Capacitance and the Parallel Plate Capacitor

In short

A capacitor is any pair of isolated conductors that can hold equal and opposite charges. Its capacitance C is defined by

C \;=\; \frac{Q}{V},

where Q is the charge on the positive plate and V is the potential difference between the plates. Capacitance is measured in farads (1 F = 1 C/V) — a huge unit; practical capacitors sit in microfarads (\muF), nanofarads (nF), or picofarads (pF).

For the parallel plate geometry — two flat conductors of area A separated by a gap d — Gauss's law gives E = \sigma/\varepsilon_0 between the plates and the field-potential relation gives V = Ed, so

\boxed{\;C_\text{parallel plate} \;=\; \frac{\varepsilon_0 A}{d}\;}.

Other standard geometries give similar formulas: for a cylindrical capacitor of length L with inner radius a and outer radius b, C = 2\pi\varepsilon_0 L/\ln(b/a); for a spherical capacitor with inner radius a and outer b, C = 4\pi\varepsilon_0\cdot ab/(b - a); for an isolated sphere of radius R, C = 4\pi\varepsilon_0 R. The key structural fact: capacitance depends on geometry (areas, distances, shapes) and the permittivity of the medium, not on the charge or the voltage. You can double the charge, but V doubles along with it, and the ratio Q/V = C stays fixed.

Slide your finger across your phone screen right now. Somewhere under the glass, a grid of microscopic transparent capacitors — one near every pixel — is watching you. Each is a tiny parallel plate: a transparent conductive film on the inside of the glass, and a ground plane just below. Your finger, when it lands, enters the field between the plates as a third conductor and changes the capacitance by about 0.1 pF. A sensitive circuit on the motherboard registers this change as a touch. That 0.1 pF of capacitance per pixel is what you are trading in every time you scroll Instagram, every time you type a message on WhatsApp. The parallel plate formula C = \varepsilon_0 A/d is what the phone's design engineers used to get the geometry right before silicon fabrication.

The same formula sits behind ISRO flight hardware, where precision reference capacitors with tolerance below 0.1% ride geostationary orbiters and sample instrument voltages. It sits behind the 360-V electrolytic capacitors in a camera flash that discharge their stored energy in milliseconds to fire the Xenon lamp. It sits behind the farad-scale ultracapacitors in Delhi Metro coaches that absorb and release energy during regenerative braking. Every one of these devices is, at its heart, a pair of conductors with a gap between them; the job of this chapter is to understand what makes the gap useful, and how the numbers on the side of a capacitor relate to its physical shape.

What a capacitor is, what capacitance measures

A capacitor is a pair of isolated conductors intended to hold equal and opposite charges. Connect the two plates to the two terminals of a battery of voltage V, and current flows briefly until the plates hold charges +Q and -Q that together produce a potential difference equal to V. At that moment the current stops (there is no complete circuit through the gap) and the capacitor is charged.

The capacitance is defined as the ratio of the charge on one plate to the resulting potential difference:

\boxed{\;C \;=\; \frac{Q}{V}\;} \tag{1}

with units farad (F), where 1 F = 1 C/V.

Why: the more charge you can dump on the plates per volt of potential difference, the better the capacitor stores charge at a given voltage. Capacitance measures precisely this — charge stored per unit potential.

Three things about (1) are worth understanding before you ever compute a specific capacitance:

C is always positive. By convention, Q is the charge on the positive plate and V is the magnitude of the potential difference; both are positive numbers.
C depends on geometry alone (for a fixed medium). You can prove this from the linearity of electrostatics: double Q, and \vec{E} doubles everywhere (by superposition), so V doubles too, and the ratio Q/V is unchanged. C is determined by the shapes of the conductors and their separation, not by how much charge happens to be on them.
The farad is huge. A 1 F capacitor charged to 1 V holds one coulomb — roughly the charge that flows through a 230-V household bulb in 4 seconds. Practical capacitors have capacitances in the \muF, nF, or pF range. A "1 F" label on a supercapacitor is a deliberate engineering feat involving enormous effective surface area.

Units and their sub-multiples

The subdivisions of the farad that you will see on datasheets:

Symbol	Name	Value
F	farad	1 F
mF	millifarad	10^{-3} F
\muF	microfarad	10^{-6} F
nF	nanofarad	10^{-9} F
pF	picofarad	10^{-12} F
fF	femtofarad	10^{-15} F

A ceramic disc capacitor used in an Arduino project is typically 100 nF. An electrolytic capacitor across a power supply is 1000 \muF. A tuning capacitor in an old transistor radio is 300 pF. A single touchscreen pixel is about 0.1 pF. A supercapacitor for regenerative braking may be 100 F.

