Parallel plate capacitor

Pull the plug on any phone charger, laptop adaptor, or camera flash and, somewhere inside, two metal plates are sitting a hair's breadth apart, doing nothing more dramatic than storing charge on their faces. That deceptively simple object — the parallel-plate capacitor — is the workhorse of the whole chapter "Electrostatic Potential and Capacitance," the geometry every other capacitor idea gets tested against. Two flat conducting plates, area $A$ each, separated by a small gap $d$ , one connected to the positive terminal of a battery and the other to the negative. Charge $+Q$ piles onto one plate, $-Q$ onto the other, a uniform electric field springs up in the gap between them, and a potential difference $V$ appears across that gap. The ratio $Q/V$ is the capacitance $C$ — and for this one geometry, that ratio turns out to depend on nothing but the plates' area, their separation, and the electric constant $\varepsilon_0$ of empty space: $C_0 = \varepsilon_0 A/d$ . Slide a slab of glass, mica or plastic into the gap and something else happens: the dielectric's own bound charges partially cancel the field inside it, the same $Q$ now produces a smaller $V$ , and the capacitance rises by a factor $K$ , the material's dielectric constant. That single fact — squeeze more capacitance out of the same plates by choosing a better insulator — is why every real capacitor you have ever touched, from a ceramic disc on a circuit board to the giant electrolytic cans inside a television, is built around a dielectric, not a vacuum gap. This article walks through where $C_0 = \varepsilon_0 A/d$ comes from, works two numbers all the way through (with and without a dielectric), and flags the traps that quietly cost marks in board exams — wrong units for $d$ , mixing up what changes when charge is held fixed versus when voltage is held fixed, and misremembering which way $K$ multiplies.

Key takeaways

A parallel-plate capacitor is two flat conducting plates of area $A$ , separated by a small gap $d$ , holding charges $+Q$ and $-Q$ .
With vacuum (or air) between the plates, the capacitance is $C_0 = \varepsilon_0 A/d$ — larger plates and a smaller gap both raise capacitance.
Filling the gap completely with a dielectric of dielectric constant $K$ multiplies the capacitance: $C = K\varepsilon_0 A/d = KC_0$ .
The physical reason is that bound charges on the dielectric's surfaces weaken the internal field, so the same charge $Q$ now needs a smaller $V$ — and $C=Q/V$ goes up.
Capacitance depends only on geometry ( $A$ , $d$ ) and the medium ( $K$ ) — never on how much charge or voltage happens to be on the capacitor at a given instant.
Dielectrics also raise the breakdown voltage a capacitor can safely hold, which is a second, independent reason real capacitors use them.
Key idea: a parallel-plate capacitor's capacitance $C_0 = \varepsilon_0 A/d$ is fixed purely by its geometry, and filling the gap with a dielectric of constant $K$ scales it up to $C = KC_0$ because the dielectric's bound charges shrink the internal field for the same free charge.

The core insight

For a parallel-plate capacitor the capacitance is a geometric property, $C_0 = \varepsilon_0 A/d$ , and inserting a dielectric that completely fills the gap multiplies it by the dielectric constant, $C = KC_0$ , because the dielectric's induced surface charges reduce the field (and hence the voltage) for the same stored charge.

Where it fits in the syllabus

The parallel-plate capacitor is the capstone worked geometry of the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), sitting right after the general definitions of capacitance and series/parallel combinations. It draws directly on the field–potential relation for a uniform field ( $V = Ed$ ) and on the field due to a charged conducting surface ( $E = \sigma/\varepsilon_0$ ) built up earlier in the chapter, and it feeds forward into the very next topic, dielectrics and how $K$ changes capacitance. Both formulas here — $C_0 = \varepsilon_0 A/d$ for vacuum and $C = K\varepsilon_0 A/d$ with a dielectric — are explicitly on the CBSE 2025-26 prescribed list and are fair game for direct numericals; the energy-stored formulas ( $U = \tfrac12 CV^2$ , and so on) are also prescribed but strictly as formulae-only, with no derivation expected, so this article keeps that boundary and does not derive them.

