Add directly for parallel; add reciprocals for series.

Parallel adds; series adds reciprocals

\[ C_{\parallel} = 12.0 + 5.30 = 17.3\,\mu\mathrm{F}, \qquad C_{\text{series}} = \left(\frac{1}{12.0} + \frac{1}{5.30}\right)^{-1} \approx 3.68\,\mu\mathrm{F} \]

The parallel result exceeds the larger capacitor; the series result is smaller than the smaller one — exactly the behaviour the formulas guarantee.

3 Energy stored and energy density

Problem. A 0.25 µF capacitor is charged to 12 V. Find the stored energy. If the gap is 1.0 mm, find the energy density in the field.

Solution. Use \(U = \tfrac{1}{2}CV^{2}\); the field is \(E = V/d\), and \(u = \tfrac{1}{2}\varepsilon_{0}E^{2}\).

U = ½CV², u = ½ε₀E²

\[ U = \tfrac{1}{2}(0.25\times10^{-6})(12)^{2} \approx 1.8\times10^{-5}\,\mathrm{J}, \qquad u = \tfrac{1}{2}(8.85\times10^{-12})\!\left(\frac{12}{10^{-3}}\right)^{2} \approx 6.4\times10^{-4}\,\mathrm{J/m^{3}} \]

The energy is genuinely spread through the field-filled volume between the plates, not stored "on" the charges themselves.

4 Inserting a dielectric (battery disconnected)

Problem. A 13.5 pF capacitor is charged to 12.5 V and the battery is removed. A porcelain slab (\(\kappa = 6.50\)) is slid in. Find the energy before and after.

Solution. Charge is fixed, so use \(U = q^{2}/2C\); the capacitance becomes \(\kappa C\).

Uᵢ = ½CV², U_f = Uᵢ/κ

\[ U_{i} = \tfrac{1}{2}(13.5\times10^{-12})(12.5)^{2} \approx 1055\,\mathrm{pJ}, \qquad U_{f} = \frac{U_{i}}{\kappa} \approx 162\,\mathrm{pJ} \]

The energy drops by a factor of \(\kappa\). The "missing" 893 pJ is the work the capacitor does pulling the slab in — the slab is sucked into the gap.

5 Energy of a charged sphere

Problem. An isolated conducting sphere of radius \(R = 6.85\,\mathrm{cm}\) carries \(q = 1.25\,\mathrm{nC}\). How much energy is stored in its field?

Solution. An isolated sphere has \(C = 4\pi\varepsilon_{0}R\); then \(U = q^{2}/2C\).

U = q² / (8πε₀R)

\[ U = \frac{q^{2}}{2C} = \frac{q^{2}}{8\pi\varepsilon_{0}R} = \frac{(1.25\times10^{-9})^{2}}{8\pi(8.85\times10^{-12})(0.0685)} \approx 1.0\times10^{-7}\,\mathrm{J} = 103\,\mathrm{nJ} \]

Even a single isolated conductor "is" a capacitor — its missing plate is the room — and it stores energy in the field stretching out around it.

Review

Chapter Summary

✓ Capacitance

\(q = CV\); \(C\) depends only on geometry. Unit: the farad, \(1\,\mathrm{F} = 1\,\mathrm{C/V}\).

✓ Four-step recipe

Assume \(q\) → Gauss' law for \(\vec{E}\) → integrate for \(V\) → \(C = q/V\).

✓ Parallel plates

\(C = \dfrac{\varepsilon_{0}A}{d}\) — grows with area, shrinks with gap.

✓ Curved geometries

Cylinder \(\dfrac{2\pi\varepsilon_{0}L}{\ln(b/a)}\), sphere \(4\pi\varepsilon_{0}\dfrac{ab}{b-a}\), isolated sphere \(4\pi\varepsilon_{0}R\).

✓ Combinations

Parallel: \(C_{\text{eq}} = \sum C_{j}\) ("par-V"). Series: \(\dfrac{1}{C_{\text{eq}}} = \sum \dfrac{1}{C_{j}}\) ("seri-q").

✓ Stored energy

\(U = \dfrac{q^{2}}{2C} = \tfrac{1}{2}CV^{2}\) — the work done to charge it.

✓ Energy density

\(u = \tfrac{1}{2}\varepsilon_{0}E^{2}\), stored in any electric field everywhere it exists.

✓ Dielectrics

\(C = \kappa C_{\text{air}}\); in fields replace \(\varepsilon_{0}\to\kappa\varepsilon_{0}\). Gauss: \(\varepsilon_{0}\oint\kappa\vec{E}\cdot d\vec{A} = q\).

Practice

Problems

Take \(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{F/m}\). For combinations, first decide whether each junction is "par-V" (same voltage) or "seri-q" (same charge); reduce the network to one capacitor, then work backwards for individual charges and voltages.

A capacitor stores 70 pC when the voltage across it is 20 V. Find its capacitance. What charge does it hold at 35 V?
A parallel-plate capacitor has circular plates of radius 8.20 cm separated by 1.30 mm. Find the capacitance, and the charge at 120 V.
A spherical capacitor has inner and outer radii 38.0 mm and 40.0 mm. Find its capacitance.
Find the capacitance of an isolated conducting sphere of radius 0.15 m. To what potential does 1.0 nC raise it?
How many 1.00 µF capacitors in parallel are needed to store 1.00 C at 110 V?
Three capacitors, 10.0 µF, 5.00 µF, and 4.00 µF, are connected in series across 100 V. Find the equivalent capacitance and the charge on each.
Repeat Problem 6 with the same three capacitors connected in parallel across 100 V. Find the charge on each.
A 2.0 µF and a 4.0 µF capacitor are connected in parallel across 300 V. Find the total energy stored.
A parallel-plate air capacitor of area 40 cm² and gap 1.0 mm is charged to 600 V. Find the capacitance, the charge, the stored energy, the field, and the energy density.
How much energy is stored in 1.00 m³ of air carrying a fair-weather field of magnitude 150 V/m?
An air-filled capacitor of capacitance 1.3 pF has its gap doubled and is then filled with wax, giving 2.6 pF. Find the dielectric constant of the wax.
A parallel-plate capacitor with plate area 0.034 m² and gap 2.0 mm is filled with a dielectric of \(\kappa = 5.5\). If the field may not exceed 200 kN/C, find the maximum stored energy.
Capacitor 1 (\(C_{1} = 3.55\,\mu\mathrm{F}\)) is charged to 6.30 V, the battery is removed, and it is connected to an uncharged \(C_{2} = 8.95\,\mu\mathrm{F}\). Find the charge on each at equilibrium.
Starting from \(q = \varepsilon_{0}EA\) and \(V = Ed\), derive \(C = \varepsilon_{0}A/d\), stating clearly where the uniform-field assumption enters.

Tip: three habits keep capacitor problems clean. First, decide at every junction whether it is par-V (parallel — same voltage) or seri-q (series — same charge); mislabelling this is the most common error. Second, reduce the whole network to a single \(C_{\text{eq}}\) before computing anything, then expand back out one step at a time. Third, when a dielectric is involved, ask first whether the battery is still connected (voltage fixed) or removed (charge fixed) — that single fact decides whether charge, voltage, and energy go up or down.