Deriving C for a parallel-plate capacitor

Take two flat, thin conductors — call them plates — of area A each, mounted face-to-face and separated by a small gap d. The gap is filled with vacuum (or air, which for most purposes behaves like vacuum with permittivity \varepsilon_0). The plate-area is large enough that we can treat the plates as effectively infinite sheets for the purpose of finding the field between them (edge effects live in the going-deeper section).

Two plates of area $A$ carry charges $\pm Q$. Between them, the field is uniform with magnitude $E = \sigma/\varepsilon_0$, where $\sigma = Q/A$.

Step 1 — field between the plates

For a single infinite plane of surface charge density \sigma, Gauss's law gives a field of magnitude \sigma/(2\varepsilon_0) on each side, pointing away from the plane (for positive \sigma). You worked this out in Applications of Gauss's Law — Spheres and Planes.

Put two such plates face to face with equal and opposite surface densities +\sigma and -\sigma. Between them, the fields from the two plates point in the same direction (from + plate to - plate) and add; outside, they cancel. So the field between the plates has magnitude

E \;=\; \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} \;=\; \frac{\sigma}{\varepsilon_0}, \tag{2}

pointing from the positive plate to the negative plate. Outside the pair (above the + plate or below the - plate), the two contributions cancel and E = 0.

Why: the +\sigma plate's field points away from it (downward below, upward above); the -\sigma plate's field points toward it (upward below, downward above). Between the plates, both fields point downward — they reinforce. Above or below the pair, they point in opposite directions — they cancel.

With \sigma = Q/A:

E \;=\; \frac{Q}{\varepsilon_0 A}. \tag{3}

Step 2 — potential difference from the field

The field is uniform between the plates (same magnitude and direction everywhere in the gap). Integrate from the negative plate (V_-) to the positive plate (V_+) against the field:

V_+ - V_- \;=\; -\int_-^+ \vec{E}\cdot d\vec{\ell}.

With \vec{E} pointing from + to - and d\vec{\ell} going the other way (from - to +), the dot product is -E\,d\ell; integrating over distance d:

V_+ - V_- \;=\; -(-E)(d) \;=\; Ed.

Why: the potential at the positive plate is higher than at the negative plate (field points from high V to low V, so V_+ > V_-). Moving from - plate to + plate gains you V = Ed in potential.

Calling this potential difference V (positive by convention):

V \;=\; Ed \;=\; \frac{Qd}{\varepsilon_0 A}. \tag{4}

Step 3 — take the ratio

From (1),

C \;=\; \frac{Q}{V} \;=\; \frac{Q}{Qd/(\varepsilon_0 A)} \;=\; \frac{\varepsilon_0 A}{d}.

\boxed{\;C_\text{parallel plate} \;=\; \frac{\varepsilon_0 A}{d}\;} \tag{5}

Why: Q cancels out of the ratio. The capacitance depends only on the geometry (A and d) and on the permittivity of the medium (\varepsilon_0), as promised. No matter what charge you put on the plates, the ratio Q/V is always \varepsilon_0 A/d.

What the formula says

Three observations about (5):

Bigger plates, more capacitance. Doubling A doubles C — the same voltage pushes twice the charge onto a plate of twice the area.
Smaller gap, more capacitance. Halving d doubles C — the charges on each plate feel more strongly the charges on the other plate, so less voltage is needed for the same amount of charge to be in equilibrium.
No dependence on charge or voltage. C is a property of the geometry, not of what you have done to the plates. Every capacitor in the world with A = 10 cm² and d = 1 mm in vacuum has capacitance C = 8.854\times 10^{-12}\times 10^{-3}/10^{-3} = 8.854 pF, regardless of whether you have put 1 nC or 1 mC on it.

A sanity check with numbers

A lab-bench parallel-plate capacitor has plates of area 0.01 m² (10 cm \times 10 cm) separated by 1 mm of air. What is its capacitance?

C \;=\; \frac{\varepsilon_0 A}{d} \;=\; \frac{8.854\times 10^{-12}\times 0.01}{10^{-3}} \;=\; 8.854\times 10^{-11}\ \text{F} \;=\; 88.54\ \text{pF}.