The physical picture

Picture the two plates edge-on: a positive sheet of charge facing a negative one, separated by the small gap $d$ . Because $d$ is small compared to the plates' width, the field lines run straight across, from the positive plate to the negative plate, and — ignoring the small fringing/bulging of field lines right at the plate edges, the standard simplification at this level — the field magnitude $E$ is the same at every point between the plates. That uniform-field picture is what lets every formula below be built from plain algebra, with no calculus.

Figure

Two parallel plates with uniform field lines running from the positive plate to the negative plate across a small gap d, area A labelled on the plates

where: $\sigma = Q/A$ is the surface charge density on each plate (charge per unit area), $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ is the permittivity of free space, $E$ is the field magnitude in the gap, $V$ is the potential difference between the plates, and $C = Q/V$ is the capacitance. The physical reading: since $E = \sigma/\varepsilon_0$ comes straight from the electrostatics of a charged conducting surface, and a uniform field gives $V = Ed$ , everything about the plates' charge, field and voltage reduces to just three numbers — $A$ , $d$ , and whatever fills the gap. When a dielectric slab of dielectric constant $K$ is slid in to completely fill the gap, its molecules polarise: each becomes a tiny dipole, and the aligned dipoles produce a thin sheet of bound (induced) charge on each face of the slab, oriented so as to partly oppose the free charge's field. The net field inside the dielectric drops to $E/K$ for the same free charge $Q$ on the plates — which, through $V = Ed$ , drops $V$ by the same factor $K$ , and through $C = Q/V$ , raises $C$ by the factor $K$ .

C = \frac{K\varepsilon_0 A}{d} = K\,C_0

where: $K$ (also called the relative permittivity, or dielectric constant) is a pure number, always greater than 1 for a real material and exactly 1 for vacuum, and $C_0$ is the vacuum capacitance of the same plates. The physical reading is simple: the dielectric never changes the geometry of the plates — it changes how easily the field "gets through" the gap, and that alone is what boosts $C$ .

The derivation

Derivation ACapacitance of a parallel-plate capacitor with vacuum between the plates

The goal is to build the formula $C_0 = \varepsilon_0 A/d$ from first principles, using only three ingredients already established earlier in the chapter — the field of a charged conducting plate, the relation between a uniform field and potential difference, and the definition of capacitance — with no calculus needed because the field between closely spaced large plates is uniform. Getting the chain of three short steps straight, in the right order, is exactly what a board-exam derivation question rewards.

Step 1. Find the field between the plates. Each plate carries a uniform surface charge density $\sigma = Q/A$ , where $Q$ is the magnitude of charge on either plate and $A$ is its area. For two oppositely charged parallel conducting plates, the fields of the two sheets add between the plates and cancel outside them, giving a uniform field of magnitude $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$ in the gap, directed from the positive plate to the negative plate, and (to a very good approximation for $d$ small compared to the plate dimensions) zero field outside.

Step 2. Convert the field into a potential difference. Because the field between the plates is uniform and the plates are flat and parallel, the standard relation for a uniform field applies directly: $V = Ed$ , where $d$ is the plate separation. Substituting the field from Step 1 gives $V = \dfrac{Qd}{\varepsilon_0 A}$ — the potential difference the battery would have to supply (or that a charged, isolated capacitor would exhibit) to hold charge $Q$ on plates of area $A$ separated by $d$ .

Step 3. Apply the definition of capacitance. By definition, $C = Q/V$ for any capacitor. Substituting the expression for $V$ from Step 2, the charge $Q$ cancels cleanly out of numerator and denominator, leaving $C_0 = \dfrac{Q}{Qd/(\varepsilon_0 A)} = \dfrac{\varepsilon_0 A}{d}$ . The cancellation of $Q$ is the physically important part: it confirms that capacitance is a fixed property of the plates' geometry and the medium between them, not of how much charge happens to be sitting on them at any instant — exactly what the definition $C=Q/V$ promises for every capacitor, and what this specific geometry now makes explicit in terms of $A$ , $d$ and $\varepsilon_0$ .

Worked examples

Board level - capacitance of a parallel-plate capacitor in vacuum

A parallel-plate capacitor has plates of area $0.02\ \text{m}^2$ each, separated by $1\ \text{mm}$ , with vacuum between them. Take $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ . Find its capacitance.

