The plate configuration
A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance. We first take the intervening medium between the plates to be vacuum; the effect of a dielectric is treated separately below. Let $A$ be the area of each plate and $d$ the separation between them. The two plates carry charges $+Q$ and $-Q$, so plate 1 has surface charge density $\sigma = Q/A$ and plate 2 has surface charge density $-\sigma$.
The defining geometric assumption is $d^2 \ll A$ — the separation is much smaller than the linear dimension of the plates. This lets us treat each plate as an infinite plane sheet of uniform surface charge density, so the field between them follows directly from the infinite-sheet result of Gauss's law. This idealisation is what makes the parallel plate capacitor analytically clean.
Field between the plates
Each charged plate, treated as an infinite sheet, produces a uniform field of magnitude $\sigma/2\varepsilon_0$ on either side, directed away from a positive sheet and toward a negative one. Adding the contributions of both plates in the three regions gives the characteristic result.
In the outer regions (above plate 1 and below plate 2) the two sheet fields point in opposite directions and cancel, so $E = 0$. In the inner region between the plates the two fields point the same way and add up:
$$ E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} $$
The direction of the field is from the positive to the negative plate. Thus the field is localised between the plates and is uniform throughout. For plates of finite area this fails near the edges, where the field lines bend outward — an effect called fringing of the field. For $d^2 \ll A$ these edge effects are ignored in regions sufficiently far from the boundary, and the interior field is taken as $\sigma/\varepsilon_0$ everywhere.
Deriving C = ε₀A/d
Since the field is uniform, the potential difference between the plates is simply the field times the separation:
$$ V = E\,d = \frac{\sigma d}{\varepsilon_0} = \frac{1}{\varepsilon_0}\frac{Q d}{A} $$
The capacitance is the ratio of charge to potential difference, $C = Q/V$. Substituting $V$ above, the charge $Q$ cancels and we are left with a purely geometric expression:
$$ C = \frac{Q}{V} = \frac{Q}{\dfrac{Qd}{\varepsilon_0 A}} = \frac{\varepsilon_0 A}{d} $$
As expected, $C$ depends only on the geometry of the system — the plate area and the separation — with $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$. For typical values $A = 1\ \text{m}^2$ and $d = 1\ \text{mm}$, this gives $C = 8.85 \times 10^{-9}\ \text{F} = 8.85\ \text{nF}$, confirming that the farad is an enormous unit; a 1 F capacitor with a 1 cm gap would need plates roughly 30 km on a side.
What capacitance depends on
The formula $C = \varepsilon_0 A/d$ encodes three dependences that NEET probes repeatedly. Capacitance scales directly with plate area, inversely with separation, and is multiplied by the dielectric constant of the medium filling the gap.
| Quantity | Relation | Effect on C |
|---|---|---|
| Plate area $A$ | C ∝ A | Double the area → double the capacitance |
| Separation $d$ | C ∝ 1/d | Halve the gap → double the capacitance |
| Medium (dielectric $K$) | C = K ε₀ A/d | Fill with dielectric $K$ → multiply by $K$ |
| Charge $Q$ / voltage $V$ | C = Q/V fixed | No effect — $Q$ and $V$ change together |
C does not depend on the charge or voltage
A common error is to assume that charging the capacitor more (larger $Q$ or larger $V$) raises its capacitance. It does not. $C = \varepsilon_0 A/d$ is fixed by geometry alone. Increasing $Q$ proportionally increases $V$, so the ratio $C = Q/V$ is unchanged.
Capacitance is a property of the conductor system, not of how much charge sits on it.
Field between plates is σ/ε₀, not σ/2ε₀
A single isolated sheet gives $\sigma/2\varepsilon_0$, but between two oppositely charged plates the two contributions add to $E = \sigma/\varepsilon_0$. The half-field appears only for one plate alone — for instance when computing the force one plate exerts on the other.
Between the plates: $E = \sigma/\varepsilon_0$. On one plate due to the other: field $= \sigma/2\varepsilon_0$.
Inserting a dielectric
When the gap is filled with a dielectric of dielectric constant $K$, the medium polarises in the applied field. The induced bound surface charges partly cancel the free charge, so the net field becomes $E = (\sigma - \sigma_P)/\varepsilon_0$, which is smaller than the vacuum field for the same free charge. A smaller field means a smaller potential difference $V = Ed$ for the same $Q$, and since $C = Q/V$, the capacitance rises. Quantitatively, the field is reduced by a factor $K$, giving:
$$ C = \frac{K\varepsilon_0 A}{d} = K\,C_0 $$
where $C_0 = \varepsilon_0 A/d$ is the vacuum value. The product $K\varepsilon_0 = \varepsilon$ is the permittivity of the medium. The dielectric also raises the maximum charge a capacitor can hold before breakdown; air, for comparison, has a dielectric strength of about $3 \times 10^{6}\ \text{V m}^{-1}$. The full polarisation mechanism and the slab-inserted-partially cases are developed in Effect of Dielectric on Capacitance.
Worked examples
A parallel plate capacitor has plates of area $1\ \text{m}^2$ separated by $1\ \text{mm}$ of vacuum. Find its capacitance.
$C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85 \times 10^{-12})(1)}{1 \times 10^{-3}} = 8.85 \times 10^{-9}\ \text{F} = 8.85\ \text{nF}.$ The small numerical value illustrates why the farad is impractically large.
The plates of the capacitor above stay connected to a battery while the separation is halved to $0.5\ \text{mm}$. What happens to the capacitance and the stored charge?
Since $C \propto 1/d$, halving $d$ doubles the capacitance to $C = 17.7\ \text{nF}$. With the battery connected, $V$ is fixed, so the stored charge $Q = CV$ also doubles. (If the capacitor were isolated, $Q$ would stay fixed and $V$ would halve instead.)
An air-filled parallel plate capacitor of capacitance $6\ \mu\text{F}$ is fully filled with a dielectric, raising the capacitance to $30\ \mu\text{F}$. Find the dielectric constant and the permittivity of the medium ($\varepsilon_0 = 8.85 \times 10^{-12}$ SI).
$K = C/C_0 = 30/6 = 5.$ Permittivity $\varepsilon = K\varepsilon_0 = 5 \times 8.85 \times 10^{-12} = 0.44 \times 10^{-10}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}.$ (This mirrors NEET 2020 Q.113.)
Parallel plate capacitor in one screen
- Each plate is treated as an infinite sheet ($d^2 \ll A$); field between plates is uniform, $E = \sigma/\varepsilon_0 = Q/\varepsilon_0 A$, directed from $+$ to $-$ plate; outside, $E = 0$.
- $V = Ed = \sigma d/\varepsilon_0$, so $C = Q/V = \varepsilon_0 A/d$ — purely geometric.
- $C \propto A$, $C \propto 1/d$; $C$ is independent of charge and voltage.
- With a dielectric: $C = K\varepsilon_0 A/d = KC_0$.
- Edge fringing is neglected for ideal plates; the farad is a huge unit ($\mu$F, nF, pF in practice).