A modest capacitor, by everyday standards. The reason real capacitors manage microfarads or more in thumb-sized packages is that they (a) roll or stack enormous sheets of very thin conductor to get large A in small volume, and (b) use a dielectric material in the gap instead of vacuum, which increases C by a factor of the dielectric constant \kappa (you will meet this in the dielectrics chapter).

Watching a parallel plate charge up

In the figure below, a battery of voltage V_0 is connected to a parallel plate capacitor at t=0. Electrons flow from the top plate to the bottom plate through the wires, leaving the top plate positive and the bottom plate negative. The potential difference rises toward V_0 and the field between the plates rises from zero to its final value E = V_0/d as the charging completes.

When the battery connects, electrons flow from the top plate (leaving it positive) through the external circuit to the bottom plate (making it negative). The field between the plates builds up until $V_\text{plate} = V_0$, at which point the charging current stops.

Exploring how C depends on d and A

For a fixed plate area $A = 0.05\ \text{m}^2$ and vacuum gap, capacitance falls as $1/d$. Move the red point to see capacitance quadruple when you halve the gap — exactly the $1/d$ scaling of equation (5).

Other standard geometries

Spherical capacitor

A sphere of radius a sits concentrically inside a hollow sphere of radius b > a. The inner sphere carries charge +Q on its surface; the outer sphere carries -Q on its inner surface. Find C.

A spherical capacitor: inner sphere of radius $a$ carrying $+Q$, outer concentric shell of radius $b$ carrying $-Q$. By Gauss's law, the field between them is that of the inner sphere alone.

Between the spheres (a < r < b), Gauss's law with a spherical Gaussian surface of radius r gives

E(r) \;=\; \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.

Why: by spherical symmetry, the field is radial and depends only on r. The Gaussian surface of radius r encloses the inner charge +Q; the outer shell's charge is outside this surface and contributes nothing to the flux.

The potential difference between inner and outer plates is

V \;=\; V(a) - V(b) \;=\; -\int_b^a E\,dr \;=\; \int_a^b E\,dr \;=\; \frac{Q}{4\pi\varepsilon_0}\int_a^b\frac{dr}{r^2} \;=\; \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right) \;=\; \frac{Q(b-a)}{4\pi\varepsilon_0 ab}.

Why: the antiderivative of 1/r^2 is -1/r. Evaluating from a to b gives -1/b + 1/a = (b-a)/(ab). The sign conventions line up because V(a) > V(b) for +Q on the inner sphere.

C_\text{spherical} \;=\; \frac{Q}{V} \;=\; \frac{4\pi\varepsilon_0 ab}{b - a}. \tag{6}

Isolated sphere limit. Take b \to \infty (the outer shell goes to infinity — equivalent to "ground at infinity"). Then ab/(b-a) \to a and

C_\text{isolated sphere} \;=\; 4\pi\varepsilon_0 R, \tag{7}

where R = a is the radius of the sphere. An isolated sphere of radius R has a well-defined capacitance, interpreted as the charge it carries divided by its potential relative to infinity. A 10 cm isolated sphere has C = 4\pi\cdot 8.854\times 10^{-12}\cdot 0.1 \approx 11.1 pF. Our planet (radius 6.37\times 10^6 m) has a "self-capacitance" of about 708 \muF — a useful number in lightning physics.

Cylindrical capacitor

A long conducting cylinder of radius a is surrounded by a concentric conducting shell of inner radius b; both have length L \gg b. Charge +Q sits on the inner cylinder, -Q on the inner surface of the outer shell.

Gauss's law with a cylindrical Gaussian surface (length L, radius a<r<b) gives, by the argument you worked through in the cylindrical-Gauss chapter,

E(r) \;=\; \frac{\lambda}{2\pi\varepsilon_0 r}, \qquad \lambda \;=\; \frac{Q}{L}.

The potential difference is

V \;=\; V(a) - V(b) \;=\; \int_a^b E(r)\,dr \;=\; \frac{\lambda}{2\pi\varepsilon_0}\int_a^b\frac{dr}{r} \;=\; \frac{\lambda}{2\pi\varepsilon_0}\ln\!\frac{b}{a} \;=\; \frac{Q}{2\pi\varepsilon_0 L}\ln\!\frac{b}{a}.