Step 1. Write down the formula for a vacuum-gap parallel-plate capacitor: $C_0 = \dfrac{\varepsilon_0 A}{d}$ .

Step 2. Convert the separation to metres before substituting — this is the single most common slip in this calculation. $d = 1\ \text{mm} = 1\times10^{-3}\ \text{m}$ , and $A = 0.02\ \text{m}^2$ is already in SI units.

Step 3. Substitute and compute: $C_0 = \dfrac{(8.85\times10^{-12})(0.02)}{1\times10^{-3}} = \dfrac{1.77\times10^{-13}}{1\times10^{-3}} = 1.77\times10^{-10}\ \text{F}$ .

Answer: the capacitance is $1.77\times10^{-10}\ \text{F}$ , or about $177\ \text{pF}$ .

Competitive level - inserting a dielectric, then comparing charge and field behaviour

The capacitor from the board-level example ( $C_0 = 1.77\times10^{-10}\ \text{F}$ ) is first charged by a $12\ \text{V}$ battery, then disconnected from the battery, and only afterward a dielectric slab of dielectric constant $K = 5$ is slid in to completely fill the gap. Find (a) the new capacitance, (b) the charge on the plates before and after inserting the slab, and (c) the new potential difference across the plates.

Step 1. New capacitance: since the slab is inserted after the battery is disconnected, the plates are isolated and $Q$ stays fixed, but $C$ still scales by the geometric relation $C = KC_0$ . $C = 5 \times 1.77\times10^{-10} = 8.85\times10^{-10}\ \text{F}$ , or $885\ \text{pF}$ .

Step 2. Charge before inserting the slab, while still connected to the $12\ \text{V}$ battery: $Q = C_0V = (1.77\times10^{-10})(12) = 2.12\times10^{-9}\ \text{C}$ . Because the battery is disconnected before the dielectric goes in, this charge has nowhere to go and stays exactly the same afterward: $Q_{\text{after}} = Q_{\text{before}} = 2.12\times10^{-9}\ \text{C}$ .

Step 3. New potential difference: with $Q$ fixed and $C$ now $5$ times larger, $V_{\text{new}} = Q/C = (2.12\times10^{-9})/(8.85\times10^{-10}) = 2.4\ \text{V}$ — exactly the original $12\ \text{V}$ divided by $K=5$ , consistent with the dielectric shrinking the field (and hence $V$ ) by a factor of $K$ for fixed charge.

Answer: $C = 8.85\times10^{-10}\ \text{F}$ ( $885\ \text{pF}$ ); $Q$ stays at $2.12\times10^{-9}\ \text{C}$ throughout (isolated plates); $V$ drops from $12\ \text{V}$ to $2.4\ \text{V}$ . Timing note: this is a two-mark-to-four-mark question depending on how many parts are asked, and it should take under three minutes. The one decision that determines everything is whether the battery stays connected (then $V$ is fixed and $Q$ rises by $K$ ) or is disconnected first (then $Q$ is fixed and $V$ falls by $K$ ) — read the question stem for "disconnected" or "removed" versus "connected throughout" before writing a single line, because the two scenarios give opposite-looking answers from the same $C = KC_0$ starting point.

Common traps

Forgetting to convert the plate separation from millimetres or centimetres to metres before substituting into $C_0 = \varepsilon_0 A/d$ — since $d$ is almost always given in mm, this is the single most common numerical slip.
Assuming capacitance itself depends on the charge $Q$ or voltage $V$ currently on the capacitor — it depends only on geometry ( $A$ , $d$ ) and the dielectric constant $K$ of the medium, exactly as the Step-3 cancellation in the derivation shows.
Believing a larger separation $d$ increases capacitance — it is the opposite, $C_0 \propto 1/d$ , so pulling the plates apart always decreases capacitance.
Confusing "battery stays connected" (voltage fixed, so inserting a dielectric raises the charge $Q$ ) with "battery is disconnected first" (charge fixed, so inserting a dielectric lowers the voltage $V$ ) — both give $C \to KC_0$ , but which of $Q$ or $V$ changes flips completely depending on this one condition in the question.
Using $K$ as a multiplier directly on charge or on voltage instead of on capacitance — the correct relation is always $C = KC_0$ ; whether $Q$ or $V$ changes, and by what factor, then follows from $C=Q/V$ , not from multiplying $K$ into $Q$ or $V$ by guesswork.

