How to Calculate the Capacitance of a Parallel Plate Capacitor

Capacitor is one of the fundamental electronic components that store electric energy. The capacity of a capacitor is measured through a parameter called capacitance, which is measured Farads. This article explores how the capacitance of the most basic type of capacitor, the parallel plate capacitor, is calculated.

Basically, the parallel plate capacitor is one of the basic electrical devices composed of two conducting plates that are parallel to each other with an insulating material, termed dielectric, sandwiched within it. In other words, it can be defined as the device that is responsible for temporarily storing electrical energy.

Also Read: Capacitance

Here’s the formula for how to calculate capacitance in parallel plate capacitors. A parallel plate capacitor exists if two conducting plates are placed parallel to one another and separated by a thin insulating material known as the dielectric. The capacitance C of A is directly proportional to the area A of the plate and inversely proportional to the separation d between the plates. This is expressed as:

C = ε ×(A / d)

where:

• C: Capacitance in Farads (F)
• ε: Permittivity of the material between the plates (F/m)
• For air or vacuum, ε = ε₀ (permittivity of free space) ≈ 8.854 × 10⁻¹² F/m
• The permittivity (ε) is a material-specific property that influences the capacitor’s capacitance. When a dielectric material with permittivity ε (greater than ε₀) fills the space between the plates, the capacitance increases.
• A: Area of each plate in square meters (m²)
• d: Distance between the plates in meters (m)

Also Read: Capacitor and Capacitance

A parallel plate capacitor is formed with two parallel conducting plates separated by a distance and are connected to a voltage source. Applying a voltage to the plates causes one plate to receive a positive charge density +? and the other an equal and opposite charge density -? .

Electric Field Calculation

• 1. Outer Region: The electric field above Plate 1 and below Plate 2 cancels out due to equal and opposite charges. Thus, E_outer = 0.
• 2. Inner Region: Between the plates, the electric fields due to the charged plates add up linearly, yielding a uniform electric field E_inner = σ/ϵ₀.

Also Read: Effect of Dielectric on Capacitance

Potential Difference Calculation

• 3. The uniform electric field between the plates generates a potential difference ‘V’ given by V = Ed, where ‘E’ is the electric field magnitude. Substituting E = σ/ϵ₀, we get V = σd/ϵ₀.

Capacitance Calculation

• 4. The capacitance ‘C’ is defined as the charge (Q) stored per unit potential difference (V), i.e., C = Q/V. For a parallel plate capacitor, Q = σA, where ‘A’ is the area of one plate.
• 5. Substituting Q = σA and V = σd/ϵ₀ into the capacitance formula, we get C = (σA)/(σd/ϵ₀).
• 6. Simplifying, we find C = (ϵ₀A)/d.

The capacitance ‘C’ of a parallel plate capacitor is directly proportional to the permittivity of free space (ϵ₀) and the area of the plates (A), and inversely proportional to the separation distance between the plates (d). This derivation provides a fundamental understanding of how capacitance is determined in such capacitors, crucial for designing electronic circuits and systems.

Key Points to Remember

• Larger plate area (A) leads to higher capacitance.
• Smaller plate separation (d) results in greater capacitance.
• The dielectric material’s permittivity (ε) significantly impacts capacitance.

Example 1. A parallel plate capacitor has plates with an area of 0.01 m² each, separated by a 0.001 m air gap. Calculate its capacitance.

Solution:

ε (air) ≈ 8.854 × 10⁻¹² F/m

A = 0.01 m²

d = 0.001 m

C = ε × A / d

= (8.854 × 10⁻¹² F/m) × (0.01 m²) / (0.001 m)

≈ 8.854 × 10⁻¹⁰ F

Example 2: A capacitor with plates of area 0.02 m² has a capacitance of 2 × 10⁻¹⁰ F. The plates are separated by a dielectric material with a permittivity of 6. Determine the distance between the plates.

Solution:

C = 2 × 10⁻¹⁰ F

ε = 6 × ε₀ (given that ε is 6 times the permittivity of free space)

A = 0.02 m²

Rearranging the formula:

d = ε × A / C

Substituting values:

d = (6 × ε₀) × (0.02 m²) / (2 × 10⁻¹⁰ F)

Note: Due to the limitations of representing very small values accurately, it’s recommended to use a scientific calculator for this calculation.

By comprehending the concept of capacitance and its calculation for parallel plate capacitors, you’ll gain a solid foundation for analyzing and designing.

It is essential to comprehend capacitance in parallel plate capacitors to develop highly effective electronic circuits. Plate area, gap between them, and dielectric properties are carefully evaluated, optimized and utilized by engineers changing how conventional capacitors work. This knowledge has been fundamental in developing electronic component-based systems suited for different applications and operational use.

Also, Check

• Capacitors in Series and Parallel
• What is the SI Unit of Capacitance?
• Miller Capacitance
• Capacitive Reactance

What is the formula for capacitance?

The formula for the capacitance C of parallel plate capacitor is C = ε₀ A / d , where ε₀ is the permittivity of free space, A is the area of each plate, and d is the separation between the plates.

How does the distance between plates affect capacitance?

As the distance between plates decreases, the capacitance increases. This is because the electric field between the plates becomes stronger, leading to a higher capacitance.

Can the capacitance of a parallel plate capacitor be negative?

The capacitance of the parallel plate capacitor can never be negative. It means the value of C can only be positive.

How does dielectric material influence capacitance?

Capacitance increases when a dielectric is introduced between the plates of a capacitor. The dielectric decreases the electric field sensed by the plates, permitting more charge to be saved at a relatively little voltage.

What happens to capacitance when plate area increases?

Capacitance across the plates increases. This is due to the fact that if the plates have more surface space for the charge to accumulate close, it has a big capacitance due to a larger plate area.