Why: the antiderivative of 1/r is \ln r. The difference \ln b - \ln a = \ln(b/a) captures the exponential nature of the 1/r field's potential drop.

C_\text{cylindrical} \;=\; \frac{Q}{V} \;=\; \frac{2\pi\varepsilon_0 L}{\ln(b/a)}. \tag{8}

This is the capacitance per unit length form you will meet in coaxial cable problems: C/L = 2\pi\varepsilon_0/\ln(b/a). A standard 50-Ω RG-58 coaxial cable, with inner conductor 0.9 mm diameter and outer braid diameter 2.95 mm (and a polyethylene dielectric, \kappa \approx 2.3), gives about 100 pF per metre — a number you will see quoted in any electronics datasheet.

Why capacitance is a property of geometry alone

You saw the calculation: in every geometry we worked out, Q canceled out of the ratio Q/V. This was not a coincidence; it is a consequence of two facts:

Electric fields are linear in the source charges. Double Q, and \vec{E} at every point doubles (by superposition).
Potential is the line integral of the field. Double \vec{E} everywhere, and V doubles.

Together: V is linear in Q, so V = (\text{something depending on geometry})\times Q, and Q/V is "one over that something" — a geometric property.

This is why you can stamp "100 nF" on a capacitor and trust it. The manufacturer never knows what voltage you will connect it to; they don't need to. The capacitance is baked into the geometry the moment the capacitor is built.

Worked examples

Example 1: Designing a parallel-plate capacitor

You want a parallel-plate capacitor of capacitance C = 500\ \text{pF} using two square aluminium plates of side 5 cm each (area A = 25\ \text{cm}^2 = 2.5\times 10^{-3}\ \text{m}^2), separated by a vacuum gap. What plate separation d is required? If the capacitor is rated to 200 V, what is the maximum charge it can hold, and is the air in the gap about to break down?

Design: pick plate separation $d$ so that $\varepsilon_0 A/d = 500$ pF. Then check that the operating field is well below the breakdown strength of air.

Step 1. Solve (5) for d.

d \;=\; \frac{\varepsilon_0 A}{C} \;=\; \frac{8.854\times 10^{-12}\times 2.5\times 10^{-3}}{500\times 10^{-12}} \;=\; \frac{2.2135\times 10^{-14}}{5\times 10^{-10}} \;=\; 4.43\times 10^{-5}\ \text{m}.

d \;\approx\; 44\ \mu\text{m}.

Why: rearranging C = \varepsilon_0 A/d to solve for d gives d = \varepsilon_0 A/C. Plug in the numbers and the gap comes out to about 44 micrometres — smaller than a typical human hair (which is about 70 \mum). That is aggressive but achievable with precision spacers.

Step 2. Maximum charge at rated voltage.

Q_\text{max} \;=\; C V_\text{max} \;=\; 500\times 10^{-12}\times 200 \;=\; 10^{-7}\ \text{C} \;=\; 100\ \text{nC}.

Why: the definition C = Q/V rearranges to Q = CV. At the rating voltage, this is the most charge the capacitor should hold.

Step 3. Check that the field stays below the breakdown of air (E_\text{breakdown} \approx 3\times 10^6 V/m).

E \;=\; \frac{V}{d} \;=\; \frac{200}{4.43\times 10^{-5}} \;\approx\; 4.5\times 10^6\ \text{V/m}.

Why: between the plates the field is uniform with magnitude V/d (from equation 4). For this capacitor, the operating field at rated voltage is about 4.5 MV/m — above the breakdown strength of air. In practice, you would either reduce the voltage rating or fill the gap with a dielectric whose breakdown strength is higher (polyethylene is about 20 MV/m; paper impregnated with oil is higher).

Result: The required gap is about 44 \mum; the maximum charge is 100 nC; and the operating field of 4.5 MV/m is above the breakdown threshold of air, so the air in the gap would ionise near the rated voltage. A practical 500 pF, 200 V capacitor uses a solid dielectric, not air.

What this shows: The parallel-plate formula is the start of a capacitor design, not the end. Once you have the geometry, you must also check the field strength against the dielectric breakdown of whatever fills the gap. This is why real capacitors use paper, ceramic, or polymer films — they sustain much larger fields before failure than vacuum or air.

