Capacitance Parallel Plate Calculator
parallel plate capacitance calculator. Get instant, accurate results. Enter values for instant results with step-by-step formulas.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
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Formula: C = ε₀εᵣA/d | Series: 1/C = Σ(1/Cᵢ) | Parallel: C = ΣCᵢ
Worked example — 88.54 pF
C = ε₀εᵣA/d | Series: 1/C = Σ(1/Cᵢ) | Parallel: C = ΣCᵢ
Capacitance of parallel plate capacitor. ε₀ = 8.854×10⁻¹² F/m.
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
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.
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.
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.
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.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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