Parallel Plate Capacitor - Formula, Definition, Derivation, FAQs

Parallel Plate Capacitor - Formula, Definition, Derivation, FAQs

Vishal kumarUpdated on 02 Jul 2025, 04:59 PM IST

A parallel plate capacitor consists of a plate connected to the positive end of a cell and another plate connected to the negative end or earthed. But let us first understand what a capacitor means. A capacitor consists of 2 conducting surfaces that are separated by a layer of an insulating medium also called a dielectric. This dielectric can be any insulating medium, the most common being parallel plate capacitors with air between the plates. The capacity of a conductor can be defined as the ratio between the charge on the conductor to its potential.

Parallel Plate Capacitor - Formula, Definition, Derivation, FAQs
Parallel Plate Capacitor

What is a Parallel Plate Capacitor?

A parallel plate capacitor consists of electrodes arranged along with some insulating material or dielectric. A parallel plate capacitor can store only a finite energy before dielectric breakdown occurs. Thus, we can say that Two parallel plates, when connected across a battery, get charged on account of an electric field existing between them and is known as a parallel plate capacitor.

Parallel Plate Capacitor Formula

The path of the electric field is defined as the path on which a positive test charge would move. The capacitance of the body is the measure of its ability to accumulate electric charge. Each capacitor has a capacitance. A typical parallel-plate capacitor has two metallic plates having area A, which are separated from each other by distance d.

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The parallel plate capacitor formula is given by:

$$
C=k \epsilon_0 \frac{A}{d}
$$
Where,
$\epsilon_0$ is the permittivity of space $\left(8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$
$k$ is the relative permittivity of dielectric material
$d$ is the separation between the plates
A is the area of plates

Also, read

Units of Capacitance

In the S.I. unit, capacitance is measured in Farad (F).
$1 \mathrm{~F}=1$ Coulomb of charge/1 volt of potential
(By equation 1) we can define 1 Farad as - the capacity of a conductor is 1 F if a charge of 1 Coulomb is required to establish a p.d. of 1 Volt between the plates.
In the C.G.S. unit, capacitance is measured in stat farad.
Dimensional formula of capacitance is $\left[M^{-1} L^{-2} T^4 A^2\right]$.

Parallel Plate Capacitor Derivation

An arrangement of parallel plates as shown in the following diagram depicts a parallel plate capacitor. This comprises two large plates aligned parallel to each other and placed at a small distance $d$ apart. Filling the gap between the plates is a dielectric material, which is shown in the dotted array. The two plates carry an equal and opposite charge.

parallel plate capacitor derivation

This shows the first plate which is charged by +Q and by -Q on the second one. Each of the plates has an area A, while d is the distance between these plates. The distance $d$ is much smaller than the area of the plates and we can write $\mathrm{d}<<\mathrm{A}$, thus the effect of the plates is considered as infinite plane sheets and the electric field generated by them is treated as that equal to the electric field generated by an infinite plane sheet of uniform surface charge density. Thus, as the plate is 1 charge Q, and as the plate has surface area A, the surface charge density may be expressed as:

$$
\sigma=\frac{Q}{A}
$$
Similarly, for plate 2 with a total charge equal to -Q and area A, the surface charge density can be given as,

$$
\sigma=-\frac{Q}{A}
$$
We divide the regions around the parallel plate capacitor into three parts, with region 1 being the area left to the first plate, region 2 is the area between the two plates and region 3 being the area to the right of plate 2.

Let us calculate the electric field in the region around a parallel plate capacitor.