Common trapCapacitance is a fixed property of the capacitor, not a snapshot of its current charge

The single biggest misconception here is treating $C = Q/V$ as if $C$ were something computed fresh from whatever charge and voltage the capacitor happens to have at a given moment — as though charging it more would somehow change its capacitance. It does not. $C_0 = \varepsilon_0 A/d$ (and $C = K\varepsilon_0 A/d$ with a dielectric) are fixed by the plates' area, their separation, and the medium filling the gap — full stop. Charge and voltage on a given capacitor can range over many values (zero when uncharged, larger when charged to a higher voltage), and $Q/V$ stays the same ratio throughout, because $Q$ and $V$ rise and fall together in direct proportion. Confusing this — for instance, saying "the capacitance went up because we charged it to a higher voltage" — is a straightforward conceptual error that costs marks even when the arithmetic elsewhere in an answer is correct.

You should be able to…

State and apply $C_0 = \varepsilon_0 A/d$ for a parallel-plate capacitor with vacuum (or air) between the plates.
Reproduce the three-step derivation of $C_0 = \varepsilon_0 A/d$ from the field of a charged plate, $V = Ed$ , and $C = Q/V$ .
State and apply $C = K\varepsilon_0 A/d = KC_0$ for a parallel-plate capacitor with a dielectric completely filling the gap.
Correctly distinguish what stays fixed (and what changes) when a dielectric is inserted with the battery connected, versus with the battery disconnected first.
Explain qualitatively, in terms of bound charges reducing the internal field, why a dielectric raises capacitance.
Solve board- and competitive-level numericals combining plate area, separation, dielectric constant, charge, and voltage.

Ask the tutor: want to work through the partial-dielectric-fill case, or connect this formula to the energy-stored expressions from the rest of the chapter — ask the AI tutor to walk through a few more examples with you.

Why does the formula only use A and d, and not the shape of the plates?

Because the derivation only ever uses the surface charge density $\sigma = Q/A$ and the uniform-field relation $V = Ed$ — neither step cares whether the plates are square, circular, or any other shape, only about their total area $A$ and how far apart they are. As long as the plates are large compared to their separation (so the field really is uniform in the middle and fringing at the edges is negligible), the exact outline of the plates never enters the formula. A circular plate and a square plate of the same area and separation have the same capacitance.

If the dielectric only partly fills the gap, does C still equal KC0?

No — $C = K\varepsilon_0 A/d$ strictly applies only when the dielectric slab fills the entire gap between the plates, touching both plates completely. If the slab only partly fills the gap (leaving some vacuum or air in series with it), the capacitor behaves like two capacitors in series — one for the dielectric-filled portion and one for the remaining gap — and you would need the series-combination formula from earlier in the chapter to combine them. That partial-fill case sits at the edge of, or just beyond, what CBSE numericals typically ask, so always check the question explicitly says the slab fills the gap completely before applying $C = KC_0$ directly.

Why do real capacitors use a dielectric instead of just leaving the gap empty?

Two independent reasons. First, for the same physical plate size and gap, a dielectric with $K > 1$ multiplies the capacitance, letting a compact component store far more charge at a given voltage than an air or vacuum gap ever could — this is why capacitors with tiny footprints can still have useful capacitance values. Second, dielectrics generally have a higher breakdown field strength than air, meaning the capacitor can be safely charged to a higher voltage before sparking (dielectric breakdown) occurs across the gap. Both effects push in the same direction: better performance from the same amount of metal.

Is the energy stored in a capacitor asked with a derivation in the CBSE exam?

No. The CBSE 2025-26 syllabus explicitly lists the energy formulas $U = \tfrac12CV^2 = \tfrac12QV = Q^2/(2C)$ as formula-only content — you are expected to know and apply them in numericals, but the derivation (which needs integrating the work done in bringing charge onto the plates bit by bit) is out of scope and will not be asked as a "derive" question. Use the formulas confidently in calculations; just do not expect, or attempt, to derive them from scratch on paper.