Capacitance Parallel Plate Calculator
Calculate parallel-plate capacitance from plate area, separation, and dielectric constant, plus combined totals for series and parallel capacitors.
Capacitance Parallel Plate Calculator
Calculator
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Formula: C = ε₀εᵣA/d | Series: 1/C = Σ(1/Cᵢ) | Parallel: C = ΣCᵢ
Worked example — 88.54 pF
Formula
C = ε₀εᵣA/d | Series: 1/C = Σ(1/Cᵢ) | Parallel: C = ΣCᵢ
Capacitance of parallel plate capacitor. ε₀ = 8.854×10⁻¹² F/m.
Worked Examples
Example 1: Air capacitor
Problem:A=0.01m², d=0.001m, εᵣ=1
Solution:C=8.854e-12×0.01/0.001=88.54 pF
Result:88.54 pF
Frequently Asked Questions
What is a parallel plate capacitor?
A parallel plate capacitor consists of two conductive plates of area A separated by distance d. Its capacitance is C = ε₀εᵣA/d, where ε₀ = 8.854×10⁻¹² F/m is the permittivity of free space and εᵣ is the relative permittivity of the dielectric material.
How do you calculate parallel plate capacitance?
Use C = ε₀εᵣA/d. For two 0.01 m² plates separated by 1 mm of air (εᵣ = 1): C = 8.854×10⁻¹² × 1 × 0.01 / 0.001 = 88.54 pF. Inserting a dielectric like glass (εᵣ ≈ 5) would increase C to about 442.7 pF.
What units does the capacitance calculator use?
Plate area is in m², plate separation in meters (m), and the result is in Farads (F), displayed in picofarads (pF) for readability. The calculator also shows stored charge Q = CV in nanocoulombs (nC) and energy U = ½CV² in nanojoules (nJ) at 5 V.
What does a dielectric do to capacitance?
A dielectric material placed between the plates increases capacitance by a factor of εᵣ (relative permittivity). Common values: vacuum = 1, air ≈ 1.0006, paper ≈ 3.5, glass ≈ 5, ceramic ≈ 10–10,000, and water ≈ 80. Higher εᵣ means more charge stored at the same voltage.
How do capacitors combine in series and in parallel?
In parallel the capacitances add, which is equivalent to increasing the plate area. In series the reciprocals add, so the total is always smaller than the smallest member; two equal capacitors in series give half of one of them, like doubling the plate separation. In a series string the voltage divides inversely with capacitance, so the smallest capacitor carries the largest share.
What is dielectric strength, and how does it set the maximum working voltage?
Dielectric strength is the electric field a material can withstand before it breaks down and conducts, quoted in volts per metre; dry air is around 3 MV/m. Since the field between the plates is V divided by d, the breakdown voltage is roughly the dielectric strength times the separation. Shrinking d to raise capacitance therefore lowers the voltage rating, which is the central trade-off in capacitor design.
How much energy does a charged parallel plate capacitor store?
Stored energy is one half C V squared, which can also be written as one half Q V or Q squared divided by twice C. A 100 pF capacitor charged to 100 V holds about 0.5 microjoules. Because energy grows with the square of voltage, doubling the applied voltage quadruples the stored energy, which is why high-voltage capacitor banks must be discharged carefully before handling.
What is the electric field between the plates?
For an ideal capacitor the field is uniform and equal to the voltage divided by the separation, so 10 V across a 1 mm gap gives 10 kV/m. It can equally be written as the surface charge density divided by the permittivity of the dielectric. This uniform field is what allows parallel plate geometry to be used as a controlled deflector in oscilloscopes and mass spectrometers.
Why does the formula assume the plate separation is small compared with the plate size?
The derivation treats the field as perfectly uniform and confined between the plates. In reality the field bulges outward at the edges, an effect called fringing, which adds a little extra capacitance. The error stays small while the separation is a tiny fraction of the plate dimensions but becomes significant as the gap approaches the plate width, where a numerical field solution is needed.
What happens to the voltage if I pull the plates apart while the charge stays fixed?
With the capacitor disconnected the charge cannot change. Capacitance is inversely proportional to separation, so doubling the gap halves C, and since V equals Q divided by C the voltage doubles. Stored energy, Q squared divided by twice C, also doubles, and that extra energy is precisely the mechanical work you do pulling the oppositely charged plates apart against their attraction.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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