Example 2: Coaxial cable capacitance

A coaxial cable used for connecting an ISRO telemetry antenna to the payload computer has an inner conductor of diameter 1.2 mm and an outer braid of inner diameter 4.0 mm. The space between is vacuum (for this problem — the real cable uses a polymer). Find the capacitance per unit length.

Coaxial geometry: inner conductor of radius $a$, outer shell of inner radius $b$. Radial field between them.

Step 1. Identify the geometric parameters.

Inner radius a = 0.6 mm = 6\times 10^{-4} m; outer radius b = 2.0 mm = 2\times 10^{-3} m.

Step 2. Apply (8).

\frac{C}{L} \;=\; \frac{2\pi\varepsilon_0}{\ln(b/a)} \;=\; \frac{2\pi\times 8.854\times 10^{-12}}{\ln(2/0.6)} \;=\; \frac{5.56\times 10^{-11}}{\ln 3.333} \;=\; \frac{5.56\times 10^{-11}}{1.204}.

\frac{C}{L} \;\approx\; 4.62\times 10^{-11}\ \text{F/m} \;=\; 46.2\ \text{pF/m}.

Why: \ln(b/a) is dimensionless and captures the geometric factor for the cylindrical geometry. The vacuum permittivity \varepsilon_0 sets the scale.

Step 3. Check with intuition.

If the cable is 10 m long, its total capacitance is about 462 pF — a number you would care about in a pulse-carrying application, where the capacitance adds to the load and slows signal risetimes. Real coaxial cables typically have dielectric inside (polyethylene, PTFE) which multiplies this by the dielectric constant \kappa \approx 2 to \kappa \approx 2.3, giving about 100 pF/m — the number you will see quoted on any RG-type cable datasheet.

Result: C/L \approx 46 pF/m for a vacuum-filled coaxial cable of these dimensions.

What this shows: The cylindrical geometry gives a logarithmic dependence of capacitance on the radius ratio, very different from the parallel-plate 1/d. Doubling b/a from 3.33 to 6.67 only drops capacitance per metre from 46 pF to about 29 pF — geometry matters, but logarithmically, not linearly.

Common confusions

"A capacitor stores charge." Only in a narrow sense. Each plate stores charge, but the two plates together have zero net charge (+Q on one, -Q on the other). What the capacitor really stores is energy — in the electric field between the plates, which you will compute in the next chapter.
"Capacitance depends on the voltage across it." No. C is a property of geometry and medium. Change the voltage and the charge changes proportionally — the ratio stays fixed.
"Bigger plates store more charge per coulomb applied." The phrase doesn't quite make sense, but the idea behind it — that bigger plates are "better" — is correct: for fixed V, larger A means more charge Q = CV.
"A parallel plate capacitor in air works perfectly up to any voltage." Wrong. Air breaks down at about 3 MV/m. As you increase the voltage, the field E = V/d eventually exceeds this threshold; the air ionises and conducts; a spark jumps across the gap and the capacitor discharges uncontrollably. The voltage rating on a capacitor's nameplate is the maximum V below which this does not happen — plus a safety margin.
"The formula C = \varepsilon_0 A/d applies even at the edges of the plates." Strictly, no. The derivation assumed the field between the plates is uniform, which is only true in the idealised infinite-plate limit. Near the edges, field lines bulge outward (fringing fields), and a little extra capacitance is stored in those fringe regions. For plates with A \gg d^2, this correction is small (a few percent). In serious calculations for physically small capacitors, you would need to correct for edge effects.
"C = \varepsilon_0 A/d with \varepsilon_0 in it means this is only for vacuum." Correct. When you fill the gap with a dielectric of dielectric constant \kappa, the capacitance becomes C = \kappa\varepsilon_0 A/d. That is the topic of the upcoming dielectrics chapter; for now, remember that (5) applies to the vacuum/air case and the dielectric generalisation is a constant factor.

If you can compute capacitances of parallel-plate, spherical, and cylindrical capacitors, you have everything you need for JEE. What follows is the edge-effect correction, the energy-density argument for where the energy is stored, and the general formulation of capacitance for conductor systems of any shape.

Edge effects and why C = \varepsilon_0 A/d is an idealisation

The derivation of (5) assumed the field between the plates is everywhere uniform with magnitude E = \sigma/\varepsilon_0. This is exactly true only for infinite plates. For plates of finite extent, the field near the edges curls outward (the field lines must close on negative charges, and those are preferentially closer to the other plate rather than far away) and the region of "uniform field" is slightly smaller than the nominal plate area.