Region I: The electric field strength produced by both the infinite plane sheets I and II is equal at all points in this region, but since both are in the opposite direction, the two forces cancel and the whole electric field is given as:

$E=\frac{\sigma}{2 \varepsilon}-\frac{\sigma}{2 \varepsilon}=0$

Region II: The electric fields produced owing to both sheets I and II in this part are directed in the same way and have the same magnitude. Hence, the overall result will be as follows,

$E=\frac{\sigma}{2 \varepsilon}+\frac{\sigma}{2 \varepsilon}=\frac{\sigma}{\varepsilon}$

Region III: Similar to the region I, here also the field strengths created due to both plane sheets I and II are equal, but their directions are opposite, resulting in the same as,

$E=\frac{\sigma}{2 \varepsilon}-\frac{\sigma}{2 \varepsilon}=0$

The electric potential difference across the capacitor can be calculated by multiplying the electric field and the distance between the planes, given as,

$$
V=E \times d=\frac{1}{\varepsilon} \frac{Q d}{A}
$$
The capacitance for the parallel plate capacitor can be given as,

$$
c=\frac{Q}{V}=\frac{\varepsilon_0 A}{d}
$$

Frequently Asked Questions (FAQs)

Q: What is meant by "capacitor leakage" and how does it affect parallel plate capacitors?
A:

Capacitor leakage refers to the slow loss of charge from a capacitor over time, even when it's not connected to a circuit. In parallel plate capacitors, this can occur due to imperfections in the dielectric material or surface conduction along the edges. Leakage reduces the effectiveness of the capacitor in storing charge for long periods.

Q: How does the capacitance of a parallel plate capacitor change in a vacuum compared to air?
A:

The capacitance of a parallel plate capacitor in a vacuum is slightly lower than in air. This is because air has a relative permittivity of about 1.00059, while vacuum has a relative permittivity of exactly 1. The difference is usually negligible for most practical purposes, which is why air is often approximated as a vacuum in calculations.

Q: What is the effect of surface roughness on the capacitance of a parallel plate capacitor?
A:

Surface roughness can increase the effective surface area of the plates in a parallel plate capacitor, potentially increasing its capacitance slightly. However, it can also lead to local field enhancements, which might lower the breakdown voltage of the capacitor. The overall effect depends on the scale of the roughness relative to the plate separation.

Q: How does the concept of "parasitic capacitance" relate to parallel plate capacitors?
A:

Parasitic capacitance refers to unintended capacitance that exists between components in an electrical circuit. In the context of parallel plate capacitors, parasitic capacitance can occur between the leads or connections of the capacitor and nearby conductors. This additional capacitance can affect the behavior of the circuit, especially at high frequencies.

Q: How does the concept of "self-capacitance" differ from the capacitance of a parallel plate capacitor?
A:

Self-capacitance is the capacitance of a single conductive object relative to infinity, while the capacitance of a parallel plate capacitor is a mutual capacitance between two conductors. Self-capacitance is typically much smaller than the mutual capacitance in a parallel plate configuration and becomes relevant mainly in antenna theory and in very high impedance circuits.

Q: What is the significance of the "quality factor" in parallel plate capacitors?
A:

The quality factor (Q factor) is a measure of the energy loss in a capacitor relative to the energy it stores. In parallel plate capacitors, a higher Q factor indicates lower energy loss and better performance, especially in AC circuits. It's particularly important

Q: What happens to the capacitance if you increase the thickness of the metal plates?
A:

Increasing the thickness of the metal plates does not significantly affect the capacitance of a parallel plate capacitor. The capacitance primarily depends on the area of the plates facing each other and the distance between them, not the plate thickness.

Q: How does temperature affect the capacitance of a parallel plate capacitor?
A:

Temperature can affect the capacitance of a parallel plate capacitor in several ways. It can cause thermal expansion of the plates, changing their area and separation. More significantly, it can affect the properties of the dielectric material, potentially changing its permittivity. The exact effect depends on the materials used.

Q: Why is it important to consider the maximum electric field strength when designing a parallel plate capacitor?
A:

Considering the maximum electric field strength is crucial in capacitor design because it determines the breakdown voltage of the capacitor. If the field strength exceeds the dielectric strength of the material between the plates, the dielectric will break down, potentially causing a short circuit and failure of the capacitor.

Q: How does the concept of electric susceptibility relate to parallel plate capacitors?
A:

Electric susceptibility (χe) is a measure of how easily a dielectric material polarizes in response to an electric field. In a parallel plate capacitor, a higher electric susceptibility of the dielectric material leads to greater polarization, which in turn increases the capacitance. It's related to the relative permittivity by εr = 1 + χe.