Quantitatively, for a square plate of side s separated by gap d with d \ll s, the fractional correction to C is of order d/s. For s = 5 cm and d = 0.1 mm (typical lab capacitor), d/s = 0.002 = 0.2\% — below the precision of most measurements, and safe to ignore. For touchscreen capacitors where individual pixel capacitances are computed to fractions of a pF, the edge correction becomes a significant design consideration and is handled numerically with finite-element solvers.

Kelvin and Maxwell worked out the leading edge correction in the late 19th century. The result for a circular plate of radius R and gap d \ll R is approximately

C \;\approx\; \frac{\varepsilon_0 A}{d} + \varepsilon_0 R\ln\frac{16\pi R}{d} - \varepsilon_0 R,

with the \ln term coming from the fringing field energy. For R = 5 cm, d = 0.1 mm, the correction is about 0.05 pF on a nominal 7 nF — truly small.

Energy density and where the energy lives

When you charge a capacitor through a potential difference V, you do work \tfrac{1}{2}CV^2 (derived in the next chapter). Where is that energy stored? Not on the plates — they are just carrying charge. The answer is: in the electric field between the plates.

The energy density of the electric field in vacuum is

u \;=\; \tfrac{1}{2}\varepsilon_0 E^2. \tag{9}

Check this for the parallel-plate capacitor: E = V/d, volume between the plates is A d, so

U \;=\; u\times\text{volume} \;=\; \tfrac{1}{2}\varepsilon_0 E^2\cdot Ad \;=\; \tfrac{1}{2}\varepsilon_0 (V/d)^2\cdot Ad \;=\; \tfrac{1}{2}\varepsilon_0 A V^2/d \;=\; \tfrac{1}{2}C V^2. \checkmark

The energy stored in the field matches \tfrac{1}{2}CV^2 exactly. This is not an accident — the energy density formula (9) is the statement of where the capacitor's energy lives. If you pull the plates apart (keeping charge constant), the volume grows, the field stays the same, and so the energy grows — that extra energy comes from the mechanical work you did pulling the plates apart. You will use this argument in the next chapter.

Capacitance for arbitrary conductor systems

For a system of N conductors, the charge on each is linearly related to the potentials of all the conductors:

Q_i \;=\; \sum_{j=1}^N C_{ij}V_j,

where C_{ij} is the capacitance matrix. The diagonal entries C_{ii} are the "self-capacitances" (how much charge builds up on conductor i when you raise its potential keeping others at zero); the off-diagonal entries C_{ij} (for i\neq j) are "mutual capacitances" (how much charge is induced on conductor i when you raise the potential of conductor j).

For a pair of conductors with one at potential +V/2 and the other at -V/2, the familiar capacitance C = Q/V is related to these matrix elements by C = C_{11} + C_{22} - 2C_{12} (or similar, depending on the convention). This reduces to the simple formula (1) in all the special cases we worked out, but the matrix view is what you need when capacitor plates are oddly shaped or when there are more than two conductors (touchscreen arrays are a practical example).

Why the farad is so huge

A 1 F capacitor charged to 1 V holds 1 coulomb — the charge that flows past a point carrying 1 A of current for 1 second. That is a lot; a lightning stroke to ground delivers roughly 5 to 30 C. What it would take to make a 1 F parallel plate capacitor in vacuum with d = 1 mm is an area of

A \;=\; C d/\varepsilon_0 \;=\; 1\times 10^{-3}/(8.854\times 10^{-12}) \;\approx\; 1.13\times 10^{8}\ \text{m}^2,

about 11 000 hectares — roughly 140 IIT Bombay campuses laid out side by side. This is why practical capacitors use tightly rolled foils and dielectric-filled gaps: to squeeze meaningful capacitance into thumb-sized packages.

Where this leads next

Capacitors in Series and Parallel — combining capacitors to build the capacitance you actually need.
Energy Stored in a Capacitor — the \tfrac{1}{2}CV^2 that sits in the field between the plates.
Dielectrics and Polarisation — how filling the gap with an insulator multiplies C by the dielectric constant \kappa.
Capacitors with Dielectrics — Advanced Problems — partially-filled capacitors, series-combined dielectric slabs, the force on a dielectric being inserted.
Gauss's Law — the field derivations used in this chapter; the sphere and cylinder capacitors both rely on Gaussian-surface arguments.